and the Dirac Equation - quantumthom.comquantumthom.com/GenRelAndDiracEqH2.pdf · 2020. 8. 17. ·...

General Relativityand the

Dirac Equation

Thomas Brennan

May 29, 2020

Preface

In this work I investigate the connections between quantum mechanics and gen-eral relativity – the topic of quantum gravity. I choose to take a first-principlesapproach and this inquiry picks up directly where Dirac and Schroedinger leftoff – the application of the general principle of relativity to the Dirac equation.I present proof of a few mathematical theorems pertaining to the way the formof Dirac equation changes, or doesn’t change, under a general transformation ofcurvilinear coordinates. Central to this is the notion that the Dirac gamma matri-ces are the unit-vectors of a flat space-time. This is something that is implicit inthe Dirac algebra and is reflected in the fundamental identity defining the Diracgamma matrices:

ηµν =12γµ, γν = γµ · γν.

ηµν is the metric of a flat space-time, and the anti-commutator is a dot product.The next step in generalizing to curvilinear coordinates is to consider other kindsof 4× 4 Dirac matrices, εµ, akin to the γ matrices, yet dependent upon time andposition that represent the basis vectors of more general metrics:

gµν =12εµ, εν = εµ · εν.

This view of the Dirac algebra opens up an alternative method for derivingthe Einstein equation that avoids the problems of non-renormalizability that comewith the usual method based on the Einstein-Hilbert action. It also reveals that the

1

2

right-hand side of the classical Einstein equation is just the leading-order term inan infinite series that we describe.

I apply this method to the well-known topics of the Schwarzschild metric aswell as hydrogen atom-type problems to show that it gives the expected results. Ialso show how the wave-function solutions to gravitational Dirac problems musteither grow (or decay) exponentially with time, but very slowly. If interpreted cos-mologically, this provides a numerical connection between the value of the Hub-ble constant, H0, the universal gravitational constant, G, and the average mass-energy density of the universe.

Contents

1 The General Principle of Relativity 51.1 Lorentz Group Transformations . . . . . . . . . . . . . . . . . . . . . 51.2 Quantum Tensor Algebra . . . . . . . . . . . . . . . . . . . . . . . . 101.3 The General Dirac Equation in Gibbs/Heaviside Notation . . . . . 131.4 Current Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5 General Covariance of Commutation Relations . . . . . . . . . . . . 171.6 The Metric Tensor and the Christoffel Symbols . . . . . . . . . . . . 19

1.6.1 The Schwarzschild Transformation . . . . . . . . . . . . . . . 211.6.2 The Schwarzschild-Dirac Problem . . . . . . . . . . . . . . . 25

2 The Free Particle 302.1 Solving the Dirac Equation as a Differential Equation . . . . . . . . 302.2 Finding the Spectrum of the Free-Particle Dirac Equation using Op-

erator Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.3 The Dirac angular momentum operator, J . . . . . . . . . . . . . . . 422.4 Angular momentum, Action and Charge . . . . . . . . . . . . . . . 45

3 The Hydrogen Atom 47

3.1 The Transformation e−i α ′2 ε . . . . . . . . . . . . . . . . . . . . . . . . 48

3.1.1 Solving for the Energy Eigenvalues . . . . . . . . . . . . . . 49

3

CONTENTS 4

4 Quantum Field Theory 544.1 Quantum Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . 544.2 The Feynman Path Integral . . . . . . . . . . . . . . . . . . . . . . . 584.3 Feynman’s Green Function Method . . . . . . . . . . . . . . . . . . . 64

5 Multiple Particles 745.1 Multiple Particles and the Principle of Superposition . . . . . . . . 74

5.1.1 Two Particles, N=2 . . . . . . . . . . . . . . . . . . . . . . . . 765.2 The Integer Normalization of the Wavefunction . . . . . . . . . . . 825.3 The Case of N = 2, Again . . . . . . . . . . . . . . . . . . . . . . . . 845.4 Principle of Superposition – Again . . . . . . . . . . . . . . . . . . . 855.5 Scattering And Feynman Diagrams . . . . . . . . . . . . . . . . . . . 86

6 The General Principle of Relativity, Part II 896.1 Newton’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . 896.2 The Quantum Maxwell Equations . . . . . . . . . . . . . . . . . . . 926.3 Conventional Quantum Gravity . . . . . . . . . . . . . . . . . . . . . 986.4 The Quantum Einstein Equation . . . . . . . . . . . . . . . . . . . . 996.5 The axiomatic basis of the General Dirac Equation . . . . . . . . . . 1036.6 The Quantum Einstein Equation . . . . . . . . . . . . . . . . . . . . 1126.7 Hubble Expansion and Quantum Gravity . . . . . . . . . . . . . . . 115

7 The Space-Time Lattice 1357.1 Discrete Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . 1357.2 The Bandwidth-limited Feynman Green’s Function . . . . . . . . . 142

8 Conclusion 146

9 The Appendices 1479.1 Appendix A: General Matrix Identities . . . . . . . . . . . . . . . . . 1479.2 Appendix B: Identities of the γ and σ matrices . . . . . . . . . . . . 148

9.2.1 The Unit Vectors in 3 and 4 Dimensions . . . . . . . . . . . . 1489.2.2 Commutation relations of γµ∂µ and γν xν in Cartesian systems1529.2.3 More basis vector commutation identities . . . . . . . . . . . 1539.2.4 The operator σL . . . . . . . . . . . . . . . . . . . . . . . . . . 1549.2.5 Identities of the Total Angular Momentum Operator, J . . . 156

9.3 Appendix C: Special Functions and their Identities . . . . . . . . . . 157

CONTENTS 5

9.3.1 The Confluent Hypergeometric Functions . . . . . . . . . . . 1579.3.2 The Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . 157

9.4 Appendix D: The Free-Particle Dirac Equation in Spherical Coordi-nates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

9.5 Appendix E: Properties of the θ+,− 2-spinors . . . . . . . . . . . . . 1639.6 Appendix F: More Tranformations . . . . . . . . . . . . . . . . . . . 166

9.6.1 The generator S = κ2 γr . . . . . . . . . . . . . . . . . . . . . . 166

9.6.2 The generator S = κ2 ε . . . . . . . . . . . . . . . . . . . . . . 166

9.6.3 The generator S = −i α2 ε . . . . . . . . . . . . . . . . . . . . . 167

9.6.4 The generator S = −i α2 γr . . . . . . . . . . . . . . . . . . . . 167

*

CHAPTER 1

The General Principle of Relativity

The General Principle of Relativity is the statement that it is possible to find atransformation of coordinates such that particles move as if they were free, albeitonly in an infinitesimally small region surrounding a particular space-time point.Our goal is to apply this idea to the Dirac equation. In this context this means thatat any point in space-time, there is a transformation of coordinates such that thewave function ψ obeys the free-particle Dirac Equation:

γµ ∂

∂xµf

ψ f =mcih

ψ f (1.1)

which can also be written:

γ0γµ ∂

∂xµf

ψ f =mcih

γ0 ψ f (1.2)

Here, the additional subscript ‘ f ’ stands for ‘free’.

1.1 Lorentz Group Transformations

Now suppose that at any point in space-time there is a relation between the boundcoordinate system and the free coordinate system and the wave functions in those

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CHAPTER 1. THE GENERAL PRINCIPLE OF RELATIVITY 7

respective systems given as:

ψ f (x f ) = eS(xb)ψb(xb) (1.3)

Where the transformation operator, S, is a four-by-four matrix taken to be a func-tion of the bound coordinates. The free coordinates, x f , are also taken to be im-plicit functions of the xb. Thus the partial derivatives transform in the usual way:

∂

∂xµf=

∂xνb

∂xµf

∂

∂xνb

(1.4)

Substituting Eqs. 1.4 and 1.3 into 1.2 initially gives:

γ0γµ ∂xνb

∂xµf

∂

∂xνb

eS(xb)ψb(xb) =mcih

γ0 eS(xb)ψb(xb).

Upon carrying thru the product rule for differentiation while remembering that Sis a function of xb, this becomes:

γ0γµ ∂xνb

∂xµf

eS ∂

∂xνb

ψb + γ0γµ ∂xνb

∂xµf

∂eS

∂xνb

ψb =mcih

γ0 eSψb.

Now multiply thru by eS†on the left:

eS†γ0γµ ∂xν

b

∂xµf

eS ∂

∂xνb

ψb + eS†γ0γµ ∂xν

b

∂xµf

∂eS

∂xνb

ψb =mcih

eS†γ0 eSψb

and then rearrange slightly, inserting a factor 1 = eS · e−S into the second term onthe left while remembering the ∂xν

b∂xµ

fterms are not four-by-four Dirac matrices, so

they commute with matrix terms:

eS†γ0γµ eS ∂xν

b

∂xµf

∂

∂xνb

ψb + eS†γ0γµ eS ∂xν

b

∂xµf

e−S ∂eS

∂xνb

ψb =mcih

eS†γ0 eSψb

Now we can identify:

eS†γ0γµ eS ∂xν

b

∂xµf= γ0γν. (1.5)


I call Eq. 1.5 the transformation law of the unit vectors, which will come in handywhen defining the quantum form of the metric tensor. If we restrict the transfor-mation eS to be a member of the general Lorentz group, that is, eS†

γ0 = γ0e−S,this becomes,

e−S γµ eS ∂xνb

∂xµf= γν. (1.6)

Thus the wave equation becomes:

γ0γν ∂

∂xνb

ψb + γ0γν e−S ∂eS

∂xνb

ψb =mcih

eS†γ0 eSψb.

Again, since we’re restricted to the general Lorentz group, the term on the rightside of the equation reduces to: eS†

γ0 eS = γ0, giving,

γ0γν ∂

∂xνb

ψb + γ0γν e−S ∂eS

∂xνb

ψb =mcih

γ0 ψb (1.7)

Since all terms are proportional to γ0, that can be factored out, giving,

γν ∂

∂xνb

ψb + γν e−S ∂eS

∂xνb

ψb =mcih

ψb. (1.8)

Eqs. 1.7 and 1.8 are the two main ways of writing what I call the general DiracEquation. Note the second term on the left hand side of these equations. This termis equivalent to a general interaction field, which results in the γµ Aµ term in theelectromagnetic case, but can also produce other interaction fields. Studying theset of possible interaction fields and the transformations which generate them is acentral goal of this work.

There is an important subtlety to point out about the transformation law of theunit vectors, Eq. 1.6. We can re-write it as

eνb,b = e−S γµ eS ∂xν

b

∂xµf

,

Since γν is a representation of the bound unit vectors, in the bound system itself.Then the representation of the bound unit vectors in the free system would be,

eνb, f =

∂xνf

∂xµb

eνb,b = e−S γµ eS. (1.9)


This implies that the norm of a bound system contravariant basis vector withitself, represented in the free system, is

eνb, f · e

νb, f = e−S γν eS · e−S γν eS = e−S γν γν eS = e−S(±1)eS = ±1, (no summation),

also has unit norm (up to a ±). So the eνb, f must truly be unit vectors, like the γν.

Likewise, by the classical definition of the transformation of covariant vectors,the covariant basis vectors must obey,

eν b =∂xα

f

∂xνb

γα.

The dot product, or norm, of one of the contravariant bound basis vectors with itscovariant dual is automatically guaranteed to have unit norm, even if the eν

b werenot true unit vectors,

eνb · eν b =

∂xνb

∂xµf

γµ ·∂xα

f

∂xνb

γα = δαµ γµ γα = γα γα = 1, (no summation). (1.10)

So we can see that the transformation for the covariant basis vectors should alsobe written,

eν b = eS γν e−S. (1.11)

This ensures that the dot product, or norm, between the contravariant and covari-ant bound basis vectors is

eνb · eν b = eS γν e−S · eS γν e−S = eS γνγν e−S, (no summation),= eS 1e−S = 1,

in agreement with Eq. 1.10.Thus, in the context of the general Dirac equation, the basis vectors in the

bound system must be unit vectors, akin to the Cartesian γν matrices, and if writ-ten as linear combinations of the γν matrices, must still have unit norm. We can,therefore, write the general Dirac equation in terms of the the unit eν

b instead ofthe γν:

eνb

∂

∂xνb

ψb + eνb e−S ∂eS

∂xνb

ψb =mcih

ψb. (1.12)

This discussion is important for understanding the transformations leading to thegeneral Dirac equation in spherical coordinates, as well as the general Dirac equa-tion for the Schwarzschild problem.


General Continuity

The first form, Eq. 1.7, is useful in deriving the general continuity equation. Sinceγν†γ0 = γ0γν, and γ0†

= γ0, taking the adjunct of 1.7 gives, at first,

∂ψ†b

∂xνγ0γν + ψ†

b∂eS†

∂xνb

e−S†γ0γν = −mc

ihψ†γ0.

Now, introducing the barred notation, ψ = γ0ψ, and moving the γ0’s left across the

middle term by using ∂eS†

∂xνb

e−S†γ0 = γ0 ∂e−S

∂xνb

eS = −γ0e−S ∂eS

∂xνb, this becomes,

∂ψb∂xν

γν − ψb e−S ∂eS

∂xνb

γν = −mcih

ψb, (1.13)

which is the adjunct of the general Dirac equation. The next step to get to thecontinuity equation is to pre-multiply Eq. 1.7 by ψ†

b , post-multiply Eq. 1.13 by ψband add the two equations, to get,

ψbγν ∂ψb∂xν

b+

∂ψb∂xν

γνψb + ψ

(γν e−S ∂eS

∂xνb− e−S ∂eS

∂xνb

γν

)ψ = 0.

Introducing the Dirac current, defined as jν = ψγνψ, this becomes,

∂jν

∂xν+ ψ

(γν e−S ∂eS

∂xνb− e−S ∂eS

∂xνb

γν

)ψ = 0, (1.14)

which is a version of the general continuity equation. Note that not all transfor-mations of the general Lorentz group preserve the form of the continuity equationin the free-particle case, ∂ν jν = 0, but include a modification term due to the po-tential ’warping’ of coordinates that we’re allowing for. If, however, we have asituation like the case of electromagnetism, where the interaction field is a purefour-vector,

γνe−S ∂eS

∂xνb= γν∆ν = γν q

ihcAν,

then ∆ν is just a c-number that can move across the γν’s, and the terms inside theparentheses in Eq. 1.14 cancel,

∂jν

∂xν+

:0ψ (γν∆ν − ∆νγν)ψ = 0.


So when the fields amount to a pure four-vector, the form of the continuity equa-tion from the free-particle case, ∂ν jν = 0, is preserved.

Now, to continue to touch base with commonly accepted results, let us forthe moment restrict the transformations to the Special Lorentz Group, so that S =

−(i/4)ωµνσµν and we still have, eS†γ0 = γ0e−S. In this case, the transformations

are no longer functions of positions, so we also have ∂eS

∂xνb= 0. Then 1.5 becomes:

e−S γµ eS ∂xνb

∂xµf= γν (1.15)

which is comparable to Bjorken & Drell, Eq. (2.12), while the Dirac equation be-comes:

γ0γν ∂

∂xνb

ψb =mcih

γ0 ψb (1.16)

the Lorentz-boosted Dirac Equation, identical in form to Eq. 1.2, as it should be. Herewe can take the subscript ‘b’ to mean ‘boosted,’ rather than ‘bound.’

1.2 Quantum Tensor Algebra

Let us elaborate on our interpretation of the γµ Dirac matrices as representationsof the four unit vectors of space-time hinted at in Eq. 1.5. Here we invoke thespirit of Oliver Heaviside and Josiah Willard Gibbs’ famous refinement of Hamil-ton’s quaternions (1, i, j, k), into modern vector notation. This ‘notational’ discov-ery resulted in the concise and illuminating way of writing Maxwell’s Equationsthat is so familiar to us today. For example, in the standard three-dimensionalGibbs/Heaviside notation, a general vector ~A is written:

~A = Ax i + Ay j + Azk

and the vector ‘del’ operator is

~∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂z

Now the conventional representation of the unit vectors taught in introductorycourses is as columns or rows, ala

i =

100

, j =

010

, k =

001


But as many who work with aircraft and spacecraft navigation or 3-d computergraphics know, it’s also possible, and computationally more efficient, to representthe unit vectors as 2× 2 square matrices. There’s also an advantage in pure al-gebraic self-consistency in doing so, as we’ll see. Even more importantly, we’llsee that writing down the unit vectors as square matrices unveils a profound, butas-yet unheralded link between Vector/Tensor Algebra and Quantum Mechan-ics. This is one of two mathematical missing links between General Relativity andQuantum Mechanics1.

Now to be specific, the Pauli matrices can be viewed as a representation of theunit vectors in three dimensions. For example, if we identify:

i = σ1 =

[0 11 0

], j = σ2 =

[0 −ii 0

], k = σ2 =

[1 00 −1

]then the familiar dot-product of vector algebra is executed as an anticommutator:

~A · ~B =12~A,~B = AxBx + AyBy + AzBz

Because the Pauli Matrices, and hence the unit vectors, obey relations like:

i · j = 12i, j = 1

2σ1, σ2 = 0,

andi · i = 1

2i, i = 1

2σ1, σ1 = 1.

But the algebraic advantage of self-consistency alluded to comes in the executionof the cross product as a commutator:

~A× ~B =12i[~A,~B] = (AyBz − AzBy)i + (AzBx − AzBx) j + (AxBy − Bx Ay)k,

which works because the unit vectors inherit the commutative properties of thePauli Matrices:

i× j =12i[i, j] =

12i[σ1, σ2] = σ3 = k

This is an advantage over the column representation because in the column repre-sentation, there is no way to actually execute the cross-product ‘on the page.’ That

1The other missing link is writing down the correct, generally covariant form of the Lagrangiandensity, which the development of this notation will help us do.


is to say, there is no way to multiply two column (or row) vectors with each otherto get another column vector of the same rank in the same space. You can writedown the symbolic algebraic relations like

i× j = k (1.17)

that the objects should conform to, but in the column (or row) representation, thisdoesn’t have any actual implementation. Meaning, there’s no way to combine or

multiply the object

100

with

010

to get

001

.

In the column representation, the cross-product between the unit-vectors can-not be written down on a two dimensional piece of paper. Now a principle assump-tion or axiom of Mathematical Physics, albeit an implicit one not spoken of fre-quently enough, is that the mathematical form of the laws of physics shall bewritten down on a two-dimensional piece of paper. You might call this the‘Chalkboard Principle.’ I believe that the implementation of Vector Algebra thatwe use in physics should conform to this principle to the utmost. The representa-tion of the unit vectors as columns fails to conform to this principle when it comesto the cross product, as to implement Eq. 1.17, you must ‘leave the page’ in asense. But in a square matrix representation the cross product can be performedon the page.

It is for this reason that the square-matrix representation of the unit vectorsis fundamentally more physical. Not only this, it also automatically and formallyunites Quantum Mechanics and General Relativity through what we can call Quan-tum Tensor Calculus. All of the vector differential equations we are familiar withfrom general relativity and electrodynamics now become quantum relations, be-cause they can all be left and right multiplied by a row or column spinor. Andnow we see the correct role reserved for columns and rows in the algebra – theseare the Dirac-spinor wavefunctions ψ and ψ†.

When we make the move from three to four space-time dimensions, the unitvectors become 4× 4 square matrices, and the four Dirac γ-matrices now fit thebill for the unit vectors.


1.3 The General Dirac Equation in Gibbs/HeavisideNotation

Now in the process of generalizing the Gibbs-Heaviside “Del” operator to 4D space-time, we are motivated to interpret the γ matrices as contravariant unit vectors,so that:

∇ = γµ ∂

∂xµb

. (1.18)

We’ve also dropped the usual arrow symbol above the ∇. (I’m certainly not thefirst person to write down ∇ = γµ∂µ in the context of the Dirac equation. Thissame notation was used by Feynman in one of his seminal papers on QuantumElectrodynamics (QED), [19], but was later supplanted in popularity by the ’slash’notation, /∂ = γµ∂µ. There are several other places in that paper where Feynmanimplies that the γµ are 4D basis vectors, but doesn’t state so explicitly. We’ll havemore to say about this along the way).

We’ll reserve the notation ~∇ to represent to 3D operator:

~∇ = σi ∂

∂xi

where the sum over i is understood to cover only the spatial dimensions, i =1, 2, 3, and the σ matrices are taken to be the 4× 4 diagonal extension of their 2× 2form:

σi4×4 =

[σi

2×2 00 σi

2×2

].

Now we can interpret the free-particle Dirac Equation as the statement that the4D gradient of the wavefunction is directly proportional to itself:

∇ψ =mcih

ψ.

As we showed in Chapter 2, regardless of whether the transformation (back tothe free system) is unitary or γ0-unitary, it is always possible to write the GeneralDirac Equation as:

γν ∂

∂xνb

ψb + γν e−S ∂eS

∂xνb

ψb =mcih

ψb

We’ve just identified the first term on the left side of the equation as the 4D Del.Now it seems natural to choose the “Delta” symbol, ∆, to represent the second


term on the left which corrects the covariant differentiation for a non-flat space-time:

∆ = γν e−S ∂eS

∂xνb

. (1.19)

In the case of the electromagnetic interaction, ∆ = qihc γµ Aµ. We introduce this no-

tation because, (1), ∆ looks an A, (2) because it nicely compliments the∇ operatorin appearance, (3), has the same units of length−1 as the ∇, and (4), in the moregeneral case of non-electromagnetic interactions such as gravity, a correction tothe ∇ differentiation will still be needed, but will not be representable by γµ Aµ,but by a more general 4× 4 Dirac operator field, which ∆ will stand for. So theGeneral Dirac Equation takes on a neat, symmetrical appearance:

(∇+ ∆)ψ =mcih

ψ (1.20)

But it’d be even nicer to find a way of writing it without the parentheses. I feelthat it’s important to take the next step, as most authors do, and define anothersymbol:

~D = ∇+ ∆. (1.21)

This is, essentially, the covariant differential operator for the Dirac space, which otherauthors often symbolize as /D. Thus the General Dirac Equation can be written:

ih ~D ψ = mc ψ. (1.22)

This is a concise and invariant way of writing the fundamental quantum waveequation which is true in any coordinate system. This can also be written,

ih γ0~D ψ = mc γ0ψ. (1.23)

Using this ~D notation makes the statement of the transformation properties ofthe Dirac equation even more concise. In the free-particle frame, it’s true that,

ψ†f γ0∇ f ψ f =

mcih

ψ†f γ0 ψ f .

Now use ψ f = eS ψb, to eliminate ψ f :

ψ†b eS†

γ0∇ f eS ψb =mcih

ψ†b eS†

γ0 eS ψb


Comparing this to Eq. 1.23, (after being pre-multiplied by ψ†b ), allows us to see

that,γ0~D = eS†

γ0∇ f eS, (1.24)

orγ0∇ f = e−S†

γ0~D e−S (1.25)

Furthermore, if the transformation, eS, is a member of the general Lorentz group,meaning, it obeys eS†

γ0 = γ0e−S, or, e−S†γ0 = γ0eS, then Eqs. 1.24 and 1.25

become,~D = e−S∇ f eS, (1.26)

or∇ f = eS ~D e−S. (1.27)

If eS and ~D have these transformation properties, then it’s guaranteed that

ψ f ∇ f ψ f =mcih

ψ f ψ f =mcih

ψbψb = ψb ~D ψb.

In line with this, we can make the statement that if any matrix, A, differentialor not, is covariant, then

ψ†b γ0Ab ψb = ψ†

f γ0A f ψ f ,

orψb Ab ψb = ψ f A f ψ f .

This should be taken as on of the most basic axioms of general relativistic quantummechanics. Now, since ψ f = eS ψb, and ψ†

f = ψ†b eS†

, this means

ψ†b γ0Ab ψb = ψ†

b eS†γ0A f eS ψb.

Thus, commuting the eS’s across the γ0’s gives,

Ab = e−S A f eS, (1.28)

orA f = eS Ab e−S. (1.29)

So we see that any matrix that transforms like Eqs. 1.28 or 1.29, transforms just like~D, Eqs. 1.26 and 1.29. We can therefore form the axiomatically invariant c-numberquantity,

ψ f A f ψ f = ψb Ab ψb. (1.30)


We can also see that products of matrices that transform like Eqs. 1.28 or 1.29 canalso form invariants:

ψb AbBb ψb = ψb e−S A f eSe−S B f eS ψb

= ψ†f e−S†

γ0e−S A f B f eS e−S ψ f = ψ f A f B f ψ f . (1.31)

This also holds true for products formed of any number of covariant matrices,ABC . . . , as well as commutators, [A, B], and anticommutators, A, B.

1.4 Current Conservation

Let’s revisit the conservation of probability currents using this Heaviside notation.The Dirac equation and its adjunct are written:(

γ0∇+ γ0∆)

ψ =mcih

γ0ψ, (1.32)

ψ†((

γ0∇)†

+(

γ0∆)†)= −mc

ihψ†γ0 (1.33)

Here (γ0∇

)†=

(γ0γµ ∂

∂xµ

)†

=∂

∂xµ γ0γµ,

is the left-acting adjunct of the del operator. The adjunct of the other term,

(γ0∆

)†=

(γ0γν e−S ∂eS

∂xν

)†

=∂eS†

∂xνe−S†

γ0γν,

is a little more straightforward to interpret since there are no ’dangling’ differ-ential operators. Now, in 1.33, left multiply the Dirac equation by ψ† and rightmultiply its adjunct by ψ and then add them. We obtain

ψ†(

γ0∇+(

γ0∇)†

+ γ0∆ +(

γ0∆)†)

ψ = 0.

Let’s get this is terms of the component derivatives,

∂

∂xµ ψγµψ + ψ(

∆ + γ0∆†γ0)

ψ = 0.


Now substitute the definition of the Dirac current, jµ = ψγµψ, to get,

∂jµ

∂xµ + ψ(

∆ + γ0∆†γ0)

ψ = 0. (1.34)

This is another version of the general continuity equation, similar to Eq. 1.14.However, here we have not assumed that ∆ is a necessarily a pure four-vector. Ifit is, then γ0∆†γ0 = −∆ and again, the terms in parentheses cancel, and ∂ν jν = 0.

We can drop the ψ’s to get a pure operator equation:

γ0∇+(

γ0∇)†

+ γ0∆ +(

γ0∆)†

= 0,

which can stand on its own without any wavefunctions pre- and post-multiplyingit. Making use of Eq. 1.21, this can also be written:

γ0~D +(

γ0~D)†

= 0 (1.35)

1.5 General Covariance of Commutation Relations

Now, we established with Eq. 1.31 that products of covariant matrices sand-wiched between ψ . . . ψ form invariants. This also must hold for commutatorsand anticommutators involving ∇ or ~D, such as:

ψ f ∇ f , A f ψ f = ψb Db, Abψb.

Now consider A f = ηµν = 12γµ, γν, the metric tensor in the free/flat space.

This is diagonal in the space-time indices, µ, ν, and in the Dirac/spinor indicesis also not only diagonal, but proportional to the number 1. When an operatoris proportional to 1 in the 4 × 4 spinor space, I also sometimes refer to it as anoperator that’s ’neutral’ in the spinor indices. Clearly, in the free/flat space, it’strue that,

ψ f ∇ f , ηµνψ f =2mcih

ψ f ηµν ψ f ,

ψ f[∇ f , ηµν

]ψ f = 0.

We can drop the ψ f . . . ψ f from both sides to get a matrix equation:

∇ f , ηµν = 2mcih

ηµν,[∇ f , ηµν

]= 0. (1.36)


However, before we can transform this to the bound frame, we must pointout that equations such as 1.30 only apply in the given form if the matrices A f orAb have internally summed four-vector indices, such as ∇ = γµ∂µ. We can findout how to form invariants from matrices which have ’unsummed’ four-vectorindices, such as γµ itself, by examining the transformation law of the four-vectorcurrent,

ψ f γµ ψ f =∂xµ

f

∂xνb

ψb γν ψb = ψb∂xµ

f

∂xνb

γν ψb = ψb γµf ,b ψb

This makes γµf ,b =

∂xµf

∂xνbγν–the representation (in the bound system) of one of the

γν vectors from the free system–invariant when sandwiched between ψ . . . ψ.Note also that if Eq. 1.6 is obeyed, which is another fundamental axiom of gen-

eral relativistic quantum mechanics, the transformation coefficients can be statedin terms of eS, e−S and the γ’s:

∂xµf

∂xνb=

12γν, e−S γµ eS,

∂xνb

∂xµf=

12γµ, eS γν e−S. (1.37)

(To derive this, start with Eq. 1.6, and use the fact that δµν = 1

2γµ, γν). Thesetransformation coefficients are, by definition, not only c-numbers, but real num-bers, and as such must be proportional to the unit matrix, 1, in the Dirac space.This will not be the case for any arbitrary eS, so Eqs. 1.37, together with eS†

γ0 =γ0e−S, put a restriction on the set of acceptable transformations and define thegeneral Lorentz group.

So, now we see that to form an invariant from the metric tensor, we must bal-ance the indices,

ψ f ηµν ψ f =∂xµ

f

∂xαb

∂xνf

∂xβb

ψb gαβ ψb

or,

ψb gαβ ψb =∂xα

b

∂xµf

∂xβb

∂xνf

ψ f ηµν ψ f ,


which gives, as a matrix equation,

gαβ =∂xα

b

∂xµf

∂xβb

∂xνf

e−Sηµν eS,

and since ηµν is neutral in the spinor indices, e−S moves across it and cancels theother eS, giving the familiar classical form of the equation,

gαβ =∂xα

b

∂xµf

∂xβb

∂xνfηµν,

and its inverse,

ηµν =∂xµ

f

∂xαb

∂xνf

∂xβb

gαβ. (1.38)

Since the transformation coefficients and ηµν are neutral in the spinor indices, thismeans that gµν must also be neutral in the spinor indices, although, dependingon the transformation, gµν generally will not be equal to ηµν in the space-timeindices, or even necessarily diagonal in µν – an important possibility. The factthat gµν is neutral in the spinor indices is essential, because it means that the entireapparatus of Riemannian tensor calculus, including quantities related to productsof gµν and the derivatives of gµν, like the Christoffel symbols, Γρ

µν, and Ricci tensor,Rµν, are well-defined in the quantum interpretation. If gµν were not neutral in thespinor indices, then the ordering of the products of gµν and its derivatives wouldmatter, and the road to connecting the classical equations of general relativity withquantum theory would be much more difficult.

1.6 The Metric Tensor and the Christoffel Symbols

In classical tensor analysis, the metric tensor is defined in terms of the basis vectorsof the space as

gµν = εµ · εν =12εµ, εν. (1.39)

Note that these εµ cannot be, in general, unit vectors like the eµ b that we dis-cussed previously, because then gµν could only be ηµν. However, we can posit aconnection between εµ and eµ b in terms of a scale factor,

εµ =√±gµµ eµ b, (no summation).


There must also be a corresponding relationship between the ’true length’ differ-entials, dxν

b , and the actual ’coordinate’ differentials, duνb , to go along with this:

dxνb =

√±gµµ duν

b , (no summation).

This construction (or hypothesis/axiom) can account for the diagonal metric ofspherical coordinates, and as we’ll see shortly, the Schwarzschild metric, whichis also diagonal. It also does not conflict with our requirement that the metricbe neutral in the spinor indices, but guarantees it. Although imposing it doeslimit the theory to the consideration of diagonal metrics. However, non-diagonalmetrics can be accounted for by positing a more general tensor of scale factors, hν

µ,such that εµ = hν

µ eν. Metrics constructed in this way will also be guaranteed to beneutral in the spinor indices. The hµ

ν is simply the matrix of coefficients relatingthe ’true length’ bound coordinates to the non-unit bound coordinates:

hµν =

∂xµb

∂uνb

. (1.40)

Now, there are two formula from classical tensor analysis that define the Christof-fel symbols. One relates the Christoffel symbols to the (non-unit) basis vectors andtheir derivatives:

Γµβν = εµ ·

∂εβ

∂uνb

. (1.41)

The other formula for the Christoffel symbols relates it to the metric tensor and itsderivatives:

Γµβν =

12

gµα

(∂gβα

∂uν+

∂gνα

∂uβ−

∂gβν

∂uα

). (1.42)

Using these two formula for the Christoffel symbols together will allow us to de-rive the quantum Einstein equation for the most general case of non-diagonal met-rics. But before we do that, let’s study the Schwarzschild metric, a diagonal metricthat comes from the solution of the classical Einstein equation in vacuum, subjectto certain boundary conditions. Even though the Schwarzschild metric comesfrom the equations of classical physics under limited circumstances, the transfor-mations that lead to its metric will still have general relevance in the quantumtheory.


1.6.1 The Schwarzschild Transformation

To illustrate and develop this method, let’s examine the Schwarzschild metric so-lution of the classical Einstein equation, with the goal of discovering the coeffi-cients of the transformation between the bound coordinates and the free coor-dinates, ∂xµ

b∂xν

f(or more precisely the ∂uµ

b∂xν

f), and the quantum transformation eS. In

spherical coordinates with the (1,−1,−1,−1) sign convention, the Schwarzschildmetric is,

gµν =

(1− 2GM

c2r ) 0 0 00 −(1− 2GM

c2r )−1 0 00 0 −r2 00 0 0 −r2sin2θ

. (1.43)

Now, let’s propose a form for the transformation by expressing the basis vectorsin the bound system, εµ, in terms of the flat-space unit vectors, γµ. The form of themetric tensor shows that the transformation mixes the time and radial coordinatesbut leaves the angular variables unaltered. Therefore, let’s propose,

ε0 = a γ0 + b γr,εr = c γr + d γ0,εθ = γθ,εθ = γφ. (1.44)

The coefficients, a, b, c, d, are assumed to be functions of (r, t) in the bound system.The definition of the metric tensor in terms of the basis vectors, Eq. 1.39, puts

the following constraints on a, b, c, d. From g0,0, we get

g0,0 =12ε0, ε0,

which gives a relation between a and b,

1− 2GMc2r

= a2 − b2.

Also, from gr,r, we get

gr,r =12εr, εr = −(1−

2GMc2r

)−1,


which means for c and d we have,

d2 − c2 = −(1− 2GMc2r

)−1.

The off-diagonal elements of gµν are zero, so from g0r, we get,

g0r =12ε0, εr = 0,

which tells usad− bc = 0.

These are three equations in four unknowns that we can recombine to eliminateb, c, d in favor of a. Up to some arbitrary sign conventions that come from takingsquare roots, the result is,

ε0 = a γ0 +

√a2 − (1− 2GM

c2r) γr,

εr = (1− 2GMc2r

)−1

(a γr +

√a2 − (1− 2GM

c2r) γ0

). (1.45)

The contravariant basis vectors can be found from εµ = gµνεν, and these are,

ε0 = (1− 2GMc2r

)−1

(a γ0 +

√a2 − (1− 2GM

c2r) γr

),

εr = −a γr −√

a2 − (1− 2GMc2r

) γ0 . (1.46)

We still need to do a little more work to discover the form of a and close theloop. We can find additional constraints by considering the Christoffel symbols.Since we already know gµν, we can use the classical definition of the Christoffelsymbols in terms of gµν and its derivatives, Eq. 1.42, to get,

Γ00r =

GMc2r2

(1− 2GMc2r )

.

Now we can compare this to the other expression for the Christoffel symbols interms of the basis vectors, Eq. 1.41. To apply this, we’ll need to calculate the


derivatives of the basis vectors. From Eq. 1.45, one of these is

∂ε0

∂r=

∂a∂r

γ0 +

(a ∂a

∂r −GMc2r2

)√

a2 − (1− 2GMc2r )

γr.

Then since Γ00r can be expressed,

Γ00r =

12ε0,

∂ε0

∂r,

we geta ∂a

∂r −(

a ∂a∂r −

GMc2r2

)(1− 2GM

c2r )=

GMc2r2

(1− 2GMc2r )

,

which is consistent for any form of a, so we must keep looking. Because Γ00r is

required to be symmetric in the lower two indices, we can also write it as

Γ00r = Γ0

r0 =12ε0,

∂εr

∂x0.

From Eq. 1.45, we have

∂εr

∂x0 = (1− 2GMc2r

)−1

∂a∂x0 γr +

a ∂a∂x0√

a2 − (1− 2GMc2r )

γ0

.

This gives us a differential equation that puts the final constraint on a,

∂a∂x0 =

GMc2r2

√a2 − (1− 2GM

c2r). (1.47)

We can solve this by proposing, as an ansatz,

a(r, t) =

√1− 2GM

c2rf (r, t),

where f (r, t) is now another undetermined function. Substituting this into Eq.1.47, we have,

∂ f∂x0 =

GMc2r2

√f 2 − 1.


This is a simple differential equation that is solved by letting f = cosh αx0. Then,since ∂ f

∂x0 = α sinh αx0, and√

f 2 − 1 = sinh αx0, we see that this solution will beconsistent if α = GM

c2r2 . We have then, for the final form of a,

a =

√1− 2GM

c2rcosh

(GMc2r2 x0

).

Thus we have completely specified the expression for the covariant basis vec-tors, εµ, in terms of the γµ:

ε0 =

√1− 2GM

c2r

(cosh

(GMc2r2 x0

)γ0 + sinh

(GMc2r2 x0

)γr

)εr =

1√1− 2GM

c2r

(sinh

(GMc2r2 x0

)γ0 + cosh

(GMc2r2 x0

)γr

). (1.48)

Likewise, the contravariant basis vectors are:

ε0 =1√

1− 2GMc2r

(cosh

(GMc2r2 x0

)γ0 + sinh

(GMc2r2 x0

)γr

)

εr = −√

1− 2GMc2r

(sinh

(GMc2r2 x0

)γ0 + cosh

(GMc2r2 x0

)γr

). (1.49)

Notice that the basis vectors turn out to depend on position, r, as well as time,x0, even though the Schwarzschild metric itself is independent of time. Because

εµ =∂uµ

b∂xν

fγν, we have found the coefficients of the Schwarzschild transformation.

We see that they are indeed c-numbers, as required, and that the stipulation thatgµν and its derivatives end up neutral in the spinor indices is satisfied.

To complete the quantum formulation of the Schwarzschild problem, we mustnow find eS. This will allow us to write down the general Dirac equation for theproblem. We will then solve that to find the stable-state eigenfunctions and energyeigenvalues of the Schwarzschild problem. We will do this by proposing a form ofthe generator, S, and see if it’s consistent with the Schwarzchild transformation,Eqs. 1.48 and 1.49. Let’s try

S = ωε,

where ε = γ0γr, and ω is a unit-less boost parameter. With this we have,

eωεγ0e−ωε = (cosh2 ω + sinh2 ω)γ0 + 2 cosh ω sinh ω γr, (1.50)


andeωεγre−ωε = (cosh2 ω + sinh2 ω)γr + 2 cosh ω sinh ω γ0, (1.51)

while the angular basis vectors are unaltered,

eωεγθe−ωε = γθ, eωεγφe−ωε = γφ.

Now, since ε0 =√

g00eωεγ0e−ωε, we can equate Eq. 1.50 with the first of Eqs. 1.48,to get: √

1− 2GMc2r

(cosh

(GMc2r2 x0

)γ0 + sinh

(GMc2r2 x0

)γr

)=

√1− 2GM

c2r

((cosh2 ω + sinh2 ω)γ0 + 2 cosh ω sinh ω γr

).

The scale factor√

g00 cancels, and this can be solved for the boost parameter, ω,by equating the coefficients of γ0, ( or γr) on each side to get,

sinh ω =

√12(cosh αx0 ± 1).

(We would also obtain the same result by setting up the equation for the radialbasis vectors, εr =

√−grr eωεγre−ωε). Since Eq. 1.48 implies the basis vectorsinitially line up with the free-space vectors at time x0 = 0, then we must select theminus sign to be consistent. Then we have,

sinh ω =

√12(cosh αx0 − 1), (1.52)

or, solving for ω,

ω(r, t) = sinh−1

√12(cosh

GMcr2 t− 1).

We see that the boost parameter, ω, depends on r, t, and thus the full transforma-tion, eωε, depends on all four variables, r, t, θ, φ.

1.6.2 The Schwarzschild-Dirac Problem

Now, from Eq. 1.12 the general Dirac equation can be written in terms of the unitvectors as

eνb

∂

∂xνb

ψb + eνb e−S ∂eS

∂xνb

ψb =mcih

ψb,


or, using the scale coefficients, Eq. 1.40, this can be written in terms of the non-unitcoordinates and the non-unit basis vectors as

ενb

∂

∂uνb

ψb + ενb e−S ∂eS

∂uνb

ψb =mcih

ψb. (1.53)

As we saw, the basis vectors in the bound system, εµb , expressed in terms of the

basis vectors of the free system, as in Eq. 1.49, depend upon time and position.However, since the metric itself does not depend on time, the basis vectors in thebound system expressed in the bound system itself will be constant in time, butmust still reflect the static metric thru Eq. 1.39. Therefore, we may write,

ε0b =

1√

g0,0γ0, εr

b =1√−gr,r

γr, εθb = γθ, ε

φb = γφ,

where γr, γθ, and γφ are the usual spherical coordinate basis vectors. For the timeand radial coordinates, the contravariant basis vectors are,

ε0b =

1√1− 2GM

c2r

γ0, εrb = −

√1− 2GM

c2rγr.

We see that at the distance of the Schwarzschild radius, r = 2GMc2 , there is a divi-

sion by zero, so at those points the basis is undefined. But, the general Dirac equa-tion can still be solved in all space through a piecemeal approach and a matchingof solutions with appropriate boundary conditions at the Schwarzschild radius.That is to say, there is a wave solution outside the so-called ’event horizon,’ as wellas a wave solution inside the event horizon. We expect the solutions in these tworegions to have different limiting properties.

Now, to begin the process of writing down the general Dirac equation, Eq.1.53, let’s first examine the ’field’ terms for the Schwarzschild generator S = ωε:

εν e−S ∂eS

∂uν= εν e−ωε ∂eωε

∂uν.

Here we’ve dropped the b subscript, since it’ll be understood that were workingin the bound basis for now. This is a sum of four components. For the time com-ponent we initially have,

ε0 e−ωε ∂eωε

c∂t=

1√

g0,0γ0 ∂ω

c∂te−ωεeωε =

1√

g0,0γ0 ∂ω

c∂t.


Then, from Eq. 1.52, together with the identities of the hyperbolic trigonometricfunctions, we calculate,

∂ω

∂t=

12

GMcr2 .

So the time component becomes,

ε0 e−ωε ∂eωε

c∂t= γ0

12

GMc2r2√

1− 2GMc2r

.

Then for the radial component we initially have,

εr e−ωε ∂eωε

∂r=

1√−gr,rγr ∂ω

∂r.

And again, from Eq. 1.52, we get,

∂ω

∂r= −GM

cr3 t.

So the radial component ends up being

εr e−ωε ∂eωε

∂r= −

√1− 2GM

c2rGMcr3 t γr.

Turning to the angular components, since ω does not depend on the angularvariables, we may treat it as a constant, although we must remember the matrixε does depend on the angles. This part of the calculation is very similar to that ofthe Hydrogen atom, and is dealt with in the Appendix, under S = κ

2 ε. Using thoseformula, and replacing κ

2 with ω, we initially have for the angular components,

γθ e−ωε ∂eωε

∂θ+ γφ e−ωε ∂eωε

∂φ=

sinh 2ω

rγ0 +

2sinh2 ω

rγr.

Then, again using Eq. 1.52 along with identities of the hyperbolic trigonometricfunctions, this becomes,

γθ e−ωε ∂eωε

∂θ+ γφ e−ωε ∂eωε

∂φ=

sinh GMcr2 t

rγ0 +

cosh GMcr2 t− 1r

γr.


Now we can bring this all together to write down the general Dirac equationfor the Schwarzschild problem:

1√1− 2GM

c2r

γ0 ∂

c∂t+

√1− 2GM

c2rγr ∂

∂r+ γθ ∂

∂θ+ γφ ∂

∂θ+ γ0

12

GMc2r2√

1− 2GMc2r

−√

1− 2GMc2r

GMcr3 t γr +

sinh GMcr2 t

rγ0 +

cosh GMcr2 t− 1r

γr ψb =mcih

ψb (1.54)

This seems unwieldy, but we can show that it results in the correct Newtonian

limit. First, let’s multiply thru by√

1− 2GMc2r , to get

γ0 ∂

c∂t+ (1− 2GM

c2r)γr ∂

∂r+

√1− 2GM

c2r(γθ ∂

∂θ+ γφ ∂

∂θ) + γ0 1

2GMc2r2

− (1− 2GMc2r

)GMcr3 t γr +

√1− 2GM

c2rsinh GM

cr2 tr

γ0 +

√1− 2GM

c2rcosh GM

cr2 t− 1r

γr ψb

=mcih

√1− 2GM

c2rψb (1.55)

In the interest of finding the Hamiltonian, let’s multiply through by ihcγ0. Let’salso use γθ ∂

∂θ + γφ ∂∂θ = − 1

h ε σLr , as well as ε = γ0γr. Then we have,

ih∂

∂t+ ihc(1− 2GM

c2r)ε

∂

∂r− icε

√1− 2GM

c2rσLr

+ ih12

GMcr2

− ih(1− 2GMc2r

)GMr3 t ε + ihc

√1− 2GM

c2rsinh GM

cr2 tr

+ ihc

√1− 2GM

c2rcosh GM

cr2 t− 1r

ε ψb

= γ0 mc2

√1− 2GM

c2rψb (1.56)

Now we can first take the limit of large c. In this case, mc2√

1− 2GMc2r → mc2 +

GMmr . The remaining terms are(

ih∂

∂t+ ihcε

∂

∂r− icε

σLr− ih

12

GMr3 tε

)ψb = γ0 (mc2 +

GMmr

)ψb (1.57)

Finally, taking the limit of large r, the leading order terms that are left are:(ih

∂

∂t+ ihcε

∂

∂r− icε

σLr

)ψb = γ0 (mc2 +

GMmr

)ψb,


which we can rearrange to give:(ih

∂

∂t− γ0 GMm

r+ ihcε

∂

∂r− icε

σLr

)ψb = γ0 mc2ψb (1.58)

Other than the leading order potential term, γ0 GMmr , this is the free-particle Dirac

equation. So we see, that in the Newtonian limit, c → ∞, and the limit of largeradial distance from the origin, the Hamiltonian is modified as,

H → Hfree + γ0 GMmr

. (1.59)

which is what we’d expect, except for one surprise: the γ0 multiplying what isotherwise the usual gravitational potential energy. Notice that this is consistentwith the classical law, H → Hfree − GMm

r , so long as the lower two components ofthe 4-spinor, the ’matter’ solution components, dominate in the limit r → ∞.

Discussion

Note to Editor: There will be a wrap-up discussion here. The material in thischapter connects up with that in Chapter 6 , and before the final version, both ofthese chapters will be re-worked or re-combined.

CHAPTER 2

The Free Particle

Now, in the previous chapter we the considered the most general formulations inthe interest of showing the full potential of the method and tore directly through tothe Schwarzschild topic. Now let’s take a step back and examine the more elemen-tary case of the free-particle Dirac equation. We’ll look specifically at the solutionsand the solution method of the free-particle Dirac equation. This is important tocover because many aspects of this solution method also apply to bound stateproblems, like the that of the hydrogen atom and gravitation. It’s also importantto go over because the free-particle solutions, which are not immediately normal-izable in the usual way, play a key role in the calculation of Feynman diagramsand the theory of quantum electrodynamics.

2.1 Solving the Dirac Equation as a Differential Equa-tion

The free-particle Dirac equation in Cartesian coordinates is

i h γµ∂µ = m c ψ.

31

CHAPTER 2. THE FREE PARTICLE 32

Through a transformation of coordinates that we demonstrate explicitly in theAppendix, this can be written in spherical coordinates as(

i h ∂0 − γ0 m c)

ψ = ε

(−i h

∂

∂r+ i

σiLi

r

)ψ (2.1)

In writing Eq. 2.1, I try to make use of as much of the traditional notation firstestablished by Dirac, [7], as possible.

Let’s discuss some of the operators and terms in 2.1. σ is the 4× 4 diagonalextension of usual 2× 2 Pauli matrices:

σi4×4 =

[σi

2×2 00 σi

2×2

].

(When the 2× 2 submatrices on the diagonal are identical, and the off-diagonalelements are null, we will sometimes freely alternate between the 2× 2 and 4× 4version, depending on context). ε, is a 4× 4, ’off-diagonal’ matrix:

ε =

[0 λ

λ 0

].

Recall that we made use of the ε matrix in the last chapter in the context of theSchwarzschild problem. The 2× 2 λ sub-matrices that make it up are

λ =

[cos θ sin θ e−i φ

sin θ ei φ − cos θ

].

I call this λ for now, because many authors have used that notation, but λ can alsobe thought of as r, the 2× 2 unit vector that points in the direction of (θ, φ). It’seasy to see that λ = r = sinθ cosφ σ1 + sinθ sinφ σ2 + cosθ σ3.

The operator σiLi, which I often just write in shorthand as σL, is equivalent to:

σiLi =

L3 L− 0 0L+ −L3 0 00 0 L3 L−0 0 L+ −L3

, (2.2)

which we also show in the Appendix. Other authors frequently write this as~σ ·~L. Idon’t like to write it as a dot product because that is a slight abuse of notation fromthe point of view of my method. In my interpretation of the γµ as basis vectors,


the 2× 2 σi are the 3D basis vectors, so it would be more appropriate if we wrote~L = σ1L1 + σ2L2 + σ3L3. But I just call it σL for short, and because of their 3DCartesian nature, there’s usually no difference between σi and σi – although therecan be in the case of the full σµν, which these σi are a subset of.

We will now construct an ansatz to solve (2.1), seeking solutions which areeigentates of H, J2, Jz. First, consider the following 2-spinor objects which can beconstructed from spherical harmonics. The first is called θ+:

θ+j,mj=

[ √j + mj Yj−1/2,mj−1/2√j−mj Yj−1/2,mj+1/2

](2.3)

And a second object that we call θ−:

θ−j,mj=

[ √j + 1−mj Yj+1/2,mj−1/2

−√

j + 1 + mj Yj+1/2,mj+1/2

](2.4)

We will see that these objects have total angular momentum j = 1/2, 3/2.... Theindex mj is half-integral and runs from mj = j, ..., 1/2,−1/2, ...− j. (Notice thatthe indices of the spherical harmonics are integral as required). (For a reference, seeBjorken and Drell, [4], pg. 53).

These 2-spinors obey the relations

σiLi θ−j = − h(j + 3/2)θ−j

andσiLi θ+j = h(j− 1/2)θ+j

Notice how the θ+,− behave slightly differently when acted upon by σiLi, al-though they are each separately eigenvectors of it – this can be verified explicityby using the representation of σiLi shown in Eq. (2.2), which we demonstrate inAppendix B.

They both have the same total J2:

J2 θ+,− = (L2 + 3h2/4 + h σiLi) θ+,− = h2 j(j + 1) θ+,−

They also both have the same value of Jz = h mj. Additionally they obey

λ θ−j =√

j+1j θ+j λ θ+j =

√j

j+1 θ−j (2.5)


That is, they are conjugate with respect to the λ operator. These last two identitiesfollow from the spherical harmonic recursion relations and are demonstrated inthe Appendix.

We find that we can construct two 4-spinor solution classes to the Dirac Equa-tion (2.1) from these:

ψej = e−iEet/h

[F−j (r) θ−jG+

j (r) θ+j

](2.6)

and

ψpj = e−iEpt/h

[F+

j (r) θ+jG−j (r) θ−j

](2.7)

where the F+,−j (r) and G+,−

j (r) are c-number coefficients which are functions ofradius only.


Let’s investigate the first solution ψej and substitute (2.6) into the Dirac equa-

tion (2.1). We get:[(Ee/c−m c)F−j (r) θ−j(Ee/c + m c)G+

j (r) θ+j

]=

[0 λ

λ 0

] [(−i h∂/∂r− i h(j + 3/2)/r) F−j (r) θ−j(−i h∂/∂r + i h(j− 1/2)/r) G+

j (r) θ+j

]which becomes[

(Ee/c−m c)F−j (r) θ−j(Ee/c + m c)G+

j (r) θ+j

]=

(−i h∂/∂r + i h(j− 1/2)/r) G+j (r)

√j

j+1 θ−j

(−i h∂/∂r− i h(j + 3/2)/r) F−j (r)√

j+1j θ+j

We can rearrange the terms and separate the matrix equation into two relations

between the coefficients of the θ+,−j :

(∂/∂r− (j− 1/2)/r) G+ =(Ee/c−m c)

√j+1

j−i h F−

(∂/∂r + (j + 3/2)/r) F− =(Ee/c+m c)

√j

j+1−i h G+

(2.8)

Now, the spherical Bessel (or Hankel) functions obey the coupled recursionrelations:

(∂/∂r− (n− 1)/r) jn−1 = −κ jn(∂/∂r + (n + 1)/r) jn = κ jn−1

(2.9)

If we make the index substitution n = j + 1/2, then these recursion relationsbecome

(∂/∂r− (j− 1/2)/r) jj−1/2 = −κ jj+1/2(∂/∂r + (j + 3/2)/r) jn+1/2 = κ jj−1/2

(2.10)

So that we can solve the system given by Eq. (2.8) by identifying

F−j = jj+1/2(κr) , G+j = ei δ jj−1/2(κr) (2.11)

(allowing for an arbitrary phase between F and G). This means that the wavenum-ber κ must obey two relations:

− κ = e−iδ(Ee/c−m c)

√j+1

j

−i h(2.12)

κ = eiδ(Ee/c + m c)

√j

j+1

−i h(2.13)


which can be equated to each other to eliminate κ, giving the following relationsatisfied by the energy Ee:

Ee = m c2 j + 1 + jei 2 δ

j + 1− jei 2 δ(2.14)

For the solution ψej to be a steady state solution, the energy Ee must be real. This

restricts δ to either δ = π or δ = π/2. In the first case δ = π we get:

Ee = 2 (j + 1/2)m c2 (2.15)

with corresponding κ from (2.12) or (2.13):

κj = −2m c

h

√j(j + 1) (2.16)

In the case δ = π/2 we get solutions with an inverse relationship betweenenergy and angular momentum. Putting δ = π/2, equations 2.14 and 2.13 give:

Ee =mc2

2j + 1, κ = −2mc

h

√j(j + 1)2j + 1

These are essentially subharmonics of the free electron. You could also call these theplanetary solutions of the free-particle Dirac equation, because like planets in oursolar system, they have generally lower frequencies and large angular momen-tum. Although the frequency (energy) tends to zero as j → ∞, the wavenumberapproaches the limit |κ| → mc

h . We will consider the physical significance of thesesolutions in later chapters.

So by solving the Dirac equation in spherical coordinates, and requiring thesolutions to be eigenstates of H, J2 and Jz, we find that the energy is quantized.By using the solutions ψ

pj , we find a similar quantization of energies, except now

with negative energy values. We can summarize the solutions for both ψej and ψ

pj

in one expression:E = ±2(j + 1/2)mc2 (2.17)

where the index j = 1/2, 3/2, 5/2... with a degeneracy of 2 j + 1 for each index j.Now we must address the discrepency between our solutions, and the tradi-

tional approach which results in the continuum of energy solutions. The begin-ning of the discrepancy comes in the choice of the definition of the θ+,− 2-spinors,


as in Eqs. 2.3, 2.4. The traditional approach, as taken in Bjorken & Drell, [8], Mes-siah, [9], or just about any other textbook on advanced quantum mechanics is todefine θ+,− like this:

θ−j,mj=

1√2j + 2

[ √j−mj + 1 Yj+1/2,mj−1/2

−√

j + mj + 1 Yj+1/2,mj+1/2

],

θ+j,mj=

1√2j

[ √j + mj Yj−1/2,mj−1/2√j−mj Yj−1/2,mj+1/2

](2.18)

These additional prefactors of 1√2j+2

and 1√2j

, result in 2-spinor objects which

are seperately prenormalized with respect to the angular variables:∫θ†θ dΩ = 1

In addition, instead of Eq. 2.5, the objects now obey a different conjugate relation-ship with respect to the λ operator:

λ θ+ = θ−, λ θ− = θ+

This in effect washes away the the j dependant factors from the problem. Further-more, instead of choosing a unitary phase factor eiδ between F and G, as in Eq.2.11, the traditional approach is to allow for a factor of arbitrary magnitude, call itv. So instead of Eqs. 2.12 and 2.13, we have:

κ =1v(E/c−m c)

i h(2.19)

κ = v(E/c + m c)−i h

(2.20)

The j dependance is gone, and the factor v, related to the relative normalizationbetween the upper two and lower two components introduces a third variable intoa set of two equations with three unkowns. The procedure continues by choosingκ to be an arbitrary parameter which can take on any continuous value from 0 to±∞ , and then solving for E and v in terms of κ. E is determined from:

E(κ) = ±√(hκ)2c2 + m2c4


A continuous function of κ which has a minimum magnitude of mc2. (This relationbetween E and κ is also satisfied by the discrete solutions, except that κ is fixed atdiscrete values).

So then, which approach is correct? One might proclaim that the traditionalapproach is to be favored because it is more in line with experimental fact. Weknow that free electrons do have continuous energies, as even the simplest cath-ode ray experiments seem to demonstrate. Furthermore, the existence of a spec-trum of heavy electrons with masses of 2m, 4m, etc. has presumably not been ob-served. But as we will see later, these heavy electron states will still have the samecharge-to-mass ratio of e/m, so they would be indistinguishable from regular elec-trons in experiments using electric and magnetic fields. Furthermore, note thatthis solution approach which results in discrete energies has been undertaken inthe rest frame of the electron. The inclusion of Lorentz boosts will still allow for acontinuum of possible momenta and total energy.

To resolve the dilemma, we must determine if the j-dependent factors whichresult from the conjugate relations between the θ+,− as in Eq. 2.5 are inherentlynecessary and natural, or if a normalization which eliminates them is correct. Toend the ambiguity, we will turn to a method of solving the Dirac equation otherthan solving it as a differential equation.

2.2 Finding the Spectrum of the Free-Particle DiracEquation using Operator Methods

Consider the operator ~r = rε. (Strictly speaking, according to our method, theradial position vector would be~r = γrr, and since ε = γ0γr = −γ0γr, it would bemore appropriate to say γ0~r = −rε. But the rε operator itself is what we’re reallyinterested in for now, so let that all be understood). Formally, the total derivativeof this with respect to time, t, is:

~r = rε + rε. (2.21)

Now we can use the Heisenberg equation of motion to calculate r and ε:

r =1ih

[H, r] , ε =1ih

[H, ε] (2.22)


From Eq. (2.1) the Hamiltonian can be written:

H = γ0 mc2 + c ε

(−i h

∂

∂r+ i

σiLi

r

)Then, noting, that that r is a diagonal scalar, we can easily calculate r as

r = 1ih [H, r] = 1

ih

[γ0 mc2 + c ε

(−i h ∂

∂r + i σi Lir

), r]

= −cε[

∂∂r , r

]= −cε

(2.23)

Now, to calculate ε, we keep in mind that ε is a full matrix operator, but a functionof the angular variables only:

ε = 1ih [H, ε] = 1

ih

[γ0 mc2 + c ε

(−i h ∂

∂r + i σi Lir

), ε]

=

:0

1ih

[−cεih ∂

∂r , ε]

+ 1ih

[iεc σi Li

r , ε]

+mc2

ih[γ0, ε

]= c

h

[ε σi Li

r , ε]

+mc2

ih[γ0, ε

](2.24)

We can now substitute the results of (2.23) and (2.24) into (2.21). Using γ0ε =−εγ0, ε2 = 1 and γ0ε− εγ0 = 2γ0ε, we obtain after some manipulation:

~r = −c− ch

γ0[ε, ε σiLi

]+

2mc2

ihrε

Rearranging this, we get an expression for the operator we call ~q:

~q = ~r + c =2mc2

ihrε− c

hγ0[ε, ε σiLi

](2.25)

We will see that this operator is equivalent in many ways to the classical fourvelocity. The first term is the zitterbewegung, which is zero when expectation valuesare taken. We will be more interested in the γ0 component in the second term,which we can call q0:

q0 =ch

(σiLi − ε σiLi ε

)(2.26)

Now let’s investigate the action of the q0 operator on the spinor solutions ofthe Dirac equation. First, consider ψe

j , (suppressing time dependence of ψ):

q0 ψej =

ch

(σiLi − ε σiLi ε

) [ F−θ−

G+θ+

](2.27)


Now, note the presence of the the operator ε =

[0 λ

λ 0

]. It is the action of λ on

the θ−,+ which is the source of the ambiguity in the solutions of Part 1. That is:

λ θ− = s θ+, λ θ+ =1s

θ− (2.28)

In the traditional approach, the factor s = 1. Earlier, we chose to let s =√

j+1j .

For now, let us leave s unspecified, and carry thru the calculation in (2.27):

q0 ψej = c

h

([−h(j + 3/2)F−θ−

h(j− 1/2)G+θ+

]− ε σiLi

[1/s G+θ−

s F−θ+

])= c

h

([−h(j + 3/2)F−θ−

h(j− 1/2)G+θ+

]− ε

[−(1/s) h(j + 3/2)G+θ−

s h(j− 1/2)F−θ+

])

= ch

([−h(j + 3/2)F−θ−

h(j− 1/2)G+θ+

]−[

h(j− 1/2)F−θ−

−h(j + 3/2)G+θ+

])

= ch

[−2h(j + 1/2) 0

0 2h(j + 1/2)

] [F− θ−

G+ θ+

]

= −2(j + 1/2)c γ0 ψej

(2.29)

Thus, if a suitable volume normalization of the free-particle states can be de-fined so that

∫ψ†ψ dV = 1, the expectation value of the four-velocity of an elec-

tron is ∫ψ†

j ~q ψj dV

=∫

ψ†j (

2mc2

ihrε− γ0q0)ψj dV

= 2(j + 1/2)c∫

ψ†j ψj dV = 2(j + 1/2)c,

Notice how the result is independent of the factor s. The double application ofthe operator ε in (2.27) results in factors of 1

s × s causing the s dependence todisappear. Since the j factors appear here in a manner independent of the valueof s, we can say that the presence of the j factors is natural and inherent.


Now let’s calculate the action of q0 on ψpj :

q ψpj = c

h(σiLi − ε σiLi ε

) [ F+θ+

G−θ−

]= c

h

([h(j− 1/2)F+θ+

−h(j + 3/2)G−θ−

]− ε σiLi

[s G−θ+

(1/s) F+θ−

])

= ch

([h(j− 1/2)F+θ+

−h(j + 3/2)G−θ−

]− ε

[s h(j− 1/2)G−θ+

−(1/s) h(j + 3/2)F+θ−

])

= ch

([h(j− 1/2)F+θ+

−h(j + 3/2)G−θ−

]−[−h(j + 3/2)F+θ+

h(j− 1/2)G−θ−

])

= ch

[2h(j + 1/2) 0

0 −2h(j + 1/2)

] [F+ θ+

G− θ−

]

= 2(j + 1/2)c γ0 ψpj

(2.30)

Again, the s factors cancel out, and now we have a result which is proportional toj + 1/2.

In general then, we have for the expectation value of ~q = ~r + c:

< ~x >= ± 2(j + 1/2)c (2.31)

That is, for the electron states ψej the positive sign pertains, while for the positron

states, the sign is negative. It seems correct then, to interpret the classical energyas E = mc < ~q >,

Ee,pj = ± 2(j + 1/2)mc2

This could also be interpreted as a spectrum of particle states with rest massesgiven by Mj = 2(j + 1/2)me

We must remember that this Ee,pj is only an expectation value – it’s not an

operator. But, the expectation value of the energy must equal the energy eigen-value of the Hamiltonian, E, if we’re considering stable states. Thus the oper-ator m~q could be interpreted as the four-velocity-based four-momentum oper-ator, which is a counterpart to the differential-based four-momentum operator,ihγµ∂µ. The zeroth-component, mcq0 is the counterpart to the Hamiltonian oper-ator H = ih∂/∂t. Classically, their expectation values must be equal, and so the


appearance of the 2(j + 1/2) factors as in Eq. 2.17 is natural and inherent. Thus,writing the θ+,− 2-spinors as they appear in Eqs. 2.3 and 2.4 is correct, and pre-normalizing them over the angular variables, as in the traditional Eq. 2.18, leadsto an inconsistency between the expectation value of energy and the eigenvalueof the Hamiltonian. This is the way things are in the free-particle case, but in thecase of bound-state problems like the Hydrogen atom, we will see that s = ±1 isthe only consistent choice.

Let’s see how this plays out further. Consider that the classical current is de-fined as

jµ = ρ xµ

We can then see that there is a correspondence with this classical definition if wepremultiply Eq. (2.31) with the electron charge e and ψ:

ρ x0 = e ψe,p q0ψe,p = ∓e 2(j + 1/2) ψe,p γ0 ψe,p

If there is a suitable normalization of ψ, then the term on the far right, whichwe know to be the Dirac current j0 = ψ γ0 ψ, underscores this correspondence,and allows us to discern that the charges of these states are quantized as Q =∓e 2(j + 1/2). Note then, that since the charge of the state, Q, is quantized in thesame way as the rest-mass, the charge-to-mass ratio is constant:

Qm

=∓2(j + 1/2) e2(j + 1/2)m0

= ∓e/m0

Which is consistent with the experimental observation that (most) electrons orpositrons have the same charge-to-mass ratio, regardless of whether they are ofthe ’heavy’ variety.

Thus the relationEe,p = ±2(j + 1/2)m0c2

stands, and the ambiguity mentioned in Part 1 is resolved. We can interpret thisas the prediction of a spectrum of heavy electrons and positrons, with inertialmasses of m = 2(j + 1/2)m0. Perhaps some of the heavy particles that have beendiscovered that have charges which are integral multiples of the electron chargebut have different charge-to-mass ratios than the electron arise from solutions ofthe general Dirac equation that break certain symmetries and result in differentcharge-to-mass ratios. There is much more work that could be done to investigatethat possibility.


Solution Table

Ee = 2(j + 1/2)mc2, Ep = −2(j + 1/2)mc2, κj = − 2mch

√j(j + 1)

ψej =

[jj+1/2(κjr)θ−jijj−1/2(κjr)θ+j

]e−i Eet

h θ−j =

[ √j−mj + 1 Yj+1/2,mj−1/2

−√

j + mj + 1 Yj+1/2,mj+1/2

]

ψpj =

[jj−1/2(κjr)θ+jijj+1/2(κjr)θ−j

]e−i

Epth θ+j =

[ √j + mj Yj−1/2,mj−1/2√j−mj Yj−1/2,mj+1/2

]

Table 2.1: A table summarizing all the solutions of the free particle Dirac equation.For each j value, there is a degeneracy of 2 j+1 .

2.3 The Dirac angular momentum operator, J

Let’s discuss another aspect of the operator ~q. We found above that the operator

~q =2mc2

ihrε− c

hγ0 [ε, εσL] ,

could be interpreted as a four-velocity. Let’s recast this in a new form using someidentities. Using the fact that ε2 = 1, the terms in the commutator can be replacedas:

[ε, εσL] = σL− εσLε.

We show in the Appendices that

σL, ε = −2hε,

which can be rearranged as

εσLε = −2h− σL,

which means that the commutator above is equivalent to

[ε, εσL] = 2 (σL + h) .

Now, Dirac defines the an operator he calls the ’total angular momentum’, [7],J, as:

JDirac = γ0 (σL + h) . (2.32)


This operator commutes with the free-particle Dirac Hamiltonian, and so must bea constant of the motion:

[H, JDirac] = 0.

The solutions of the free-particle Dirac equation listed above are eigenstates of thisoperator:

JDiracψej = h(j + 1/2)ψe

j , JDiracψpj = −h(j + 1/2)ψp

j .

From this we see that the eigenvalues of J are positive for the particle solutions, ψej ,

and negative for the antiparticle solutions, ψpj . Usually total angular momentum

is considered to be a positive magnitude, h√

j(j + 1), and in the non-relativisticSchrodinger approach, the solutions are taken to be eigenstates of J2, such thatJ2ψ = h2 j(j + 1). This is a key point where the Dirac J operator is different, andthe J2 operator of the non-relativistic theory is not exactly the same thing as thesquare of the J Dirac operator:

J2ψej = h2(j + 1/2)2ψe

j = h2(j2 + j + 1/4)ψej =

(h2 j(j + 1) + h2/4

)ψe

j

The two differ by h2/4, a constant term which will be the same for any angu-lar momentum state and is proportional to the identity matrix. We see then thatthe square of the Dirac and Schrodinger total angular momentum operators arerelated as:

J2Dirac = J2

Schrodinger +h2

41

Because the difference between these squared operators is proportional to theidentity matrix, an eigenvector of J2

Dirac will also be an eigenvector of J2Schrodinger,

and their eigenvalues always differ by h2/4. Thus, requiring the solutions of theDirac equation to be eigenvectors of the un-squared operator, JDirac, means thatthey will also be eigenvectors of the squared operators, J2

Dirac and the more famil-iar J2

Schrodinger. However, because the Dirac operator is ’un-squared’ it contains alittle more information in the fact that its eigenvalues can be positive or negative.

Now that we understand the Dirac operator, J, we can see that the four-velocityoperator ~q can be written:

~q =2mc2

ihrε− 2c

hJ.


We’ll be interested in the commutation relations that this operator forms withthe ∇ operator. In spherical coordinates, the ∇ operator is written:

∇ = γ0 ∂

c ∂t+ γ0ε

(∂

∂r− 1

hσLr

),

which we can also write as:

∇ = γ0 ∂

c ∂t+ γ0ε

(∂

∂r− 1

hγ0 J − h

r

).

We can now prove an interesting identity:

12ih∇,~q = γ0 mc2.

To do so, calculate:

∇,~q = ∇,2mc2

ihrε− 2c

hJ = 2mc2

ih∇, rε − 2c

h∇, J

Now, assuming these anti-commutators are acting on eigenstates of J, and∇, suchthat Jψ = ∓h(j + 1/2)ψ and ∇ψ = mc

ih ψ, it’s easy to see that

−2ch∇, J = −2c

h(∇J + J∇)

= −2ch

(mcih· ∓h(j + 1/2) +∓h(j + 1/2) · mc

ih

)= −2mc2

ih· ∓(j + 1/2).

For the other term,

2mc2

ih∇, rε = 2mc2

ih

(γ0 +

1h

2J)=

2mc2

ih

(γ0 +∓(j + 1/2)

).

So the sum of the two becomes:

2mc2

ih∇, rε− 2c

h∇, J = 2mc2

ih

(γ0 +∓(j + 1/2)

)− 2mc2

ih·∓(j+ 1/2) =

2mc2

ihγ0.

Thus we have proved the identity:

12ih∇,~q = γ0mc2. (2.33)


We see that this is the quantum version of the familiar formula from classical spe-cial relativity:

pµdxµ

dτ= mc2,

and it underscores the idea that ~q can be interpreted as the quantum four-velocityoperator.

2.4 Angular momentum, Action and Charge

Let’s continue to study the operator J = γ0(σiLi + h

)Earlier in this chapter, we

found that the positive-energy solutions (interpreted as electrons), which we’ll

now call ψ+, are written as ψ+ =

[gθ+

i f θ−

]. These obey

Jψ+ = +(j + 1/2)hψ+,

and have positive energy eigenvalues.

Now consider the other solution, ψ− =

[gθ−

i f θ+

]. These obey

Jψ− = −(j + 1/2)hψ−

Notice that they have negative eigenvalues of J, (although they can be constructedwith positive energy eigenvalues). Labelling this operator as ’J’ is a conventionthat goes back to Dirac, who interpreted it as the ’total angular momentum’ op-erator. But, as we mentioned before, total angular momentum is usually under-stood to have only positive values. Furthermore, total angular momentum statesare also usually understood to be eigenvalues of the operator J2 such that J2ψ =j(j + 1)h2ψ. But with J defined as above, we find that J2ψ = (j(j + 1) + 1/4) h2 ψ.However, it does commute with the Hamiltonian, [J, H] = 0, so its eigenstates aresimultaneous with those of H. Because of its ability to be positive or negative,instead of calling this operator the total angular momentum, it might be more ap-propriate to call it the total action, or even the free action. It has units of action,and in the semi-classical old quantum theory, action is quantized in units of nh,where n can be a positive or negative integer. Requiring the expectation-valueaction-integral of J around a closed angular path to be quantized,∫

ψ† J · d~θ ψ = nh.


serves to define those classical paths. There is much more that can be done toexplore exactly how the quantized action theorems of the old quantum theory arisefrom the Dirac wave-theory.

Lastly, note that if a total state consists of a superposition of two states of op-posite total action, J, ψtotal = ψ+

j + ψ−j , then the expectation value of J will bezero: ∫

ψ†total J ψtotal dV = 0,

since ψ+j and ψ−j are orthogonal. But stable states of matter are known to have

zero overall electric charge. Therefore total action and total electric charge areessentially the same thing. So it might be useful to interpret the electric chargeoperator, Q, as

Q =eh

J.

Discussion

Stable states of matter tend to have an expectation value of zero for total actionor total charge, although stable charged states are also possible. These states canbe constructed from a linear superposition of the different orthogonal solutions ofthe free-particle Dirac equation, shown in Table 2.1. In the absence of any inter-action potentials, the orthogonality of these solutions ensures a zero probabilityof transitions occurring. The presence of potential fields will create a couplingbetween states, allowing transitions to occur in certain circumstances. Dependingon the details of the interaction field that binds the particles, the orthogonalityof the spherical harmonics will still leave many transitions forbidden. Hence itis possible to ’pile’ many of these solutions on top of each other, about the sameorigin of coordinates.

Note to Editor: More discussion and tidying up of this chapter is planned.

CHAPTER 3

The Hydrogen Atom

Our goal in this chapter will be to derive the energy spectrum of the Hydrogenatom based upon the methods outlined in the previous chapters. We will showhow the accepted spectrum of the Hydrogen atom can be derived from a transfor-mation of the general Dirac equation.

When the Dirac equation is solved in practice, it is usually solved in a sphericalcoordinate system. It may be that this is an additional fundamental physical prin-ciple: whenever possible, nature tends to choose a spherical coordinate system forthe description of stable states which are eigenstates of both the Hamiltonian, H,and the total angular momentum operator, ~J. In spherical coordinates, the covari-ant ∇ operator is written:

∇ = γ0 1c

∂

∂t+ γr ∂

∂r+ γθ ∂

∂θ+ γφ ∂

∂φ

This can be recast in terms of the orbital angular momentum operators, Li, (whichamount to derivatives w.r.t. the angles) and the Pauli matrices, σi as:

∇ = γ0 1c

∂

∂t+ γr ∂

∂r− γr 1

hσiLi

r

48

CHAPTER 3. THE HYDROGEN ATOM 49

3.1 The Transformation e−i α ′2 ε

We must first find a transformation that gives us the Dirac equation for the hydro-gen atom. Let’s try:

S = −iα ′

2ε

With this, S† = −S, so this choice of S generates a transformation which is unitary.(Note thought, that this transformation is not a part of the general Lorentz grounp,since eS†

γ0 6= γ0e−S). Then we also have an identity:

eS = e−i α ′2 ε = cos

α ′

2− i ε sin

α ′

2

and

e−S = ei α ′2 ε = cos

α ′

2+ i ε sin

α ′

2The field generated by this transformation will be calculated from:

γ0∆ = γ0γµe−S ∂eS

∂xµ =

*0

e−S 1c

∂eS

∂t+

*0

γ0γre−S ∂eS

∂r+ γ0γθe−S ∂eS

∂θ+ γ0γφe−S ∂eS

∂φ

The first two terms on the right are zero because eS does not depend on r or t sinceε is a function of only of (θ, φ). On carrying thru and making use of the cyclicr, θ, φ identities from the Appendix, we find:

γ0∆ = −isinα ′

r− 2

sin2 α ′2

rε (3.1)

Or, multiplying through by γ0 to get ∆ itself:

∆ = −iγ0 sinα ′

r− 2

sin2 α ′2

rγ0ε (3.2)

From this it’s clear that ∆ + ∆† = 0, since this is a unitary transformation. We cancall it the unitary Coulomb field, even though it differs from the usual Coulomb’sLaw in the term proportional to sin2 α ′

2 . Interestingly, we’ll find that the final ex-pression for the energy eigenvalues is identical to the traditional result, despitethis additional term which ends up dropping out.


Making more use of the ε operator, the differential operator γ0∇ can be writ-ten:

γ0∇ =∂

c ∂t+ ε

(∂

∂r− 1

hσLr

)so that the full Dirac equation, in operator form, becomes:

∂

c ∂t+ ε

(∂

∂r− 1

hσLr

)− i

sinα ′

r− 2

sin2 α ′2

rε =

mcih

γ0

Let’s multiply through by ihc, to get

ih∂

∂t+ ihc ε

∂

∂r− ic ε

σLr

+ hcsinα ′

r− 2ihc ε

sin2 α ′2

r= mc2 γ0

Now remembering that the solutions will be eigenstates of the Hamiltonian, H =ih ∂

∂t , as well as the total angular momentum, J = γ0 (σL + h), we finally have forthe Dirac hydrogen Hamiltonian:

H = −ihc ε∂

∂r+ ic ε

γ0 J − hr

− hcsinα ′

r+ 2ihc ε

sin2 α ′2

r+ mc2 γ0 (3.3)

3.1.1 Solving for the Energy Eigenvalues

Now I should mention that I am following a method of solution very similar tothat described in Bethe and Salpeter’s, The Quantum Mechanics of One and Two Elec-tron Atoms, [2]. Another resource on the derivation of the Dirac hydrogen atomthat I highly recommend is available on the web at quantummechanics.ucsd.edu,[3]. The only difference between the derivation presented here, and the traditionalmethod described in these two references is the presence of the one additionalterm in the Hamiltonian:

2ihc εsin2 α ′

2r

,

and the merely formal replacement:

sinα ′

r→ α

r.

Continuing with the derivation, let’s propose that the solution can be writtenas:

ψ = e−iEt/h[

gθ+

i f θ−

]

http://quantummechanics.ucsd.edu/ph130a/130_notes/node501.html


In this case, J ψ = (j + 1/2)h ψ.Now, as we saw in Ch. 2, it’s always possible to find a representation of the θ

2-spinors such that

λθ+ = sθ−

λθ− =1s

θ+

which preserves the unitary nature of the ε matrix which is built from the λ’s.We’ll use the representation where s = −1, and this is the representation usedin the traditional derivations such as [3]. This means that the operation of the ε

matrix is equivalent to

εψ =

[0 λ

λ 0

] [gθ+

i f θ−

]=

[−i f θ+

−gθ−

].

Using this, and remembering that Jψ = (j + 1/2)hψ, Eq. 3.3 unfolds into thefollowing real, coupled differential equations, in the one variable, r:(

Ehc− mc2

hc+

sinα ′

r

)g =

(− ∂

∂r− j + 3/2

r+

2sin2 α ′2

r

)f

(Ehc

+mc2

hc+

sinα ′

r

)f =

(∂

∂r− j− 1/2

r−

2sin2 α ′2

r

)g

Continuing to follow the standard solution method, let f = 1r F and g = 1

r G.We then have, after some minor rearrangement:(

∂

∂r+

j + 1/2r

−2sin2 α ′

2r

)F =

(mc2 − E

hc− sinα ′

r

)G(

∂

∂r− j + 1/2

r−

2sin2 α ′2

r

)G =

(mc2 + E

hc+

sinα ′

r

)F

Now let k1 = mc2+Ehc , k2 = mc2−E

hc , and make the variable substitution ρ =√k1k2r. The coupled equations become:(

∂

∂ρ+

j + 1/2ρ

−2sin2 α ′

2ρ

)F−

(√k2

k1− sinα ′

ρ

)G = 0(

∂

∂ρ− j + 1/2

ρ−

2sin2 α ′2

ρ

)G−

(√k1

k2+

sinα ′

ρ

)F = 0


The next step is to propose a more specific form for F and G in terms of expo-nentially damped power series:

F = e−ρ∞

∑m=0

am ρs+m, G = e−ρ∞

∑m=0

bm ρs+m, (3.4)

where s is a non-integral exponent.After substituting these into the coupled differential equations, we find the

following equations for the coefficients of the m-th integral power of each sum:

−am +

(s + m + 1 + (j + 1/2)− 2sin2 α ′

2

)am+1 −

√k2

k1bm + sinα ′bm+1 = 0

−bm +

(s + m + 1− (j + 1/2)− 2sin2 α ′

2

)bm+1 −

√k1

k2am − sinα ′am+1 = 0

(3.5)

Now, the series terminates on the left at m = 0, so a−1 and b−1 must be zero.Letting m = −1 in Eq. 3.5, this implies(

s + (j + 1/2)− 2sin2 α ′

2

)a0 + sinα ′b0 = 0 (3.6)

−sinα ′a0 +

(s− (j + 1/2)− 2sin2 α ′

2

)b0 = 0 (3.7)

Linear algebra requires that the determinant of these coefficients vanish. This putsa restriction on the non-integral power, s:

s = ±√(j + 1/2)2 − sin2α ′ + 2sin2 α ′

2(3.8)

In the traditional solution method, the + sign solution is chosen:

s =√(j + 1/2)2 − sin2α ′ + 2sin2 α ′

2. (3.9)

This forces the wave-function to be normalizable and finite at the origin.Let’s combine the recursion formulae, 3.5, into a single equation by multiply-

ing the first equation in the pair by√

k1k2 and the second by k2, and then subtract-ing the second from the first. The am and bm terms drop out, and we find that the


terms bm+1 and am+1 are proportional to each other:

bm+1 =

√k2k1

sinα ′ +(

s + m + 1 + (j + 1/2)− 2 sin2 α ′2

)√

k2k1

(s + m + 1− (j + 1/2)− 2 sin2 α ′

2

)− sinα ′

am+1 (3.10)

This shows that if the series in am and bm terminate, they must do so at the samepower, call it m = nr. Then in equations 3.5, anr+1 = bnr+1 = 0, and 3.5 becomes:

−anr −

√k2

k1bnr = 0

−bnr −

√k1

k2anr = 0

These are both the same condition on the coefficients of the terminal power:

bnr = −

√k1

k2anr (3.11)

Now let’s use 3.11 in equation 3.10, considering m + 1 = nr, this becomes:

−

√k1

k2anr =

√k2k1

sinα ′ +(

s + nr + (j + 1/2)− 2 sin2 α ′2

)√

k2k1

(s + nr − (j + 1/2)− 2 sin2 α ′

2

)− sinα ′

anr

Now the anr ’s drop out, and we obtain a condition relating k1 and k2 to the integersnr and j + 1/2:√

k1

k2sinα ′ −

(s + nr − (j + 1/2)− 2 sin2 α ′

2

)

=

√k2

k1sinα ′ +

(s + nr + (j + 1/2)− 2 sin2 α ′

2

)Let’s transform this again by multiplying thru by

√k1k2. After rearranging this

becomes:

2(

s + nr − 2 sin2 α ′

2

)√k1k2 = (k1 − k2) sinα ′


Recalling the definitions of k1 and k2, we find that√

k1k2 =√

m2c4−E2

hc and k1− k2 =2Ehc , so this becomes:

2(

s + nr − 2 sin2 α ′

2

) √m2c4 − E2

hc=

2Ehc

sinα ′

Cancelling factors of 2 and hc and then squaring both sides:(s + nr − 2 sin2 α ′

2

)2 (m2c4 − E2

)= E2sin2α ′

Now use our choice of s =√(j + 1/2)2 − sin2α ′ + 2sin2 α ′

2 :(√(j + 1/2)2 − sin2α ′ +

2 sin2 α ′

2+ nr −

2 sin2 α ′

2

)2 (m2c4 − E2

)= E2sin2α ′

Note how the second order terms have cancelled – this will lead to a energy eigen-value equation that’s identical in form to the traditional result, despite the slightlydifferent Hamiltonian that we started with. This can now be rearranged andsolved for the energy eigenvalues:

E =mc2√

1 + sin2α ′

(nr+√

(j+1/2)2−sin2α ′)2

(3.12)

Note that if we identify α = sinα ′, we have the accepted solution for the energyspectrum of the H-atom. Thus, we’ve shown that the fine structure constant isa parameter of the transformation that relates the wave-functions of the boundsystem to that of the free system in accord with the general principle of relativity.

Discussion

Note to Editor: There’s another transformation, generated by S = −i α2 ε, the same

transformation we discussed in relation to the Schwarzschild problem, that resultsin a spectrum of energy eigenvalues very similar to that of the hydrogen atom, andcan be solved by a method almost identical to what we used above. That will beadded to a re-worked version of this chapter.

CHAPTER 4

Quantum Field Theory

From the point of view of the quantum field theory (QFT) approach based uponLagrangian methods, it has been shown that certain theories of quantum gravityare not renormalizable, [18]. To understand what this means, we need to take atour through quantum field theory.

4.1 Quantum Electrodynamics

First, let’s look at where QFT succeeds most prominently: the description of elec-tromagnetic interactions between charged particles – a theory is known as quan-tum electrodynamics, or QED.

The most popular formulation of QED centers around Lagrangian methodsand the idea of second quantization. The QED Lagrangian density can be written:

LQED = ψ(

ihc~D−mc2)

ψ +1

16πFµνFµν (4.1)

Here I’ve used my notation for the covariant derivative, ~D, which is essentiallythe same as /D, and we’re working in Gaussian units, with all the factors of h andc included explicitly. In this case,

~D = ∇+ ∆ = γµ∂µ +q

ihcγµ Aµ.

55

CHAPTER 4. QUANTUM FIELD THEORY 56

Remember also that the Maxwell stress-energy tensor Fµν is expressed in terms ofthe four-vector potential, Aµ, as:

Fµν = ∂µ Aν − ∂ν Aµ, (4.2)

so that there are really only two independent fields in 4.1, Aµ and ψ.The dynamical field equations, which in the case of QED are the Dirac equation

and Maxwell’s equations, can be derived from 4.1 by applying the canonical Euler-Lagrange prescription. For a Lagrangian density, L, which is a function of a fieldwith multiple components φk and the space-time derivatives of that field, ∂αφk,that prescription is:

∂α∂L

∂(∂αφk)=

∂L∂φk

. (4.3)

It’s important to note that the field indices in 4.3, k, can be either the four spinorindices of the Dirac field, ψ, or the four-vector indices on Aµ. When we get to thecase of gravitational Lagrangians, we’ll even see how the indices k will stand forthe sixteen tensor indices of the field gµν.

For now, let’s first see how Maxwell’s equations are derived from 4.1 by writ-ing the Euler-Lagrange equation as:

∂α∂LQED

∂(∂α Aρ)=

∂LQED

∂Aρ. (4.4)

For the right-hand side of 4.4 , we have

∂LQED

∂Aρ= qψγρψ.

To calculate the left-hand side, we find first, that:

∂L∂(∂α Aρ)

=1

4πηαβηρσFβσ,

so the left-hand side of 4.4, is,

∂α∂L

∂(∂α Aρ)=

14π

ηαβηρσ∂αFβσ.

Equating the left and right sides of 4.4, then gives:

14π

ηαβηρσ∂αFβσ = qψγρψ.


To make contact with the classical form of Maxwell’s equations, let’s raise andlower a few indices and get rid of the η’s, giving:

14π

∂βFβα = qψγαψ.

If we identify Jα = qcψγαψ, then this becomes,

∂βFβα =4π

cJα,

which is the classical inhomogeneous Maxwell equation. The conservation of thecurrent, ∂α Jα = 0, follows directly from this because of the antisymmetric defini-tion of Fµν in 4.2. The homogeneous Maxwell equation

∂αFβγ + ∂βFγα + ∂γFαβ = 0,

is also automatically satisfied because of the definition of the antisymmetric Fµν,and does not come from the application of the Euler-Lagrange prescription. Sothere we have Maxwell’s equations.

The Dirac Equation can also be derived by applying 4.3 to 4.1, but this timeby taking partial derivatives with respect to the Dirac field spinor components,ψk, and the space-time derivatives of those components, ∂αψk. We must point outthat in applying the Euler-Lagrange prescription, ψ and ψ are formally treated asindependent fields. Let’s deal with variations in ψ, treating ψ as a constant. Theright-side of the Euler-Lagrange formula gives:

∂L∂ψk

=∂

∂ψk∑ij

(ihc ψi γ

µij ∂µψj + q ψi γ

µij Aµ ψj −mc2 δij ψiψj

)= ∑

iq ψi γ

µik Aµ −mc2 ψk

To get the left side, first calculate:

∂L∂(∂αψk)

=∂

∂(∂αψk)∑ij

(ihc ψi γ

µij ∂µψj + q ψi γ

µij Aµ ψj −mc2 δij ψiψj

)= ∑

iihc ψiγ

αik.


Then for the left-side of the Euler-Lagrange formula, we have:

∂α∂L

∂(∂αψk)= ∑

iihc ∂αψi γα

ik

Equating the left and right sides gives:

∑i

ihc ∂αψi γαik = ∑

iq ψi γ

µik Aµ −mc2 ψk

or,∑

i

(ihc ∂αψi γα

ik − q ψi γµik Aµ

)= −mc2 ψk.

We can dispense now with the explicit spinor indices, and write this as:

ihc ∂αψ γα − q ψ γα Aα = −mc2 ψ,

Which we recognize as the adjoint of the Dirac equation. Remembering that ψ =ψ† γ0, we can take a transpose conjugate, and then cancel the γ0’s, to get:

ihc γα∂α ψ + q γα Aα ψ = mc2 ψ,

which is the Dirac equation. (Note that we could have also obtained the Diracequation by considering variations in ψ and its derivatives, while treating ψ as theconstant – this would have given us the Dirac equation directly, without havingto take a transpose conjugate in the end).

So we see how both Maxwell’s equations and the Dirac equation follow fromthe definition of the QED Lagrangian density in Eq. 4.1, and the formal applica-tion of the canonical Euler-Lagrange prescription, Eq. 4.3. But other than justify-ing Eq. 4.1, which in a way is just an ansatz that leads to the equations of motionthat we already knew, this formal procedure doesn’t teach us anything new on itsown.

But now that we have the form of the QED Lagrangian density, transition am-plitudes for various electromagnetic processes can be calculated from it. In termsof the Lagrangian density, L, an amplitude, iM, can be defined as,

− iM = eih∫L d4x = e

ih∫(ψ(ihc~D−mc2)ψ+ 1

16π FµνFµν) d4x (4.5)

Note also that in Eq. 4.5, the Dirac field ψ (or ψ) is interpreted as an particle cre-ation (or destruction) operator acting on a Fock space vacuum ket, ala ψ|0 > –


this is what is known as the second quantization in conventional QFT. The electro-magnetic field γµ Aµ is also related to the creation or destruction of photons as asecond quantized operator.

When we expand 4.5 to the the first few orders of ex, (and neglecting the termsinvolving Fµν...), we get:

−iM =∫

d3x ψ†outψin +

ih

∫ψ(

ihc~D−mc2)

ψ d4x

+12!

(ih

∫ψ(

ihc~D−mc2)

ψ d4x)·(

ih

∫ψ(

ihc~D−mc2)

ψ d4x′)+ . . .

(4.6)

If we time-order the expansion, (see [23], pg. 85), we have:

−iM =∫

d3x ψ†outψin +

ih

∫ t

t0

dt∫

xψ(

ihc~D(t)−mc2)

ψ d3x

+

(ih

∫ t

t0

dt1

∫x

ψ(

ihc~D(t1)−mc2)

ψ d3x)·(

ih

∫ t1

t0

dt2

∫x′

ψ(

ihc~D(t2)−mc2)

ψ d3x′)

+ . . . (4.7)

This amplitude, or matrix element, can be shown to give rise to the Feynmanrules of QED, and its calculations of scattering cross-sections that agree with ex-perimental results to a high degree of accuracy. I find this experimental agreementto be as convincing an argument for the validity of the QED results as the fact thatthis Lagrangian and the formal Euler-Lagrange procedure gives the correct equa-tions of motion. (We should point out that Feynman himself originally discoveredand derived his formulation of QED, valid to any order of approximation, basedon a Green’s function method, [19], [20], not the Lagrangian procedure – but he didprove the equivalence of the two methods.)

4.2 The Feynman Path Integral

As part of the formalism of conventional QFT, and specifically, in QED, the Diracfields ψ and ψ, and the electromagnetic fields Aµ are reinterpreted as operators ina larger, more abstract Fock-space. This is what is referred to as the second quanti-zation. There are details involving Fourier decomposition in terms of positive andnegative frequencies, but ψ is essentially a fermion particle-creation operator, and


ψ is a fermion anti-particle-creation operator. Depending on their ordering in eachterm in the series, Eq. 4.6, ψ and ψ can also act as annihalation operators, mean-ing that they can destroy or ’uncreate’ particles or anti-particles. The fields Aµ

are operators that can create or destroy photons when acting on Fock-space kets,although the photon is its own anti-particle. For example, to create a state fromthe vacuum containing one electron,

ψ|0 >= |1electron >,

or to create one electron and one photon from the vacuum,

Aµψ|0 >= |1electron, 1photon >,

And so on. There are many other details involving the correct time-ordering ofthe operators and all kinds of possibilities involving the interchange of identicalparticles that must be also be accounted for to correctly apply this procedure. Wewill address these details when we calculate some specific scattering amplitudeexamples in a bit, but this is the general idea.

Now, the amplitude for transition to a ’final’ Fock-space ket from an earlier,’initial’ ket is given by a version of the Feynman path integral:

< Ψ f |Ψi >=∫DψDψDAµ e

ih∫L d4x =

∫DψDψDAµ e

ih∫(ψ(ihc~D−mc2)ψ+ 1

16π FµνFµν) d4x

(4.8)Here, the notation

∫DψDψDAµ signifies a sum over all possible configurations

of the fields ψ, ψ, and Aµ. This is an integration on top of the 4D volume inte-gration also occurring in the exponential. A rigorous mathematical interpretationof the meaning of

∫DψDψDAµ is challenging, but in practice this integration is

parametrized in terms of a less-esoteric integration in the Fourier-transform k-space over all possible momenta and frequencies of the waves ψ, ψ and Aµ.

Now the QED Lagrangian density, LQED, can be separated into a sum of freefield terms and an interaction term:

LQED = L0 + Lint

whereL0 = ψ f

(ihγµ∂µ −mc2

)ψi +

116π

FµνFµν

andLint = qψ f γνψi Aν (4.9)


Thus, the the exponentials in 4.8 can be factored into a product of terms involvingthe free fields and the interaction terms,

eih∫L d4x = e

ih∫L0 d4x e

ih∫Lint d4x.

With this in mind, the Fock-space transition amplitude, Eq. 4.8, can be normalizedwith respect to the amplitude for no interaction:

< Ψ f |Ψi >=

∫DψDψDAµ e

ih∫L d4x∫

DψDψDAµ eih∫L0 d4x

(4.10)

Now, in the interaction Lagrangian factor, 4.9, the electromagnetic potentialfield, Aν, can be expressed in terms of a current, jν, which is the source of thatfield, and an electromagnetic Green’s function as,

Aν(x) =4π

c

∫G(x, x′) jν(x′) d4x′.

Since Aν is a solution of Maxwell’s equations, we must have ∂2Aν = 4πc jν, (in the

Lorentz gauge). This means G(x, x′) must satisfy the delta-function condition

∂2G(x, x′) = δ4(x− x′). (4.11)

Since the delta-function can be represented in a Fourier expansion as

δ4(x− x′) =1

(2π)4

∫e−iqµ(xµ−x′µ) d4q,

it’s easy to see that a Green’s function that satisfies 4.11, can be represented as

G(x, x′) =1

(2π)4

∫ e−iqµ(xµ−x′µ)

qνqνd4q. (4.12)

Thus, Aν(x) can be written as

Aν(x) =4π

c

∫ 1(2π)4


qνqνd4q jν(x′) d4x′ (4.13)

Now here comes an important intuitive leap that (I’ve heard) goes back toHeisenberg, yet is not often highlighted enough in modern textbooks on QFT.


The electromagnetic field, Aν, that effects the behavior of the particle which is thesubject of consideration (and represented by the Dirac field, ψ,) is itself producedby a current jν that comes from the other particle of the interaction. If we let φ

represent the Dirac field of this other particle, then the current jν from this otherparticle should be represented as

jν(x′) = q′c φ f γν φi. (4.14)

Notice that this current involves an initial and final spinor, not one and the samespinor, as a Dirac current is more often written. When we look at the interactionLagrangian term, this creates an important balance between ψ and φ, as both havefinal and initial states in the overall picture, and both should have equal footingas subjects of consideration. Thus, when we substitute Eqs. 4.14 and 4.13 into theexpression for the interaction Lagrangian, Eq. 4.9, we have,

Lint(x) = qψ f (x)γνψi(x)4π

c

∫ 1(2π)4


qνqνd4q q′c φ f (x′) γν φi(x′) d4x′.

In the final expression for the scattering amplitude, Eq. 4.6, the variable x is alsointegrated over all space-time. Thus, the first-order scattering amplitude for goingfrom an initial state represented by two spinors, ψi, φi, to a final state representedby ψ f , φ f , is,

iM = qq′ (4π)1

(2π)4

∫x

∫x′

∫q

ψ f (x)γνψi(x)e−iqµ(xµ−x′µ)

qνqνφ f (x′)γνφi(x′) d4q d4x′ d4x.

(4.15)Now, here’s another major assumption: Suppose that all of the initial and final

spinors are free-particle plane-waves, normalized to a delta-function,

ψi(x) = ψ(1)e−ik1µxµ,

φi(x′) = φ(2)e−ik2µx′µ ,

ψ f (x) = ψ(3)e−ik3µxµ,

φ f (x′) = φ(4)e−ik4µx′µ .

Here we use the notation, ψ(1), for example, to represent a constant four-componentspinor ’coefficient’, representing a state carrying spin, s1, as the k-dependance hasbeen factored out into the term e−ik1µxµ

. Thus, the terms in Eq. 4.15, will become,

ψ f (x)γνψi(x) = ψ(3)γνψ(1)e−i(k1µ−k3µ)xµ,


andφ f (x′)γνφi(x′) = φ(4)γνφ(2)e−i(k2µ−k4µ)x′µ .

Making these substitutions into Eq. 4.15 allows us to integrate out the space-timevariables x and x′:

iM = qq′ (4π)1

(2π)4

∫x

∫x′

∫q

ψ(3)γνψ(1)e−i(k1µ−k3µ)xµ e−iqµ(xµ−x′µ)

qνqν

× φ(4)γνφ(2)e−i(k2µ−k4µ)x′µ d4q d4x′ d4x.

Regrouping terms in the exponentials around x and x′, this becomes,

iM = qq′ (4π)1

(2π)4

∫q

ψ(3)γνψ(1)1

qνqνφ(4)γνφ(2)

×∫

e−i(k1µ−k3µ−qµ)xµd4x

∫e−i(k2µ−k4µ−qµ)x′µ d4x′ d4q.

Inserting the definition of the delta-function, this becomes

iM = qq′ (4π)1

(2π)4

∫q

ψ(3)γνψ(1)1

qνqνφ(4)γνφ(2)

× (2π)4δ4(k1µ − k3µ − qµ) (2π)4δ4(k2µ − k4µ + qµ) d4q.

Finally, carrying out the integral over q-space, we can eliminate the first of thedelta-functions, implying qµ = k1µ − k3µ, in the other delta-function, giving,

iM = (2π)4δ4(k1µ + k2µ− k3µ− k4µ) ψ(3)γνψ(1)4πqq′

(k3µ − k1µ)(k3µ − k1

µ)φ(4)γνφ(2).

Now, we can see that even this first-order scattering amplitude is divergent, (sinceit’s proportional to an overall delta-function). It turns out that the second-orderterm is divergent to a second power of infinity, and so on for third and higher-order terms. Before we deal with higher-order terms, we point out that it is stan-dard practice to ’cancel’ the overall delta-function (and its (2π)4 coefficient) inthis first-order term. To get this even more into the standard form, also insert agµν term to raise the index on the second gamma matrix, and rearrange slightly toget,

M = 4πqq′ ψ(3)γνψ(1)−igµν

(k3µ − k1µ)2 φ(4)γνφ(2) (4.16)


We’ve also employed the shorthand, (k3µ − k1µ)(k3µ − k1

µ) = (k3µ − k1µ)2.

Let’s look at the specific case of electron-electron scattering, represented bythe two first-order Feynman diagrams in Fig. 4.1. In this case the charges q =q′ = −e. Furthermore, since both are electrons, which are identical fermions, ananti-symmetrization of the amplitude represented in Eq. 4.16 must be performedto be in accord with the exclusion principle1. Because the two electrons are iden-tical, the sequence of events leading up to the emergence of one electron havingmomentum (and spin) k3, s3 and the other having momentum and spin k4, s4 areindistinguishable. Thus to get the total amplitude, the amplitude for the secondprocess must be subtracted, (or added, with a − sign by the exclusion principle),from the amplitude for the first:

M = 4πe2 ψ(3)γνψ(1)−igµν

(k3µ − k1µ)2 φ(4)γνφ(2)

− 4πe2 φ(4)γνψ(1)−igµν

(k4µ − k1µ)2 ψ(3)γνφ(2) (4.17)

!k1

!k2

!k3

!k4

!k1

!k2

!k3

!k4

Figure 4.1: The two Feynman diagrams for electron-electron scattering.

1By the way, this is a hint that there is a principle of superposition at work. We will deal withthis more in the next chapter.


4.3 Feynman’s Green Function Method

For comparison, let’s to review the formulation of Feynman’s Green function/scatteringseries method of QED as first derived by Feynman himself in his seminal papersof 1949, [19], [20]. I like this formulation of QED because it does not require theideas of a ’Fock space’ or second quantization – it doesn’t contradict those ideas,it just sidesteps them. Bjorken and Drell’s ’Relativistic Quantum Mechanics,’ [4],presents a good distillation of Feynman’s original argument, and here I closelyfollow the development they present in Chapter 6 of their text, while filling insome details that they may have omitted for the sake of being concise. Also, it’simportant to note that renormalization can be applied to this method, showingthat the Fock space and second quantization methods are also not essential forbuilding a renormalized theory.

The Feynman Green’s function(s) are defined by the integral equations:

θ(t′ − t)ψ+(x′, t′) = i∫

d3x G+(x′, t′, x, t)ψ+(x, t)

θ(t− t′)ψ−(x′, t′) = i∫

d3x G−(x′, t′, x, t)ψ−(x, t) (4.18)

This is essentially a statement of Huygen’s principle. For positive frequency wavesthat travel forwards in time, ψ+, the wave at any point, x′, at a later time t′ > t,is the superposition of wavelets arriving from distant points, x, emitted at earliertimes t. The Green’s function, G+(x′, t′, x, t), which we can abbreviate as G+(x′, x),describes how those wavelets propagate. Likewise, G− propagates negative fre-quency waves that travel backwards through time, to the present moment t′ < t,from wavelets at future times, t. Note also that in 4.18, the Green’s functions canin general be either interaction or non-interaction, free-particle Green’s functions.

First, let’s examine the forward-time Green’s function, G+(x′, x). The first ofEq’s 4.18, implies that

(ih∂

∂t′− H′0)θ(t

′ − t))ψ+(x′) = i∫

d3x(ih∂

∂t′− H′0)G

+0 (x′, x)ψ+(x).

Where H′0 = γ0mc2 − ihγ0γi ∂∂x′i is the free-particle Hamiltonian. Thus, if we re-

quire,

(ih∂

∂t′− H′0)G

+0 (x′, x) = hδ(t′ − t)δ3(x′ − x), (4.19)


Then the first of Eq.’s 4.18, implies,

ihδ(t′ − t)ψ+(x′) = i∫

d3xδ(t′ − t)hδ3(x′ − x)ψ+(x) = ihδ(t′ − t)ψ+(x′),

which is consistent if G+(x′, x) satisfies Eq. 4.19, which we can rewrite more ex-plicitly as,

(ih∂

∂t′+ ihγ0γi ∂

∂x′i− γ0mc2)G+(x′, x) = hδ(t′ − t)δ3(x′ − x).

Divide through by hc and multiply both sides by γ0, to get

(i∇′ − mch)G+ = γ0δ4(x′ − x).

Now define SF(x′, x)γ0 = G+(x′, x). This must then obey,

(i∇′ − mch)SF(x′, x) = δ4(x′ − x). (4.20)

This is a nice equation because the space and time variables appear in a balanced,covariant way. If we then analyze the negative-frequency Green’s function in thesecond of Eqs. 4.18 in a similar way, we find that

G−(x′, x) = −SF(x′, x)γ0,

and SF(x′, x) again ends up needing to satisfy Eq. 4.20 to be consistent. Thus,finding a solution to Eq. 4.20 will give us both G+ and G−, so Eq. 4.20 is taken tobe the defining property of the Feynman Green’s functions.

Now, to get an explicit expression for SF(x′, x), it’s useful to turn to Fourieranalysis. Let

SF(x′, x) = SF(x′ − x) =∫ d4k

(2π)4 e−ikµ(x′µ−xµ)SF(k)

Using this in Eq. 4.20, we have, on the left-hand side,

(i∇′ − mch)SF(x′ − x) =

∫ d4k(2π)4 e−ikµ(x′µ−xµ)(γµkµ −

mch)SF(k).

Thus, if

SF(k) =1

γµkµ − mch

,


then 4.20 will be satisfied. As long as kµkµ 6= m2c2

h2 , the inverse of the operatorγµkµ − mc

h , is well-defined, and we can write,

SF(k) =γµkµ +

mch

kµkµ − m2c2

h2

.

Now, the Dirac equation, (ih∇+

qc~A)

ψ = mcψ

can be rearranged as(ihc∂0 − H0)ψ = −qγ0 ~Aψ, (4.21)

where H0 = γ0mc2 − ihcγ0γi∂i is the free Hamiltonian. Assume that the electro-magnetic potential field, ~A, only acts for a short time between t1−∆t1 and t1. Thusthe total wave at the moment t1 is then ψ(x1, t1) = φ(x1, t1) + ∆ψ(x1, t1). Here,φ is a solution of the free-particle Dirac equation, and ∆ψ is a small modificationdue to the brief presence of the field. Thus, 4.21, becomes,

(ihc∂0 − H0) (φ + ∆ψ) = −qγ0 ~A(φ + ∆ψ),

which becomes,

*0(

ih∂

∂t− H0

)φ +

(ih

∂

∂t− H0

)∆ψ = −qγ0 ~Aφ− qγ0 ~A∆ψ.

The first term on the left cancels to zero because φ is a free-particle solution. Whatremains can be integrated from t1 − ∆t1 to t1:∫ t1

t1−∆t1

(ih

∂

∂t− H0

)∆ψ dt =

∫ t1

t1−∆t1

(−qγ0 ~Aφ

)dt +

∫ t1

t1−∆t1

(−qγ0 ~A∆ψ

)dt

which becomes,

ih∆ψ(x1, t1)−

:0ih∆ψ(x1, t1 − ∆t1)−

:0H0∆ψ∆t1 = −qγ0 ~Aφ∆t1 −

:0

qγ0 ~A∆ψ∆t1.

The second term on the left was cancelled because the at the moment t1− ∆t1, theinteraction was still zero, so there would be zero alteration to the wavefunction


at that time and ∆ψ(x, t1 − ∆t1) must be zero. The other terms were cancelledbecause they are all second-order in the small quantities. Thus, to first order,

∆ψ(x, t1) = −qih

γ0 ~A φ ∆t1 (4.22)

Now consider the total wave at a later time, t′ > t1 when the interaction is nolonger taking place. It would simply be written,

ψ(x′, t′) = φ(x′, t′) + ∆ψ(x′, t′).

If we want to know the value of ∆ψ(x′, t′) at a time t′ > t1, we can propagate itforwards from ψ(x1, t1) using the free-particle Green’s function, G0, since there isno interaction between t1 and t′:

∆ψ(x′, t′) = i∫

d3x1 G0(x′, t′, x1, t1)∆ψ(x1, t1).

Using Eq. 4.22, this becomes,

∆ψ(x′, t′) = i∫

d3x1 G0(x′, t′, x1, t1) (−qih) γ0 ~A φ(x1, t1)∆t1

But since φ is a free-particle wave, it can be propagated from any time t < t′ to tas,

φ(x′, t′) = i∫

d3x G0(x′, t′, x, t) φ(x, t).

Thus, the total wave at a time t′ > t1 > t after the interaction is over is,

ψ(x′, t′) = i∫

d3x G0(x′, t′, x, t) φ(x, t)+ i∫

d3x1 G0(x′, t′, x1, t1) (−qih) γ0 ~A φ(x1, t1)∆t1.

Using the free-particle Green’s function propagation one more time to expressφ(x1, t1) in terms of the earlier φ(x, t) in the last term on the right,

ψ(x′, t′) = i∫

d3x G0(x′, t′, x, t) φ(x, t)

+ i∫

d3x1 G0(x′, t′, x1, t1) (−qih) γ0 ~A i

∫d3x G0(x1, t1, x, t) φ(x, t)∆t1

= i∫

d3x[

G0(x′, t′, x, t) + c∆t1i∫

d3x1G0(x′, t′, x1, t1)(−qihc

)γ0 ~AG0(x1, t1, x, t)]

φ(x, t)


Thus, to first order in small quantities, the interaction Green function, G1, expressedin terms of the free-particle Green’s function, G0, is,

G1(x′, x) = G0(x′, x) + c∆t1i∫

d3x1G0(x′, x1)(−qihc

)γ0 ~A(x1)G0(x1, x)

(We’ve simplified the notation here and dropped some of the t′’s, t1’s and t’s andimplicitly include them with their x′’s x1’s and x’s where its unambiguous). So,we’ve shown that

ψ(x′) = i∫

d3x G1(x′, x)φ(x).

Now let’s consider how the above expressions appear when we include twosequential interactions occurring at t1 and t2, where t′′ > t2 > t′ > t1 > t. Afterthe first interaction takes place, the total wave at the time t′ will be as before,

ψ1(x′, t′) = i∫

d3xG1(x′, x)ψ0(x)

Then, just after the second interaction takes place between t2 − ∆t2 and t2, anadditional wave arises,

∆ψ2(x2, t2) = −qih

γ0 ~A(x2, t2)ψ1(x2, t2)∆t2

So that the new total wave at the moment, t2, is,

ψ2(x2, t2) = ψ1(x2, t2) + ∆ψ2(x2, t2)

= ψ1(x2, t2)−qih

γ0 ~A(x2, t2)ψ1(x2, t2)∆t2

= i∫

d3xG1(x2, x)ψ0(x)− qihc

γ0 ~A(x2, t2) c∆t2 i∫

d3xG1(x2, x)ψ0(x)

After the second interaction has taken place, ψ2 can be propagated to the finaltime t′′ using the free-particle Green’s function,

ψ(x′′) = i∫

d3x2G0(x′′, x2)ψ2(x2)

= i∫

d3x2G0(x′′, x2)

(i∫

d3xG1(x2, x)ψ0(x)− qihc

γ0 ~A(x2, t2) c∆t2 i∫

d3xG1(x2, x)ψ0(x))

Using the expression above to replace G1, and using the fact that

i∫

d3x2G0(x′′, x2)G0(x2, x) = G0(x′′, x),


and taking the limit ∆ti → 0, while allowing the number of interaction momentsto go to infinity, we have, to second order in (presumably small) q

hc ,

G2(x′′, x) = G0(x′′, x)− qhc

∫d4x1G0(x′′, x1)γ

0 ~A(x1)G0(x1, x)

+ (qhc

)2∫

d4x2

∫d4x1G0(x′′, x2)γ

0 ~A(x2)G0(x2, x1)γ0 ~A(x1)G0(x1, x)

(4.23)

But remember, to get the the final wave at the final moment of consideration,t′′, we still require one more volume integration,

θ(t′′ − t)ψ(x′′) = i∫

d3x G2(x′′, x)ψ0(x),

= i∫

d3x G0(x′′, x)ψ0(x, t),

+ i∫

d3x (− qhc

)∫

d4x1G0(x′′, x1)γ0 ~A(x1)G0(x1, x)ψ0(x),

+ i∫

d3x (qhc

)2∫

d4x2

∫d4x1G0(x′′, x2)γ

0 ~A(x2)G0(x2, x1)γ0 ~A(x1)G0(x1, x)ψ0(x)

(4.24)

Now we are finally able to calculate the probability amplitude for scatter-ing. Doing this brings us back to one of the most elementary ideas of quantum-mechanics – that the amplitude for a transition from an initial state to a final stateis equal to the integral over all space of the inner product of the two waves at themoment of interest, t′′:

−iM =∫

d3x′′ψ†out(x′′, t′′)ψin(x′′, t′′)

This gives us yet another volume integration. The outgoing final wave, ψout, doesnot require any propagation, as, we assume know its form at the moment t′′, butwe use Eq. 4.24, for ψin, since it’s been ’altered’ by the interactions that havehappened since it was ’prepared’ at the time t as ψ0(x, t):

− iM = i∫

d3x′′ψ†out(x′′, t′′)G0(x′′, x)

∫d3xψ0(x, t)

− i(qhc

)∫

d3x′′ψ†out(x′′, t′′)

∫d4x1G0(x′′, x1)γ

0 ~A(x1)∫

d3xG0(x1, x)ψ0(x)

+i(qhc

)2∫


∫d4x2

∫d4x1G0(x′′, x2)γ

0 ~A(x2)G0(x2, x1)γ0 ~A(x1)

∫d3xG0(x1, x)ψ0(x).


Now we can simplify this and perform the∫

d3x volume integrals on the right bymaking use of the original definition of the Green’s function, G0,:

−iψ0(x′′, t′′) =∫

d3xG0(x′′, x)ψ0(x, t),

and−iψ0(xi, ti) =

∫d3xG0(xi, x)ψ0(x, t),

so we have,

− iM =∫

d3x′′ψ†out(x′′, t′′)ψ0(x′′, t′′)

− (qhc

)∫


∫d4x1G0(x′′, x1)γ

0 ~A(x1)ψ0(x1, t1)

+(qhc

)2∫


∫d4x2

∫d4x1G0(x′′, x2)γ

0 ~A(x2)G0(x2, x1)γ0 ~A(x1)ψ0(x1, t1).

We can simplify this yet again and perform the∫

d3x′′ integrations on the left. Italso follows from the definition of G0 that,∫

d3x′′ψ†out(x′′, t′′)G0(x′′, xi) = iψ†(xi, ti).

Thus,

− iM =∫

d3x′′ψ†out(x′′, t′′)ψ0(x′′, t′′) (4.25)

− i(qhc

)∫

d4x1ψ†out(x1, t1)γ

0 ~A(x1)ψ0(x1, t1)

+i(qhc

)2∫

d4x2

∫d4x1ψ†

out(x2, t2)γ0 ~A(x2)G0(x2, x1)γ

0 ~A(x1)ψ0(x1, t1). (4.26)

This is the scattering series of QED. Since we’ve taken this out to second order,mathematical induction allows us to infer how the pattern continues for the higher-order terms. Every possible Feynman diagram for the interaction of two fermionscorresponds to one term of this series. In other words, applying the Feynman rulesof QED to a correctly drawn diagram gives you one of the amplitude terms fromthis series, because this series is where the Feynman rules originally come from.


Now, let’s switch to our notation ∆ = qihc

~A, replace ψ0 with ψin, and replacethe unnecessary x′′’s with x’s,

− iM =∫

d3xψ†out(x, t)ψin(x, t)∫

d4x1ψ†out(x1, t1)γ

0∆(x1)ψin(x1, t1)

−i∫

d4x2

∫d4x1ψ†

out(x2, t2)γ0∆(x2)G0(x2, x1)γ

0∆(x1)ψin(x1, t1).

If ψin and ψout are free-particle plane-waves, then the first term is a delta-function.Also, replace G0γ0 with SF, and ψ†γ0 with ψ,

− iM =∫


d4x1ψout(x1, t1)∆(x1)ψin(x1, t1)

−i∫

d4x2

∫d4x1ψout(x2, t2)∆(x2)SF(x2, x1)∆(x1)ψin(x1, t1).

Now, because ψin is a plane wave, and (∇ − mcih )ψin = 0, you can insert a

∇ − mcih alongside the ∆ in front of the ψin wherever it appears without chang-

ing the value of the integrals. Looking at the third term, use the fact that (∇2 −mcih )SF(x2, x1) = δ4(x2 − x1) to insert a ∇2 − mc

ih − δ4(x2 − x1) in front of theSF(x2, x1) in the third term without changing the value of the integral, to get:

− iM =∫


d4x1ψout(x1, t1)(∇+ ∆(x1)−mcih

)ψin(x1, t1)

− i∫

d4x2

∫d4x1ψout(x2, t2)

× (∇+ ∆(x2)−mcih− δ4(x2 − x1))SF(x2, x1)(∇+ ∆(x1)−

mcih

)ψin(x1, t1).

Next, since we’ve assumed that t2 is strictly greater than t1 in the nested, time-


ordered integrals, δ4(x2 − x1) = 0, so that can be dropped, to give,

− iM =∫


d4x1ψout(x1, t1)(∇+ ∆(x1)−mcih

)ψin(x1, t1)

− i∫

d4x2

∫d4x1ψout(x2, t2)

× (∇+ ∆(x2)−mcih

)SF(x2, x1)(∇+ ∆(x1)−mcih

)ψin(x1, t1).

Then, using Di = ∇i + ∆i, we have,

− iM =∫


d4x1ψout(x1, t1)(D1 −mcih

)ψin(x1, t1)

−i∫

d4x2

∫d4x1ψout(x2, t2)(D2 −

mcih

)SF(x2, x1)(D1 −mcih

)ψin(x1, t1).

Now, we’ve discussed the representation of SF(x′, x) as a Fourier integral:

SF(x′, x) =1

(2π)4

∫d4k e−ikµ(x′µ−xµ) γµkµ − mc

ih

kµkµ − m2c2

h2

.

But there’s another way to represent SF in terms of the plane-wave solutions ofthe free-particle Dirac equation.

Discussion

Now we’ve set the stage for the discussion of the problems of conventional quan-tum gravity. Like the Fµν field of QED, an action, (or matrix element), for quan-tum gravity can be defined for the free gravitational field. Through the applicationof the Euler-Lagrange minimization procedure this can be shown to give rise tothe classical equation of motion (the Einstein equation) for the gravitational field.Let’s show how this procedure begins.

In classical general relativity, the Einstein-Hilbert action is,

S =∫ ( c4

16πG(R− 2Λ) + LM

)√−g d4x,


and the associated Lagrangian density is

LGR =c4

16πG(R− 2Λ) + LM)

√−g. (4.27)

In these, g is the determinant of the space-time metric, gµν, R = gµνRµν is the Ricciscalar, and Λ is an (optional) cosmological constant. LM is the Lagrangian densitydescribing the matter-fields, which is ultimately related to the mass-energy tensor,Tµν. It’s lengthy, so I won’t go through every step here, but the application of theformal Euler-Lagrange procedure, Eq. 4.3, for each field component φk = gµν

and their space-time derivatives leads to the Einstein equation of classical generalrelativity,

Rµν −12

gµνR + Λgµν =8πG

c4 Tµν,

where we can interpret, as a matter of definition,

Tµν ≡ −2δLM

δgµν+ gµνLM.

Although the Lagrangian and second quantization procedure outlined abovesuccessfully describes QED, when applied to gravity in an analogous manner,problems arise. As many have pointed out, [18], the Lagrangian density for classi-cal gravitational interactions does not lead to a renormalizable (i.e. useful) theoryof quantum gravity, so a different approach is needed.

In Chapter 6, we will show how to derive both Maxwell’s equations and theEinstein equation directly from the Dirac equation itself, without any reference tothe free-field terms.

Note to Editor: This chapter is a little too long and will be tightened up.

CHAPTER 5

Multiple Particles

5.1 Multiple Particles and the Principle of Superposi-tion

Consider a system of N particles. What I call the “direct coordinates,” are thespace-time coordinates xν

n that point to the centroid of the “wavelet” of each par-ticle. The total wavefunction is simply a superposition of each of the individualwavelets:

ψtotal(x) = ψ1(x− x1) + ψ2(x− x2) + . . . , (5.1)

where the four-vector nature of the coordinates xν is implicitly understood.Note here that I’m totally breaking away from the dogma of the direct product

wavefunction:ψtotal(x1, x2, . . . ) = ψ1(x1)⊗ ψ2(x2)⊗ . . .

which will show to be wrong, or at least superfluous, since it doesn’t lead to anyresults that can’t also be arrived at from 5.1 together with the general Dirac equa-tion. The principle of superposition is not only simpler, but it’s also more successfulat describing experimental fact. Think of how molecular bonding is described toa fair degree of accuracy by defining bonding and anti-bonding orbitals whichare a linear combination of atomic orbitals. Think also of the simple cathode ray

75

CHAPTER 5. MULTIPLE PARTICLES 76

experiments which show that the interference patterns observed result from thesimple linear addition of all of the innummerable individual electron waves. Incontrast to this, since the solutions of the Dirac equation are 4-spinors, think ofthe 16-spinor that would result from the direct product of only two spinors, orthe absurd complexity resulting from the direct product of N spinors. Instead, wepropose a more unified and simpler idea: That there is a single wavefunction, ψ,describing the entire physical universe, which is a 4-spinor represented as either acolumn or row vector, and it can be written as superposition of separate solutionsof the Dirac Equation as shown in Eq. 5.1.

I’ll deal with the refutation of the direct-product in more detail after we estab-lish a few more things. For now let’s consider something which everyone agreeson: The “exchange coordinates” of the system consist of 1 center of mass coordi-nate, ~RCM, and N − 1 reduced mass coordinates, ~ρn, for a total of N coordinates.The center of mass coordinate is defined as:

~RCM =1M

N

∑n=1

mn~rn, (5.2)

where M is the total mass of the system, M = m1 + m2 + · · ·+ mN.But the definition of the reduced mass coordinates will vary, depending upon

the quantum state that the combined system collapses or scatters into after inter-action sets in. One possibility is a simple sequential definition;

~ρn =~rn+1 −~rn. (5.3)

In the case of just two particles, there is only one reduced mass coordinate, ~ρ =~r2 −~r1, so this sequential definition works. But with N interacting particles, thereare at most (N − 1)!/2 different ways one could define the remaining N − 1 ex-change coordinates that describe the system. In general though, we can assumethat reduced mass coordinates, a.k.a. the remaining N − 1 exchange coordinates,must be a simple linear combination of all of the direct coordinates:

~ρn =N−1

∑i=1

αn,i~ri (5.4)

To make the algebra more concise, let’s consider the CM coordinate ~RCM tosimply be the Nth coordinate ~ρN. Then if we assume that this matrix α is invert-ible, the original direct coordinates can be reexpressed in terms of the exchange


coordinates:

~rn =N

∑j=1

α−1n,j~ρj (5.5)

This means that the total wavefunction becomes:

ψtotal(~x) = ψ1(~x−N

∑j=1

α−11,j~ρj) + ψ2(~x−

N

∑j=1

α−12,j~ρj) + . . . (5.6)

!r1!r2

!RCM

Figure 5.1: After condensation occurs, the number of dynamic coordinatesneeded to describe the system drops – here there are only two. (The CM coor-dinate is not a dynamic coordinate).

5.1.1 Two Particles, N=2

To make these ideas easier to digest for the time being, let’s restrcit our consider-ation to the case of just two particles. Then the two exchange coordinates are

~ρ =~r2 −~r1,


and~RCM =

1m1 + m2

(m1~r1 + m2~r2).

And the total wavefunction becomes:

ψtotal(~x, t) = ψ1(~x−~r1(t)) + ψ2(~x−~r2(t))

Note here that the direct coordinates must be functions of time, (think of the orbitof the Moon around the Earth). But let’s reexpress~r1 and~r2 in terms of ~ρ and ~R:

~r1 = ~R− m2

M~ρ, (5.7)

~r2 = ~R +m1

M~ρ. (5.8)

So that the total wavefunction is:

ψtotal(~x, t) = ψ1(~x− (~R− m2

M~ρ)) + ψ2(~x− (~R +

m1

M~ρ))

Going back to the touchstone idea that this paper is inspired by, the generalprinciple of relativity applied to the Dirac equation, we assert that the wavefunctiondescribing each particle, in the free coordinate space, x, obeys the free particle Diracequation:

ih γµ ∂

∂xµ ψ1(x− x1) = m1c ψ1(x− x1),

ih γµ ∂

∂xµ ψ2(x− x2) = m2c ψ(x− x2).

where m1 and m2 are the masses of the two particles under consideration, and c isthe speed of light in vacuum. That’s another surprising result of this application ofthe principle of relativity: even wavefunctions describing massive, complex parti-cles like the proton, or the Earth - Moon system, when differentiated in an inertial,un-warped reference frame, are proportional to the inertia of the free particle.

Now consider this purely formal change of variables which pivots on the ar-guments of the wavefunctions (x− x1), and (x− x2):

∂

∂xψ1(x− x1) = −

∂

∂x1ψ1(x− x1),

∂

∂xψ2(x− x2) = −

∂

∂x2ψ1(x− x2).


!R

⊗

!r2

!r1

!ρ

Figure 5.2: The reduced mass coordinate vector,~ρ, points from particle 1 to particle2 and passes through the center-of-mass.


In this sense then, the two wavefunctions must also obey:

− 1m1

γµ ∂

∂xµ1

ψ1(x− x1) =cih

ψ1(x− x1), (5.9)

− 1m2

γµ ∂

∂xµ2

ψ2(x− x2) =cih

ψ2(x− x2). (5.10)

Now let’s simply add the two equations:

− 1m1

γµ ∂

∂xµ1

ψ1(x− x1)−1

m2γµ ∂

∂xµ2

ψ2(x− x2) =cih

ψ1(x− x1) +cih

ψ2(x− x2)

For now, suppressing the arguments, (x − x1), (x − x2), which are understood,and also rearranging slightly, this becomes:

γµ

(− 1

m1

∂

∂xµ1

)ψ1 + γµ

(− 1

m2

∂

∂xµ2

)ψ2 =

cih

(ψ1 + ψ2) (5.11)

Now, let’s look at the linear relationship between the direct coordinates andexchange coordinates, reflected in 5.8, with an eye towards changing partial dif-ferential variables via the chain rule of calculus. Look at the contravariant differ-entials (changes in distance):

dρν = dxν2 − dxν

1 (5.12)

anddRν =

1m1 + m2

(m1dxν1 + m2dxν

2)

But remember that the center of mass, RCM, is a constant, so any partial differenti-ation must be such that RCM is held constant, so dR = 0. This implies:

0 =1

m1 + m2(m1dxν

1 + m2dxν2)

or thatm1dxν

1 = −m2dxν2

This means that Eq. 5.12 may be written as either:

dρν = dxν2 +

m2

m1dxν

2 =

(1 +

m2

m1

)dxν

2 =

(1

m2+

1m1

)m2 dxν

2


in terms of dx2, or in terms of dx1 as:

dρν = −m1

m2dxν

1 − dxν1 = −

(1 +

m1

m2

)dxν

1 = −(

1m1

+1

m2

)m1 dxν

1

Letting 1µ = 1

m1+ 1

m2, where µ is known as the reduced mass, we then have the

following relations between differentials:

µ dρν =m2 dxν2, (5.13)

µ dρν = −m1 dxν1. (5.14)

This makes it clear that:

1m1

∂

∂xν1= − 1

µ

∂

∂ρν,

1m2

∂

∂xν2= +

1µ

∂

∂ρν.

Notice the difference in sign, which has its origin in the original definition of thereduced mass coordinate, ρ = x2 − x1.

Now, 5.11 becomes:

γν

(1µ

∂

∂ρν

)ψ1 − γν

(1µ

∂

∂ρν

)ψ2 =

cih

(ψ1 + ψ2)

which becomes:γν 1

µ

∂

∂ρν(ψ1 − ψ2) =

cih

(ψ1 + ψ2) (5.15)

Again, going back to 5.10, subtracting, instead of adding the two equations, ulti-mately results in:

γν 1µ

∂

∂ρν(ψ1 + ψ2) =

cih

(ψ1 − ψ2) (5.16)

By again adding or subtracting, 5.15 and 5.16, then slightly rearranging, we obtain:

ih γν ∂

∂ρνψ1 = +µc ψ1

ih γν ∂

∂ρνψ2 = −µc ψ2


The signs must be swapped one more time to obtain the correct orientation of thevariable ρ, because of the pivot from x to x1 and x2. Then we have

ih γν ∂

∂ρνψ1 = −µc ψ1

ih γν ∂

∂ρνψ2 = +µc ψ2

Showing that both ψ1 and ψ2 separately obey almost the same equation, the onlydifference being that ψ1 has a negative reduced mass. Because of this, it will alsohave a negative energy spectrum.

More about the Direct Product Error.

Now notice how the familiar definition of the reduced mass as

1µ=

1m1

+1

m2

emerged from this application of the principle of superposition, without any useof direct product wavefuntions. Part of what has tempted others (and myself)into considering direct-product wavefunctions in the first place is the unfortunatemathematical coincidence that for second derivatives:

−h2

2m1~∇2

1 +−h2

2m2~∇2

2 =−h2

2M~∇2

R +−h2

2µ~∇2

ρ

or thatp2

12m1

+p2

22m2

=p2

µ

2µ+

p2M

2M

And thus the total Hamiltonian can be written as a sum of two terms, either interms of the direct coordinates as Htotal = H1 + H2, or in terms of the exchangecoordinates as Htotal = Hµ + HM. (In order to arrive at the above relations, onealso assumes that the variable describing the center-of-mass, R, is allowed to varyduring partial differentiations.)

But this is only true for second derivatives. We must remember that the SchrodingerEquation is a non-relativistic approximation to a fuller relativistic theory, wherethe Hamiltonian is actually an infinite series of higher powers of p or d/dx. In


the non-relativisitc Schroedinger theory, one can write a direct-product wave-function,

ψ(R, ρ) = ψ(R) · ψ(ρ)and show that these separate factors obey:(

p2µ

2µ+ V(ρ)

)ψ(ρ) = Eµ ψ(ρ)

andp2

M2M

ψ(R) = ECMψ(R)

with Etotal = Eµ + ECM. But because this coincidence does not apply to the higherorder derivatives, this direct product recipe does not obey Lorentz invariance andso is not applicable to the Dirac Equation. But the method based upon the prin-ciple of superposition outlined above obeys Lorentz invariance and one better: itobeys the general principle of relativity.

5.2 The Integer Normalization of the Wavefunction

In the free-particle system, the total wave function is in general a sum of separateprimitive waves. These primitive waves are the fundamental quantum excitationsof the positron and electron, normalized to ±1:∫

ψ†i γ0ψi dV = ±1

with the value of +1 representing the positron and the value of −1 the electron.The total wave function, ψtot is a sum of orthogonal primitive waves according

to the principle of superposition:

ψtot, f = ψ1, f + ψ2, f + · · · = ∑i

ψi, f

which implies that the total wave, in some free reference frame must be normal-ized as: ∫

ψ†f γ0ψ f dVf = Q,

where Q = 0,±1,±2 . . . . This normalization is essentially the overall charge. It isof course possible to have a total wave which has zero overall charge.


Restating this in terms of the bound wave, ψb = eSψ f , we have∫ψ†

b eS†γ0eSψb dVf = N.

The integrand here is a hybrid, since ψb is a function of the xb, whereas the volumeelement, dVf , is still parametrized in terms of the x f . Since the x f are implicitfunctions of the xb, we might be tempted to relate the two through the ClassicalJacobian determinant:

dVf =

∣∣∣∣∂x f

∂xb

∣∣∣∣ dVb (5.17)

...Now, from classical relativity, we know that the momentum of the center-of-

mass obeys:PµPµ = M2c2

We can derive this fact from the principle of superposition. If the total wavefunc-tion is:

ψtotal(x) = ψ1(x− x1) + ψ2(x− x2) + . . . ,

Then,

ih∇ψtot = ih∇ (ψ1(x− x1) + ψ2(x− x2) + . . . )= m1c ψ1(x− x1) + m2c ψ2(x− x2) + . . .

If as stated above, we assume that the different partial waves, ψ1, ψ2, are or-thogonal and each normalized to 1, then the expectation value of the total mo-mentum satisfies:

〈P〉 =∫

ψ†tot ih∇ψtot dV

=∫ (

ψ†1 + ψ†

2 + . . .)(m1c ψ1(x− x1) + m2c ψ2(x− x2) + . . . ) dV

= m1c + m2c + · · · = Mc,

as it should, if we interpret

〈P〉 =√

PµPµ.

This confirms the principle that each partial wave ψi should be separately nor-malized as above. So we see that the principle of superposition combined withthe Dirac Equation produces the correct results regarding the total center-of-massand reduced masses.


5.3 The Case of N = 2, Again

In the two particle case...The wave equation is solved in the reduced mass sys-tem, and the solutions that result are functions of the reduced mass coordinate,ρ, which is the argument of the function, i.e. ψ( )ρ. But how do we relate thesesolutions obtained in the reduced mass system to the free particle coordinate, x,in order to write down the principle of superposition?

ψtot(x) = A1 ψ1(α1 x)ρ + A2 ψ2(α2 x)ρ (5.18)

ihγµ ∂

∂xµ ψtot(x) = ihγµ ∂

∂xµ A1 ψ1(α1 x)ρ + ihγµ ∂

∂xµ A2 ψ2(α2 x)ρ

ihγµ ∂

∂xµ ψ1(α1 x)ρ = α1ihγµ ∂

∂ρµ ψ1( )ρ = α1µcψ1( )ρ

ihγµ ∂

∂xµ ψ1(α2 x)ρ = α2γµ ∂

∂ρµ ψ2( )ρ

ihγµ ∂

∂xµ ψtot(x) = α1µcA1ψ1(α1 x)ρ + α2µcA2ψ1(α2 x)ρ

α1µ = m1, α2µ = m2

and remember,µ =

m1m2

m1 + m2, M = m1 + m2

so,

α1 =Mm2

, α2 =Mm1

For normalization over d3x, we must have,

A1 = α3/21 , A2 = α3/2

2

Thus

ψtot(x) =(

Mm2

)3/2

ψ1(Mm2

x) +(

Mm1

)3/2

ψ2(Mm1

x) (5.19)

And this ensures:∫ψ†

tot

(ihγµ ∂

∂xµ

)ψtot dV = (m1 + m2)c = Mc


5.4 Principle of Superposition – Again

Consider the description of the total wavefunction in a free reference frame whichis at rest with respect to the center-of-mass of the system of particles under consid-eration. If this system is in a stable, periodic state, then the most general way towrite the principle of superposition is:

ψtot(x) = ∑i

e−i Eih t · eSi(ρi(x)) · ψi(ρi(x))

This system will have a period of major conjunction, Tmajor, if:

Ei

h· Tmajor = 2πni

for each partial wave, ψi, where ni is an integer.So now I’ve partially followed-through on my promise to complete the de-

velopment and exposition of the quantum theory of sound in terms of the Diracequation which I made in my previous book, Understanding Sonoluminescece, [12].

One of the key theoretical developments in that book that led to useful resultswas the use of the principle of superposition to describe a sound wave propagatingin a gas:

ψtotal(x, t) = e−i mc2h tψgas(x, t) + e−iωstψsnd(x, t). (5.20)

Here, ψgas is the combined wavefunction of all of the gas molecules under consid-eration, m is the mass of each gas molecule1, ωs is the frequency of the imposedsound wave, and ψsnd is the fermion wave representing the sound disturbance.We will see that the reduced mass, µ, of the sound wave is much smaller than themass of an electron, so sound waves are nearly massless. Most importantly, andsurprisingly, c, is not the speed of light, but rather the speed of sound in the gas.

The normalization integral of the total wave over the volume containing thegas is written:∫

ψ†totalψtotal dV =

∫ (ei mc2

h tψ†gas + eiωstψ†

snd

)·(

e−i mc2h tψgas + e−iωstψsnd

)dV

=∫ (

ψ†gasψgas + e−iωstei mc2

h tψ†gasψsnd + e−i mc2

h teiωstψ†sndψgas + ψ†

sndψsnd

)dV

1For a pure gas composed of only one type of molecule, m is the mass of one molecule – for agas mixture, m becomes the average molecular mass


The condition of the integer quantization is that the total wave-function mustbe normalized to an integer number of disturbances, which is to say,∫

ψ†totalψtotal dV = Ntotal (5.21)

And, without the addition of the sound wave, we require that the backgroundwave be normalized such that:∫

ψ†gasψgas dV = Nmolecules.

5.5 Scattering And Feynman Diagrams

Now that we’ve discussed the meaning of the reduced-mass coupled state a bit,we can use that to help understand a subtlety in the calculation of probabilitiesfrom Feynman diagrams.

One step in the calculation of scattering probabilities from Feynman diagramsthat used to mystify me is the prescription that to get the final transition probabil-ity, you have to ’square the amplitude’ before using it to calculate cross-sectionsor lifetimes, usually symbolized as |M|2. But the calculation ofM comes from avolume integral of ψ†ψ, which is already a square amplitude, and is thus alreadya probability, so why do you have to square it again? The answer is that a com-plete (tree-level) Feynman diagram with incoming and outgoing particles actuallyrepresents a sequence of two events: The first event is when the uncoupled incom-ing plane waves enter into a coupled, reduced mass state represented by just thephoton and vertices in a traditional diagram, but is more accurately thought of asa short-lived state where the particles move (classically) along hyberbolic paths,under the influence of the photon field, as shown in Fig. 5.3. The second event iswhen the particles leave the coupled state and enter into a final outgoing state oftwo uncoupled plane waves. Basic probability theory, then, tells us that the totalprobability for this sequence of two events is the product of the two probabilitiesfor each individual event.

Let the incoming state consisting of the superposition of two plane waves berepresented as,

ψin = ψ1(~k1, x, t) + ψ2(~k2, x, t).

Let the intermediate, coupled, reduced mass state be represented, as in Eq. 5.19,


by,

ψc =

(Mm2

)3/2

ψ1(Mm2

x) +(

Mm1

)3/2

ψ2(Mm1

x).

Let the final, outgoing state consisting of the superposition of two plane wavesbe,

ψout = ψ3(~k3, x, t) + ψ4(~k4, x, t).

The first event is the transition from the incoming plane-wave state, ψin, to thecoupled, reduced-mass state ψc. The probability ’amplitude’ for this is,

−iM1 =∫

d3x ψ†c ψin.

Notice that, unlike the probability density of a single particle wave, ψ†ψ, whichmust be real, this quantity is, in general, complex, and that’s why it’s more cor-rectly called a ’probability amplitude.’ The second event is the transition fromthe coupled state ψc to the final outgoing plane-wave state, ψout. The probabilityamplitude for that is,

−iM2 =∫

d3x ψ†outψc.

The total probability for this sequence of two events is the product of the twoprobabilities:

|Mtotal| = |M1| · |M2|.Because of the symmetry, in the center-of-mass frame, between the incoming andoutgoing plane wave states and the intermediate coupled state, whose hyperbolicpaths must asymptotically approach and link up to the straight paths of the in-coming and outgoing states , it can be shown that

|M1| = |M2| = |M|,

so,|Mtotal| = |M|2.

Thus, I believe it’s a slight abuse of language to call M an ’amplitude’ – it’sactually a probability for half the event, and the final ’square amplitude,’ |M|2,is also a probability, just for the full sequence of events. This is a fine point that’sglossed over in most field-theory textbooks, but it’s one that needs to be con-sciously addressed to fully understand the calculation of observable scatteringfluxes from Feynman diagrams.


Figure 5.3: When a scattering interaction occurs between two particles, they mo-mentarily enter into a ’coupled’ state described by a single reduced-mass wave-function ψ(ρ, t).


Discussion

Note to editor: More discussion and tidying up of this chapter is intended.

CHAPTER 6

The General Principle of Relativity, Part II

In this chapter, we continue the development begun in chapter 2, and use themethods of quantum tensor calculus to explore more of the implications of thegeneral Dirac equation.

6.1 Newton’s Second Law

We can use the methods of quantum tensor calculus to derive two different ver-sions of Newton’s 2nd law – one that applies to waves, ψ carrying an overallcharge, and one that applies to waves, ψ carrying no overall charge, (which resultfrom the superposition of waves having zero charge overall).

Charged Waves

The general Dirac Hamiltonian may be written as

H = γ0mc2 + γ0ihcγi ∂

∂xi − ihcγ0~∆.

We will see that we should interpret ~p = ihσi ∂∂xi to be the 3D momentum vector

object – because the Pauli sigma matrices are the 3d unit vectors, so ~p = p1 i +

91

CHAPTER 6. THE GENERAL PRINCIPLE OF RELATIVITY, PART II 92

p2 j + p3k. Notice also that it is block diagonal, unlike the operator ihγi ∂∂xi , which

couples upper and lower spinor components to each other. Now, the Heisenbergequation of motion for any operator O, is

dOdt

=1ih

[H, O]

So this tells us that

~p =1ih

[γ0mc2 + ihcγ0γi ∂

∂xi − ihcγ0~∆, ihσi ∂

∂xi

]Analyzing the first two terms in the commutator, we have[

γ0mc2, ihσi ∂

∂xi

]= 0,

for the first term, and [ihcγ0γi ∂

∂xi , ihσi ∂

∂xi

]= 0,

for the second term. So we’re left with

~p =

[σi ∂

∂xi , ihcγ0~∆]

. (6.1)

This is the general quantum version of Newton’s Second Law.If we define the 3d del operator as ~∇ = σi ∂

∂xi , and interpret the potential (en-ergy), Φ, to be Φ = −ihcγ0~∆, then this means,

~p = −[~∇, Φ

].

In the case that Φ is a scalar potential, (meaning only containing terms propor-tional to 1, and none involving things like γi or σi), then this becomes the familiar

~p = −~∇Φ.

Or the force vector is the negative gradient of the potential, and the time rate ofchange of momentum is equal to this force.

Now let’s apply this to the case of the hydrogen atom, to see that it gives ex-pected results. For the H-atom, we found in Chapter 6 that the transformationgenerated by S = − i

2 α′ε resulted in the correct energy level spectrum. We also


discovered that the boost parameter, α′ is related to the fine-structure constant α,as α = sinα′. Furthermore, since the fine-structure constant, α, is small, then sinα′

is also small. So, to lowest order in the small parameters,

ihcγ0∆ = ihc

(−i

sinα′

r− 2

sin2 α′2

rε

)≈ hcsinα′

r=

e2

r,

(where, in the last step, we made the replacement, hcsinα′ = e2). Thus, Eq. 6.1,becomes

~p =

[σi ∂

∂xi , ihcγ0~∆]=

[σi ∂

∂xi ,e2

r

].

If we change to spherical coordinates, then only the partial derivative with respectto r is involved, and we have,

~p =

[r

∂

∂r,

e2

r

]= − e2

r2 r.

This is indeed Coulomb’s Law for the attractive force between the positively chargednucleus and negatively charged electron:

~F = − e2

r2 r.

If we wanted to get an exact analytic expression for the force, including all ordersof the fine-structure constant, we could also find that from Eq. 6.1.

Uncharged waves

Now let’s consider the operator γ0~P = −ihγ0γi∂i.

ddt

(γ0~P

)=

1ih

[γ0mc2 + γ0~pc− ihcγ0~∆, γ0~p

]Which becomes

ddt

(γ0~P

)= −2mc2γi

∂

∂xi − ihc[

γ0γi ∂

∂xi , γ0~∆]

orddt

(γ0~P

)= −2mc2~∇3 − ihc

[γ0~∇3, γ0~∆

]


This is the version of Newton’s 2nd law that applies to hydrodynamics. Theleft side is the time rate of change of momentum density, which can also be calledforce per unit volume. The first term on the right of the equality is equivalent to thepressure gradient, since pressure variations are proportional to density variationsvia the bulk modulus, κ.

6.2 The Quantum Maxwell Equations

The classical Maxwell Equations, in Gaussian units, in vacuo, and in manifestlycovariant component form, are:

∂φµν

∂xν=

4π

cJ µ

∂φµν

∂xσ+

∂φνσ

∂xµ +∂φσµ

∂xν= 0. (6.2)

Here, the electromagnetic field tensor, φµν, is defined in terms of the four-vectorpotential, Aµ, as

φµν =∂

∂xνAµ −

∂

∂xµ Aν,

and the electromagnetic current four-vector, J µ, is assumed to obey the continuityequation,

∂J µ

∂xµ = 0. (6.3)

The first of Eqs. 6.2 is known as the inhomogeneous Maxwell equation while thesecond is called the homogeneous Maxwell equation. Remember that if we are for-mulating this in a flat space, then gµν = ηµν, and

φµν = ηµαηνβφαβ, Aµ = ηµν Aν, etc.

We will show how these classical equations can be derived from the generalDirac Equation. Now, the General Dirac Equation is,

(∇+ ∆)ψ =mcih

ψ.

When this is also written in manifestly covariant form, showing all the indices, itis, (

γµ ∂

∂xµ + γµe−S ∂eS

∂xµ

)ψ =

mcih

ψ,


We identify the field

∆ = γµe−S ∂eS

∂xµ = γµ∆µ.

This last equality assumes that the vector field ∆ is composed of four four-vectorcomponents, each defined as

∆µ = e−S ∂eS

∂xµ .

These are components that transform as a four-vector under the Lorenz group, butthey are also 4× 4 Dirac matrices that have the ability to act as matrix operatorson the wave ψ:

∆µ ψ = . . . ,

just as the γµ do. We can take derivatives of ∆µ, but we must be careful to applythe product rule for differentiation correctly:

∂

∂xν∆µψ =

∂∆µ

∂xν+ ∆µ

∂

∂xνψ.

Thus∂∆µ

∂xν+ ∆µ

∂

∂xνψ =

∂

∂xνe−S ∂eS

∂xµ ψ

or∂∆µ

∂xν+ ∆µ

∂

∂xνψ =

(∂e−S

∂xν

∂eS

∂xµ + e−S ∂2eS

∂xµ∂xν

)ψ + e−S ∂eS

∂xµ

∂

∂xνψ.

Then the terms involving direct derivatives on ψ cancel from both sides, and wehave the well-defined operator equation

∂∆µ

∂xν=

∂e−S

∂xν

∂eS


∂xµ∂xν

From this we can form the anti-symmetric tensor operator, Φµν,

Φµν =∂∆µ

∂xν− ∂∆ν

∂xµ .

If the order of differentiation in the terms like e−S ∂2eS

∂xµ∂xν can be interchanged, thenthose terms cancel, and we have

Φµν =∂e−S

∂xν

∂eS

∂xµ −∂e−S

∂xµ

∂eS

∂xν(6.4)


Assuming a Cartesian metric, let’s raise both of the indices using ηµν,

Φµν = ηµαηνβ

(∂e−S

∂xβ

∂eS

∂xα− ∂e−S

∂xα

∂eS

∂xβ

)Now, to get towards Maxwell’s equations, let’s take a another derivative, for nowover an unduplicated index, σ, (using the same product rule argument as above):

∂Φµν

∂xσ= ηµαηνβ

(∂2e−S

∂xσ∂xβ

∂eS

∂xα+

∂e−S

∂xβ

∂2eS

∂xσ∂xα− ∂2e−S

∂xσ∂xα

∂eS

∂xβ− ∂e−S

∂xα

∂2eS

∂xσ∂xβ

)From this we can see, from inspection of the indices, that the operator Φµν simul-taneously obeys the following two equations,

∂Φµν

∂xµ =4π

cJ ν

∂Φµν

∂xσ+

∂Φνσ

∂xµ +∂Φσµ

∂xν= 0, (6.5)

where the current J ν is defined as:

4π

cJ ν = ηµαηνβ

(∂2e−S

∂xµ∂xβ

∂eS

∂xα+

∂e−S

∂xβ

∂2eS

∂xµ∂xα− ∂2e−S

∂xµ∂xα

∂eS

∂xβ− ∂e−S

∂xα

∂2eS

∂xµ∂xβ

).

(6.6)Again, by inspection of the indices, we see that the (operator) current J ν mustobey the continuity equation:

∂J ν

∂xν= 0, (6.7)

and then also that∂2Φµν

∂xµ∂xν= 0.

Remember that all of Eqs. 6.5 through 6.7 are implicitly operator equations thatcan act on the right on ψ, or on the left on ψ†.

We can clearly see how Eqs. 6.5 are identical in form to their classical formu-lation, Eqs. 6.2, and that the quantum continuity condition, Eq. 6.7, is the same asits classical equivalent, Eq. 6.3. Thus we have another example of a tried and truequantum principle that goes back to Heisenberg – that quantum equations can beformed from their classical counterparts by simply promoting the variables andsymbols to quantum operators. However, exactly what version of the classical


equations to use can be difficult to determine beforehand. Apparently, the mani-festly covariant formulations, always preferred by Einstein, are the correct versionto apply the procedure to. But we have accomplished much more than just a for-mal promotion of classical field variables to quantum operators. We’ve shownthat the fields ∆µ of the general Dirac equation automatically obey Maxwell’sequations. We’ve also shown how the current, J ν, must, as a consequence, obeycontinuity – in contrast, in the classical formulation, the continuity of J ν mustbe assumed as a supplemental condition to the Maxwell equations. Most impor-tantly, we’ve also shown how the electromagnetic vector potential field, Aµ, andthe electromagnetic current, J ν, both have a common origin in the generator ofthe transformation, S, that relates the solutions of the general Dirac equation tothe free-particle reference frame. That is to say,

ec

Aµ = ∆µ = e−S ∂eS

∂xµ ,

and

4π

cJ ν = ηµαηνβ

(∂2e−S

∂xµ∂xβ

∂eS

∂xα+

∂e−S

∂xβ

∂2eS

∂xµ∂xα− ∂2e−S

∂xµ∂xα

∂eS

∂xβ− ∂e−S

∂xα

∂2eS

∂xµ∂xβ

).

Without this quantum formulation based upon the general Dirac equation, thereis no classical equation that points out the common origin of Aµ and J ν. In theclassical formulation the currents and fields are logically separate entities that arecoupled by the (empirically derived) Maxwell equations. Here, then, we haveaccomplished a unification from the first principle of the general Dirac equation.We’ve also shown how there can be a simpler path to discovering the analytic formof the fields and currents – instead of having to solve a differential equation likeMaxwell’s equations, if we know the form of the generator, S, then we instantlyhave the analytic forms of Aµ and J ν in hand.

Another important thing to point out is that the quantum electromagnetic cur-rent operator, J ν, is not the same thing as the Dirac probability current, jµ,

jµ = ψ† γ0γµ ψ,

since, as it is sandwiched between a ψ† . . . ψ, the Dirac current only has the rankof a complex c-number, not a 4× 4 matrix operator like J ν. However, we can as-sociate the Dirac current with the operator γ0γµ, which becomes the Dirac currentonce so sandwiched. We will find, that for certain transformation generators, S,


the J ν, as defined in Eq. 6.6, simplifies down to terms proportional to the opera-tors γµ or γ0γµ.

One other note on the operator ∆µ, or Aµ. In the foundational published workson QED, like Julian Schwinger’s early articles, [24] and [25], the electromagnetic

field, Aµ(x), and it’s derivatives, ∂Aµ(x)∂xν , are considered operators. Equations like[

Aµ(~r, t),1c

∂

∂tAν(~r′, t)

]|Ψ >= ihc ηµν δ3(~r−~r′)|Ψ >,

can appear, where |Ψ > is considered an abstract ket in a Fock space. These typesof ’Fock space’ conditions and operator identities are used in many calculationsto good effect, and come into play particularly in higher-order diagrams, wheremultiple powers of time-ordered operators arise. Although, as I’ve said, I believethat the Fock space formalism is superfluous in some ways, these type of operatoridentities still have a valid meaning when we consider Aµ, (or ∆µ), in terms ofit’s relation to the transformation eS. As we found above in the derivation of thequantum Maxwell equations,

∆µ = e−S ∂eS

∂xµ ,

∂∆µ

∂xν=

∂e−S

∂xν

∂eS


∂xµ∂xν.

In the electromagnetic case, ∆µ itself is usually proportional to the 4× 4 identity

in the Dirac space, 1, but its derivative, ∂∆µ(x)∂xν is not necessarily proportional to 1,

but is a more general 4× 4 operator in the Dirac space. Thus, meaningful com-mutation and anti-commutation relations with these operators can be formed andused to manipulate time-ordered products, but there is no need to view them asanything more than 4× 4 Dirac operators.

Let’s look at how the quantum Maxwell equations relate to the Hydrogen

atom. In Chapter 3, we found that the transformation e−i α′2 ε led to the accepted en-

ergy spectrum of the hydrogen atom. The field that arose from this transformationwas:

∆ = iγ0 sinα′

r− γ0ε

2sin2 α′2

r.

Now, γ0ε = γr = γi xi

r , so this field can be stated in Cartesian form as,

∆ = γµ∆µ = iγ0 sinα′

r− γi 2 sin2 α′

2 xi

r2 .


Then, remembering ∆µ = −eihc Aµ, the Aµ components are:

A0 =hce

sinα′

r, Ai = i

hce

2 sin2 α′2 xi

r2 .

Using the fact that sinα′ = α = e2

hc , this can also be written:

A0 =er

, Ai = ie√

1− α2 xi

r2 . (6.8)

This is the electromagnetic field associated with the positively charged nucleus.Now let’s calculate the current, J ν, associated with this field. We could do thisusing Eq. 6.6, the most logically fundamental definition of the current based onthe transformation, eS, but, we’ll have an easier time if we use the inhomogeneousMaxwell equation, Eq. 6.5, since we also already know ∆µ, and from ∆µ we canget Φµν. The inhomogeneous Maxwell equation gives:

4π

cJ ν = ηνβ

(∂

∂x0 Φ0,β −∂

∂xi Φi,β

)= η0ν

(− ∂2

∂xi2∆0

)= η0ν ∂2

∂xi2−i sin α′

r

From this we see that the spatial components, J i, are zero and only J 0 is nonzero.Using, ∂2

∂xi21r = −4π δ3(~r), we get

J 0 = i sinα′ δ3(~r).

Finally, let J0 = ihc−eJ 0 to get a quantity with the units electromagnetic current

density, (because multiplying ∆µ by this factor gives you Aµ), and use sinα′ = e2

hc .We get, for the charge density,

ρ(~r) = J0/c = e δ3(~r).

This is the charge density of the positively charged proton in the nucleus of thehydrogen atom, not the charge density associated with the electron itself, whichwould be negative. This makes sense, since the positively charged nucleus is clas-sically identified as the source of the field that binds the electron and is the subjectof the Maxwell equations. Thus, the electromagnetic vector potential, Aµ, given in

Eq. 6.8, which we derived from the quantum transformation, e−i α′2 ε, and yields the

correct energy spectrum of the hydrogen atom, is also consistent with the quan-tum Maxwell equations, Eq. 6.5, as well as with classical intuition.


6.3 Conventional Quantum Gravity

Now that we’ve shown how Maxwell’s equations can be derived as a direct con-sequence of the Dirac equation itself, we can see how the other derivation basedupon the Euler-Lagrange reduction of the free-field electromagnetic Lagrangian,which we went over in Chapter 4, could be viewed as superfluous. This might bea major point where conventional QFT and QED goes wrong –the inclusion of thefree-field operator 1

16π FµνFµν as part of the Lagrangian density in the QED Matrixelement of the form of Eq. 4.5. But this error has no impact on the calculation ofscattering amplitudes for Feynman diagrams that contain ingoing and outgoingFermions, since the amplitude for each of those diagrams come from a term in theQED scattering series, Eq. 4.26, and none of the terms of that series come fromthe free-field electromagnetic Lagrangian. Even non-’tree-level’ diagrams that in-clude intermediate virtual particle-antiparticle pairs come from a term in Eq. 4.26.

Applying the Euler-Lagrange method to this term and the coupling term ψγµ Aµψ

results in a derivation of Maxwell’s equations, but I believe this may be a temp-tation that leads us astray. One can derive any equation of motion by positingthe right Lagrangian, but that could just be a mathematical ad-hoc coincidence thatdoesn’t have any bearing in physical reality. I call this kind of erroneous deriva-tion a Lagrungian formulation. As I just showed, you can derive Maxwell’s equa-tions directly from the Dirac equation itself, so the Maxwell Lagrangian approachis superfluous. There is no need to include the Fµν field, or any other fields otherthan the Dirac field ψ in the Lagrangian or the matrix elements −iM that comefrom that Lagrangian. The field Aµ is still present, but it is simply the covariantcorrection to differentiation for the coordinate system in which the particle ex-ists and in which ψ comes as a solution to the Dirac equation, and Aµ, or moregenerally, ∆µ, describes both electromagnetic and gravitational interactions.

I take the view that electromagnetic fields do not exist as entities that are in-dependent from material particles. Photons are not things, they are relationshipsthat arise between things. The Dirac field is the only field, and the amplitudeterms for all interactions between observable particles follow only from the gen-eral Dirac Lagrangian density, L =

√g ψ(

ihc~D−mc2)

ψ from which follows theDirac equation. This applies not only to electromagnetic interactions, but also togravitational interactions.

This position, if correct, solves the biggest problem with conventional quan-tum gravity, non-renormalizability, by simply obviating it. If the free-field term


involving the Fµν field in QED is not needed to describe anything observable inelectrodynamics, then action terms involving the free gravitational field wouldalso be unneeded. As we mentioned at the end of Chapter 4, even though theEinstein-Hilbert action, which includes the free-field terms, can be shown to giverise to the classical equation of motion for the gravitational field, the Einsteinequation, through the formal application of the Euler-Lagrange minimization pro-cedure, that doesn’t mean it’s correct or relevant to physics. Especially if the Ein-stein equation can be shown to follow directly from the group-theory propertiesthat lead to the Dirac equation.

6.4 The Quantum Einstein Equation

We just showed how to derive Maxwell’s equations without any need for the free-field Lagrangian. Now let’s prove the assertion we made in the preface and inChapter 1 – that the Einstein equation can be derived from first principles in termsof the 4× 4 Dirac basis vectors, εµ, which are akin to the γµ Dirac matrices of aflat space. This will also be done without any need for the Einstein-Hilbert action.This form can be called the quantum Einstein equation, and we will be able toshow that it reduces to the classical Einstein equation in the leading-order terms.

The metric tensor in a generally curvilinear space, gµν, is expressed in terms ofthe 4× 4 basis vectors, εµ :

gµν =12εµ, εν.

A first-principles parallel displacement argument relates the derivatives of thecovariant and contravariant basis vectors to the Christoffel symbols:

∂εµ

∂uν= Γσ

µνεσ,∂εα

∂uσ= −Γα

ρσερ. (6.9)

Eq. 6.9 can be solved for the Christoffel symbols by taking a dot-product (anti-commutator):

Γαµν =

12εα,

∂εµ

∂uν = 1

2εα,

∂εν

∂uµ . (6.10)

giving an expression for the Christoffel symbols in terms of these basis vectorsand their derivatives.

Now, the differential of a position vector is expressed d~r = εµduµ. Because aposition vector points from the origin of the coordinate system to a point in space,


the change in the position vector around a parallelogram along the leg duµduν

must be the same as along the leg duνduµ, as you arrive at the same point in theend regardless of the leg of travel. Thus the order of partial differentiation in thesecond derivatives of a position vector can be interchanged, ∂2~r

∂uµ∂uν = ∂2~r∂uν∂uµ . And

therefore we know that ∂εµ

∂uν = ∂εν∂uµ , so the Christoffel symbols must be symmetric

in the lower two indices.However, the order of partial derivatives cannot usually be interchanged as in

the case of a position vector. In the case of any other vector, parallel displacementof a vector around different legs of a parallelogram results in different changesto the components of the vector. And this is also the case for the basis vectorsthemselves, so the ordering of the second derivatives matters. In the one case wehave,

∂2εµ

∂uα∂uν=

(∂Γλ

µν

∂uα+ Γρ

µνΓλρα

)ελ, (6.11)

and in the other case,

∂2εµ

∂uν∂uα=

(∂Γλ

αµ

∂uν+ Γρ

αµΓλρν

)ελ. (6.12)

We see that these two quantities are not, in general, equal.The Ricci tensor, Rµν, derived from the Riemann tensor, Rα

βγδ, is traditionallyexpressed in terms of the Christoffel symbols, Γα

µν as,

Rµν =∂Γα

µν

∂uα−

∂Γααµ

∂uν+ Γβ

µνΓαβα − Γβ

µαΓαβν. (6.13)

Now an equivalent expression for the Ricci tensor is proved:

Rµν =12εα,

(∂2εµ

∂uα∂uν−

∂2εµ

∂uν∂uα

). (6.14)

To see this, replace the second partials of the basis vectors using 6.11 and 6.12, andthen follow through on the anticommutator dot-product with εα. The traditionalexpression for the Ricci tensor, Eq. 6.13, results, proving the equivalence of ournew expression, 6.14.

Eq. 6.14 is nice because it lends to an intuitive understanding of the Ricci ten-sor, Rµν, as a quantity related to the second difference of a basis vector, εµ around


two different legs of a parallelogram, duαduν vs. duνduα, dotted into each of thefour εα’s and summed over α. It’s also simpler in appearance than the traditionalexpression. Most importantly for our purposes, it relates the Ricci tensor to the4× 4 basis vectors which appear in the General Dirac Equation and hints at thenext step in making the connection to quantum theory.

Using the same techniques a related expression for the more general Riemanntensor (from which the Ricci tensor follows), can also be derived:

Rασρβ =12εα,

(∂2εβ

∂uσ∂uρ −∂2εβ

∂uρ∂uσ

). (6.15)

Also, by making use of the identity δαβ = εα · εβ, and the fact that the second

derivatives of δαβ commute (and vanish, since it’s constant), yet another expression

for the Ricci tensor, (with a different pattern of raised and summed indices), canbe derived:

Rµν =12εµ,

(∂2εα

∂uα∂uν− ∂2εα

∂uν∂uα

). (6.16)

Let’s discuss some of the basics of the transformation theory of the Dirac equa-tion again. In locally orthonormal coordinates, the Dirac equation is written,(

eµ∂µ + eµ∆µ

)ψ =

mcih

ψ,

where

∆µ = e−S ∂eS

∂xµ = −∂e−S∂xµ eS.

The last equality can be made because ∂µ(e−SeS) = 0.Now to make direct contact with quantum theory and the Dirac equation. Re-

call from Chapter 1, that for a locally orthogonal coordinate system which relatesback to a flat space where the free-particle Dirac equation is satisfied, the basisvectors can be written:

εα =1√

gααeSγαe−S. (6.17)

Here the γα are the basis vectors in the free system, and eS is the fundamentalquantum transformation. The coefficients, 1√

gαα, are the scale factors, which as of

Chapter 1 were introduced as an additional hypothesis and were not yet derivablefrom the eS. As we continue the process of arriving at the quantum Einstein equa-tion, we will see that the scale factors will arise naturally from the eS and not as


a separate hypothesis. Now continuing from Eq. 6.17, we can derive expressionsfor the derivatives. Initially, we have,

∂εα

∂uρ = −12(gαα)

−3/2 ∂gαα

∂uρ eSγαe−S + (gαα)−1/2 ∂eS

∂uρ γαe−S + (gαα)−1/2eSγα ∂e−S

∂uρ .

Then, using Eq. 6.17, this can be rearranged into:

∂εα

∂uρ =∂ log (gαα)−1/2

∂uρ εα +∂eS

∂uρ e−Sεα + εαeS ∂e−S

∂uρ .

Now, because ∂∂uρ (eSe−S) = ∂

∂uρ (1) = 0, we can make the replacement,

∂eS

∂uρ e−S = −eS ∂e−S

∂uρ ,

and then define an (intermediate) symbol, Pρ ≡ ∂eS

∂uρ e−S, giving,

∂εα

∂uρ =∂ log (gαα)−1/2

∂uρ εα +[Pρ, εα

].

Applying these techniques again allows us to calculate the second derivatives ofthe basis vectors to get

∂2εα

∂uα∂uν− ∂2εα

∂uν∂uα=

[(∂Pν

∂uα− ∂Pα

∂uν+ PνPα − PαPν

), εα

].

Then reintroducing the definition of Pν in terms of the transformation eS, we have,

∂Pν

∂uα− ∂Pα

∂uν+ PνPα − PαPν =

∂2eS

∂uα∂uνe−S − ∂2eS

∂uν∂uαe−S ≡ Qνα.

The matrix Qνα is explicitly antisymmetric, and is directly related to eS, e−S,and their second derivatives. So, returning to the expression for the Ricci tensor,Eq. 6.16, we have shown,

Rµν =12εµ, [Qνα, εα].

Now, Rµν must be symmetric in µ, ν, but it must also end up being neutral inthe spinor indices, since as we showed in Chapter 1, gµν must be neutral in the


spinor indices. Not any arbitrary Qνα will result in this, so not any arbitrary eS

will give an acceptable Rµν. For a given eS, if it turns out that

Qνα = iq(u)σνα = −12

q(u) [εν, εα] ,

where q(u) is a coefficient that’s is a function of the coordinate position compo-nents, uµ, then Rµν will be guaranteed to be symmetric and neutral in the spinorindices.

6.5 The axiomatic basis of the General Dirac Equation

When Dirac originally derived the equation that bears his name, as covered in hisfamous text, [1], his logical starting point or first principle was the covariance ofcurrent conservation:

∂µ jµf = 0,

where the conserved current, in a flat-space, is

jµf = ψ f γµψ f .

Requiring this to be true allows one to propose the Dirac equation, as an ansatz,and show that if ψ f satisfies the free-particle Dirac equation, then the current, jµ

f ,so defined, is automatically conserved.

A similar argument should also be the logical starting point in deriving theGeneral Dirac equation in a non-flat space. In this case, we let

jµb =√

gψbεµψb, (6.18)

where the current is now appended by a factor of√

g, giving it the status of atensor density. Then the conservation law

∂

∂xµb

jµb = 0,

will be covariant across all reference frames, warped or flat. The current jµb will

also transform as a contravariant four-vector that can be related back to a flat spaceas

jµb =

∂xµb

∂xνf

jµf .


This factor of√

g cannot be overlooked, and its inclusion highlights a subtleproperty of the general Lorentz group. That is, a transformation, eS, must satisfy:

eS†γ0 =

√gγ0e−S. (6.19)

ore−S†

γ0 =1√

gγ0eS. (6.20)

That is, commuting eS†or e−S†

across γ0 brings out a factor of√

g, or 1√g .

For the the scalar, ψψ, to be invariant in the expected way, it must also beappended by a factor of

√g:

√gψbψb =

√gψ f e−S†

γ0e−Sψ f =√

gψ f1√

gγ0eSe−Sψ f = ψ f ψ f .

Using these preliminary axioms we can derive the General Dirac Equation.First, if the free particle Dirac equation is satisfied in the flat-space:

γ0γµ ∂

∂xµf

ψ f =mcih

γ0ψ f ,

Then continuity will be satisfied in the free frame, ∂µ jµf = 0, as well as in the bound

frame. And from the form of the free-particle Dirac equation in the flat space,we can find the equation satisfied by the wave in the bound system, the GeneralDirac equation: this is the application of the general principle of relativity. Usingψ f = eSψb, and ∂

∂xµf=

∂xνb

∂xµf

∂∂xν

b, this at first becomes:

γ0γµ ∂xνb

∂xµf

∂

∂xνb

eSψb =mcih

γ0eSψb.

Carrying through the partial derivatives on the left, remembering that eS(xb) is ageneral transformation that depends upon the coordinates, xµ

b , we then have,

γ0γµ ∂xνb

∂xµf

∂eS

∂xνb

ψb + γ0γµ ∂xνb

∂xµf

eS ∂

∂xνb

ψb =mcih

γ0eSψb.


Next, multiply both sides by eS†:

eS†γ0γµ ∂xν

b

∂xµf

∂eS

∂xνb

ψb + eS†γ0γµ ∂xν

b

∂xµf

eS ∂

∂xνb

ψb =mcih

eS†γ0eSψb.

Now, commute eS† across γ0, using the defining group property of a generalLorentz transformation, Eq. 6.19:

√gγ0e−Sγµ ∂xν

b

∂xµf

∂eS

∂xνb

ψb +√

gγ0e−Sγµ ∂xνb

∂xµf

eS ∂

∂xνb

ψb =mcih√

gγ0e−SeSψb.

We see that the√

g factor cancels from both sides–we can also now cancel the γ0

from both sides, and use e−SeS = 1 on the right side, to get:

e−Sγµ ∂xνb

∂xµf

∂eS

∂xνb

ψb + e−Sγµ ∂xνb

∂xµf

eS ∂

∂xνb

ψb =mcih

ψb.

Remembering that ∂xνb

∂xµf

is not a 4× 4 Dirac matrix, so eS can be moved across it,

and again using eSe−S = 1, to insert an eS in the first term on the left, we have:

e−SγµeS ∂xνb

∂xµf

e−S ∂eS

∂xνb

ψb + e−SγµeS ∂xνb

∂xµf

∂

∂xνb

ψb =mcih

ψb.

Now we can recognize what we must take as a basic transformation axiom of thegeneral Dirac theory, (and quantum gravity):

ενb,b = e−SγµeS ∂xν

b

∂xµf

. (6.21)

where ενb,b is the representation of the unit vectors of the bound frame, in the

bound frame itself. This could be taken to be the γµ, which is what is done in thecase of a simple flat-space Lorentz boost between flat-space frames. In the case ofa warped-space bound frame, we could still impose the condition that εν

b,b = γν,since, according to the general principle of relativity, it’s always possible to find acoordinate system which is locally orthonormal. But that would be more restric-tive than we need to be. In the most general relativistic case, we won’t be workingin coordinates that are locally orthonormal.


Thus, the general Dirac equation is written,(εµ ∂

∂xµb+ εµe−S ∂eS

∂xµb

)ψb =

mcih

ψb.

This is the same as what we wrote in Chapter 1, but now we’ve fully taken into ac-count the properties of the general Lorentz transformation, including the factor of√

g that was needed to preserve general continuity, which we glossed over. Eventhough the factor of

√g does not appear explicitly in the general Dirac equation,

it is still implicitly present in the transformation eS. Given this, we can proposethat the most general Lorentz boost can be represented as:

eS = g1/4 e−i4 ωµνσµν , (6.22)

where ωµν are boost parameters that are functions of the coordinates xb, and thewarped-space Pauli matrices, σµν = i

2

[εµ, εν

], are defined in terms of the basis

vectors in the same way they are in a flat-space. We would then also have,

e−S = g−1/4 ei4 ωµνσµν . (6.23)

The form of the transformation in Eq. 6.22 and its inverse, Eq. 6.23 satisfy theconditions defining a general Lorentz boost, Eqs. 6.19 or 6.20. We must also pointof that the metric tensor, gµν, and its determinant, g, is a function of the boostparameters, ωµν. It’s specific functional dependence on ωµν can be worked outfrom the fact that,

gµν =∂xα

f

∂xµb

∂xβf

∂xνb

ηαβ,

And the fact that the transformation coefficients,∂xα

f

∂xµb

can also, in principle, be

expressed in terms of ωµν. Thus any general Lorentz transformations as expressedin Eqs. 6.22 and 6.23, is entirely specified by the boost parameters, ωµν. Thusall properties of the metric in the bound frame, including its scale factors, can beexpressed in terms of the boost parameters, ωµν, which relate it to a local flat-space. Remember, this local flat-space may not be accessible to observers andexperimenters physical reality (actually, it never truly is), but it still must exist asa mathematical reality, if the general principle of relativity is valid.

If we assume that the εµ retain some of the key properties of the γµ regard-ing transpose conjugation and commutation across γ0, namely, γ0εi = −εiγ0 for


spatial basis vectors, so εi†γ0 = γ0εi, and the like, this ensures that the definingproperty of the general Lorentz group, Eq. 6.19, is satisfied. Note that here, andthroughout, ε0 is not necessarily the same thing as γ0. Also note that, when wewrite ψb = ψ†

b γ0, it is written in terms of γ0, not ε0. Eqs. 6.22 and 6.23, can beuseful when considering specific problems, such as the Schwarzschild case.

Now, the εµb,b are, in general, functions of position that need to satisfy εµ · εν =

gµν. What can we learn about their spatial derivatives? If we choose to workin a locally orthonormal system, then, formally, since they have no dependenceon the bound coordinate, xµ

b , their partial derivatives would be zero. However,since we would be working in a system which is not actually flat, we know that inthe absolute sense as seen from the point of view of the flat-space related to thatbound-space by the boost, the orientation and magnitude of the bound coordinatebasis vectors will be different at xµ

b + δxµb than they are at xµ

b . Thus, to calculatethe partial derivatives of ε

µb,b, we should represent them in the free space before

taking differentials:

εµb, f =

∂xµf

∂xνb

ενb,b =

∂xµf

∂xνb

∂xνb

∂xαfe−SγαeS = δ

µα e−SγαeS.

So,ε

µb, f = e−SγµeS,

is the representation of the basis vectors of the bound system in the free system,and γµ are the unit basis vectors of the free system in the free system. Now wemay take the differentials with respect to the bound-space coordinates,

δεµb, f = ε

µb, f (xb + δxb)− ε

µb, f (xb) =

(∂e−S

∂xαb

dxαb

)γµeS + e−Sγµ

(∂eS

∂xαb

dxαb

),

remembering that eS depends upon xb but γµ does not. Now use 1 = eSe−S tointroduce factors where needed:

δεµb, f =

(∂e−S

∂xαb

dxαb

)eSe−SγµeS + e−SγµeSe−S

(∂eS

∂xαb

dxαb

).

Now we can transform this differential back to the bound system:

δεµb,b =

∂xµb

∂xνfεν

b, f =

(∂e−S

∂xαb

dxαb

)eS ∂xµ

b∂xν

fe−SγνeS +

∂xµb

∂xνfe−SγνeSe−S

(∂eS

∂xαb

dxαb

).


Referring back to Eq. 6.21, we can eliminate factors in favor of εµb,b:

δεµb,b =

(∂e−S

∂xαb

dxαb

)eSε

µb,b + ε

µb,b e−S

(∂eS

∂xαb

dxαb

).

The variations dxαb are arbitrary, so we may take them to only involve one coordi-

nate variation at a time, to give us the partial derivative of the bound basis vectorsin the bound system:

∂εµ

∂xαb=

∂e−S

∂xαb

eSεµb,b + ε

µb,be−S ∂eS

∂xαb

.

Since ∂∂xα

b(e−SeS) = ∂

∂xαb(1) = 0, it’s true that

∂e−S

∂xαb

eS = −e−S ∂eS

∂xαb

,

so the partial derivative becomes,

∂εµ

∂xαb= −e−S ∂eS

∂xαb

εµb,b + ε

µb,be−S ∂eS

∂xαb

,

or∂εµ

∂xαb=

[εµ, e−S ∂eS

∂xαb

],

and we’ve dropped the b, b subscript, since now it can be understood to be repre-sented entirely in the bound system. Thus we’ve proved the identity

∂εµ

∂xνb= [εµ, ∆ν] , (6.24)

Where, ∆µ = e−S ∂eS

∂xµb, are the same fields that appear in the General Dirac equa-

tion. This derivation was based upon the first principle of the general principle ofrelativity, and so must be true for any reference frame that can be related back toa flat-space where the free-particle Dirac equation is satisfied and the continuityequation is also satisfied.

From fundamental tensor-calculus, we also know that the derivatives of thebasis vectors are related to the Christoffel symbols,

∂εµ

∂xνb= −Γµ

νσεσ. (6.25)


This must still be true in the quantum theory, as it comes from a basic geometricargument on the parallel displacement of vectors. Thus, equating the two pro-vides a connection between the fields, ∆µ, and the Christoffel symbols,

[εµ, ∆ν] = −Γµνσεσ. (6.26)

This is similar to a formula derived by E. Schrodinger in his famous paper ’Diracelectron in the gravitational field’, [36]. Except that he wrote it as

∂εµ

∂xνb= [εµ, ∆ν]− Γµ

νσεσ.

He was very close to the correct formula, but writing it like this obscures a keyinsight. Schrodinger’s version states that the derivative of the basis vectors, andhence the warping of coordinates, has two separate contributions which must beadded: one from the EM fields via ∆µ, and another from the gravitational fieldsvia Γµ

νσ. He underscores this in the text and he seemed unsatisfied that there isn’tany deeper link between ∆µ and Γµ

νσ. The source of Schrodinger’s error was thathis formula didn’t come from the general principle of relativity leading to thegeneral Dirac equation via eS, but rather was confused by the pre-suppositionthat gravitational fields do not naturally appear in the Dirac equation. The correctversion, Eq. 6.26, shows that the EM field components, ∆µ, are connected to thegravitational field via the Christoffel symbols, Γµ

νσ. Thus gravitational and elec-tromagnetic fields are ultimately unified, and have as their common origin thetransformation, eS, that relates the wavefunction in the bound system to that in afree system, ala, ψ f = eSψb.

We can derive an additional formula. Take another partial derivative of bothsides of Eq. 6.26 and use Eq. 6.24 to substitute derivatives of εµ on the left, andEq. 6.25 in the same way on the right. Finally, interchange the free indices α and ν

to get a second equation and take the difference of the two to get:[εµ,(

∂∆ν

∂xα− ∂∆α

∂xν+ ∆α∆ν − ∆ν∆α

)]= −Rµ

βανεβ, (6.27)

Where Rµβαν is the Riemann tensor as it is traditionally defined in terms of the

Christoffel symbols and their derivatives.Both equations 6.26 and 6.27 can be rearranged to solve for the ∆µ and their

derivatives in terms of the Christoffel symbols or the Riemann tensor using the


same procedure. In both equations, the term on the right is a linear combinationof basis vectors. For the commutator term on the left to also turn out to be propor-tional to a linear combination of basis vectors, the object being commuted with εµ

must be proportional to the σµν. So, in Eq. 6.26, let,

∆ν = ∆ikν σik, (6.28)

where ∆ikν are c-number coefficients. (Before going any further it’s worth noting

that, because of the antisymmetric form of σik, this is equivalent to,

∆ν =12

(∆ik

ν − ∆kiν

)σik, (6.29)

so ∆ν really only depends on the difference, ∆ikν − ∆ki

ν ). Now, substituting Eq. 6.28into Eq. 6.26, on the left side, we have,

[εµ, ∆ν] = ∆ikν [εµ, σik] .

The term in the commutator on the right can be recast using fundamental identi-ties. First expanding it out using the definition of σik:

[εµ, σik] =i2[εµ, [εi, εk]] =

i2[εµ, εiεk]−

i2[εµ, εkεi] .

Then using an identity for a commutator involving products of matrices, (Eq. 9.4in the Appendix),

[A, BC] = B, AC− BC, A,

We have,[εµ, εiεk] = εi, εµεk − εiεk, εµ = 2δ

µi εk − 2δ

µk εi.

and, likewise,[εµ, εkεi] = 2δ

µk εi − 2δ

µi εk.

Thus,[εµ, σik] = 2i

(δ

µi εk − δ

µk εi)

.

We can then insert a dummy index to combine the εk and εi into a single term:

[εµ, σik] = 2i(δ

µi δ

ρk − δ

µk δ

ρi)

ερ.


This proves our assertion that if ∆µ is composed of σik, then the result of the com-mutator will be proportional to the εµ as required. So now we have the equation,

[εµ, ∆ν] = ∆ikν (2i)

(δ

µi δ

ρk − δ

µk δ

ρi)

ερ.

Now we can insert this into 6.26 to get:

∆ikν (2i)

(δ

µi δ

ρk − δ

µk δ

ρi)

ερ = −Γµνσεσ.

Now, dot both sides with εα, to get

∆ikν (2i)

(δ

µi δα

k − δµk δα

i)= −Γµ

νσgσα,

or (∆µα

ν − ∆αµν

)=

i2

Γµνσgσα.

So we’ve shown that the antisymmetric difference of the coefficients is propor-tional to Christoffel symbols. Inserting this into Eq. 6.29, we’ve now shown that,

∆µ =i4

gkσΓiµσσik. (6.30)

The same technique can be used to rearrange Eq. 6.27. Let,

Qνα =∂∆ν

∂xα− ∂∆α

∂xν+ ∆α∆ν − ∆ν∆α.

Then for equation Eq. 6.27 to be consistent, Qνα must be composed of σik,

Qνα = Qikνασik =

12

(Qik

να −Qkiνα

)σik.

The procedure continues as before. We find that,

Qνα =i4

Riβανgβkσik (6.31)

We can rearrange this by raising and lowering some indices:

Qνα =i4

Rikανσik.


Now, Rikαν is antisymmetric under interchange of ik, and να, and symmetric underinterchange of kα and iν. Let’s first swap να so that those indices appear in thesame order on both sides,

Qνα = − i4

Rikνασik. (6.32)

Then, using the antisymmetric property of Rikνα in ik and the definition of σik interms of εi, εk, this can be shown to be equivalent to,

Qµν =14

Rikµνεiεk. (6.33)

6.6 The Quantum Einstein Equation

First we write the second-order general Dirac equation as,

εµ(∂µ + ∆µ

)εν (∂ν + ∆ν)ψb = −

m2c2

h2 ψb,

or,

εµDµ ενDν ψb = −m2c2

h2 ψb,

where Dµ = ∂µ + ∆µ. Because the partial derivative of the εµ can be written interms of a commutation relation, Eq. 6.24, we can use that to ’move’ the second εν

across the first Dµ to get:

εµενDµDν ψb = −m2c2

h2 ψb,

just as can be done with the second-order free-particle Dirac equation. Now, thesecond-order covariant derivative terms, DµDν, are components of a second-ranktensor. And as is the case with any second-rank tensor, it can be written as a sumof symmetric and anti-symmetric parts:

DµDν =12Dµ, Dν+

12[Dµ, Dν

].

Thus, the second-order Dirac equation becomes:

εµεν

(12Dµ, Dν+

12[Dµ, Dν

])ψb = −

m2c2

h2 ψb. (6.34)


For now, let’s call the symmetric part:

Tµν =12Dµ, Dν, (6.35)

where the ’overbar’ indicates that it’s not quite yet the mass-energy tensor, Tµν,but we will find that it can be related to it. Looking more closely at the antisym-metric part, we have, at first:[

Dµ, Dν

]= ∂µ∂ν + ∂µ∆ν − ∆ν∂µ − ∂ν∆µ + ∆µ∂ν − ∂ν∂µ + ∆µ∆ν − ∆ν∆µ.

In the second-order general Dirac equation, the partial derivative terms above,like ∂µ∂ν, act directly on the wavefunction to the right, and so just give you thesecond derivative of each spinor component after acting. Since each spinor com-ponent is assumed to be a differentiable, continuous function, the order of partialdifferentiation can be interchanged, so those terms cancel. Then what remains is,[

Dµ, Dν

]=[∂µ, ∆ν

]−[∂ν, ∆µ

]+ ∆µ∆ν − ∆ν∆µ.

The commutators of the partials with ∆µ are simply the derivatives of ∆µ, so thisbecomes, [

Dµ, Dν

]=

∂∆µ

∂xνb− ∂∆ν

∂xµb+ ∆µ∆ν − ∆ν∆µ = Qµν,

and we showed above that Qµν can be expressed in terms of the Riemann tensor inEq. 6.33. Using all of this, we can now re-write the second-order Dirac equation,Eq. 6.34, as

εµεν

(Tµν +

116

Rikµνεiεk)

ψb = −m2c2

h2 ψb.

Now let’s exploit the symmetry of the Riemann tensor under exchange of the firstand last indices, iν. We can write,

Rikµνεµενεiεk =12

(Rikµνεµενεiεk + Rνkµiε

µεiενεk)

(6.36)

=12

Rikµνεµ(

ενεi + εiεν)

εk = gνiRikµνεµεk

But, that brings in the Ricci tensor, Rkµ = gνiRikµν, so

Rikµνεµενεiεk = Rkµεµεk = Rµνεµεν.


Thus, the second-order Dirac equation becomes,

εµεν

(Tµν +

18

Rµν

)ψb = −

m2c2

h2 ψb.

Now, because Rµν and Tµν are symmetric in µν, this can be rearranged as,

gµν

(8Tµν + Rµν +

2m2c2

h2 gµν

)ψb = 0.

Note also that this sum is essentially the trace of the tensor

8Tµν + Rµν +2m2c2

h2 gµν.

The fact that the trace of a tensor is zero does not allow you to conclude that eachcomponent of the tensor is zero. So we must keep looking...

We’ve shown that

Qνα =∂∆α

∂xν− ∂∆ν

∂xα+ ∆ν∆α − ∆α∆ν,

Using the fact that,

∆ν = e−S ∂eS

∂xν= −∂e−S

∂xνeS,

It’s simple to show that,

Qνα = e−S(

∂2eS

∂xν∂xα− ∂2eS

∂xα∂xν

).

Recalling Eqs. 6.22 and 6.23, it can be shown that the factors of√

g in the full eS

transformation drop out of the final expression for Qνα, giving,

Qνα = e−Σ(

∂2eΣ

∂xν∂xα− ∂2eΣ

∂xα∂xν

).

where Σ = i4 ωµνσµν. Recalling Eq. 6.32 that relates Qνα to the Riemann tensor, we

now have shown,

i4

Rikνασik = e−Σ(

∂2eΣ

∂xν∂xα− ∂2eΣ

∂xα∂xν

). (6.37)


This is an important result, because it relates the boost parameters ωµν–the param-eters that ’generate’ or define the boost into the warped coordinate system–to theRiemann tensor which characterizes the geometry of that warped coordinate sys-tem. In a sense we’ve now ’closed the loop’, and even though it doesn’t look likeit just yet, this is essentially the quantum Einstein equation. If the boost is char-acterized by a small parameter, then the lowest-order terms from the expressionabove will give the classical Einstein equation. Let’s continue...

Here is the result. We find that,

Rµν =2πG

c4 ρc2(

ωσµωνσ

)−1+O(ω4) + . . . (6.38)

Where ωµν are the parameters of the boost leading to the general Dirac equation.Therefore, we can interpret

T∗µν = −14

ρc2(

ωσµωνσ

)−1...

...more to come...

6.7 Hubble Expansion and Quantum Gravity

In addition to investigating the Einstein equation and its quantum equivalent,more can be learned about the theory of quantum gravity and how it relates toHubble expansion by considering the equation of continuity and the conserva-tion of probability current. One reason that gravitational fields must exist is tooffset, or ’balance’ the probability flux created by the expansion of the universalwavefunction.

Now, going back to the principle of relativity as described in Chapter 1, theuniversal wave function in the local bound system must be related to the universalwave function in a local free system by:

ψ f = eS(xb)ψb

Hubble expansion implies that this universal wavefunction is growing expo-nentially, and that the transformation generator, S, must include a slowly growingtime dependent term:

S =12

H0t + ...


For now I’ve left the other terms unspecified to focus on the effect that the timeexpansion term has. We can see that with an exponentially growing wavefunction,the expectation value of the distance between two points must grow as:

〈x〉 =∫

ψ†f x ψ f dVf =

∫ψ†

b e12 H0t x e

12 H0tψb dVf = eH0t

∫ψ†

f (t0) x ψ f (t0) dVf

(6.39)where the volume elements and expectation values are evaluated in the free-particle reference frame. Therefore

〈x(t)〉 = eH0t 〈x0〉 ,

so that the expectation value of the distance between any two points grows expo-nentially. Taking the first time derivative of this gives us the classical version ofthe Hubble Law:

dxdt

= H0 eH0t 〈x0〉or

v(t) = H0 x(t)

and that the further apart those two points were originally, the faster the reces-sional velocity between them will be. This is how Hubble growth is incorporatedinto quantum theory. We can also see that since its an exponential law, the rate ofexpansion will also be accelerating with time:

a(t) =dvdt

= H20 x(t)

Discussion

Note to Editor: This chapter, along with Chapter 1, form the heart of this book.There will be more discussion here, along with more of the details above for thederivation of Eq. 6.38 from Eq. 6.37.

February 22, 2020

It occurred to me that the most general transformation, eS, should involve trans-lations as well as rotations/boosts. The generator of the translation transforma-tions is the canonical momentum, pν, ala Sp =

∫dxµ pµ. The generator of the


boosts/rotations is Jµν = Lµν + Sµν. The most general transformation would in-volve both translation and boost/rotation. This could be built up infinitessimaly.

Sj =∫

dθµνJµν = (j + 1/2)h

Jµν = Lµν + Sµν

Sp =∫

dxµ pµ = nph

pµ =∂

∂xµ + ∆µ

eStot = eSp+Sj

∂εµ

∂xν=

[εµ, e−S ∂eS

∂xν

]= −Γµ

νσεσ

∂eS

∂xν=

∂

∂xν

(1 +

∫dxµ pµ +

12

∫dxµ pµ

∫dxµ pµ + . . .

)= pν +

12

pν (∫

dxβ pβ) +12(∫

dxα pα) pν + . . .

Therefore,

e−S eS

∂xν=

(1−

∫dxµ pµ + . . .

)(pν +

12

pν (∫

dxβ pβ) +12(∫

dxα pα) pν + . . .)

= pν − (∫

dxµ pµ) pν +12

pν (∫

dxβ pβ) +12(∫

dxα pα) pν

= pν +12

pν (∫

dxβ pβ)−12(∫

dxα pα) pν

= pν +12

[pν,∫

dxβ pβ

]+ . . .

Therefore, [εµ,

12

[pν,∫

dxβ pβ

]+ . . .

]= −Γµ

νσεσ


Therefore,

∂

∂xα[

εµ,12

[pν,∫

dxβ pβ

]+ . . .

] = −

∂Γµνβ

∂xαεβ + Γµ

νσΓσαβεβ

Therefore,

[

∂εµ

∂xα,

12

[pν,∫

dxβ pβ

]+ . . .

]+

[εµ,

12

[∂pν

∂xα,∫

dxβ pβ

]+ . . .

]+

[εµ,

12[pν, pα] + . . .

] = −

∂Γµνβ

∂xαεβ + Γµ

νσΓσαβεβ

Therefore, to lowest order in h (and/or dτ), (which I have to double-check),[εµ,

12[pν, pα]

]= −

∂Γµνβ

∂xαεβ + Γµ

νσΓσαβεβ

Now, even though [ε, pν] = 0 normally, that only holds if εµ and pν are evalu-ated at the same moment in time. At different times they don’t commute, so theabove commutator is not automatically zero. Interchanging the free indices α andν, we obtain a second equation,[

εµ,12[pα, pν]

]= −

∂Γµαβ

∂xνεβ + Γµ

ασΓσνβεβ

Then, taking the difference of the two,

[εµ, [pν, pα]] = −∂Γµ

νβ

∂xαεβ + Γµ

νσΓσαβεβ +

∂Γµαβ

∂xνεβ − Γµ

ασΓσνβεβ

A little more analysis could turn this into the Einstein equation...Before doing that, I need to introduce h properly, as in eS → e

ih S, and be careful

about operators evaluated at different moments in time. Perhaps also using thealternative definition of ex as the product of a sequence of operators at differenttimes as

ψ(x, t+ 2dt) = eih Sψ(x, t) =

(1 +

ih

∫dxβ pβ(t + dt)

)(1 +

ih

∫dxα pα(t)

)ψ(x, t)


February 25, 2020

I need to re-visit the basics of the propagator and field theory. Huygen’s Principleis the starting point:

θ(t′ − t)ψ+(x′, t′) = i∫

d3x G+(x′, t′, x, t)ψ+(x, t)

θ(t− t′)ψ−(x′, t′) = i∫

d3x G−(x′, t′, x, t)ψ−(x, t)

Let

ε0H(x′, t′) =(

mc2 − ihcε0∆0 − ihcεi(∂

∂x′i+ ∆i(x′, t′))

)Thus, if ψ+(x′, t′) is a solution of the general Dirac equation at time t′, then the leftside of the forward-time Huygen’s principle is:

ε0(

ih∂

∂t′− H(x′, t′)

)θ(t′ − t)ψ+(x′, t′) = ih ε0 δ(t′ − t)ψ+(x′, t′)

To be consistent, the right hand side of the equation implies,

ε0(

ih∂

∂t′− H(x′, t′)

)G+(x′, t′, x, t) = h ε0 δ(t′ − t)δ3(x′ − x).

Divide through by hc, and use the definition of H(x′, t′) to get,(iD′ − mc

h

)G+(x′, t′, x, t) =

1c

ε0 δ(t′ − t)δ3(x′ − x).

or, (iD′ − mc

h

)G+(x′, t′, x, t) = ε0 δ4(x′ − x).

Now ε0(x′, t′) is a function of the primed space-time variables, and because

[εµ, ∂ν + ∆ν] = 0

...


February 26, 2020

I’ve proved that[εµ, ∂ν + ∆ν] = 0.

What, then, is [εµ, ∂ν + ∆ν

]=?

Is it also = 0 ? ... No, it’s not. Starting from [εµ, ∂ν + ∆ν] = 0, and using εµ =gµαεα, you can show that,

[εβ, ∂ν + ∆ν

]= gµβ

∂gµα

∂xνεα.

Using the parallel displacement argument, δ(gµν Aµ Aν) = 0 and that δAν =Γα

νσ Aαdxσ, you can relate this to the Christoffel symbols, as[εβ, ∂ν + ∆ν

]= −

(δα

µδλβ + gµβgλα

)Γµ

λν εα.


February 28, 2020

Rµν is only symmetric if,

∂gσα

∂xβ

∂gσα

∂xµ −∂gσα

∂xµ

∂gσα

∂xβ= 0

Proof. Since,

Γσµσ =

12

gσα ∂gσα

∂xµ .

Using this to calculate∂Γσ

µσ

∂xβ and∂Γσ

βσ

∂xµ , you find that for these two terms to be equal,the above must be satisfied.

Thus, the Einstein equation is only strictly true and consistent if the above issatisfied.

March 3, 2020

Using the identities, [εµ, σνα

]= 2i

(gµνδ

ρα − gµαδ

ρν

)ερ

as well as,ενα = gνα − iσνα

You can show that

ωµνωαβσµνσαβ = 3 ωµνωµν −ωµνωαβσνασβµ.

This also holds for any cyclic permutations, µναβ→ ναβµ→ αβµν, and so on.Another identity that I derived back around 12/28/2019: Using σµν = i

2 [εµ, εν]

and ∂εµ

∂xα = [εµ, ∆α], you can show that,

∂σµν

∂xα= [σµν, ∆α]

Continuing in that manner, you can show,

∂2σµν

∂xα∂xβ− ∂2σµν

∂xβ∂xα=

[σµν,

∂∆β

∂xα− ∂∆α

∂xβ+ ∆α∆β − ∆β∆α

]


March 4, 2020

But,

Qαβ =∂∆β

∂xα− ∂∆α

∂xβ+ ∆α∆β − ∆β∆α

So,∂2σµν


∂xβ∂xα=[σµν, Qαβ

](6.40)

Now, another expression for Qαβ, is

Qαβ = −e−Σ(

∂2eΣ

∂xα∂xβ− ∂2eΣ

∂xβ∂xα

), (6.41)

from Chapter 7, where Σ = − i4 ωµνσµν. I’ve been having a hard time trying to sum

the series,

eΣ = 1− i4

ωµνσµν +12

(− i

4ωµνσµν

)(− i

4ωαβσαβ

)+ . . . (6.42)

But, I don’t have to sum that series, I need to sum the one above it, Eq. 6.41.Using Eq. 6.42 in Eq. 6.41, we find,

− e−Σ(

∂2Σ∂xα∂xβ

− ∂2Σ∂xβ∂xα

)= − i

4ωµν

(∂2σµν


∂xβ∂xα

)+

12(

i4)2ωµνωλκ

[(∂2σµν


∂xβ∂xα

), σλκ

]+O(ω3) + . . .

Then, using Eq. 6.40 to replace terms on the right side, and Eq. 6.41, on the leftside, we have

−Qαβ = − i4

ωµν

[σµν, Qαβ

]+

12(

i4)2ωµνωλκ

[[σµν, Qαβ

], σλκ

]+O(ω3) + . . .

or,

−Qαβ =

[− i

4ωµνσµν, Qαβ

]− 1

2

[− i

4ωλκσλκ,

[− i

4ωµνσµν, Qαβ

]]+O(ω3) + . . .

Now, comparing this to the well-known formula,

e−A B eA = B− [A, B] +12!

[A, [A, B]]− 13!

[A, [A, [A, B]]] + . . .


or,

B− e−A B eA = [A, B]− 12!

[A, [A, B]] + . . .

But, before we can use that, we have to use

Qαβ = − i4

Rikαβ σik

and the well-known formula (pg. 39 of Peskin & Schroeder),[σµν, σik

]= 2i

(gνiσµk − gµiσνk − gνkσµi + gµkσνi

)


March 6, 2020

Redo of proof of

Qνα =i4

Riναk σik + Fνα

where, Fνα is a c-number field, and

Qνα =∂∆ν

∂xα− ∂∆α

∂xν+ ∆α∆ν − ∆ν∆α

Starting from,∂εµ

∂xν= [εµ, ∆ν] = −Γµ

νσ εσ,

take another derivative of both sides. This gives,

∂2εµ

∂xα∂xν=

[∂εµ

∂xα, ∆ν

]+

[εµ,

∂∆ν

∂xα

]= −∂Γµ

νσ

∂xαεσ − Γµ

νσ∂εσ

∂xα

Then replacing derivatives of εµ on the left using ∂εµ

∂xα = [εµ, ∆α], and on the rightwith ∂εσ

∂xα = −Γσαρερ, you get,

εµ∆α∆ν − ∆αεµ∆ν − ∆νεµ∆α + ∆ν∆αεµ +

[εµ,

∂∆ν

∂xα

]= −∂Γµ

νσ

∂xαεσ + Γµ

νρΓρασεσ

Then, in the above, interchange the free indices α and ν to obtain a second equa-tion, and then subtract the second equation from the first to get,[

εµ,∂∆ν

∂xα− ∂∆α

∂xν+ ∆α∆ν − ∆ν∆α

]=

(−∂Γµ

νσ

∂xα+

∂Γµασ

∂xν+ Γµ

νρΓρασ − Γµ

αρΓρνσ

)εσ.

Now, using P. Johnson’s definition of the Riemann Tensor, from his Eq. 5.53,(swapping his indices for mine, ρ→ µ, λ→ α, µ→ ν, ν→ σ),

Rµανσ = ∂νΓµ

ασ − ∂αΓµνσ + Γρ

ασΓµνρ − Γρ

νσΓµαρ,

we have now shown,[εµ, Qνα] = Rµ

ανσ εσ.

Now, propose,

Qνα = Qikνα σik + Fνα =

12

(Qik

να −Qkiνα

)σik + Fνα (6.43)


Therefore,[εµ, Qνα] = Qik

να [εµ, σik] .

But, fundamental identities can be used to show,

[εµ, σik] = (2i)(δ

µi δ

ρk − δ

µk δ

ρi)

ερ.

so,Qik

να [εµ, σik] = Qikνα (2i)

(δ

µi δ

ρk − δ

µk δ

ρi)

ερ = (2i)(Qµρ

να −Qρµνα

)ερ.

Thus,(2i)

(Qµρ

να −Qρµνα

)ερ = Rµ

ανσ εσ.

Then, dotting both sides with ερ, and dividing both sides by 4i, and letting ρ→ kand µ→ i we have,

12

(Qik

να −Qkiνα

)= − i

4gkσ Ri

ανσ.

Comparing this to the proposed form, Eq. 6.43, and lowering and raising someindices, we have,

Qνα = − i4

Riανkσik + Fνα.

Finally, use the fact that Riανk = −Riναk, to get,

Qνα =i4

Riναk σik + Fνα

Q.E.D.


March 10, 2020

R =14

ωαµωiµ Rαi

or,

4 = ωαµωiµ

1R

Rαi

Let,

Rαi =1R

Rαi

so,4 = ωαµωi

µ Rαi.

But,4 = RαiRαi = δα

α

Therefore,Rαi = ωαµωi

µ. (6.44)

or, multiplying through by R,

Rαi = Rωαµωiµ (6.45)

Now, the Einstein Eq. is, (using P. Johnson’s sign conventions),

Rµν − 12

gµνR = −κTµν,

whereκ =

8πGc4 .

This implies,R = κT

The Einstein Eq. can also be written,

Rµν = −κT∗µν (6.46)

where T∗µν = Tµν − 12 gµνT. It’s also the case that T∗ = −T.

Now, referring back to Eq. 6.45, this implies,

Rωαµωiµ = −κT∗αi


or

ωµρωνρ = −8πG

c41R

(Tµν − 1

2gµνT

)Thus, if the classical Einstein Eq. is solved, and both R and Tµν are known orgiven, the boost parameters can be found...Also, this shows that they’re small.

Going back to the Einstein Eq. in it’s alternate form, Eq. 6.46, this can also bewritten,

Rµν = T∗µν

Comparing to Eq. 6.44, this tells us,

T∗µν = ωµρωνρ

or,

ωµρωνρ = − 1

T

(Tµν − 1

2gµνT

)But, since T = ρc2,

ωµρωνρ = −

(1

ρc2 Tµν − 12

gµν

).

So the boost parameters are related to the four-velocities of the matter, since,

Tµν = ρdxµ

dτ

dxν

dτ.

March 28, 2020

Now,

∆µ = e−S ∂eS

∂xµ .

But sinceeS = g1/4e−

i4 ωµνσµν

= g1/4eΣ

andΣ = − i

4ωµνσµν

Therefore,

∆µ = −14

∂ ln g∂xµ + e−Σ ∂eΣ

∂xµ .


So let

∆µ = e−Σ ∂eΣ

∂xµ .

Now,

e−Σ ∂eΣ

∂xµ = e−Σ ∂

∂xµ eΣ − ∂

∂xµ

And since

e−ABeA = B− [A, B] +12!

[A, [A, B]]− 13!

[A, [A, [A, B]]] + . . . ,

Therefore,


∂xµ = −[

Σ,∂

∂xµ

]+

12!

[Σ,[

Σ,∂

∂xµ

]]− 1

3!

[Σ,[

Σ,[

Σ,∂

∂xµ

]]]+ . . .

and since −[Σ, ∂

∂xµ

]= ∂Σ

∂xµ , therefore,


∂xµ =∂Σ∂xµ −

12!

[Σ,

∂Σ∂xµ

]+

13!

[Σ,[

Σ,∂Σ∂xµ

]]+ . . . (6.47)

Now,∂Σ∂xα

= − i4

∂ωµν

∂xασµν − i

4ωµν

∂σµν

∂xα,

but,∂σµν

∂xα= [σµν, ∆α] = [σµν, ∆α] .

So,∂Σ∂xα

= − i4

∂ωµν

∂xασµν − i

4ωµν [σ

µν, ∆α] .

Then, putting this back into Eq. 6.47:

∆α =− i4

∂ωµν∂xα σµν− 1

2!

[− i

4 ωρσσρσ,− i4

∂ωµν∂xα σµν

]+ 1

3!

[− i

4 ωζησζη ,[− i

4 ωρσσρσ,− i4

∂ωµν∂xα σµν

]]+...

− i4 ωµν[σµν,∆α]− 1

2! [−i4 ωρσσρσ,− i

4 ωµν[σµν,∆α]]+ 13! [−

i4 ωζησζη ,[− i

4 ωρσσρσ,− i4 ωµν[σµν,∆α]]]+...

most explicitly. Then, replacing terms with Σ = − i4 ωµνσµν, and letting ∂αΩ =

− i4

∂ωµν

∂xα σµν to make it more concise for more algebraic manipulation, we have,

∆α = ∂αΩ− 12!

[Σ, ∂αΩ] +13!

[Σ, [Σ, ∂αΩ]] + . . .

+ [Σ, ∆α]−12!

[Σ, [Σ, ∆α]] +13!

[Σ, [Σ, [Σ, ∆α]]] + . . .


Or,

∆α − [Σ, ∆α] +12!

[Σ, [Σ, ∆α]]−13!

[Σ, [Σ, [Σ, ∆α]]] + . . .

= ∂αΩ− 12!

[Σ, ∂αΩ] +13!

[Σ, [Σ, ∂αΩ]] + . . .

But, the left hand side of this is just:

∆α − [Σ, ∆α] +12!

[Σ, [Σ, ∆α]]−13!

[Σ, [Σ, [Σ, ∆α]]] + · · · = e−Σ∆αeΣ

For the right hand side, let

Λα = ∂αΩ− 12!

[Σ, ∂αΩ] +13!

[Σ, [Σ, ∂αΩ]]− 14!

[Σ, [Σ, [Σ, ∂αΩ]]] + . . . (6.48)

So,e−Σ∆αeΣ = Λα

or,∆α = eΣΛαe−Σ

Now, out to fourth order in ω, it can be shown that,

eΣΛαe−Σ = ∂αΩ− 12!

[Σ, ∂αΩ] +13!

[Σ, [Σ, ∂αΩ]] +14!

[Σ, [Σ, [Σ, ∂αΩ]]] + . . .

although, it’s apparently not the case that eΣΛαe−Σ = Λα, since looking at Eq.6.48, the fourth order term is slightly different. So, therefore we’ve shown that

∆α = ∂αΩ− 12!

[Σ, ∂αΩ] +13!

[Σ, [Σ, ∂αΩ]] +14!

[Σ, [Σ, [Σ, ∂αΩ]]] + . . .

Now, returning to the fact that

∆α =i4

gijΓiασσjσ

we have thus shown

Γµανσµν = −∂ωµν

∂xασµν +

12

∂ωµν

∂xα

[− i

4ωρσσρσ, σµν

]+ . . .


Since all terms in the series are proportional to σµν, that can be factored out. Tosecond order, we have,

Γµαν = −∂ωµν

∂xα− gβρ ∂ωµβ

∂xαωρν + . . .

This can be re-arranged to eliminate gβρ. By going back to the explicit form justafter Eq. 6.47, and raising and lowering appropriate indices,

Γµαν = −∂ω

µν

∂xα−

∂ωµβ

∂xαω

βν + . . .

Thus, specifying, ωµν, the transformation generators for the boost between the free

and warped space, specifies the Christoffel symbols of the warped space. Also,since Γµ

αν has to be symmetric in αν, this puts a constraint on ω:

∂ωµν

∂xα=

∂ωµα

∂xν

Now we can express the Ricci tensor, Rµν, also in terms of ωµν.

Now, since the Ricci tensor can be expressed in terms of the Christoffel symbolsas (using P. Johnson’s sign convention),

Rµν = ∂ρΓρµν − ∂µΓρ

ρν + ΓσµνΓρ

ρσ − ΓσρνΓρ

µσ

Using the above expression for Γµαν in terms of ω

µν, we find

Rµν = −2∂ω

ρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + . . .

Thus the Ricci tensor is also entirely determined by the ωµν. Since the Einstein

equation can be written

Rµν = −8πGc4 T∗µν

We can thus interpret the right-hand side of the Einstein equation, when writtenin terms of ω

µν and its derivatives, as the stress-energy tensor.

−8πGc4 T∗µν = −2

∂ωρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + . . .


or

T∗µν =c4

4πG

∂ωρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + . . .

The left hand side of the Einstein equation, when expressed in terms of gµν and itsderivatives, will be a different way to express Rµν.

So, for a given mass-energy distribution, there will boost, ω, resulting in thatdistribution, specified as:

∂ωρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + · · · = 4πG

c4 T∗µν

So the Einstein equation follows as a consequence of the General Dirac Alge-bra, and the properties of the general Lorentz boost leading to the general DiracEquation. Thus the Einstein-Hilbert Lagrangian method is superfluous.

April 8, 2020

The goal is to prove:∂εµ

∂xα=[εµ, ∆α

], (6.49)

And,∂σ ν

µ

∂xα=[σ ν

µ , ∆α

]. (6.50)

Now, because of the first principle, that

εµb,b · εν b,b = δ

µν ,

then, since

εµb,b =

∂xµb

∂xαfe−SγαeS,

It must be true that,

εν b,b =∂xβ

f

∂xνb

e−SγβeS.

Thus,

εµ b, f =∂xα

b

∂xµf

εα b,b =∂xα

b

∂xµf

∂xβf

∂xαb

e−SγβeS = δβµe−SγβeS = e−SγµeS.


so we’ve proved,εµ b, f = e−SγµeS.

Therefore,

δεν b, f =∂e−S

∂xαb

dxαb γνeS + e−Sγν

∂eS

∂xαb

dxαb

=∂e−S

∂xαb

eSdxαb e−SγνeS + e−SγνeSe−S ∂eS

∂xαb

dxαb

= −dxαb e−S ∂eS

∂xαb

e−SγνeS + e−SγνeSe−S ∂eS

∂xαb

dxαb .

Now,

δεν b,b =∂xµ

f

∂xνb

δεµ b, f

Therefore,

δεν b,b = −dxαb e−S ∂eS

∂xαb

∂xµf

∂xνb

e−SγµeS +∂xµ

f

∂xνb

e−SγµeSe−S ∂eS

∂xαb

dxαb .

but, as we established,

εν b,b =∂xµ

f

∂xνb

e−SγµeS,

so,

δεν b,b = −dxαb e−S ∂eS

∂xαb

εν b,b + εν b,be−S ∂eS

∂xαb

dxαb .

Thus, we’ve shown,∂εν b,b

∂xαb

=

[εν b,b, e−S ∂eS

∂xαb

]So we proved,

∂εν

∂xα= [εν, ∆α] .

QED for Eq. 6.49.


An interesting consequence, or corollary of Eq. 6.49 , is that it puts an addi-tional symmetry restriction on the Christoffel symbols, that we don’t usually getin classical tensor calculus. Namely, it follows that,

Γρµα = −Γµαρ (6.51)

This will end up being a restriction on ω νµ .

Now to prove Eq. 6.50. It begins with,

σ νµ =

i2[εµ, εν

],

then taking the derivative,

∂σ νµ

∂xα=

i2

[∂εµ

∂xα, εν

]+

i2

[εµ,

∂εν

∂xα

]=

i2[[

εµ, ∆α

], εν]+

i2[εµ, [εν, ∆α]

]= −

[∆α, σ ν

µ

]So we’ve proved,

∂σ νµ

∂xα=[σ ν

µ , ∆α

]QED on Eq. 6.50.

Now, the reason for deriving these identities, especially Eq. 6.50 , is for takingderivatives of the generator of the transformation, Σ,

Σ = − i4

ωµνσµν = − i4

ωµνσ ν

µ ,

since it will be more convenient to write ω and σ as mixed tensors. Thus,

∂Σ∂xα

= − i4

∂ωµν

∂xασ ν

µ +− i4

ωµν

∂σ νµ

∂xα.

Therefore,∂Σ∂xα

= − i4

∂ωµν

∂xασ ν

µ +− i4

ωµν

[σ ν

µ , ∆α

].

Thus, the work that we entered under March 28, 2020 (actually it was March30), holds, and we have, to second order in ω,

Γµανσ ν

µ = −∂ωµν

∂xασ ν

µ +12

∂ωµν

∂xα

[− i

4ω

ρσσ σ

ρ , σ νµ

],


and therefore,

Γµαν = −∂ω

µν

∂xα−

∂ωµβ

∂xαω

βν

So, the derivation we did that day for the Ricci tensor also holds,

Rµν = −2∂ω

ρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + . . .

Thus the Ricci tensor is clearly symmetric in µν as it should be, and the interpre-tation of the right-hand-side of this as the tensor T∗µν, related to the stress-energytensor also holds. Thus, we’ve derived the Einstein equation,

Rµν = −8πGc4 T∗µν.

where4πG

c4 T∗µν =∂ω

ρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + . . .

April 9, 2020

So now, this result,

Rµν = −2∂ω

ρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + . . .

or4πG

c4 T∗µν =∂ω

ρβ

∂xµ

∂ωβρ

∂xν+O(ω3) + . . .

makes sense, because the boost generators, or more specifically, their first deriva-tives, can be interpreted as a four-velocity or momentum. As in the case of aspecial Lorentz boost, those parameters end up being ’rapidities’ directly relatedto the four-velocity of the boosted frame. Now, Einstein proposed that the stress-energy tensor should be written

Tµν = ρdxµ

dτ

dxν

dτ,

and this is consistent in many ways. It causes the Einstein equation to reduce tothe Poisson equation, ∇2φ = −4πGρ, in the non-relativistic, small-field limits.

CHAPTER 7

The Space-Time Lattice

My goal in this chapter is to show that the cutoff procedure of renormalizationis equivalent to requiring that the Dirac wavefuntion, ψ, is normalizable over adiscrete space-time lattice.

7.1 Discrete Fourier Transform

Consider a 1D lattice of points in space (or time), with a lattice spacing, dx, (whichwe will see later will be related to the particle mass as dx ∝ h

2mc ). Let ψ [x]be a function of the discrete variable [x] = ndx, where n is an integer, n =0,±1,±2, . . . . Then, a discrete Fourier transform of some function, ψ, can be de-fined as:

ψ(k) = ∑[x]

e−ik[x] ψ [x] . (7.1)

Note that this transformed function is periodic over an interval 2K = 2πdx , such

that ψ(k + 2K) = ψ(k). In this sense one could say that this is a bandwidth-limitedFourier transform, since all of the ’information’ encapsulated by the function iscontained in an interval of size 2K, such as −K < k < +K. The original function

137

CHAPTER 7. THE SPACE-TIME LATTICE 138

ψ [x] can then be recovered from this:

ψ [x] =1

2K

∫ K

−Kψ(k)eik[x]dk

= ∑[y]

ψ [y]1

(2K)

∫ K

−Keik[x−y]dk

= ∑[y]

ψ [y] δ [x− y] .

where,

δ [x− y] =1

2K

∫ K

−Keik[x−y]dk =

sinK[x− y]K[x− y]

=

1, if [x] = [y]0, otherwise

Also, it’s easy to see that if the function is normalized over an infinite, discretelattice in x-space, then it’s also normalized over a continuous, finite interval ofsize 2K:

∑[x]

ψ∗ [x]ψ [x] =1

2K

∫ K

−Kψ∗(k)ψ(k)dk,

which is called a Parseval condition. This region, k ∈ [−K, K] can also be called aBrillouin zone, as it is in the context of solid state physics.

In this simple demonstration, we witness a general mathematical theorem atplay: A function defined over an infinite, discrete lattice in x-space is periodic (orhas a cutoff) in k-space. The inverse of this theorem is also true: if a function isperiodic (or has a cutoff) in k-space, then it is defined (and normalizable) overan infinite lattice in x-space. There aren’t any radical new ideas here–I’m justpointing out a basic aspect of Fourier transform mathematics: an infinite latticein one space corresponds to a continuous, periodic interval in the transformspace.

Now here’s a quantum leap in creative, lateral thinking that goes back toClaude Shannon: A continuous, bandwidth-limited interpolation in x-space can beinferred by letting [x]→ (x), due to the well-defined nature of eikx for continuousarguments:

ψ(x) =1

2K

∫ K

−Kψ(k)eikxdk =

12K ∑

[y]

∫ K

−Kψ[y]eik(x−[y])dk,

= ∑[y]

sinK(x− y]K(x− y]

ψ[y]. (7.2)


This is known as Shannon’s sampling theorem, [28]. Claude Shannon was an elec-trical engineer and he originally derived this theorem in the context of analyzingdigital discrete-time signals sent over telegraph lines. It continues to be an es-sential mathematical tool in modern digital signal processing and information theory.Note also that the interpolated value of ψ(x) in between the lattice (or sample )points depends upon a sum of the values of ψ at every other discrete point on theinfinite lattice. This is an important echo (or foreshadowing) of Feynmann’s sumover all paths, which will see is closely related.

A key aspect of discrete Fourier/Shannon analysis is the definition of the delta-function that emerges:

δ(x− y] =1

2K

∫ K

−Keik(x−y]dk. (7.3)

In terms of this delta function, the interpolation of Shannon’s sampling theoremis simply

ψ(x) = ∑[y]

δ(x− y]ψ[y].

We see that, unlike the case of the usual continuous delta-function, this bandwidth-limited delta-function does not diverge and is finite everywhere (if the k-spaceintegral is real). In this sense, it’s more like a Kronecker delta, and it is exactly aKronecker delta if the first argument, x, in (x− y] is discrete. Otherwise it behavesas a kernel for a ’perfect’ interpolation between lattice points. Perfect, so long as theinput function, ψ(x) – the signal in Shannon’s telegraphy theory – is intrinsicallybandwidth-limited.

In the transform k-space, there’s also a delta-function, but that one is continu-ous and periodic:

δ(k− q) =1

2K ∑[x]

e−i(k−q)[x] (7.4)

This diverges if k = q, but its integral over the first Brillouin zone is unity:∫2K

δ(k− q)dk =∫

2K

12K ∑

[x]ei(k−q)[x]dk = 1.

To verify this, just examine the integral of this k-space delta over an interval of


length 2K, containing the point k = q:∫ K

−Kδ(k− q)dk =

∫ K

−K

12K ∑

[x]e−i(k−q)dk

=1

2K ∑[x]

eiq[x] 1−i[x]

(e−iK[x] − eiK[x])

= ∑[x]

eiq[x] sinK[x]K[x]

= 1.

The last step in the equality holds since sinK[x]K[x] = 1 when [x] = 0, and is zero for

all other points on the lattice. Note also the cardinality, (in the sense of Cantor), ofthe divergence of δ(k− q) when k = q:

δ(0) = ∑[x]

e−i0[x] = 1 + 1 + 1 + · · · = ℵ0.

The cardinality of the infinity is of the order of ℵ0, since 1 + 1 + 1 + . . . is thenumber of integers that exist – an infinity which is not as ’large’ as ℵ1. This meansthat the k-space continuum is not as dense as the usual continuous number line oforder ℵ1, as it is mapable to the integers and so is only a continuum of the rationalnumbers. Thus we have refined our understanding of the relationship betweenthe ’amount’ of information contained in the first Brillouin zone of the periodic,continuous k-space and the corresponding infinite, discrete x-space lattice: thefunctions ψ[x] or ψ(k) defined in either region contain mapping information oforder ℵ0. This is important because it will allow us to be more precise about thedivergences encountered during renormalization, when we get to that shortly.

Lorentz-Invariance on a Lattice

So why don’t we simply plow forward and reformulate of all of quantum theoryin terms of a 4D Cartesian discrete space-time lattice? Well, there are some subtlechallenges involved, but the main issue with a Cartesian lattice is that it lacksrotational and Lorentz invariance: obviously, if you rotate a cubic lattice by anarbitrary angle, the points on the rotated lattice won’t line up with points on theoriginal lattice. Furthermore, in the transformed k-space, the edges of the 4D cubic


region will overlap parts of adjacent Brillouin zones, as cubes are not rotationallyinvariant, and the limits of integration in k-space will involve nested functions –those could be complicated but would be well-defined and calculable in principle.

However, choosing the right lattice geometry, like a 4D ’hybrid’ spherical lat-tice instead of a Cartesian one, will make the calculations much more straightfor-ward1. This hybrid spherical lattice will be continuous and periodic in the angular(θ, φ) coordinates, and discrete in the radial and time coordinates (hence the term’hybrid.’) We anticipate that the 4D hybrid spherical delta function should looksomething like this:

δ4(x− x′) ≈ 1r2 δ(ct− ct′]δ(r− r′]δ(φ− φ′)δ(cosθ − cosθ′) (7.5)

Because the two angular coordinates are continuous and periodic over 2π, the k-space transform of the angular coordinates will be an infinite, 2D discrete latticewith indices l, m, (or j, mj), which as we just mentioned above, is a general prop-erty of Fourier transforms. Thus the θ and φ delta functions can be represented interms of an infinite, discrete series of spherical harmonics:

δ(φ− φ′)δ(cosθ − cosθ′) =∞

∑l=0

l

∑m=−l

Y∗lm(θ′, φ′)Ylm(θ, φ).

A mathematical implication here is that physically measurable angles are lim-ited to the set of rational divisions of the interval (0, 2π), even though π itself isan irrational number. This must be so, since all physical measurements which canbe written down as a number must terminate at some decimal point and henceare rational numbers. We are mortal experimenters who only have finite time andenergy available to make measurements – thus all measurable quantities must beeither rational numbers or integers. The correct formulation of quantum mechan-ics must reflect this ’metaphysical’ fact on some level.

Bandwidth-Limited Theory of Bessel Functions

Before we can get a more rigorous fix on Eq. 7.5, we need to understand thebandwidth-limited theory on cylindrical and spherical lattices. This begins withthe fundamentals of Bessel series and Bessel transforms. Let

f (k) =∞

∑m=1

cνm Jν(ανmkκ), (7.6)

1Personal communications, M.R. Hamilton.


where ανm is the mth zero of Jν. Since,∫ κ

0Jν(ανm

kκ)Jν(ανn

kκ) k dk =

0, if n 6= mκ2

2 [Jν+1(ανm)]2 , if n = m,

then we may solve Eq. 7.6 for the coefficients of the series,

cνn =2

κ2 [Jν+1(ανn)]2

∫ κ

0f (k)Jν(ανn

kκ) k dk. (7.7)

So far, the value of the inverse-length constant, κ, is unspecified – we’ll get to that.Now, comparing Eq. 7.6 with Eq. 7.1, we see that they are homologous in form,

and are simply applications of the same principle of orthogonal decompositionusing different basis functions: the basis of eikx in the former, and the basis ofJν(ανm

kκ ) in the latter. The ανm are the zeros of Jν and are a discrete lattice of points

– although they are not equally spaced as in the case of the eikx basis. The ανm arealso unit-less, but we can associate each zero with a point [m]ν measured in unitsof length such that κ[m]ν = ανm. Thus, we can replace cνm with ψ[m]ν, (whichactually looks better when written as ψν[m]) , and replace f (k) with ψν(k) in Eq.7.6,

ψν(k) = ∑[m]

ψν[m]Jν(ανmkκ). (7.8)

Here, even though [m] is not integral, it is discrete, and the sum is understood torange over the entire infinite set of [m]. Now we can rewrite Eq. 7.7 as

ψν[m] =2

κ2 [Jν+1(κ[m]ν)]2

∫ κ

0ψν(k)Jν(ανm

kκ) k dk,

and then use Eq. 7.8 to replace ψν(k), to give,

ψν[m] = ∑[r]ν

2κ2[Jν+1(κ[m]ν)]2

∫ κ

0Jν(k[r]ν)Jν(k[m]ν) k dk ψν[r].

If, for the meantime, we continue to restrict [m]ν to be one of the points on thelattice of the zeroes of Jν, then the Jν+1(κ[m]ν) term in the denominator can bereplaced with Jν+1(κ[r]ν), without altering the completeness and orthogonalityproperties of the series representation. That is, on the lattice points, it’s also truethat,

ψν[m] = ∑[r]ν

2κ2[Jν+1(κ[r]ν)]2

∫ κ

0Jν(k[r]ν)Jν(k[m]ν) k dk ψν[r].


Now we are in a position to perform the same leap to a continuous variable asShannon did to get Eq. 7.2, and let [m]→ (m):

ψν(m) = ∑[r]ν

2κ2[Jν+1(κ[r])]2

∫ κ

0Jν(k[r]ν)Jν(km) k dk ψν[r]. (7.9)

We can be sure that we are never dividing by zero above, since the Jν+1(κ[r]ν) inthe denominator is evaluated at one of the zeros of Jν, not one of the zeroes ofJν+1. The zeroes of Jν+1 always occur at points in between successive zeroes of Jν,so Jν+1 is guaranteed to be non-zero at the zeroes of Jν. Thus, we have deriveda well-defined representation of the bandwidth-limited delta-function/kernel ofthe radial variable on a cylindrical lattice:

δ(m− r] =2

κ2[Jν+1(κ[r])]2

∫ κ

0Jν(k[r]ν)Jν(km) k dk. (7.10)

The same approach can be used in the case of a spherical lattice in spherical co-ordinates. The cylindrical Bessel functions are replaced with the spherical Bessel

functions, jn(x) =√

2πx Jn+1/2(x), (and these have the same sets of zeros). Eq. 7.8,

becomes,

ψn(k) = ∑[m]

ψn[m]jn(αnmkκ).

The ’coefficients,’ ψn[m] become:

ψn[m] =2

κ3 [jn+1(κ[m]n)]2

∫ κ

0ψn(k)jn(ανm

kκ) k2 dk,

The continuous spherical interpolation then becomes,

ψn(m) = ∑[r]n

2κ3[jn+1(κ[r])]2

∫ κ

0jn(k[r]n)jn(km) k2 dk ψν[r]. (7.11)

Thus the bandwidth-limited delta function/kernel for the radial variable on aspherical lattice is:

δ(m− r] =2

κ3[jn+1(κ[r])]2

∫ κ

0jn(k[r]n)jn(km) k2 dk. (7.12)

and Eq. 7.11 can be written

ψn(m) = ∑[r]n

δ(m− r]ψn[r]


7.2 The Bandwidth-limited Feynman Green’s Function

So far this is nice, but we haven’t quite yet made the jump to mathematical physics.Physics usually focuses on a kernel which is not a delta-function, but a Green’sfunction, G(x, x′), that is closely related to a delta function. For example, in electro-static or Newtonian gravitational analysis, the Green’s function satisfies∇2G(x, x′) =−4πδ3(x − x′). Green’s functions also play a central role in Feynman’s formula-tion of quantum field theory. The ’forward-time’ and ’backward-time’ spinors ψ+

and ψ− obey

θ(t− t′)ψ+(x, t) = i∫

d3x′ G+(x, x′, t, t′)ψ+(x′, t′),

θ(t′ − t)ψ−(x, t) = i∫

d3x′ G−(x, x′, t, t′)ψ−(x′, t′),

which is essentially Huygen’s principle. We showed in the previous chapter howthis pair leads to the single condition,

(i∇− mch)SF(x, x′) = δ4(x− x′), (7.13)

defining, SF, the Feynman Green’s function.In the discrete x-space, Huygen’s principle for the forward and backward-time

waves is a sum over the discrete radial and time lattice variables and a continuousintegral over the angular variables:

θ(t− t′]ψ+(r, θ, φ, t) = i ∑j

∑[r′]j,[t′]j

∫[r′]2dΩ′ G+

j (r, θ, φ, t, [r′], θ′, φ′, [t′])ψ+j [r′, t′](θ′, φ′),

θ[t′ − t)ψ−(r, θ, φ, t) = i ∑j

∑[r′]j,[t′]j

∫[r′]2dΩ′ G−j (r, θ, φ, t, [r′], θ′, φ′, [t′])ψ−j [r

′, t′](θ′, φ′),

We point out that this is a sum over finer and finer lattices, with each lattice in-dexed by j. But in a moment we’ll pivot to an analysis in k-space, where insteadof sums over finer and finer lattices, there will be a sum of integrals over largerand larger Brillouin zones.

Now consider the k-space/momentum-space representation of the FeynmanGreen’s functions in terms of unnormalized (but renormalizable), free-particleCartesian plane waves in the traditional theory. As Bjorken and Drell show, [4], if


it’s required that Eq. 7.13, is satisfied, then the solution can be represented as

SF(x− x′) = −iθ(t− t′)∫

d3p2

∑r=1

ψrp(x)ψr

p(x′) + iθ(t′ − t)∫

d3p4

∑r=3

ψrp(x)ψr

p(x′),

where the momentum-space integrals go from −∞ to ∞. This expansion is interms of Cartesian plane waves, but it instructs us how we might construct anansatz for a bandwidth-limited version of same thing in spherical coordinates:

SF(x− x′) = −i ∑j,mj

∫ ωj

0dω

∫ κj

0k2dk e−iω(t−t′) Se,j

F (ω, k)ψej,mj

(kr, θ, φ)ψej,mj

(kr′, θ′, φ′)

+ i ∑j,mj

∫ ωj

0dω

∫ κj

0k2dk eiω(t−t′) Sp,j

F (ω, k)ψpj,mj

(kr, θ, φ)ψpj,mj

(kr′, θ′, φ′).

(7.14)

Here, each Se,jF (ω, k) or Sp,j

F (ω, k) is a diagonal 4 × 4 matrix that can be chosenso that Eq. 7.13 is satisfied, and ψe

j,mjand ψ

pj,mj

are the free-particle solutions wederived in Chapter 2, and summarized in Table 2.1.

Looking more closely at the ψψ matrices formed from the outer product of twospinors, we have for ψeψe, for example:

ψej,mj

(kr, θ, φ, t)ψej,mj

(kr′, θ′, φ′, t′)

=[

jj+1/2(kr)jj+1/2(kr′)θ−j (θ, φ)θ−†j (θ′ , φ′) ijj+1/2(kr)jj−1/2(kr′)θ−j (θ, φ)θ+†

j (θ′ , φ′)

ijj−1/2(kr)jj+1/2(kr′)θ+j (θ, φ)θ−†j (θ′ , φ′) −jj−1/2(kr)jj−1/2(kr′)θ+j (θ, φ)θ+†

j (θ′ , φ′)

].

This is a 4× 4 matrix composed of 2× 2 submatrices, one of which is, for example,

θ−j (θ, φ)θ−†j (θ′, φ′)

=

[(j−mj + 1)Yj+1/2mj−1/2

(θ, φ)Y∗j+1/2mj−1/2(θ′ , φ′) −

√(j−mj + 1)(j + mj + 1)Yj+1/2mj−1/2

(θ, φ)Y∗j+1/2mj+1/2(θ′ , φ′)

−√(j−mj + 1)(j + mj + 1)Yj+1/2mj+1/2

(θ, φ)Y∗j+1/2mj−1/2(θ′ , φ′) (j + mj + 1)Yj+1/2mj+1/2

(θ, φ)Y∗j+1/2mj+1/2(θ′ , φ′)

]Thus, for the full Feynman Green’s function, weNow, using the results for the free-particle solutions in spherical coordinates

from Chapter 2, it’s straightforward to show that:

(i∇− mch)e−iωtψe

j,mj(kr, θ, φ)

=

ωc + k

√j

j+1 −mch 0

0 −(ωc + k

√j+1

j + mch )

e−iωtψej,mj

(kr, θ, φ)


and,

(i∇− mch)eiωtψ

pj,mj

(kr, θ, φ)

=

−(ωc + k

√j+1

j + mch ) 0

0 ωc + k

√j

j+1 −mch

eiωtψpj,mj

(kr, θ, φ)

Thus, if we let

Se,jF (ω, k) =

ωc + k

√j

j+1 −mch 0

0 −(ωc + k

√j+1

j + mch )

−1

,

and

Sp,jF (ω, k) =

−(ωc + k

√j+1

j + mch ) 0

0 ωc + k

√j

j+1 −mch

−1

Then

Se,jF (ω, k)(i∇− mc

h)e−iωtψe

j,mj(kr, θ, φ) = e−iωtψe

j,mj(kr, θ, φ)

Sp,jF (ω, k)(i∇− mc

h)eiωtψ

pj,mj

(kr, θ, φ) = eiωtψpj,mj

(kr, θ, φ)

Discussion

This bandwidth-limited analysis shows that we can view the full ℵ1 Green’s func-tion as composed of an (infinite series) sum of ℵ0 Green’s functions. Likewise,since the delta function results from the action of i∇− mec

h on each Green’s func-tion component, the total ℵ1 delta-function is composed of an (infinite series)sum of ℵ0 delta functions. We showed above how each ℵ0 Green’s function iscomposed of spinor basis functions from a finite region of k-space (a Brillouinzone), which are also the heavy-electron solutions of the free-particle Dirac equa-tion we studied in Chapter 3, having the sequence of masses mj = (2j + 1)me.In frequency-space each Brillouin zone extends over a region from ω = 0 toω = (2j + 1)mec2

h , and the integral over the radial κj k-space variable goes fromκ = 0 to κ = 2mec

h

√j(j + 1). We can represent this as

SF,ℵ1(x− x0) = SF,2me(x− x0) + SF,4me(x− x0) + · · ·


Each successive term on the right, SF,2(j+1/2)me(x− x0) corresponds (or gives riseto) Feynman diagrams having more and more intermediate virtual particles (orintermediate states of heavier and heavier mass). Under many circumstances,the contribution of higher-mass terms to the amplitude diminishes so they can beneglected in an approximations, but they must be included to represent the fullℵ1 Green’s function. Since each SF,2(j+1/2)me(x− x0) satisfies

(∇− mech

)SF,2(j+1/2)me(x− x0) =

This is easiest to picture in one-dimension, but in four dimensions is essentiallythe same idea. The ℵ1 delta-function is defined in terms of a Fourier integral overall space:

δℵ1(x− x0) =1

2π

∫ ∞

−∞eik(x−x0) dk

While an ℵ0 delta-function is defined in terms of an integral over a bandwidth-limited region:

δℵ0(x− x0) =1

2K

∫ K

−Keik(x−x0) dk

The full ℵ1 delta-function can be represented by an infinite series summation ofbandwidth-limited, ℵ0 delta-functions:

δℵ1(x− x0) =∞

∑i=1

δℵ0,i(x− x0).

Note to Editor: This topic of this chapter seems like it’s not directly connectedto the topic of quantum gravity. It’s more of a side discussion and it is connectedto the topic of renormalization, and follows up on the work in Chapter 2.

CHAPTER 8

Conclusion

There will be a conclusion...

148

CHAPTER 9

The Appendices

These appendices consist of a collection of identities and other derivations, mostof which are used in the main part of this text.

9.1 Appendix A: General Matrix Identities

For any matrix operators, A, B, C, differential or not, the following commutatoridentities hold:

[AB, C] = A [B, C] + [A, C] B (9.1)[AB, C] = AB, C − A, CB (9.2)[A, BC] = B [A, C] + [A, B]C (9.3)[A, BC] = B, AC− BC, A (9.4)

These anti-commutator identities are also true:

AB, C = A [B, C] + A, CB (9.5)AB, C = AB, C − [A, C] B (9.6)A, BC = [A, B]C + BA, C (9.7)A, BC = A, BC− B [A, C] (9.8)

149

CHAPTER 9. THE APPENDICES 150

Another important identity, for any two operators, A, B, differential or not,(from Messiah, vol. 1, pg. 339):

eiABe−iA = B + i [A, B] +i2

2![A, [A, B]] +

i3

3![A, [A, [A, B]]] + . . .

Let A→ iA, then this becomes:

e−A B eA = B− [A, B] +12!

[A, [A, B]]− 13!

[A, [A, [A, B]]] + . . .

ore−A B eA = B + [B, A] +

12!

[[B, A] , A] +13!

[[[B, A] , A] , A] + . . .

And a version of this that we may find useful:

e−S∇ eS = ∇+ [∇, S] +12!

[[∇, S] , S] +13!

[[[∇, S] , S] , S] + . . . (9.9)

9.2 Appendix B: Identities of the γ and σ matrices

9.2.1 The Unit Vectors in 3 and 4 Dimensions

The Pauli matrices are the correct quantum representation of the 3D cartesian unitvectors, meaning i = σ1, j = σ2 and k = σ3. Therefore the quantum representationof the spherical coordinate unit vectors is:

r = sinθ cosφ i + sinθ sinφ j + cosθ k =

[cosθ sinθ e−iφ

sinθ eiφ −cosθ

](9.10)

θ =∂r∂θ

= cosθ cosφ i + cosθ sinφ j− sinθ k =

[−sinθ cosθ e−iφ

cosθ eiφ sinθ

]φ =

1sinθ

∂r∂φ

= −sinφ i + cosφ j =[

0 −ie−iφ

ieiφ 0

]This really makes it clear how the SU(2) Pauli matrices very precisely encapsulatethe 3D unit vector algebra, and why the imaginary number i should not be purgedfrom the algebra.


These inverse identities are also useful:

σ1 = i = sinθ cosφ r + cosθ cosφ θ − sinφ φ (9.11)

σ2 = j = sinθ sinφ r + cosθ sinφ θ + cosφ φ

σ3 = k = cosθ r− sinθ θ

The r, θ, φ inherit the cyclic nature of the Pauli matrices:

r θ = i φ

θ φ = i r

φ r = i θ

And of course, r, θ, φ inherit the anticommutative and commutative properties ofthe σ1, σ2, σ3:

12

r, θ= 0, etc.

and12i[r, θ]= φ, etc.

This last one is a representation of the cross-product identities which the unit vec-tors must obey e.g. , r× θ = φ, etc.

Using the 3D spherical unit vectors, we can express the 4D contravariant Lorentzγ matrices in the spherical coordinate system as

γ0 =

[1 00 −1

]

γr =

[0 r−r 0

]γθ =

1r

[0 θ

−θ 0

]γφ =

1rsinθ

[0 φ

−φ 0

]And the covariant Lorentz γ matrices in the spherical coordinate system are:

γ0 = γ0 =

[1 00 −1

]


γr =

[0 −rr 0

]γθ = r

[0 −θ

θ 0

]γφ = rsinθ

[0 −φ

φ 0

]These satisfy

12γµ, γν = 1 · gµν

12γµ, γν = 1 · gµν

12γµ, γν = 1 · δµ

ν

The spherical representation of the γ matrices is Lorentz covariant, althoughwe must remember that specifying a spherical center in one Lorentz frame breaksa certain symmetry: it singles out the reference frame in which the Dirac Equationis solved as the "home" coordinate system – when boosted to another frame, thespherical symmetry is lost, although the Lorentz covariance is maintained.

We must also note that these γr matrices are not the same thing as the 3D unitvectors. The representations of the 3D unit vectors in the full 4× 4 representationare simply the diagonal extension of the 2× 2 representations, i.e:

r =[

r 00 r

]But we must remember that the 3D unit vectors are not Lorentz covariant, mean-ing they don’t transform as four-vectors the way the γ matrices do, as in Eq. 1.5.In fact, they are pseudo-vectors, which are actually tensors, and transform like theσµν which they’re built from. (Remember, σ3 = σ12 = 1

2i[γ1, γ2], etc.) This is an

important point we must bear in mind when we attempt to construct tensors suchas the 3D volume element: In the x, y, z coordinates this is

dV3D = d~z · (d~x× d~y) =12

σ3,

12i

[σ1, σ2

]dx dy dz (9.12)


To derive certain identities we will need to calculate the partial derivatives ofthe r, θ and φ unit vectors. Based on 9.11 we have:

∂r∂θ

= θ =

[−sinθ cosθ e−iφ

cosθ eiφ sinθ

]∂r∂φ

=

[0 −i sinθ e−iφ

i sinθ eiφ 0

]= sinθ φ

∂θ

∂θ=

[−cosθ −sinθ e−iφ

−sinθ eiφ cosθ

]= −r

∂θ

∂φ=

[0 −i cosθ e−iφ

i cosθ eiφ 0

]= cosθ φ

∂φ

∂θ= 0

∂φ

∂φ=

[0 −e−iφ

−eiφ 0

]= −cosφ i− sinφ j = − 1

sinθ

(r− cosθ k

)Using these, we can derive all partial derivatives of the γµ in spherical coordi-

nates. For γ0:∂γ0

∂xµ = 0 ∀ µ.

For γr:∂γr

∂x0 = 0,∂γr

∂r= 0,

∂γr

∂θ=

∂

∂θ

[0 −rr 0

]=

[0 −θ

θ 0

]=

1r

γθ,

∂γr

∂φ=

∂

∂φ

[0 −rr 0

]= sinθ

[0 −φ

φ 0

]=

1r

γφ.

For γθ:∂γθ

∂x0 = 0,

∂γθ

∂r=

[0 −θ

θ 0

]=

1r

γθ,

∂γθ

∂θ= r

[0 r−r 0

]= −r γr,


∂γθ

∂φ= r cosθ

[0 −φ

φ 0

]= cotθ γφ

For γφ:∂γφ

∂x0 = 0,

∂γφ

∂r= sinθ

[0 −φ

φ 0

]=

1r

γφ,

∂γφ

∂θ= r cosθ

[0 −φ

φ 0

]= cotθ γφ,

∂γφ

∂φ= rsinθ

[0 − ∂

∂φ φ∂

∂φ φ 0

]= r

[0 r− cosθ k

−(r− cosθ k) 0

]= −r γr + r cosθ γ3,

which can also be written:

∂γφ

∂φ= −r sin2θ γr − sin θ cos θ γθ

9.2.2 Commutation relations of γµ∂µ and γνxν in Cartesian sys-tems

We will find it necessary to calculate various commutation and anticommutationrelations of the operators γµ∂µ and γν xν. Assume a flat Cartesian metric, andthat the coefficients xν have no dependance on (x, t), only proper time, τ, so that[∂µ, xν

]= 0.

γµ∂µ, γν xν = γµ, γνxν∂µ = 2 δµν xν∂µ = 2 xµ∂µ[

γµ∂µ, γν xν]= [γµ, γν] xν∂µ = 2i σµ

ν xν∂µ

Slightly rearranging and grouping these as a pair for reference, we have,

12γµ∂µ, γν xν = xµ∂µ (9.13)

12[γµ∂µ, γν xν

]= i σµ

ν xν∂µ (9.14)

The first of these, Eq. 9.13 is essentially the convective derivative. The second ofthese, Eq. 9.14, if set equal to zero, is equivalent to the classical law, ~p × ~x = 0,


which is equivalent to the conservation of orbital angular momentum, and that’sthe same thing as Kepler’s Second Law.

Pursuing a related pair of identities, first use the matrix identity, Eq. 9.6, to get,

γ0γν, γµ = γ0γν, γµ −[γ0, γµ

]γν = 2γ0 δ

µν + 2i σµ0 γν.

Then, use the matrix identity, Eq. 9.1 to show,[γ0γν, γµ

]= γ0 [γν, γµ] +

[γ0, γµ

]γν = 2iγ0σν

µ + 2iσ0µγν.

Thus, rearranging these last two slightly, and grouping them as a pair for refer-ence, we have,

12γµ, γ0γν = γ0 δ

µν + i σµ0 γν (9.15)

12

[γµ, γ0γν

]= i γ0σµ

ν − i σµ0γν (9.16)

9.2.3 More basis vector commutation identities

In the case of general curvilinear coordinates, the free-space orthonormal basisvectors, γµ, give way to the εµ basis vectors. The εµ are not necessarily orthogonalor normal, but they still obey

εµ · εν =12εµ, εν = gµν, εµ · εν =

12εµ, εν = δ

µν .

The tensor σµν is also defined in terms of these, as it is in a flat-space:

σµν =i2[εµ, εν] , σ

µν =

i2[εµ, εν] .

From these, and the commutation and anti-commutation relations for com-pound products of matrices, identitis. 9.1 through 9.8, we can derive more. Con-sider, [

ελ, εµεν

]= εµ, ελεν − εµελ, εν = 2δλ

µ εν − 2δλν εµ,

by 9.4, and likewise, [ελ, ενεµ

]= 2δλ

ν εµ − 2δλµ εν.


Thus, [ελ,[εµ, εν

]]= 4

(δλ

µ εν − δλν εµ

),

so, [ελ, σµν

]=

[ελ,

i2[εµ, εν

]]= 2i

(δλ

µ εν − δλν εµ

).

Introducing an additional dummy index gives a more symmetric-looking version,[ελ, σµν

]= 2i

(δλ

µ δρν − δλ

ν δρµ

)ερ. (9.17)

Similarly, by using identities 9.7 and 9.8 , it is shown that,

ελ,[εµ, εν

] = εν

[ελ, εµ

], (9.18)

which is equivalent to the familiar vector identity ~A · (~B× ~C) = ~C · (~A× ~B).

9.2.4 The operator σL

Throughout this work we employ the shorthand σL, or σiLi for the operator whichother authors often symbolize as~σ ·~L. It is more explicitly written as:

σiLi = σ1L1 + σ2L2 + σ3L3 = σxLx + σyLy + σzLz (9.19)

This is a cartesian coordinate representation of σL. But let’s use some identities toget a representation in spherical coordinates. We have a spherical representationof Lz,

Lz = −ih∂

∂φ. (9.20)

As well as a spherical representation for the orbital angular momentum raising,L+, and lowering L− operators:

L± = Lx ± iLy = h e±iφ(± ∂

∂θ+ icotθ

∂

∂φ

).

These can be rearranged to give spherical representations of Lx and Ly:

Lx = ih(

sinφ∂

∂θ+ cotθ cosφ

∂

∂φ

), (9.21)


and

Ly = −ih(

cosφ∂

∂θ− cotθ sinφ

∂

∂φ

). (9.22)

Now we can use Eqs. 9.20, 9.21 and 9.22 with Eq. ?? to express Eq. 9.19 inspherical form. The result is:

σL = ih(

θ1

sinθ

∂

∂φ− φ

∂

∂θ

)(9.23)

Here, θ and φ are, as usual, the 4× 4 diagonal extensions of the 2× 2 forms.But we also could have read this off from the two equivalent expressions of the

∇ operator in spherical coordinates, derived by other means in Appendix D:

∇ = γ0 ∂

∂x0 + γr(

∂

∂r− 1

hσLr

)and

∇ = γ0 ∂

∂x0 + γr ∂

∂r+ γθ ∂

∂θ+ γφ ∂

∂φ

and thus equating:

−1h

γr σLr

= γθ ∂

∂θ+ γφ ∂

∂φ

Then multiplying through by γr, we have

1h

σLr

= γrγθ ∂

∂θ+ γrγφ ∂

∂φ,

Since γr2 = −1. Then using

γrγθ =

[0 r−r 0

]· 1

r

[0 θ

−θ 0

]=

1r

[−rθ 0

0 −rθ

]= −i

1r

[φ 00 φ

]= −i

1r

φ,

and

γrγφ =

[0 r−r 0

]· 1

rsinθ

[0 φ

−φ 0

]=

1rsinθ

[−rφ 0

0 −rφ

]= i

1rsinθ

[θ 00 θ

]= i

1rsinθ

θ,

We have,1h

σLr

= −i1r

φ∂

∂θ+ i

1rsinθ

θ∂

∂φ.


After cancelling the 1/r’s and rearranging, this becomes:

σL = ih(

θ1

sinθ

∂

∂φ− φ

∂

∂θ

),

The same as Eq. 9.23. This shows how consistent the methods of Quantum TensorCalculus are, and the validity of interpreting the Pauli σ matrices as unit vectorsas in Eqs. 9.11 and ??.

9.2.5 Identities of the Total Angular Momentum Operator, J

Here’s another one relating the total angular momentum operator J to the orbitalangular momentum components:

J = γ0 (σL + h)

where σL is shorthand for σiLi, which contains an implicit sum over i = 1, 2, 3.Starting from:

σL, λ = −2hλ

andσL, ε = −2hε

It can be shown that[J, ε] = 0,

and [J, γ0

]= 0,

and [J, γ0ε

]= 0.

The following two identities will be important in proving how the ‘old’ quan-tum theory arises from the ‘new’ quantum theory:

12ih∇, J = mc J

and12[ih∇, J] = 0


9.3 Appendix C: Special Functions and their Identi-ties

9.3.1 The Confluent Hypergeometric Functions

The solution of Coulomb-Dirac equation involves the confluent hypergeometricfunctions. Lucy Joan Slater’s Confluent Hypergeometric Functions, [30] is the defini-tive work on these special functions, filled with proofs of many identities andrecursion formulae – some of those results are quoted in the more well-knownreference text of Abramovitz and Stegun, [31]. As Bethe shows explicitly in [11],the confluent hypergeometric functions, M(α, β, γ), are the radial eigenfunctionsof the Coulomb Dirac equation, (after factoring out an overall exponential andsome other minor transformations). These can be shown to obey the followingcoupled, first-order differential relations:

M′(−n′ + 1, b, ρ)

= M(−n′ + 1, b, ρ) +−n′ + 1− b

ρ

(M(−n′ + 1, b, ρ)−M(−n′, b, ρ)

),

M′(−n′, b, ρ) =−n′

ρ

(M(−n′ + 1, b, ρ)−M(−n′, b, ρ)

). (9.24)

Transforming the Coulomb Dirac equation in spherical coordinates leads to thisidentity. Intriguingly, this exact identity does not explicitly appear in [30] or [31],(nor anywhere else I’ve been able to find), but the identities they do provide allowone to prove that the confluent hypergeometric functions solve the Coulomb Diracequation – hence they must obey the above coupled differential equations whichare equivalent to the radial portion of Coulomb Dirac equation. A pure mathe-matical proof of this identity, not making reference to the physical circumstancesof the Hydrogen atom, can be extracted from the work in Chapter 3, although it’snot critical.

9.3.2 The Bessel Functions

Here we discuss a collection of useful identities satisfied by the cylindrical andspherical Bessel functions, Jν(x), and jn(x). Some of these may also apply to theclosely related Neumann and Hankel functions where indicated.


As we saw in Chapter 2, the spherical Bessel (and Hankel or Neumann) func-tions that arise in the solution of the free-particle Dirac equation obey a set of twocoupled differential equations:(

∂

∂r− n− 1

r

)jn−1(κr) = −κ jn(κr),(

∂

∂r− n + 1

r

)jn(κr) = κ jn−1(κr). (9.25)

These differential-recursion relations are the free-particle analogues of 9.24. I be-lieve that there must be a pure mathematical theorem that connects the Besselfunctions to the confluent hypergeometric functions and their respective coupleddifferential-recursions shown above. It would be nice to see the proof of that, butit’s not essential. There are many identities, some of which are cited in [31] and[10], that connect the Bessel-type functions to the hypergeometric-type functionsunder various circumstances, so the existence of a general theorem connecting9.24 and 9.25 would be expected.

Summation and Integration Identities

These identities relate discrete sums to continuous integrals of Bessels functions,and are useful in defining bandwidth-limited delta functions and Green’s func-tions. As far as I know, these only apply to the Bessel functions, but not to theNeumann or Hankel functions.

From Arfken and Weber, pg. 650, (1995), [10]: let

f (x) =∞

∑n=1

an Jm(αmnx),

where αmn is the nth zero of Jm, and m = 0, 1, . . . . Then it can be shown that∫ 1

0[ f (x)]2 xdx =

12

∞

∑n=1

a2n [Jm+1(αmn)]

2 .

From Dominici et al, (2012), [29]:∫ ∞

0

Jν(at)Jν(bt)t

dt =∞

∑n=0

εnJν(an)Jν(bn)

n,


where, a, b ∈ [0, π], εn = 12 when n = 0, and εn = 1 for n ≥ 1, and ν = 1

2 , 32 , . . . .

Also, from [29], if the integration goes from −∞ to ∞, the εn = 1 for all n:∫ ∞

−∞

Jν(at)Jν(bt)t

dt =∞

∑n=−∞

Jν(an)Jν(bn)n

,

still with a, b ∈ [0, π], and ν = 12 , 3

2 , . . . .Also, [29] proves an identity for general powers of t and n,∫ ∞

0

Jµ(at)Jν(bt)tµ+ν−2k dt =

∞

∑n=0

εnJµ(an)Jν(bn)

nµ+ν−2k ,

which is true if 0 < b < a < π, Re(µ + ν− 2k) > −1, and k is an integer.Dominici et al, [29], also prove some theorems that are useful in developing the

bandwidth-limited theory of Green’s functions and delta functions. The ’Newto-nian kernel,’ K(~r,~r′), can be expressed as a series,

K(~r,~r′) =1

|~r−~r′| =∞

∑l=0

l

∑m=−l

Y∗lm(θ, φ)Ylm(θ′, φ′)Kl(r, r′), (9.26)

=∞

∑n,l=0

l

∑m=−l

φ∗nlm(~r)φnlm(~r′) (9.27)

where

Kl(r, r′) = 4π∫ ∞

0

Jl+1/2(kr)Jl+1/2(kr′)k√

rr′dk.

and

φnlm(~r) =

√4πεn

rnJl+1/2(rn)Ylm(θ, φ).

We can replace cylindrical Bessel functions with spherical, so,

φnlm(~r) =

√4πεn

rnJl+1/2(rn)Ylm(θ, φ) =

√8εn jl(rn)Ylm(θ, φ)

thus,

K(~r,~r′) =∞

∑n,l=0

l

∑m=−l

8εnY∗lm(θ, φ)Ylm(θ′, φ′)jl(rn)jl(r′n), (9.28)


and Kl(r, r′), is

Kl(r, r′) =∞

∑n=0

8εn jl(rn)jl(r′n).

Now, the ’Newtonian kernel,’ is otherwise known as the Green’s function ofthe 3D radial wave equation, and it satisfies,

∇2 1|~r−~r′| = −4πδ3(~r−~r′). (9.29)

With this, we can foresee that there must be an equivalent expansion series for the3D delta function:

δ3(~r−~r′) =∞

∑n,l=0

l

∑m=−l

δl(r, r′)Y∗lm(θ, φ)Ylm(θ′, φ′).

We’ll be able to get an explicit form for δl(r, r′) by applying ∇2 to 9.28. Now, inspherical coordinates,

∇2 (Ylm(θ, φ)jl(kr)) =(

d2

dr2 +2r

ddr− l(l + 1)

r

)Ylm(θ, φ)jl(kr) = k2Ylm(θ, φ)jl(kr).

It’s also the case that Y∗lm = −Yl,−m, so,

∇2 (Y∗lm(θ, φ)jl(nr)) = n2Y∗lm(θ, φ)jl(nr).

Thus, from 9.28, and 9.29, we have

δ3(~r−~r′) = − 14π

∞

∑n,l=0

l

∑m=−l

n28εnY∗lm(θ, φ)Ylm(θ′, φ′)jl(rn)jl(r′n).

and,

δl(r, r′) =∞

∑n=0

n2 8εn jl(rn)jl(r′n)

9.4 Appendix D: The Free-Particle Dirac Equation inSpherical Coordinates

To demonstrate the transformations leading to Eq. (2.1), we will use the propertiesof the γ and σ matrices and their identities. We will work in the representation of


the γ used by Bjorken and Drell [8]:

γ0 =

[1 00 −1

]γi =

[0 σi

−σi 0

]γ5 =

[0 11 0

](9.30)

and use the typical representation of the Pauli matrices:

σ1 =

[0 11 0

]σ2 =

[0 −ii 0

]σ3 =

[1 00 −1

](9.31)

The Pauli matrices obey the identities

σ1 σ2 = i σ3 σ3 σ1 = i σ2 σ2 σ3 = i σ1

σ2 σ1 = −i σ3 etc.(9.32)

so therefore the γ obey the identities:

γ1 γ2 = −i σ3 γ3 γ1 = −i σ2 γ2 γ3 = −i σ1

γ2 γ1 = i σ3 etc.(9.33)

Here, the σ are 4× 4 block-diagonal matrices formed from the 2× 2 Pauli matrices.The γ and σ also have the properties:

(σi)2 = 1 (γ0)2 = 1 (γi)2 = −1 (9.34)

Now consider the following:

(γj xj)(γi ∂i) = (γ1)2 x1∂1 + γ1 γ2 x1∂2 + ...+ (γ2)2 x2∂2 + γ2 γ1 x2∂1 + ...

which by virtue of the identities (9.33) and (9.34) becomes

(γj xj)(γi ∂i) = −xi∂i + i σ1 (x3∂2 − x2∂3) + ...= −r ∂/∂r + σiLi/h

so that we have established the identity

(γj xj)(−i h γi ∂i) = i h r∂

∂r− i σiLi

which we will find convinient to recast as(γ0 γj xj

r

)(−i h γ0 γi ∂i

)= −i h

∂

∂r+ i

σiLi

r(9.35)


When we write the γj xj object explicitly as

γj xj =

0 0 x3 x1 − i x2

0 0 x1 + i x2 −x3

−x3 −x1 + i x2 0 0−x1 − i x2 x3 0 0

(9.36)

it becomes apparent that (γj xj

)2= −r2 14×4

so that we can make the following statement about the inverse matrix(γj xj

r

)−1

= −(

γj xj

r

)(9.37)

Then, using (9.37) with (9.35) we obtain

− i hγ0 γi ∂i =γ0 γj xj

r

(−i h

∂

∂r+ i

σiLi

r

)(9.38)

so that the Dirac equation(i h ∂0 −mc γ0

)ψ = −i h γ0 γi ∂i ψ

becomes (i h ∂0 −mc γ0

)ψ = ε

(−i h

∂

∂r+ i

σiLi

r

)ψ

This establishes Eq. (2.1), with the object ε defined as

ε ≡ γ0 γj xj

r(9.39)

When the explicit form of γj xj from (9.36) is combined with the definition of thespherical coordinates

x1 = r sin θ cos φ

x2 = r sin θ sin φ

x3 = r cos θ


We find that the explicit form of ε becomes

ε =

0 0 cos θ sin θ e−i φ

0 0 sin θ ei φ −cos θ

cos θ sin θ e−i φ 0 0sin θ ei φ −cos θ 0 0

(9.40)

Note how in the definition of the ε object, Eq. 9.39 and in other relations, thesummation indicies j are both raised in the γjxj object. In a simple flat spacemetric, this does not create a problem.

9.5 Appendix E: Properties of the θ+,− 2-spinors

Here we will establish the properties of the θ+,− 2-spinors when acted upon byλ and σiLi. We will make use of the the properties of the angular momentumoperators and spherical harmonics:

L+ Yl,m =√(l −m)(l + m + 1)Yl,m+1

L− Yl,m =√(l + m)(l −m + 1)Yl,m−1

(9.41)

σiLi θ+j =

[Lz L−L+ −Lz

] [ √j + mj Yj−1/2,mj−1/2√j−mj Yj−1/2,mj+1/2

]

= h[ √

j + mj(mj − 1/2)Yj−1/2,mj−1/2 +√

j−mj

√(j−mj)(j− 1/2−mj − 1/2 + 1)Yj−1/2,mj−1/2√

j + mj

√(j− 1/2−mj + 1/2)(j− 1/2 + mj − 1/2 + 1)Yj−1/2,mj+1/2 − (mj + 1/2)

√j−mj Yj−1/2,mj+1/2

]= h

[ ((mj − 1/2)

√j + mj +

√j−mj

√(j + mj)(j−mj)

)Yj−1/2,mj−1/2(√

j + mj

√(j−mj)(j + mj)− (mj + 1/2)

√j−mj

)Yj−1/2,mj+1/2

]= h

[ √j + mj(mj − 1/2 + j−mj)Yj−1/2,mj−1/2√j−mj(j + mj −mj − 1/2)Yj−1/2,mj+1/2

]= h (j− 1/2) θ+j

(9.42)


σiLi θ−j =

[Lz L−L+ −Lz

] [ √j−mj + 1 Yj+1/2,mj−1/2

−√

j + mj + 1 Yj+1/2,mj+1/2

]

= h[ √

j−mj + 1(mj − 1/2)Yj+1/2,mj−1/2 −√

j + mj + 1√(j + 1/2 + mj + 1/2)(j + 1/2−mj − 1/2 + 1)Yj+1/2,mj−1/2√

j−mj + 1√(j + 1/2−mj + 1/2)(j + 1/2 + mj − 1/2 + 1)Yj+1/2,mj+1/2 + (mj + 1/2)

√j + mj + 1 Yj+1/2,mj+1/2

]

= h

[ ((mj − 1/2)

√j−mj + 1−

√j + mj + 1

√(j + mj + 1)(j−mj + 1)

)Yj+1/2,mj−1/2(√

j−mj + 1√(j−mj + 1)(j + mj + 1) + (mj + 1/2)

√j + mj + 1

)Yj+1/2,mj+1/2

]

= h[ √

j−mj + 1(mj − 1/2− j−mj − 1)Yj+1/2,mj−1/2√j + mj + 1(j−mj + 1 + mj + 1/2)Yj+1/2,mj+1/2

]= −h (j + 3/2) θ−j

(9.43)Now use the following identities to calculate the action of λ:

cos θ Yl,m =√

(l−m+1)(l+m+1)(2l+1)(2l+3) Yl+1,m +

√(l−m)(l+m)(2l−1)(2l+1) Yl−1,m

sin θ eiφ Yl,m = −√

(l+m+1)(l+m+2)(2l+1)(2l+3) Yl+1,m+1 +

√(l−m)(l−m−1)(2l−1)(2l+1) Yl−1,m+1

sin θ e−iφ Yl,m =√

(l−m+1)(l−m+2)(2l+1)(2l+3) Yl+1,m−1 −

√(l+m)(l+m−1)(2l−1)(2l+1) Yl−1,m−1

(9.44)


λ θ−j =

[cosθ sinθ e−iφ

sinθ eiφ −cosθ

] [ √j−mj + 1 Yj+1/2,mj−1/2

−√

j + mj + 1 Yj+1/2,mj+1/2

]

=

√j−mj + 1

√ (j + 1/2−mj + 1/2 + 1)(j + 1/2 + mj − 1/2 + 1)

(2j + 2)(2j + 4)Yj+3/2,mj−1/2 +

√(j + 1/2−mj + 1/2)(j + 1/2 + mj − 1/2)

(2j)(2j + 2)Yj−1/2,mj−1/2

−√

j + mj + 1

√ (j + 1/2−mj − 1/2 + 1)(j + 1/2−mj − 1/2 + 2)

(2j + 2)(2j + 4)Yj+3/2,mj−1/2 −

√(j + 1/2 + mj + 1/2)(j + 1/2 + mj + 1/2− 1)

(2j)(2j + 2)Yj−1/2,mj−1/2

√

j−mj + 1

−√

(j + 1/2 + mj − 1/2 + 1)(j + 1/2 + mj − 1/2 + 2)

(2j + 2)(2j + 4)Yj+3/2,mj+1/2 +

√(j + 1/2−mj + 1/2)(j + 1/2−mj + 1/2− 1)

(2j)(2j + 2)Yj−1/2,mj+1/2

+√

j + mj + 1

√ (j + 1/2−mj − 1/2 + 1)(j + 1/2 + mj + 1/2 + 1)

(2j + 2)(2j + 4)Yj+3/2,mj+1/2 +

√(j + 1/2−mj − 1/2)(j + 1/2 + mj + 1/2)

(2j)(2j + 2)Yj−1/2,mj+1/2

=

[ ((j−mj + 1)

√j+1/2+mj−1/2

(2j+2)(2j) + (j + mj + 1)

√j+1/2+mj−1/2

(2j+2)(2j)

)Yj−1/2,mj−1/2(

(j−mj + 1)

√j+1/2−mj−1/2

(2j)(2j+2) + (j + mj + 1)

√j+1/2−mj−1/2

(2j+2)(2j)

)Yj−1/2,mj+1/2

]

=

√2j+22j√

j + mj Yj−1/2,mj−1/2√2j+2

2j√

j−mj Yj−1/2,mj+1/2

=√

j+1j θ+j

(9.45)Because λ2 = 1 this then implies that

λ θ+j =

√j

j + 1θ−j (9.46)

It can also be shown that:

θ θ+j = −iφ θ−j (9.47)

and:

θ θ−j = −iφ θ+j (9.48)


9.6 Appendix F: More Tranformations

As we showed in Chapter 5, the transformation generator, S = −i α2 ε, generates

a unitary transformation that produces the Coulomb-Dirac equation and the fa-miliar spectrum of the hydrogen atom. Let’s explore a few other closely relatedtransformations.

9.6.1 The generator S = κ2γr

Consider the generatorS =

κ

2γr

and its transformation operator

eS = eκ2 γr

= cosκ

2+ γrsin

κ

2

If κ is assumed real, then this transformation is unitary, since (eκ2 γr

)† = e−κ2 γr

.Following Eq. 1.19, the field generated by this transformation is

∆ = γµe−S ∂eS

∂xµ =sinκ

r+ 2

sin2 κ2

rγr

The resulting alteration to the probability (hydraulic) current is:

γ0∆ + (γ0∆)† = 2γ0 sinκ

r+ 4γ0γr sin2 κ

2r

So this transformation, although unitary, does not preserve the form of the localcontinuity equation, ∂µ jµ = 0.

9.6.2 The generator S = κ2ε

For the generator S = κ2 ε, the transformation operator is

eS = eκ2 ε = cosh

κ

2+ εsinh

κ

2

(Remember that ε = γ0γr ).


The alteration to the Dirac equation is

∆ =sinhκ

rγ0 +

2sinh2 κ2

rγr

The alteration to the hydraulic current is

γ0∆ + (γ0∆)† =2sinhκ

r+

4sinh2 κ2

rε

So this transformation is not unitary, and it also does not preserve local continuity.

9.6.3 The generator S = −i α2 ε

The generator S = −i α2 ε generates the transformation

eS = e−i α2 ε = cos

α

2− iεsin

α

2

This is the transformation we used in Chapter 5 that leads to the Coulomb-Diracequation and the spectrum of the hydrogen atom.


∆ = −isinα

rγ0 −

2sin2 α2

rγr,

and the alteration to the hydraulic current is

γ0∆ + (γ0∆)† = −4sin2 α

2r

ε.

So this transformation, although it is unitary, does not preserve local hydrauliccontinuity.

9.6.4 The generator S = −i α2 γr

The generator S = −i α2 γr generates the transformation

eS = e−i α2 γr

= coshα

2− iγrsinh

α

2



∆ = isinhα

r+

2sinh2 α2

rγr

and the alteration to the hydraulic current is

γ0∆ + (γ0∆)† =4sinh2 α

2r

ε.

This transformation in non-unitary and does not preserve local hydraulic conti-nuity.

Here’s a table to summarize these results.

S eS ∆ γ0∆ + (γ0∆)† Unitary? Local Cons? Global Cons?

I. κ2 γr cos κ

2 + γrsin κ2 ∆ = sinκ

r +2sin2 κ

2r γr 2sinκ

r γ0 +4sin2 κ

2r ε Yes No No

II. κ2 ε cosh κ

2 + εsinh κ2 ∆ = sinhκ

r γ0 +2sinh2 κ

2r γr 2sinhκ

r +4sinh2 κ

2r ε No No No

III. −i α2 ε cos α

2 − iεsin α2 ∆ = −i sinα

r γ0 − 2sin2 α2

r γr −4sin2 α2

r ε Yes No Yes

IV. −i α2 γr cosh α

2 − iγrsinh α2 ∆ = i sinhα

r +2sinh2 α

2r

4sinh2 α2

r ε No No Yes

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[3] quantummechanics.ucsd.edu – An excellent web resource outlining the tra-ditional solution of the Dirac hydrogen atom.

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171

http://quantummechanics.ucsd.edu/ph130a/130_notes/node501.html

BIBLIOGRAPHY 172

[10] George B. Arfken and Hans J. Weber. Mathematical Methods for Physicists.Fourth Edition. Academic Press, 1995.

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