Junior Honours Quantum Mechanics

Chris Stock (cstock@ed.ac.uk)

September 21, 2015

1 Introduction

1.1 Outline

This course will involve a short introduction on the breakdown of classical theory followed by a seriesof lectures on the wave mechanics approach to quantum theory. The second part of the course willdiscuss Dirac notation and applications of this using matrix mechanics. Understanding the techniquesapplied to matrix mechanics is key to this course. The final part of the class will involve an introduc-tion to perturbation theory and applications to the multielectron atoms.

The outline of the class is listed below.

1) Introduction- what is wrong with classical theory?2) Wave mechanics - introduction, “postulates”, Schrodinger equation, and introduction to angularmomentum3) Applications to scattering4) The Hydrogen atom5) Simple Harmonic Oscillator solved using Hermite polynomials6) Matrix mechanics - Dirac notation7) Observables and operator algebra (CSCO-Complete Set of Commuting Observables)8) Ladder operators and the harmonic oscillator revisited9) Angular momentum10) Spin operators11) The two level problem12) Time-independent perturbation theory13) Application of perturbation theory to the multielectron atom

1.2 References

[1] C. Cohen-Tannoudji, D. Diu, and F. Laloe, “Quantum Mechanics” (Hermann, Paris, France, 1973).(very formal treatment of the mathematical basis for quantum mechanics with emphasis on matrixmechanics)

[2] L. Pauling, and E.B. Wilson, “Introduction to Quantum Mechanics with Applications to Chem-istry” (Dover, New York, 1935).(good presentation of the solution to the hydrogen atom, perturbation methods, and the harmonicoscillator using wave mechanics)

[3] M.A. Morrison, T.L. Estle, and N.F. Lane, “Quantum States of Atoms, Molecules, and Solids”(Prentice-Hill, New Jersey, 1976).(excellent presentation and discussion of some key aspects to atomic physics and also perturbationmethods)

1

[4] B. R. Judd, “Operator Techniques in Atomic Spectroscopy” (Princeton University Press, New Jer-sey, 1998).(a short book providing a compact discussion of angular momentum and operators for crystal fieldanalysis)

[5] L. D. Landau and E. M. Lifshitz, “Quantum Mechanics” (Elsevier Science, Oxford, 1977).(advanced treatment of quantum mechanics with little use of Dirac Notation)

[6] A. Messiah, “Quantum Mechanics” (Dover, New York, 1958).(a very good treatment of both matrix and wave mechanics)

[7] J. Hardy, “Quantum Physics” (Notes from second year quantum physics).(These notes provide a starting point for this course and some reference will be made. The two text-books used to formulate these notes by French and Taylor and Eisberg also provide useful startingreferences.)

[8] J. A. Peacock “Fourier Analysis” (Notes from current third year course).(The techniques developed in this course will be referred to and are fundamental to many of the resultsdescribed here.)

2 What is wrong with classical mechanics? (Messiah -Chapter 1, Cohen-Tannoudji-Chapter 1)

There are many examples and illustrations where experiments clearly show that classical theory isincomplete and something else is required to describe real systems. Probably most people are familiarwith the “UV catastrophe” problem and the photoelectric effect (summarized by the relation E =hν = hω) which clearly show the quantized nature of light (see notes on the photoelectric effect byHardy). In parallel to these, the discovery of fundamental particles such as the electron highlight thequantized nature of matter. It therefore is clear that a new theory is needed and must treat matterand light on an equal footing.

While these experiments highlight the failure of continuous classical theory, I would like to focuson two experiments that describe what properties should exist in a “new theory”.

2.1 Experiment 1: Double slit experiment

There are several conclusions we can draw from this experiment. First, while the photoelectric effecttells us that light should be considered in terms of discrete units called photons, the interference effectillustrates the underlying wave nature. The theory that we develop must have “wave-particle duality”.While most people are familiar with the double slit experiment resulting in interference patterns withlight, a similar effect can be observed with particles such as electrons. An electron diffraction pattern isshown in Fig. 2 which similar to that of light. This indicates that electrons can be behave qualitativelylike waves, while the photoelectric effect shows that light can behave like particles.

2.1.1 de Broglie relation

We have shown above the need for both a wave and particle description of light based upon ourknowledge of the photoelectric effect (see J. Hardy’s notes) and Young’s double slit experiment. Thewave particle duality is formalized through the de Broglie relation which is written as follows.

λ =2π

k=h

p(1)

The energy of the particle is related to the frequency by the Einstein relation.

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Figure 1: Figure taken Cohen-Tannoudji, Quantum Mechanics.

E = hω (2)

Recall that ω ≡ 2πν.

2.1.2 Experiment 2: Malus’ law

Consider an experiment consisting of a polarized plane monochromatic light wave incident on a de-tector.

Classically, the light wave takes the following form,

~E(~r, t) = E◦ei(kz−ωt)p (3)

After passing through the analyzer, the plane wave is polarized along a particular direction,

~E′(~r, t) = E′◦ei(kz−ωt)x (4)

The intensity of this transmitted wave is proportional to |E′◦|2 and is given by Malus’ law -I ′ = I cos θ where θ is the angle between x and |p|. This law is difficult to understand at a quantumlevel when the incident beam is so weak that the photon will either cross the polarization analyzeror be absorbed (we cannot detect fractions of photons). The result does make sense if we consider alarge number of photons incident on the analyzer where the detected intensity will be proportional toN cos2 θ photons. There are several results.

The analyzer only chooses particular chosen or “eigen” (we will discuss more why we use thisword later) results. This makes sense if we think about the analyzer only pointing along two possible

3

Figure 2: Figure taken from B. C. Hall, Quantum Theory for Mathematicians.

directions, but if it is an angle (like in Fig. 3) we can think about the result in terms of probabilities.This experiment highlights the need to consider probabilities in the new theory and to find theseprobabilities, we decomposes the particles into a linear combination of various eigenstates or chosenstates. This idea is often referred to as spectral decomposition.

The final interesting result from this experiment is that the measurement disturbs the system beingprobed. We will come back to this point later. A final point that needs to be made is that measuringdevice is linear in terms of E and note that I ∝ E2. This is particularly important in the context ofthe double slit experiment.

2.2 Correspondence principle and cleaning up classical mechanics

In the discussion above we have discussed two examples (Young’s double slit experiment and Malus’law) where classical theory seems to be incompatible with experimental results. However, classicaltheory does work (and is very successful) so our new quantum theory must be equivalent to classicaltheory in some limit. A natural question to ask is - Can we clean up classical mechanics by notingthat in the “classical limit” our new theory should reduce the familiar classical results?

3 Wave mechanics - postulates and the Schrodinger equation

The main conclusion from the experiments discussed in the previous section that a new theory mustbe linear and obey the superposition principle. Note the similarity between these physical constraintsand the definitions above for a linear operator acting on a vector.

Lets describe some postulates that we have learned from these two experiments to create a formu-lation of quantum mechanics:

4

Figure 3: Figure taken from Cohen-Tannoudji, Quantum Mechanics, vol 1.

(i) “The quantum state of a particle such as the electron is characterized by a wave functionψ(~r, t), which contain all of the information it is possible to obtain about the particle.”

(ii) “ψ(~r, t) is interpreted as a probability amplitude of the particle’s presence. Since the pos-sible positions of the particle form a continuum, the probability dP (~r, t) of the particle being attime t, in a volume element d3r = dxdydz situated at the point ~r must be proportional to d3r andtherefore infinitesimal. |ψ(~r, t)|2 is then interpreted as the corresponding probability density withdP = C|ψ(~r, t)|2d3r, where C is a normalization constant.”

(iii) “The principle of decomposition applies to the measurement of an arbitrary physical quantity:1) The result found must belong to a set of eigen results a; 2) With each eigenvalue a is associatedan eigenstate, that is, an eigenfunction ψ(~r). This function is such that, if ψ(~r, t0) = ψa(~r) (where t0is the time at which the measurement is performed), the measurement will always yield a; 3) For anyψ(~r, t), the probability Pa of finding the eigenvalue a for a measurement at the time t0 is found bydecomposing ψ(~r, t) in terms of the functions ψa(~r):”

ψ(~r, t0) =∑a

caψa(~r) (5)

then,

Pa =|ca|2∑a |ca|2

(6)

(the presence of the denominator ensures that the total probability is equal to 1)

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(iv) The equation describing the evolution of the function ψ(~r, t) remains to be written. It ispossible to introduce it in a very nature way, using the Planck and de Broglie relations. Based on theexperiments discussed above, the equation must be linear and homogeneous and hence we write thefollowing,

Hψ = Eψ (7)

Given the correspondence principle discussed above, we can postulate what the form of this equa-tion is based on the result that any theory we write must cross over to the classical limit and also

must be obeyed by plane wave solutions of the form ψ(~r, t) ∝ eiEhteikx. We therefore expect the time

dependence to be linear and equal to E → −ih ∂∂t . Also, given that classically we expect the energy

of a free particle of momentum p to be p2

2m which (using the de Brogile relations) equal to h2k2

2m wetherefore make the association p→ ih∇. Therefore, the Schrodinger equation for a free particle wouldbe

− h2

2m∇2ψ(~r, t) = ih

∂

∂tψ(~r, t) (8)

If we have a particle in a potential V (~r, t), then classically we would write the energy as the sum of thekinetic energy and the potential energy. Adding this term in gives us the full Schrodinger equation.

− h2

2m∇2ψ(~r, t) + V (~r, t)ψ(~r, t) = ih

∂

∂tψ(~r, t) (9)

This is the full equation which we will be solving for a series of cases during the wave mechanics partof this course. To motivate this equation, we have associated various classical observable quantitieswith operations based on the correspondence principle requiring us to retain the solutions found withclassical dynamics. For example, we associate momentum p of a particle with the operation −ih∇and the energy with the time derivative −ih ∂

∂t .

4 Particles in a time independent scalar potential

In this section, we will solve the Schrodinger equation for several one-dimensional potential wells andestablish a procedure for solving these problems in general. In the previous section, we argued that ournew theory must be based upon a linear homogeneous equation based upon superposition arguments(Young’s experiment) and spectral decomposition (Malus’ Law). We then invoked the fact that anytheory we write should give the classical result as a particular limit and also we noted that it should beconsistent with our knowledge of plane wave solutions based on the photoelectric effect and classicalelectromagnetic theory. We then wrote the following wave equation (Schrodinger’s equation).

ih∂

∂tψ(~r, t) = − h2

2m∇2ψ(~r, t) + V (~r)ψ(~r, t) (10)

In this section, we will study solutions of this wave equation for certain “toy” models includingthe infinitely deep well, the Dirac well, and the harmonic well. We will then go onto apply thesetechniques to solving the hydrogen atom.

For the remainder of this section on wave mechanics, we will be studying potentials that are timeindependent implying that V is only a function of ~r as written above. Therefore, using the separationof variables method for solving different equations we can start of with the following solution of Eqn.12.

ψ(~r, t) = φ(~r)e−iEht (11)

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substituting this back into the Schrodinger equation above gives the time independent Schrodingerequation

Eφ(~r) = − h2

2m∇2φ(~r) + V (~r)φ(~r) (12)

Solving this for a series of potentials will be the topic of the next few sections.

4.1 Square well with infinite walls (Messiah-Chapter 3, Hardy - Chapter 9.1.2)

To start our study of special potentials applying Quantum wave mechanics, we start with the infinitewell. This was done previously in J. Hardy’s course, but it is instructive to review the potential andthe techniques for solving it before studying other problems.

V (x) =

{0 if |x| < a

2∞ if |x| ≥ a

2

We can start off by noting that solution of ψ for |x| ≥ a is 0. There is no possibility that that theparticle can be found within the walls. Within the region |x| < a, the Schrodinger equation can bewritten as,

d2ψ(x)

dx2+ εψ = 0 (13)

The solution has been worked out by Hardy and we will not go into the details here but rather tostate that it is,

ψ(x)n = sin(nπax), n = even (14)

ψ(x)n = cos(nπax), n = odd (15)

(16)

with energy eigenvalues of

εn =n2π2

a2(17)

Note that when n = 0 there is no particle in the box. This is a valid solution of the Schrodingerequation above and corresponds to zero energy. When there is a particle in the box it must have aground state energy given by n = 1 in the above formula. We will return to this concept of zero pointenergy later in the harmonic oscillator.In the exercises, you will be investigating the finite depth well and connecting the solution with theDirac potential discussed in the next section.

4.2 Some special functions- the Dirac Delta and Heaviside functions(Cohen-Tannoudji-Appendix-2)

For the remainder of this course, we will require some understanding of several functions with uniqueproperties. In particular we will make use of the “δ-function. As pointed out by Cohen-Tannoudji,this is not mathematically strictly a function but rather a distribution. We will refer to it as a functionas dictated by convention.

Consider a function δε(x) with the following definition

δε(x) =

{1ε if −ε2 < x < ε

20 if |x| > ε

2 .

7

If we consider the integral∫∞−∞ δ

εf(x)dx ∼ f(0)∫∞−∞ δ

ε(x)dx = f(0). The smaller ε, the bet-ter this approximation. We therefore examine limε→0 and define the δ function by the relation∫ +∞−∞ f(x)δ(x)dx = f(0). In general we can write the following,

∫ +∞

−∞δ(x− α)f(x)dx = f(α). (18)

It follows from this that∫∞−∞ δ(x)dx = 1. There are a number of ways of defining the δ function and

for simplicity we have chosen 1/ε. Other examples which approach this when ε approaches zero are12εe−|x|/ε, 1

πε

x2+ε2, and 1

πsin(x/ε)

x . A more comprehensive list can be found in Cohen-Tannoudji (volume2 and appendix 2).

Another definition of the δ-function can be formulated in terms of the step or “Heaviside” function.

Θ(x) =

{1 if x > 00 if x < 0.

It can be shown, taking our definition of δ above in terms of ε, that δ-function is the derivative ofthe Heaviside function. These two “functions” are very important and will be used in our study ofpotentials. Further properties of the δ function will be derived in the problem set.

4.3 Dirac well (Davies-Chapter 3, Cohen-Tannoudji- Complement K1)

A very interesting potential which as many applications in scattering and also molecular physics is theDirac potential. This is also a useful potential for establishing techniques in analyzing any problemdealing with a potential well.

V (x) = −αδ(x) (19)

where α is a positive real number. The Schrodinger equation then becomes the following

−h2

2m

d2ψ(x)

dx2− αδ(x)ψ(x) = Eψ(x) (20)

First consider the two limiting cases where x 6= 0 and hence the equation reduces to −h2

2md2ψ(x)dx2 =

Eψ(x). The solution to this is,

ψ(x) = Ae−βx +Be+βx, x 6= 0 (21)

where A and B need to be determined and β =√−2mE/h. Imposing well defined properties of ψ (ie

finite at all values of x) fixes the solution to be the following

ψ(x) =

{Beβx if x < 0Ae−βx if x > 0.

and imposing the fact that ψ must be continuous gives us A = B (take the limit as x→ 0 from bothpositive and negative sides and set them equal). We now deal with the x=0. At this point, we havethe above ordinary differential equation for ψ(x) containing the special function δ(x). Integrating witha region of ±ε around x = 0 gives,

∫ +ε

−ε

(−h2

2m

d2ψ(x)

dx2− αδ(x)ψ(x)

)dx =

∫ +ε

−εEψ(x)dx (22)

and integrating

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−h2

2m

(limε→0+

dψ(x)

dx− limε→0−

dψ(x)

dx

)− αψ(0) = E

∫ +ε

−εψ(x)dx (23)

Lets now remember that ψ(x) must be continuous (as well as being finite as used above already). Thecontinuity property of ψ(x) implies that lim|ε|→0

∫ +ε−ε ψ(x)dx ≡ 0 (identically). Therefore, the right

hand part of the equation above is 0. We now have,

(limε→0+

dψ(x)

dx− limε→0−

dψ(x)

dx

)= −2mα

h2 ψ(0) (24)

Using our solutions for ψ(x) at |x| > 0, we find that by taking derivatives that 2Aβ = 2mαh2 ψ(0), and

continuity implies limx→0 ψ(x) = A implying β = mαh . There is there only one bound state, or energy

eigenvalue, to this potential, namely

E = −mα2

2h2 . (25)

To complete this off we still need to find A and this can be derived by noting∫ +∞−∞ ψ(x)dx ≡ 1. Doing

the algebra gives A =√mαh . We can then write the solution of ψ compactly as

ψ(x) =

√mα

he−

mαh2 |x| (26)

This solution will also be derived on the problem set where you will take a well of finite width andthen take the width to zero studying how many energy eigenvalues there are as a function of width.

4.4 Harmonic Oscillator (Pauling - Chapter 3, Landau- Chapter 23)

Consider a potential energy of the form V (x) = 12mω

2◦x

2 where x is the displacement of the particle ofmass m from the equilibrium position x = 0. The classical energy of this system is written as follows,

E =p2

2m+

1

2mω2◦x

2. (27)

Using the correspondence principle discussed above, we can associate E with the Hamiltonian operatorH and the moment p with its corresponding operator −ih d

dx . We can then write the Schrodingerequation as follows,

−d2ψ(x)

dx2+ α2x2ψ(x) = λψ(x) (28)

or

d2ψ(x)

dx2+(λ− α2x2

)ψ(x) = 0. (29)

We would like to find solutions for ψ(x) that are physical and are suitable everywhere (ie for x inthe range of −∞ to +∞). We could use our knowledge of special functions to solve this, howeverit is useful to approach this from a different standpoint starting from finding the solution in certainlimits and then perform a power series expansion. We are searching for finite solutions which aresingle valued over this region (well behaved functions). First, consider the situation with |x| is large

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in the limit where α2x above dominates and λ is negligible by design. In this limit we want to findthe solution to the following equation,

d2ψ(x)

dx2+ α2x2ψ(x) = 0. (30)

We can guess the solution to this equation as ψ = e±α2x2

and this can be verified by direct substitutionand also remembering we are studying the solution at large |x| (try to verify this and remember that

we are working in the limit of large x). We note that of the two solutions e+α2x2

is not satisfactorygiven the fact that it diverges. Using the fact that we found a good solution at large |x|, we can nowproceed to derive solutions by introducing a power series in x and deriving the coefficients.

Define a trial wavefunction ψ(x) = e−α2x2f(x). Lets list the derivatives here,

dψ

dx= −αxf exp

[−α

2x2]

+ f ′ exp[−α

2x2]

(31)

d2ψ

dx2=(−αf + α2x2f − αxf ′ + f ′′ − αxf ′

)exp

[−α

2x2]

(32)

=(α2x2f − αf − 2αxf ′ + f ′′

)exp

[−α

2x2]

Substituting this into Eqn. 30 gives the following differential equation for f

f ′′ − 2αxf ′ + (λ− α)f = 0 (33)

To make the connection with special functions in a few steps lets make the substitution ξ =√αx and

replace the function f(x) with H(ξ) to get the following differential equation,

d2H

dξ2− 2ξ

dH

dξ+

(λ

α− 1

)H = 0 (34)

We now represent H(ξ) as a power series

H(ξ) =∑n

anξn = a0 + a1ξ + a2ξ

2 + ... (35)

At this point we have made an important assumption regarding the properties of H as we have assumedthat this series converges for all ξ and that the function is well defined. Physically, we know this hasto be the case for the solutions to make sense and to satisfy our overall constraint that the integral ofthe probability density must be finite :

∫∞−∞ |ψ(x)|2dx = 1.

By now differentiating H(ξ) and substituting the derivatives back into Eqn. 34 we come up witha rather messy expression which contains powers of ξ. However, for this equation to hold for all ξ,we can equate coefficients of terms with the same power of ξ and hence derive the following recursionrelation (this will be done more explicitly in the problem set). We therefore get the following recursionformula,

(n+ 1)(n+ 2)an+2 +

(λ

α− 1− 2n

)an = 0. (36)

We are basically done as we have derived a general solution ψ(x) for the harmonic oscillator in terms

of functions of the form ψ(x) = f(x)e−α2x2

with f(x) being a power series. There is however a seriousproblem. While we assumed a few steps ago that the series f(x) converged for all x, this is not generally

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Figure 4: Figure taken Pauling, Introduction to Quantum Mechanics.

true in the recursion relation in Eqn. 36 for all values of λ (the energy values) and α (potential wellcurvature). The function is square integrable if f(x) is truncated and does not contain an infinitenumber of terms in its series expansion. Note that the constraint that we have imposed on ψ(x) ismuch more strict than just convergence, we require the following to be true for our solutions.

∫ ∞−∞|ψ(x)|2dx = 1. (37)

It can be seen from Eqn. 36 that the series expansion is finite (and hence converges ∀x) if

(λ

α− 1− 2n

)= 0 (38)

or the energy values have the following discrete spectra,

λ = (2n+ 1)α. (39)

This condition can be expressed as the following,

En =

(n+

1

2

)hν◦ (40)

and this should be compared with the result derived from the photoelectric effect which states thatE = nhν. The big difference here is the presence of a zero point energy when n = 0. This resulthas experimentally been confirmed and manifests as a weak attraction between two parallel plates ina vacuum. This effect only results from “vacuum” fluctuations and is known as the Casimir effectafter the person who predicted it in 1948 (see Phys. Rev. Foc. 2 28 (1998) for a nice discussion andintroduction).

Graphically, examples of the wavefunctions are plotted in Fig. 4 and 6. It can be seen that thenumber of zeroes (or nodes) in the wavefunctions increases with energy and n with the ground stateplotted in Fig. 5. An extreme case is plotted for n = 10 in Fig. 6.

One final note is that we have avoided using special functions here and instead of relied on thepostulates of our new quantum wave theory which includes imposing well defined properties on ψ(x)

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Figure 5: Figure taken Pauling, Introduction to Quantum Mechanics.

Figure 6: Figure taken Pauling, Introduction to Quantum Mechanics.

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(like convergence). The functions we have written down H(ξ) are known as Hermite polynomialsand the properties of these will be studied in problem sets or they can be investigated by readingthe references noted above. In summary, using the solutions in the limit of large |x| we were able toderive a general solution for the harmonic oscillator. By stipulating the physical criterion that thewavefunction must be finite and single valued, we found that naturally we had to impose a cutoff inthe series expansion which resulted in the presence of discrete energy values. In the end, we found thepresence of a zero point energy value and have noted that this agrees with experiments.

We will return to the harmonic oscillator problem later and solve it using operator techniques andintroduce ladder operators important for studies in quantum optics and field theory.

5 Operators and angular momentum (Landau-Chapter 4)

Before solving the hydrogen atom we need to develop some formalism to describe and understandangular momentum. As noticed in classical theory, the symmetry of central force problems (wherethe potential only depends on r) angular momentum plays an important role because it is a conservedquantity. In classical theory, angular momentum is defined as,

~L = ~r × ~p (41)

It is helpful, particularly for the hydrogen atom, to first consider the classical case of a rigid rotor.Classically, a fixed particle of mass m with a fixed distance r has the following expression for theenergy.

E =L2

2mr2(42)

Recall from our formulation of the Schrodinger equation, we associated various “observable” quantitieswith various mathematical operations or “operators”. As noted previously that while we are usingthese terms loosely at the moment, we will formally define them in the matrix mechanics section ofthis course where we introduce Dirac notation. From the correspondence principle, we associate eachclassical quantity with one of these operations and in particular note that P → −ih∇. We thereforewrite the Schrodinger equation (time independent) for a quantum rigid rotor as,

Eψ =L2

2mr2ψ (43)

It is clear from this Hamiltonian that an equivalent problem is to solve the equation λσ = L2σ. Once wehave the solution to this, we have the solution to the rigid rotor equation above. We can write the totalangular momentum L2 ≡ L2

x+L2y+L2

z, making the transformation to spherical coordinates ((x, y, z)→(r sin θ cosφ, r sin θ sinφ, r cos θ)) and transforming the ∇ function into spherical coordinates, we getthe following equation in terms of θ and φ,

L2σ = λσ (44)

−(

1

sin2 θ

∂2

∂φ2+

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

))σ = λσ

We can solve this equation through separation of variables where we write σ(θ, φ) = Θ(θ)Φ(φ). Thisthen leads to the following equation after some rearrangement

1

Φ

∂2

∂φ2Φ +

sin θ

Θ

∂

∂θ

(sin θ

∂

∂θ

)Θ + sin2 θλ = 0 (45)

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Notice the first term only depends on φ and is equal to terms independent of φ. For this to be validfor all φ this term must be equal to a constant. We therefore write,

d2Φ

d2φ= −a2Φ (46)

The solution to this is Φ = 1√2πeiaφ. For this function to be periodic in φ (that is to be single valued

for example φ=0 and φ=2π are the same) a must be an integer m = 0,±1,±2, .... Note that thefactor

√2π is included in the solution of Φ to ensure that the function is normalized, that is that∫

dφΦ∗Φ = 1. Substituting this solution for Φ back into our equation above, gives the followingexpression only in terms of θ.

−m2

sin2 θ+

1

Θ

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)Θ + λ = 0 (47)

The next step is to solve the equation which depends on θ. In order to solve this, lets first introducea substitution to a new independent variable z = cos(θ) which varies over the range z = [−1, 1]. Toreplace Θ(θ) by the function

P (z) = Θ(θ), (48)

we need the following Jacobian

dΘ

dθ=dP

dz

dz

dθ= −dP

dzsin θ (49)

and the fact that sin2 θ = 1−z2. Putting this all together gives the following new differential equation,

d

dz

((1− z2)

dP (z)

dz

)+

(β − m2

1− z2

)P (z) = 0 (50)

We can solve this equation using the same method developed for the solution of the harmonic oscillatorabove. This is actually quite tedious in this case and the solution is outlined very nicely by Paulingand also in Eisberg used in the second year course. We will just note that the solution to this equationis the associated Legendre functions Pml (cos θ):

Pml = (1− z2)|m|/2d|m|

dz|m|Pl(z) (51)

where Pl(z) are the ordinary Legendre polynomials. A general expression for these are given byRodrigues’ formula

Pl(x) =1

2l!

dl

dxl(x2 − 1)l (52)

The important thing to note here is that we have a second integer which enters in based on a recursionrelation and reasoning similar to that in the harmonic oscillator. Recall that in the harmonic oscillatorwe required the series expansions to be finite at large x and ensured this by truncating them. If wewere to go through the entire expansion here we would find a problem at z = ±1 requiring a similartruncation. The overall angular solutions to this problem are call spherical harmonics and have theform,

14

Figure 7: Figure taken Morrison, Estle, and Lane, Quantum States of Atoms, Molecules, and Solids.

Y ml (θ, φ) ∝ Pml (cos θ)eimφ (53)

Out of this truncation in the series expansion, we have found a new integer l with λ = l(l + 1)where l and m are connected by l = 0, 1, 2, 3... and m = −l...+ l. We can therefore write the followingsolution to the original problem,

L2σ = λσ → L2Y ml = l(l + 1)Y m

l (54)

To understand the physical significance of this, it is interesting to note what the object Lz is equalto. Using our concept of correspondence, this can be associated with the operation −i∂∂φ . Acting thison our angular solution gives a factor of m. We can then write the following equation,

LzYml = mY m

l (55)

15

Figure 8: Figure taken Pauling, Introduction to Quantum Mechanics.

The angular dependence in real space is plotted in Fig. 7. The ground state (l = 0) has no nodes inthe angular dependence and as the energy increases the number of nodes increases. Three dimensionalcontours for l = 0 and l = 1 are shown in Fig. 8.Therefore m is associated with the z − axis projection of the total angular momentum.

6 Hydrogen atom (Morrison, Pauling-Chapter 18, Messiah-Chapter 9, Landau-Chapter 5)

To conclude the first section of course dealing with wave mechanics based on applications of Schrodingerequation, we discuss the solution to the Hydrogen atom. This is one of the few exactly solvable prob-lems in quantum mechanics and of importance to quantum states in molecules. The underlyingequation can be written as,

H = − h2

2m∇2 + V (r) (56)

where the potential V (r) takes the Coulomb form of − e2

r for the hydrogen atom which consists of apositive proton being circled by a negatively charged electron. For the moment we will only assumethat V (r) only depends on r and investigate solutions. There are several ways of solving this outlinedin different books. All books use separation of variables and write separate equations for the angularand radial components. Pauling solves for three equations (r, φ, θ) while Messiah and Landau takeadvantage of the angular momentum operators discussed in the previous section. We will follow thislater approach as it involves slightly less work and highlights the physical significance of the result.However, we will exploit Pauling’s technique (used for the harmonic oscillator above) to solve theradial part of the equation.

Given the symmetry of the problem (note the equation does not explicitly depend on angles θ orφ), it is convenient to switch to spherical coordinates where,

16

∇2 =1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2(57)

the equation of motion then becomes,

1

r2

∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

∂2

∂ψφ2+

2m

h2 [E − V (r)] = 0 (58)

If we introduce the operator L2 we can reduce this to

h2

2m

(− 1

r2

∂

∂r

(r2∂ψ

∂r

)+L2

r2

)+ V (r)ψ = Eψ (59)

We know classically that angular momentum is conserved during motion in a central potential. Weshall consider stationary states in which the angular moment l and the component m have definitevalues. These values will then fix the angular dependence. We therefore will perform separation ofvariables to write ψ in terms of an angular and a radial part as follows.

ψ = R(r)Ylm(θ, φ) (60)

Since L2Yl,m = l(l + 1)Ylm we obtain for the radial function R(r),

1

r2

d

dr

(r2dR

dr

)− l(l + 1)

r2R+

2m

h2 [E − V (r)]R = 0 (61)

Let’s solve the radial part and set V (r) = − e2

r and therefore specialize the problem to that of thehydrogen atom. Following Pauling, lets consider cases where the energy is insufficient to ionize theatom and hence consider negative energy values. It is helpful to group together the constants usingthe following substitutions,

α2 = −2mE

h2 (62)

λ =me2

hα(63)

and the new independent variable ρ = 2αr. The wave equation then becomes,

1

ρ2

d

dρ

(ρ2dS

dρ

)+

(−1

4− l(l + 1)

ρ2+λ

ρ

)S = 0 (64)

where 0 ≤ ρ < ∞ and note that we have set S(ρ) = R(r) and computed the appropriate Jacobian.Just like the case of the simple harmonic oscillator above, lets consider the case at large r when thepotential is weak and the terms ∝ 1

r are small. The equation then becomes,

d2S

dρ2=

1

4S (65)

with solutions S = e±ρ2 of which only S = e

−ρ2 is satisfactory based on the fact that wavefunction

must be finite for all r. As in the simple harmonic oscillator, we now assume that the solution to thisODE has the form,

17

S(ρ) = e−ρ2 F (ρ) (66)

The equation satisfied by F (ρ) can be derived to be,

d2F

dρ2+

(2

ρ− 1

)+

(λ

ρ− l(l + 1)

ρ2− 1

ρ

)F = 0 (67)

where 0 ≤ ρ <∞. The coefficients of dFdρ and F possess singularities at the origin. To ensure that the

solutions are well behaved and finite for all r, we make the additional substitution F (ρ) = ρsL(ρ) inwhich L(ρ) is the following power series which a non vanishing constant term.

L(ρ) =∑ν

aνρν (68)

where a0 6= 0. Taking derivatives, we have the following,

dF (ρ)

dρ= sρs−1L+ ρ

dL

dρ(69)

and

d2F (ρ)

dρ2= s(s− 1)ρs−2L+ 2sρs−1dL

dρ+ ρs

d2L

dρ2(70)

Putting this all together, our ODE for the radial part of the wavefunction becomes,

ρs+2d2L

dρ+ 2sρs+1dL

dρ+ s(s− 1)ρsL+ 2ρs+1dL

dρ+ 2sρsL− ... (71)

ρs+2dL

dρ− sρs+1L+ (λ− 1)ρs+1L− l(l + 1)ρsL = 0

By design, L begins with terms a0 and therefore the coefficients of ρs must vanish and for this tohappen [s(s− 1) + 2s− l(l+ 1)] = 0. This gives us a relations for s implying that s = +l or −(l+ 1).Of the two solutions −(l + 1) does not give acceptable solutions and we can therefore write

F (ρ) = ρlL(ρ) (72)

and further substitution gives

ρd2L

dρ2+ [2(l + 1)− ρ]

dL

dρ+ (λ− l − 1)L = 0 (73)

We now explicitly introduce the power series presented above (L(ρ) =∑

ν aνρν) and again the coeffi-

cients must vanish individually. We then come up with the following recursion relations.

(λ− l − 1)a0 + 2(l + 1)a1 = 0 (74)

(λ− l − 2)a1 + (4(l + 1) + 2) a2 = 0 (75)

(λ− l − 3)a2 + (6(l + 1) + 6) a3 = 0 (76)

or for general coefficients of ρν

18

(λ− l − 1− ν)aν + (2(ν + 1)(l + 1) + ν(ν + 1)) aν+1 = 0 (77)

It can be argued, like in the case of the harmonic oscillator, that for any values of λ and l the serieswhose coefficients are determined by this formula must be truncated. For very large values of ν, thesuccessive terms of an infinite series are two large and do not converge. For the series to break off weneed the following relation

(λ− l − 1− n′) = 0 (78)

which implies λ = n where n = n′ + l + 1. n′ is called the radial quantum number and n the totalquantum number. The allowed radial function can then be written (finally),

R(r) = e−ρ2 ρlL(ρ) (79)

in which L(ρ) is defined by the recursion formula above with λ = n. These functions are specialfunctions and known as associated Laguerre functions.

Putting this all together in our expression earlier for λ gives the following for the energy eigenvalues(after dealing with the constants),

En = −13.6eV1

n2(80)

Except for n = 1, each energy level is degenerate being represented by more than one independentsolution of the wave equation. If we introduce quantum numbers n, l,m as subscripts, the wavefunctions we have found as acceptable solutions of the wave equation are,

ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ) (81)

The wavefunctions corresponding to distinct sets of values for n, l,m are independent with the allowedvalues being determined as follows,

m = 0,±1,±2, ... (82)

l = |m|, |m|+ 1, |m|+ 2, ...

n = l + 1, l + 2, l + 3, ...

or

n = 1, 2, 3, ... (83)

l = 0, 1, 2, ..., n− 1

n = −l,−l + 1, ...− 1, 0, 1, ...+ l − 1,+l (84)

There are consequently (2l + 1) independent wave functions with given values of n and l.There are are few qualitative things to note about the solutions. Figures 9 and 10 show plots of

the radial wavefunctions as a function of ρ (which is proportional to r). Note that as n increases,the number of nodes increases just like in the harmonic oscillator and what was previously done withparticle in the box. Note that the ground state has no nodes.

19

Figure 9: Figure taken Pauling, Introduction to Quantum Mechanics.

20

Figure 10: Figure taken Pauling, Introduction to Quantum Mechanics.

21

7 Formal presentation of wave mechanics

In the previous sections we have been studying applications of the Schrodinger equation to scattering,the harmonic oscillator, and the hydrogen atom. We took a probabilistic interpretation of the wavefunction ψ(~r, t) and in particular noted that |ψ(~r, t)|2d3r represented the probablity of find at a timet the particle in a volume d3r. From this application, we have gained some intuition on what definesa general wavefunction ψ(~r, t). Here we summarize properties that we have found and generally mustexist in real physical systems.

1)∫d3r|ψ(~r, t)|2=1: This makes sense as the probability cannot diverge and for the single particle

problems discussed above, the particle must exist somewhere in space.2) ψ(~r, t) should be continuous and have continuous derivatives (except where there is a disconti-

nuity in the potential V -remember the δ function potential).Applying these “rules” helped us in solving the Dirac potential, the harmonic oscillator, the quantumrigid rotor, and finally the hydrogen atom.

7.1 A little bit of mathematics

Before developing a general theory of quantum mechanics, it is timely and important to review somelinear algebra. Particularly, what defines a vector. In most early mathematics courses a vector isdefined as something with magnitude and direction. This is completely wrong and is an unfortunatesimplification.

7.2 Vector space

The previous section concludes the wave mechanics part of this course and we will now enter intothe formal construction of quantum mechanics through the use of techniques from linear algebra.Given our experience with potential wells and the discussion in the previous sections, we now wishto formulate a general theory involving the set of square-integrable functions and have the propertythat they are defined everywhere and infinitely differentiable. Lets define F the set of wave functionscomposed of sufficiently regular functions that meet these criterion. While this is a slightly vaguedefinition at the moment, we will make this more precise in the coming sections.

One can go through all of the axioms that define a vector space and formally prove that ourproperties (postulates above) derived from fairly simple physical arguments defines a vector space.For example, if two functions ψ1(~r) and ψ2(~r) ∈ F , then ψ(~r) = λ1ψ1(~r) + λ2ψ2(~r) also ∈ F for λ1

and λ2 being two arbitrary complex numbers. Note that given the linear and homogeneous nature ofthe Schrodinger equation, it should be clear that ψ is a solution. We could also show that ψ is squareintegrable as required.

7.2.1 General definition of a Vector space

An unfortunate misconception is that a vector is an object that has direction and magnitude. Thisdefinition on its own is only true for a small subset of examples and a vector space is something muchmore. I will cite the definition of a vector space below taken from L. Smith Linear Algebra (Springer,New York, 1998).

A vector space is a set V, whose elements are called vectors, together with two operations. Thefirst operation, called vector addition, assigns to each pair of vectors A and B a vector denoted asA+B, called their sum. The second operation, called scaler multiplication, assigns to each vector Aand number r a vector denoted by rA. The two operatios are required to have the following properties:

AXIOM 1: A+B=B +A for each pair of vectors A and B (commutative law of vector addition).AXIOM 2: (A+B) + C = A+ (B + C) for each triple of vectors A, B, and C (associative law of

vector addition).AXIOM 3: There is a unique vector, called the zero vector, such that A+ 0 = A for every vector

A.

22

AXIOM 4: For each vector there corresponds a unique vector −A such that A+ (−A) = 0.AXIOM 5: r(A+B) = rA+ rB for each pair of real number and r and each pair of vectors A and

B.AXIOM 6: (r + s)(A) = rA+ sA for each pair of real numbers, r and s, and each vector A.AXIOM 7: (rs)(A) = r(sA) for each pair of real numbers and r and s and each vector A.AXIOM 8: For each vector A, 1A = A.

Most people are familiar with vectors spaces defined in R2 (two dimensional cartesian space) andR3 (three dimensional cartesian space) which easily satisfy all of the axioms listed above, but it isinteresting to think about what is NOT a vector space. For example, the set (red, green, blue) is nota vector space as there is no −green. An interesting example of a vector space is the memory bits 0and 1 and the operations associated with them. Finally, think about the properties of the functions∈ F that are solutions of the Schrodinger equation (take for example the Hermite polynomial basedsolutions for the harmonic oscillator) and try to prove to yourself that they do satisfy the axioms listedabove for a vector space.

7.3 Inner product

We will now define several general properties in vector spaces of interest. Lets define the scaler productfor a pair (φ(~r) and ψ(~r) ∈ F . This operation is similar to the “dot product” that we are familiarwith in vectors spaces ∈ R2 and R3. We can associate a complex number (denoted for now as (φ, ψ))equal to.

(φ, ψ) ≡∫d3rφ∗(r)ψ(r) (85)

Note that this integral always converges if φ and ψ ∈ F . There are some fairly straightforwardproperties worth noting about this scaler product which is similar to the “dot” product you havedefinitely encountered previously.

(φ, ψ) = (ψ, φ)∗ (86)

(φ, λ1ψ1 + λ2ψ2) = λ1(φ, ψ) + λ2(φ, ψ) (87)

(λ1φ1 + λ2φ2, ψ) = λ∗1(φ1, ψ) + λ∗2(φ2, ψ) (88)

If (φ, ψ) = 0, then φ(~r) and ψ(~r) are said to be orthogonal. Examples of this can be taken fromthe previous sections. For example, in our study of the harmonic oscillator above we found a seriesof discrete energy levels described by the Hermite polynomials Hn(x). For Hn(x) and Hm(x) it wasshown in the exercises that the

∫ +∞−∞ H∗nHmdx=0 (for m 6= n). These solutions are therefore orthogonal

and ∈ our new set of functions F . Similar properties can be derived for the quantum rigid rotor (Y ml )

and the hydrogen atom (Legendre polynomials etc.). As a final note, we also have the property that(ψ,ψ) =

∫d3r|ψ(~r)|2 is a real and positive number, and is zero only when ψ(~r) = 0.

7.4 Linear operators

A linear operator A is, by definition, “a mathematical entity which associate every function ψ(~r) ∈ Fanother function ψ′(~r), the correspondence being linear.”

Formally,

ψ′(~r) = Aψ(~r) (89)

and for two complex numbers λ1 and λ2,

A[λ1ψ1(~r) + λ2ψ2(~r)] = λ1Aψ1(~r) + λ2Aψ2(~r) (90)

23

Lets consider some simple examples of linear operators. For example the parity operator P behaves asfollows Pψ(~r) = ψ(−~r). The position operator Xψ(x, y, z) = xψ(x, y, z), and the differential operatorthat differentiates with respect to x Dxψ(x, y, z) = ∂

∂xψ(x, y, z). Going back above we can find thatour operators Lx,y,z and Px,y,z we used for the quantum rotor and the hydrogen atom are also linearoperators with similar definitions.

Consider the product of operators A and B, their product is defined by,

(AB)ψ(~r) = A[Bψ(~r)] (91)

In general, AB 6= BA (!!!). Lets define the following commutator

[A,B] = AB −BA (92)

Above, we talked a lot about X and P . Do the operators X and P actually commute? Lets computethe commutator acting on a wavefunction ψ(x) ∈ F .

[X,Px]ψ(x) = −ih[x∂

∂x− ∂

∂xx]ψ(x) = −ih

(x∂

∂xψ(x)− ∂

∂x(xψ(x))

)= ihψ(x) (93)

based on the product rule of calculus. Since this is true ∀ψ(x) ∈ F , then [X,Px] = ih. In the exercises,you will compute similar commutators for the angular momentum operators that we used in our studyof the quantum rotor.

7.5 Discrete orthonormal basis in F

We now discuss some properties and definitions of bases in F . It is first important to quantify whatwe mean as orthonormal in general. Consider a countable set of functions of F labelled by a discreteindex i. These could be our Hermite polynomials we found in the harmonic oscillator or the functionsY ml we found for the quantum rigid rotor. In general, consider the set {u1, u2, ....} ∈ F . The set is

orthonormal if

(ui, uj) =

∫d3ru∗i (r)uj(r) = δij (94)

where δij is defined as follows

δij =

{1 if i = j0 if i 6= j

The set of {u1, u2, ....} ∈ F constitute as basis if every function ψ(~r) ∈ F can be expanded in one andonly one way in terms of the ui(~r):

ψ(~r) =∑i

ciui(~r) (95)

We now have the results that (ui, ψ) =∑

j cj(ui, uj) =∑

j cjδij = ci. Therefore, ci = (ui, ψ) =∫d3ru∗i (r)ψ(r). If we have two functions φ and ψ ∈ F - φ(~r) =

∑i biui(~r) and ψ(~r) =

∑j cjuj(~r).

Then (φ, ψ) =∑

i,j b∗i cjδij =

∑i b∗i ci and in particular (ψ,ψ) =

∑i |ci|2. These relations are again

generalizations of the familiar relations that we have worked with in vector spaces in R2 and R3.

24

7.6 The closure relation

The relation stated above - (ui, uj) =∫d3ru∗i (r)uj(r) = δij implicitly implies that the set {ui(~r)} are

normalized to 1. Based on this, we can define the closure relation by noting

ψ(~r) =∑i

ciui(~r) =∑i

(ui, ψ)ui(~r) =∑i

(∫d3r′u∗i (~r

′)ψ(~r′)

)ui(~r) (96)

=

∫d3r′ψ(~r′)

(∑i

u∗i (~r′)ui(~r)

)

For this to be true, then the following relation must hold,

∑i

ui(~r′)u∗i (~r) = δ(~r − ~r′) (97)

This is known as the closure relation. If an orthonormal set satisfies the closure relation, it constitutesa basis.

7.7 Bases not belonging to F

Above we discussed the properties of the set of functions {ui(~r)} which are composed of squareintegrable functions. It can also be useful to introduce bases of functions not belonging to F , but interms of which any wavefunction ψ(~r) can be expanded. We will consider two examples.

7.7.1 Plane waves

The basis consider above, {ui(~r)} consisted of square integrable functions. It is sometimes useful toconsider bases that do not belong to this set, but that a wavefunction can be expressed. We have seenthis in our analysis of the free particle Schrodinger equation where we found it helpful to considerfunctions which are not square integrable (∼ eikx). We found in the first workshop that we couldconstruct sensible solutions which are square integrable by creating wave packets. You have seen inyour Fourier analysis course the idea of Fourier transforms ψ(p) of ψ(x),

ψ(x) =1√2πh

∫ ∞−∞

dpψ(p)eipx/h (98)

ψ(p) =1√2πh

∫ ∞−∞

dxψ(x)e−ipx/h

For example, consider a wavefunction defined as vp(x) = 1√2πh

eipx/h - vp(x) is a plane wave with wave

vector p/h. You might recall that we used these plane wave solutions to motivate the Schrodingerequation in the first few sections of this class. Note that the integral of |vp(x)|2 diverges and thereforethis function cannot belong to F . Lets denote {vp(x)} as the set of all plane waves, that is, of allfunctions corresponding to the continuous index p (which varies over −∞ to +∞). We can write thefollowing in analogy with our discussion of the set {ui(~r)} given above.

ψ(x) =

∫ ∞−∞

dpψ(p)vp(x) (99)

ψ(p) = (vp, ψ) =

∫ ∞−∞

dxv∗p(x)ψ(x) (100)

therefore,

25

(ψ,ψ) =

∫ ∞−∞

dp|ψ(p)|2 (101)

This continuous basis also obeys closure

∫ ∞−∞

vp(x)v∗p(x′) =

1

2πh

∫dpeip(x−x

′)/h = δ(x− x′) (102)

We now calculate the following inner product.

(vp, vp′) =1

2πh

∫dxeix(p′−p)/h = δ(p− p′) (103)

While the continuous and discrete basis (discussed earlier) may appear quite different, all of therelations carry over by making the following correspondence,

i←→ ~p (104)∑i

←→∫d3p (105)

δij ←→ δ(~p− ~p ′) (106)

7.7.2 Delta function basis

As a second (and final) example of a continuous basis consider the set of functions of ~r {ξ~r0(~r)} labeledby the continuous index ~r0 defined by,

ξ~r0(~r) ≡ δ(~r − ~r0) (107)

We can show that these have similar closure relations as the continuous plane wave basis discussedabove. In particular, all of our original relations for our original discussions of square integrablefunctions ∈ F with the following associations.

i←→ ~r (108)∑i

←→∫d3r0 (109)

δij ←→ δ(~r0 − ~r0′) (110)

These are similar to the association made above for the plane wave solutions. We will explore theproperties of this example of a continuous basis if there is time, however we will make frequent use ofthe plane wave basis outlined above.

7.7.3 Generalization - Continuous orthonormal bases

Generalizing the discussion above, we can define a continuous orthonormal basis as a set of functionsof ~r, wα(~r), labeled by a continuous index α, which satisfy the orthonormalization

(wα, wα′) =

∫d3r w∗α(~r)wα′(~r) = δ(α− α′) (111)

and closure

26

∫dα wα(~r)w∗α(~r ′) = δ(~r − ~r ′) (112)

Therefore, again we make the following correspondence which applies to a continuous orthonormalbasis.

i←→ α (113)∑i

←→∫dα (114)

δij ←→ δ(α− α′) (115)

7.8 Isn’t this messy and annoying (yet)?

In the previous sections we used the physical fact that∫d3r|ψ(~r, t)|2=1 to develop a wave vector

formalism of quantum mechanics and defined some important mathematical operators (for bases ∈ Fand not). We then discussed linear operators and some simple examples. Based on this, we talkedabout several basis states included discrete and continuous bases. We found that these could bebased upon a number of different parameterizations and used the examples of spatial and momentumcoordinates. This description of quantum mechanics is quite cumbersome and in general is problemspecific (example real vs momentum space). In particular, the procedure of establishing a basis andthen projecting it onto some coordinate system can be tedious (as evidence above in our discussionof the harmonic oscillator, the quantum rotor, and the hydrogen atom). It would really be nice if wecould develop a mathematical framework which is arbitrary and applies to all of these basis states andalso allow us to quickly and easily rederive the relations stated above for various basis states. This iswhat Dirac notation aims to do.

8 Simplifying life - Dirac notation and state space (Cohen-Tannoudji-Chapter 2,Messiah-Chapter 7)

While it is useful to project a our state onto a particular basis, it is rather cumbersome and problemspecific. To get around this, we will introduce Dirac notation and the idea of a state space. We willstart of by stating that each quantum state of a particle can be characterized by a state vector. Weare going to now define the notation and rules in state space. In particular, lest consider a set of statesin E which are a subspace of F . Lets discuss the rules of vector calculation in E .

8.1 Definition of Kets and Bras

As a matter of notation, any element or vector that lives in E is called a ket vector, or for the remainderof this course, a ket. It is represented by the symbol |〉. For example - |ψ〉. In particular, for any ψ(~r)∈ F , there is a |ψ〉 ∈ E .

We now define the scaler product. With each pair of kets |φ〉 and |ψ〉, we associate a complexnumber which is their scaler product (|φ〉, |ψ〉). To write this out in Dirac notation we need to introducethe concept of dual space and hence the “bra” vector. The scaler product is a linear function whichassociates a complex number with every ket. It can be shown that the set of linear functions definedon the kets |ψ〉 ∈ E constitues a vector space which is called the dual space of E and we will symbolizethis by E∗. Any element, or vector, of the space E∗ is called a bra vector, or just a bra. We symbolizeit as 〈 |.

Note, linear functional and linear operator must not be confused (!!). In both cases, the operationis linear but a functional associates each ket with a complex number while the latter associates anotherket. Associated with every ket in E there is a bra in E∗. We define a linear function that associates (ina linear way) with each ket |ψ〉 ∈ E a complex number which is equal to the scaler product (|φ〉, |ψ〉).Let 〈φ| be this linear functional defined as,

27

〈φ|ψ〉 = (|φ〉, |ψ〉) (116)

Note that that the product is antilinear

(λ1|φ1〉+ λ2|φ2〉, |ψ〉) = (λ∗1〈φ1|+ λ∗2〈φ2|)|ψ〉 (117)

Therefore, the correspondence between kets and bras is analogous to the correspondence between wavefunctions of wave mechanics and their complex conjugate. We have now introduced two notations forthe scaler product (|φ〉, |ψ〉) and 〈φ|ψ〉. From now on, we will use the later notation (〈φ|ψ〉) and dropthe former. For completeness, we now list (in Dirac notation) the properties of the scaler product.

〈φ|ψ〉 = 〈ψ|φ〉 ∗ (118)

〈φ|λ1ψ1 + λ2ψ2〉 = λ1〈φ|ψ1〉+ λ2〈φ|ψ2〉 (119)

〈λ1φ1 + λ2φ2|ψ〉 = λ∗1〈φ1|ψ〉+ λ∗2〈φ2|ψ〉 (120)

and we finally have 〈ψ|ψ〉 is real, positive, and zero if and only if |ψ〉 = 0. Note that two kets |ψ1〉and |ψ2〉 are orthogonal if 〈ψ1|ψ2〉 = 0.

9 Linear operators

A linear operator associates with every |u〉 ∈ E , another ket |v〉 ∈ E with the correspondence beinglinear,

A|u〉 = |v〉 (121)

A (λ1|ψ1〉+ λ2|ψ2〉) = λ1A|ψ1〉+ λ2A|ψ2〉

The product of two operators is defined as follows,

(AB)|ψ〉 = A(B|ψ〉) (122)

That is, B first operators on |ψ〉 to give the ket B|ψ〉, A then operates on this new ket B|ψ〉. Thesedefinitions are exactly analogous to the definitions in wave mechanics discussed above when we wereconsider wavefunctions ψ(~r). Like above, the commutator is defined as [A,B] = AB − BA. Recallthat in general AB 6= BA.

It can be seen that the notation can be quite convenient. For example the inner product between|φ〉 and |ψ〉 is simply 〈φ|ψ〉 (noting the use of the dual space and one to one correspondence betweenkets and bras). There needs to be some caution associated with the notation as seen by consideringthe following,

|ψ〉〈φ| (123)

This corresponds to an operator which can be seen by acting it on an arbitrary ket |χ〉. The orderof symbols is critical! Only complex numbers can be moved about without direct consequences. Forexample, |ψ〉λ = λ|ψ〉 and 〈ψ|λ = λ〈ψ|. To see how this notation works in a bit more detail, letsconsider the following “projector” operator

Pψ = |ψ〉〈ψ| (124)

28

and apply it to an arbitrary ket |φ〉

Pψ|φ〉 = |ψ〉〈ψ|φ〉 (125)

Therefore, this operation gives a ket proportional to |ψ〉 with the coefficient of proportionality 〈ψ|φ〉being the scaler product between the two kets |ψ〉 and |φ〉. Therefore, this operator projects onto aparticular ket |ψ〉. This is confirmed by the fact that acting this operator twice is the same as once.For example, P 2

ψ = PψPψ = |ψ〉〈ψ|ψ〉〈ψ| = |ψ〉〈ψ| = Pψ. We can then project from one subspace of Eto another through use of a generalization of this projector operator

Pq =

q∑i=1

|φi〉〈φi| (126)

Acting this on an arbitrary |ψ〉 gives the linear superposition of the projection of |ψ〉 onto thevarious |φi〉. We can use this to general the closure relation for a discrete basis as,

∑i=1 |ui〉〈ui| =

1 where the sum is done over the entire subspace spanned by |ui〉 and for a continuous basis as∫dα|uα〉〈uα| = 1. Here we have imposed the orthogonality relation 〈ui|uj〉 = δij and 〈uα|uα′〉 =

δ(α− α′) for a continuous basis.

10 Hermitian conjugate

We have only discussed the operation of linear operators on kets |ψ〉. From the one to one corre-spondence between bras and kets, we can derive an analogous conjugation relation between linearoperators. Let A be a linear operator and let |v〉 be the ket conjugate to the bra 〈u|A. As notedabove, |v〉 depends antilinearly upon the bra 〈u|. This linear correspondence defines a linear operatorwhich is termed Hermitian conjugate operator of A or sometimes called the adjoint operator of A andis symbolised by A†. Hence |v〉 = A†|u〉.

An important property of Hermitian operators can be derived from noting that from the propertiesof scaler products 〈u|v〉 = 〈v|u〉∗. Therefore,

〈ψ|A†|φ〉 = 〈φ|A|ψ〉∗ (127)

There is a number of properties of the Hermitian conjugate including (A†)† = A, (λA)† = λ∗A†, andbased on the linear nature of this operation (A+B)† = A† +B†. As a final note, 〈A†φ| = 〈φ|(A†)† =〈φ|A and 〈A†φ|ψ〉 = 〈φ|Aψ〉.

We now calculate (AB)†. Consider the ket |φ〉 = AB|ψ〉 ≡ A|χ〉, with |χ〉 = B|ψ〉. Therefore,〈ψ|(AB)† = 〈χ|A† = 〈ψ|B†A†. This gives us the result that (AB)† = B†A†. Note the reversalin the order of the operators in the expression. Moreover, the Hermitian conjugate of the operator(|u〉〈v|)† = |v〉〈u|. This can be proven by operating this on an arbitrary |φ〉 and |ψ〉 - 〈ψ|(|u〉〈v|)†|φ〉 =[〈φ|(|u〉〈v|)|ψ〉]∗ = 〈φ|u〉∗〈v|ψ〉∗ = 〈ψ|v〉〈u|φ〉. Since this is true for ∀ |ψ〉 and |φ〉 ∈ E we have(|u〉〈v|)† = |v〉〈u|.

We can summarize the operation of taking the Hermitian conjugate as follows (from Messiah-Chapter 7): ”replace everywhere the numbers by their complex conjugate, bras by the conjugate ketsand vice versa, the operators by their Hermitian conjugates, and reverse in each term the order of thevarious symbols occurring there, that is to say the order of the bras, kets, and operators.”

10.1 Hermitian operators

By definition, we refer to an operator as Hermitian if A = A†. Just to note (for terminology) thatanti-Hermitian is defined as I = −I†. Note the product of two Hermitian operators is not necessarilyHermitean since (HK)† = KH. HK is only Hermitian if an only if H and K commute.

29

10.2 Unitary operators

An operator U is unitary if it is the inverse of its own Hermitian conjugate: UU † = U †U = 1. Thesewill be the subject of the problem set.

11 Observables (Cohen-Tannoudji-Chapter 2, Messiah-Chapter 7)

11.1 Eigenvalues and vectors of an operator

We state that |ψ〉 is an eigenvector (or eigenket) of the linear operator A if

A|ψ〉 = λ|ψ〉 (128)

where λ is a complex number. The eigenvalue λ is called nondegenerate when its corresponding eigen-vector is unique to within a constant factor. However, if there exist a least two linearly independentkets which are eigenvectors of A with the same eigenvalue, this eigenvalue is said to be degenerate.For example, if λ is g − fold degenerate, there correspond to it g independent kets |ψi〉 (i = 1, 2, ...g)such that

A|ψi〉 = λ|ψi〉 (129)

We can find the eigenvalues and eigenvectors from the characteristic equation Det[A− λI] = 0.There are several important properties of the eigenvalues of a Hermitian operator (that is when

A = A†).

Property 1: The eigenvalues of a Hermitian operator are real. This follows if we take the definition ofthe eigenvalue problem above and therefore write 〈ψ|A|ψ〉 = λ〈ψ|ψ〉. We further note by the definitionof a Hermitian operator, 〈ψ|A|ψ〉∗ = 〈ψ|A†|ψ〉 = 〈ψ|A|ψ〉. Since this is a real number and 〈ψ|ψ〉 is areal number, then λ, the eigenvalue, must be a real number.

Property 2: Two eigenvectors of a Hermitian operator corresponding to two different eigenvalues areorthogonal. This derives from considering two eigenvectors |ψ〉 and |φ〉 of the Hermitian operator A.

A|ψ〉 = λ|ψ〉 (130)

A|φ〉 = µ|φ〉 (131)

Since A is Hermitian, we can further note that 〈φ|A = µ〈φ|. Then multiplying the eigenvalue equationsabove, we obtain,

〈φ|A|ψ〉 = λ〈φ|ψ〉 (132)

〈ψ|A|φ〉 = µ〈ψ|φ〉 (133)

Subtracting then gives the (µ − λ)〈φ|ψ〉 = 0. Since λ and µ are different eigenvalues by definition,then φ and ψ are orthogonal.

11.2 Definition

Consider a Hermitian operator A and kets |u〉 such that A|uk〉 = a|uk〉 and therefore 〈uj |A = a′〈uj |.Lets take the case that the subspace spanned by {ui} forms a complete orthonormal basis such thatany ket |ψ〉 can be described as a linear combination of these orthonormal vectors. By definition, theHermitian operator A is an observable if this orthonormal system of vectors forms a basis in the statespace. This can be expressed formally through the closure relation.

30

∞∑n=1

|ui〉〈ui| = 1 (134)

This mathematical relation applies to to a discrete basis. We can generalize it to the case of acontinuous basis.

∫dα|α〉〈α| = 1 (135)

The ultimate problem of knowing if a given Hermitian operator is an observable is in most cases avery difficult mathematical problem.

11.3 Commuting Observables (Cohen-Tannoudji-Chapter 2, Messiah - Chapter 8)

We now prove three useful theorems applying to commuting observables.

Theorem 1: If two operators A and B commute, and if |ψ〉 is an eigenvector of B, B|ψ〉 is also aneigenvector of A, with the same eigenvalue.

Proof: Take A|ψ〉 = a|ψ〉 and apply B to both sides to give B (A|ψ〉) = B (a|ψ〉). By definition Aand B commute implying A (B|ψ〉) = a (B|ψ〉). Therefore B|ψ〉 is an eigenvector of A with eigenvaluea.

Theorem 2 : If two observables A and B commute, and if |ψ1〉 and |ψ2〉 are two eigenvectors of A withdifferent eigenvalues, the matrix element 〈ψ1|B|ψ2〉 is zero.

Proof: 〈ψ1|(AB −BA)|ψ2〉=0 ; by the assumption that A and B commute. Therefore, using theeigenvalues we get the relation (a1 − a2)〈ψ1|B|ψ2〉 = 0. Given that a1 6= a2, 〈ψ1|B|ψ2〉 = 0.

Theorem 3 : If two observables A and B commute, one can construct an orthonormal basis of the statespace with eigenvectors common to A and B. The converse is also true that “if two observables Aand B possess a common basis, they commute.”

This follows from theorem 1 and theorem 2.

11.4 CSCO - Complete set of commuting observables

Consider an observable A and a basis of E composed of eigenvectors |u〉 of A. If none of the eigenvaluesof A are degenerate, the various basis vectors of E can be labelled by the eigenvalue an. In other words,there exists only one basis of E formed by the eigenvectors of A. It is then said that the observable Aconstitutes, by itself, a complete set of commuting observales (CSCO). The key issue in any quantummechanics problem, is to come up with a complete set of commuting observables (CSCO).

The concept of a CSCO is quite abstract. Lets apply this to the concept to the quantum rigidrotor discussed early on in this course where the Hamiltonian H = αL2. We found the eigenfunctionswere the spherical harmonics which required two quantum numbers to uniquely identify them - l,m.We can then label the eigenstates as |l,m〉 using Dirac notation where L2|l,m〉 = l(l + 1)|l,m〉 andLz|l,m〉 = m|l,m〉. We note here that the operator L2 by itself does not form a CSCO because thereis a degeneracy associated with each eigenvalue l. The combination of L2 and Lz does form a CSCO.The example of the harmonic oscillator we studied earlier is a bit more trivial as each eigenstate ofH is non degenerate and therefor H forms a CSCO for the one dimensional harmonic oscillator weinvestigated early on in this course.

To define this more formally, “a set of observables A, B, C ... is called a complete set of commutingobservables if 1) all of the observables A, B, C ... commute by pair; 2) specifying the eigenvalues ofall the operators A, B, C ... determines a unique common eigenvector.”

31

A central goal to every physical problem is the determination of a CSCO.

12 Postulates of Quantum Mechanics (Cohen-Tanoudji-Chapter 3)

First postulate: “At a fixed time t0, the state of a physical system is defined by specifying a ket|ψ(t0)〉 belonging to the state space E .”

Second postulate: “Every measureable physical quantity A is described by an operator A acting inE ; this operator is an observable.”

Third postulate: “The only possible result of the measurement of a physical quantity A is one ofthe eigenvalues of the corresponding observable A.”

Fourth postulate: “When the physical quantity A is measured on a system in the normlizedstate |ψ〉, the probability P (an) of obtaining the eigenvalue an of the corresponding observable Ais: P (an) =

∑gni=1 |〈uin|ψ〉|2, where gn is the degree of degeneracy of an and {|uin〉}” (i=1,2,...gn) is an

orthonormal set of vectors which forms a basis in the eigensubspace En associated with the eigenvaluean of A.

Fifth postulate: “If the measurement of the physical quantity A on the system in the state |ψ〉 givesthe result an, the state of the system immediately after the measurement is the normalized projectionPn|ψ〉/

√〈ψ|Pn|ψ〉 of |ψ〉 onto the eigensubspace associated with an.”

Sixth postulate: “The time evolution of the state vector |ψ(t)〉 is governed by the Schrodingerequation ih d

dt |ψ(t)〉 = H(t)|ψ(t)〉 where H(t) is the observable associated with the total energy of thesystem.”

13 Physical consequences of the postulates and our new theory

Before going to to apply this new theory of quantum mechanics to specific problems, we discuss somegeneral physical consequences of the theory we have developed. We first discuss the notion of conservedquantities and then uncertainty relations.

13.1 Conserved quantities

In this section, we discuss an important physical consequence of the Schrodinger equation in terms ofwhat formally defines a conserved quantity. Consider an arbitrary linear operator A, we can write theexpectation value of this operator in general as follows,

〈A〉(t) ≡ 〈ψ(t)|A(t)|ψ(t)〉 (136)

Differentiating and using the product rule of calculus

d

dt〈ψ(t)|A|ψ(t)〉 =

(d

dt〈ψ(t)|

)A(t)|ψ(t)〉+ 〈ψ(t)|A(t)

(d

dt|ψ(t)〉

)+ ... (137)

〈ψ(t)|∂A∂t|ψ(t)〉 (138)

Recall the Schrodinger equation above ih ddt |ψ(t)〉 = H(t)|ψ(t)〉 gives the following,

ihd

dt〈A〉 = 〈[A,H]〉+ ih〈∂A

∂t〉 (139)

32

Therefore, a constant of motion is an observable A which does not depend explicitly on time andwhich commutes with H. Therefore, for A to be a constant of motion 1) ∂A

∂t = 0 and 2) [A,H]=0.Lets consider this in terms of the quantum rotor discussed above where H = αL2. Since L2

commutes with the Hamiltonian and does not explicitly depend on time, it is a conserved quantity.Likewise, since H commutes with itself, energy is conserved quantity.

13.2 The uncertainty relations

Two observables, A and B are said to be compatible if they can be simultaneously determined. Theprecise basis of this derives from theorem derived above where if |ψ〉 is a eigenstate of B, then B|ψ〉is an eigenstate of A. Note the converse is not (!!) true in general. If A and B do not commute,in general a state cannot be a simultaneous eigenvector of these two observables. The are said to beincompatible.

This leads to some important physical results regarding operators that do not commute. Forexample, this result seems to imply that measuring the position X, will introduce an uncertainty inthe momentum P since these two operators do not commute. It is particularly interesting to thinkabout the case of the quantum rigid rotor solved above. Recall that [L2, Lx,y,z] = 0 implying thatwe can measure the total angular momentum and only on spatial projection. We cannot know, forexample, both the eigenstates of Lz and Lx.

To quantify this, we will “derive” the position-momentum uncertainty relation. We will show that,in general, that if two observables A and B satisfy the equation,

[A,B] = ih (140)

then the product of their root-mean-square deviations will always remain greater than h2/2.

∆A ∆B ≥ h

2(141)

By definition,

∆A = (〈A2〉 − 〈A〉2)12 (142)

∆B = (〈B2〉 − 〈B〉2)12

Lets define the following observables

A = A− 〈A〉 (143)

B = B − 〈B〉 (144)

(145)

Therefore,

∆A = ∆A = 〈A2〉12 (146)

∆B = ∆B = 〈B2〉12 (147)

Assume a dynamical state of the system is defined by the ket |u〉 normalized to unity. From Schwarzinequality to the vectors A|u〉 and B|u〉:

(∆A)2(∆B)2 ≡ 〈u|A2|u〉〈u|B2|u〉 ≥ |〈u|AB|u〉|2 (148)

Separating AB into Hermitean and anti Hermitian

33

AB =AB + BA

2+AB − BA

2=AB + BA

2+ih

2(149)

By direct substitution we can separate real and imaginary parts

〈u|AB|u〉 = 〈AB + BA

2〉+

ih

2(150)

We now rewrite the Schwarz inequality

(∆A)2(∆B)2 ≥ 〈AB + BA

2〉2 +

h2

4(151)

We therefore have the following relation we wanted to show,

∆A ∆B ≥ h

2(152)

There in the case of position and momentum where [X,P ] = ih, measuring one of these observablesintroduces and uncertainty in the other.

14 Ladder operators and the Harmonic oscillator (Messiah-Chapter 12,Cohen-Tannoudji-Chapter 5)

Previously we solved the Schrodinger equation for the harmonic oscillator in representation space.This was slightly cumbersome and involved the derivation of Hermite polynomials and recursionrelations. Using Dirac notation we can solve this problem without projecting onto a particular basis(namely the position representation used previously) and solve the problem more generally and moredirectly by considering operators acting in state space. In this section we revisit this with operatortechniques and introduce ladder (or creation and annihilation operators). We rewrite the Hamiltonianfor quantum-mechanical problem of a particle of mass m in a one dimensional harmonic well.

H =P 2

2m+

1

2mω2X2 (153)

Now, P and X are operators and are connected via the commutation relation [X,P ] = ih. Theproblem is of fundamental importance as it appears in the study of Quantum Electrodynamics and alsoQuantum Field theory. Before studying the solution to this Hamiltonian using operator techniques,we first make the following substitutions to remove constants and simplify the notation.

X =

√mω

hX (154)

P =1√mhω

P (155)

With these substitutions the commutator [X, P ] = i. The Hamiltonian can also be written as, H =

hωH, with H = 12

(X2 + P 2

). We therefore seek solutions to the eigenvalue problem H|ψν〉 =

εν |ψν〉. If X and P were just numbers and this was a classical problem, we could write X2 + P 2 as(X + iP )((X − iP )), but X and P do not commute so these two expression are not equivalent. We

34

can simplify the problem considerably by defining the following operators which mimic this step wewould have done in more classical theory where these two variables commute.

a =1√2

(X + iP

)(156)

a† =1√2

(X − iP

)(157)

Note that while X and P are Hermitian, a and a† are not, but are adjoints of each other. Noting that[a, a†] = 1

2 [X + iP , X − iP ] = i2 [P , X]− i

2 [X, P ], gives the following relation

[a, a†] = 1 (158)

To see how this simplifies the notation we calculate a†a,

a†a =1

2

(X − iP

)(X + iP

)=

1

2

(X2 + P 2 + iXP − iP X

)=

1

2

(X2 + P 2 − 1

)(159)

Comparing this with our expression above for H = 12

(X2 + P 2

)gives the following,

H = a†a+1

2(160)

or

H = aa† − 1

2(161)

.Note that the fact that X and P do not commute results in this extra zero point energy term of12 . This key aspect of our new theory of quantum mechanics gives us this fundamentally differentexpression and new results like the Casimir effect discussed above.

Lets now define the operator N

N = a†a (162)

The operator is Hermitian since N † = a†(a†)†

= a†a = N . Moreover, we can then now write

H = N +1

2(163)

therefore, the eigenvectors of N are the eigenvectors of H and vice versa. This may give the impressionof an endless series of operator redefinitions to simplify H, but to determine how this helps us withthe goal of solving this Hamiltonian and determining the spectrum in state space, we need to derivea few properties of N , a, and a†. We first note that [N, a] = −a and [N, a†] = a†. To solve for thespectrum of the harmonic oscillator in state space we therefore need to determine the eigenvectorsand values of the operator N . We therefore seek solutions N |φiν〉 = ν|φiν〉. We consider a theoremoutlined in Cohen-Tannoudji and in Messiah.

35

14.1 Properties of a and N

Theorem If |φiν〉 is an eigenvector of the operator N with the corresponding eigenable ν, then thefollowing are true.

Lemma 1 : “The eigenvalues ν of the operator N are positive or zero.”

PROOF: This can be shown by noting |(a|φiν〉

)|2 = 〈φiν |a†a|φiν〉 ≥ 0. We then note that

〈φiν |a†a|φiν〉 = 〈φiν |N |φiν〉 = ν〈φiν |φiν〉. Since 〈φiν |φiν〉 is positive it therefore follows that ν >≥ 0.This is what we wanted to show.

Lemma 2 : (properties of the vector a|φiν〉) “Let |φiν〉 be a non-zero eigenvector of N with eigenvalueν (as stated above), we shall prove that (i) if ν = 0, the ket a|φiν〉 is zero and (ii) if ν > 0, the keta|φiν〉 is a non-zero eigenvector of N with the eigenvalue ν − 1.”

PROOF: (i) Recall from above that 〈φiν |a†a|φiν〉 = ν〈φiν |φiν〉, therefore the square of the normof vector a|φiν〉 is zero if (and only if) this vector is zero. If ν = 0 is an eigenvalue of N , alleigenvectors associated with this eigenvalue must satisfy the relation a|φiν〉 = 0. This implies thata†a|φiν〉 = N |φiν〉 = 0. Therefore, any vector which satisfies a|φiν〉 = 0 is an eigenvector of N witheigenvalue ν = 0.

We have just worked out what the ground state is of N and hence H. Note that ν must be zeroor positive and this there implies that the ground state of H = N + 1

2 has a zero point ground stateenergy of 1

2 . Now that we have sorted out the ground state (and derived an interesting result pointto the presence of a zero point energy), lets try and work out the excited state spectrum.

PROOF: (ii) Lets assume that ν > 0 to begin to determine what the excited spectrum looks like.Applying what we know about commutators of a and a† we get the following,

[N, a]|φiν〉 = −a|φinu〉 = (Na− aN) |φiν〉 = Na|φiν〉 − aN |φiν〉 = Na|φiν〉 − νa|φiν〉 (164)

⇒ −a|φiν〉 = Na|φiν〉 − νa|φiν〉

Rearranging gives the following,

N(a|φiν〉

)= (ν − 1)

(a|φiν〉

)(165)

Therefore a|φiν〉 is an eigenvector of N with eigenvalue ν − 1. This is a nice relation and seems toimply a has the effect of lowering the energy, but we have not determined the spectrum or the detailsof our state space vectors |φiν〉. To carry on, lets determine some properties of a†.

Lemma 3 : (properties of the vector a†|φiν〉) “Let |φiν〉 be a non-zero eigenvector of N with eigen-value ν (as stated above), we shall prove that (i) the ket a†|φiν〉 is nonzero and (ii) the ket a†|φiν〉 is aeigenvector of N with the eigenvalue ν + 1.”

PROOF: (i) Recall that [a, a†] = 1 ⇒ aa† − a†a = 1 ⇒ aa† = 1 + a†a (from the commutatorrelations discussed above). To prove this lemma we now note the following

|a†|φiν〉|2 = 〈φiν |a†a|φiν〉 (166)

= 〈φiν |(1 + a†a)|φiν〉 = (1 + ν)〈φiν |φiν〉

By lemma 1 (proved above), ν is always postive or zero and therefore a†|φiν〉 must be always nonzero.Note, this distinguishes a†|φiν〉 from a|φiν〉 which can be zero. Lets now look at how a†|φiν〉 responds

36

to the operator N .

Proof: (ii) The proof is similar and relies on the commutator [N, a†] = a†.

[N, a†]|φiν〉 = a†|φinu〉 =(Na† − a†N

)|φiν〉 = Na†|φiν〉 − a†N |φiν〉 = Na†|φiν〉 − νa†|φiν〉 (167)

⇒ a†|φiν〉 = Na†|φiν〉 − νa†|φiν〉

N(a†|φiν〉

)= (ν + 1)

(a†|φiν〉

)(168)

Therefore, the effect of a† appears to be to raise the energy. Note, we have not assumed anything aboutν or |φiν〉. We now have the necessary properties and tools to work out what the energy spectrum ofN is and hence the hamiltonian H.

14.2 Spectrum and basis of N and hence H

Lets summarize what we just did in the previous section of formal operator algebra. We have rewritten the Hamiltonian in terms of the operator a, and a†. We then noted that since H = a†a + 1

2 ,

that we mind as well simplify this one more step and define N = a†a and write H = N + 12 . Since

eigenvectors of N are also eigenvectors of H, we then studied the properties of a and a†.Given the theorem above applying to a|φiν〉 whose eigenvalue of N is ν − 1, we can apply this

theorem to a2|φiν〉, a3|φiν〉, etc. The implication of the theorem is that the eigenvalue spectrum ofthese is ν − 2, ν − 3 etc. However, the set of eigenvalues has a lower limit (also from the theoremabove) and that limit is 0. The implication of this fact is that ν must be an integer n.

Conversely, we can also apply this logic to a†|φiν〉 and derive the eigenvectors of repeated operationsof a† are ν+1,ν+2 etc. The integer nature of the eigenvalues is easier to see from the repeated actionof a†. We know that the ground state (from the theorem above) has an eigenvalue of ν=0. Actinga† again gives an eigenvalue of 1 and again 2 etc. This technique is actually quite general and willcome up again particularly with angular momentum. In general, one can determine the ground state(or conversely the highest energy state) and through repeated operations of a or a† obtain all of theeigenvalues and eigenvectors. This will be the topic of an exercise session.

In conclusion, the spectrum of eigenvalues of N is formed by the set of non-negative integers.Moreover, by repeated action of a and a† on one of them, all of the eigenvectors can be obtained.Given the eigenvalues are integers, we now denote (for simplicity) the eigenvectors to be proportionto |n〉 = |0〉, |1〉, |2〉, |3〉.... with eigenvalues of N equal to 0, 1, 2, 3, .... Given that a tends to lower theenergy spectra it is referred to as the annihilation operator and since a† has the effect of raising it iscalled the creation operator. We can work back our previous results in terms of Hermite polynomialsby projecting these states onto the position representation |x〉. While this sounds rather abstract, itis straightforward and is a topic of the exercises. We can therefore take our new basis states, whichexist in state space, and project them onto any representation we like. This illustrates the eleganceof using Dirac notation over our previous techniques of working in the position representation andsolving rather difficult ODE’s and PDE’s.

Lets put these results together - given that H = hωH = hω(N + 1

2

), we can write the eigenvalue

spectrum as

En = hω

(n+

1

2

), n = 0, 1, 2, 3... (169)

Note that since none of these eigenvectors are degenerate (also part of the exercises), the operatorN (or H) form a CSCO. Therefore, once we have determined the properties of these eigenvectors and

37

eigenvalues, we have formally solved this problem. While we have solved for the eigenvalue spectrum,we now must solve for the details of the eigenvectors. We will do this in the |n〉 representation.

The vector |0〉 associated with n = 0 is a vector that satisfies a|0〉 = 0. According the theorem(namely lemma 3), the vector |1〉 corresponds to n = 1 and is proportional to a†|0〉:

|1〉 = c1a†|0〉 (170)

where c1 is some normalization constant which will now be determined. By imposing that the eigen-vectors are normalized, we can determine what c1 is as well as cn.

〈1|1〉 = |c1|2〈0|aa†|0〉 = |c1|2〈0|1 + a†a|0〉 (171)

Since |0〉 is a normalized eigenvector, this implies that |c1|2 = 1 and that c1=1. We can now proceedto get c2

〈2|2〉 = |c2|2〈1|aa†|1〉 = |c2|2〈1|1 + a†a|1〉 = 2c22 (172)

We therefore have,

|2〉 =1√2a†|1〉 =

1√2

(a†)2|0〉 (173)

We can keep doing this for |3〉 etc.. and we would find that cn = 1√n

. We then write down the fully

normalized eigenvectors of the harmonic oscillator in the |n〉 basis.

|n〉 =1√n!

(a†)n|0〉 (174)

Since H is Hermitian, the kets |n〉 corresponding to different values of n are orthogonal. Since eachone is normalized as discussed above we have the following relation

〈m|n〉 = δm,n (175)

Since H is observable (by assumption) the set of |n〉 obey the following closure relation.

∑n

|n〉〈n| = 1 (176)

Using the normalization above we also note for completeness the following properties of the creationand annihilation operators.

a|n〉 =√n|n− 1〉 (177)

a†|n〉 =√n+ 1|n+ 1〉

Note that the observables X and P can be written in terms of a and a† and therefore we can switchbetween the number representation |n〉 and the position representation |x〉. This will be explored morein the exercises.

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15 Angular momentum revisited - a general theory (Cohen-Tannoudji - Chapter6, Messiah-Chapter 13)

We now revisit the angular momentum problem in quantum mechanics, but apply our use of raisingand lowering operators to solve for the eigenvectors and values. Recall the fundamental formula forangular momentum,

~L = ~R× ~P (178)

Since we know the commutation relations for R and P observables, we can calculate the commutatorrelations for the operators Lx, Ly, and Lz. For example the commutator [Lx, Ly]:

[Lx, Ly] = [Y Pz − ZPy, ZPx −XPz] = [Y Pz, ZPx] + [ZPy, XPz] (179)

and since Y Pz commutes with XPz, and ZPy with ZPx. We then have,

[Lx, Ly] = Y [Pz, Z]Px +X[Z,Pz]Py = −ihY Px + ihXPy = ihLz (180)

We can repeat this for all possible iterations and we would get the following

[Lx, Ly] = ihLz , [Ly, Lz] = ihLx , [Lz, Lx] = ihLy (181)

We have therefore established the commutation relations for the components of the angular mo-mentum. We can adopt a more general point of view and define an angular momentum J as any setof three observables Jx, Jy, and Jz which satisfies,

[Jx, Jy] = ihJz , [Jy, Jz] = ihJx , [Jz, Jx] = ihJy (182)

A rather compact way of writing this is in terms of the completely antisymmetrical tensor εijk -[Ji, Jk] = iεijkJ

k (recall that repeated indices are summed over). We can also introduce the totalangular momentum operator

J2 = J2x + J2

y + J2z (183)

which commutes with all three components of J

[J2, Ji] = 0, i = x, y, z (184)

This will be explicitly shown in the exercises using the same method as done above for the case of orbitalangular momentum. We now seek out the eigenvalue spectra and eigenvectors of angular momentum.We have a problem though that the individual components do not commute and therefore we cannothave simultaneous eigenvectors. We will therefore seek eigenvectors and eigenvalues common to J2

and Jz. To determine this, the procedure is actually quite similar to that of the harmonic oscillatoroutlined above.

It is convenient to introduce two new operators that are linear combinations of Jx and Jy.

J+ = Jx + iJy (185)

J− = Jx − iJy

39

Similar to the operators a† and a of the harmonic oscillator, J± are not Hermitian but they are adjointsof each other. In the rest of this discussion we will only use the operators J±, Jz, and J2 and thereforewe note the following relation,

J2 =1

2(J+J− + J−J+) + J2

z (186)

and therefore,

J−J+ = J2 − Jz(Jz + h) (187)

J+J− = J2 − Jz(Jz − h).

It is fairly straightforward to show the following commutator relations (based on the definitions pre-sented above).

[Jz, J+] = hJ+ , [Jz, J−] = −hJ− , [J+, J−] = 2hJz (188)

and

[J2, Jz] = [J2, J−] = [J2, Jz] = 0 (189)

The eigenvalues of J2 are positive and this can be seen by noting that it is the sum of positive definiteHermitian operators. Since J2 commutes which of the components of J , one can form a complete setof common eigenvectors of J2 and one of its components, Jz for example. We will write the eigenvaluesof J2 as j(j + 1) from now on with j ≥ 0.

Let |jm〉 be an eigenvector of J2 and Jz with eigenvalues j(j + 1)h2 and mh respectively. Wetherefore write,

J2|j,m〉 = j(j + 1)h2|j,m〉 (190)

and

Jz|j,m〉 = mh|j,m〉. (191)

Lets consider J±, we now wish to find the properties of the operation J±|j,m〉. As with the case ofthe ladder operators above, we first wish to consider |J+|j,m〉|2 and therefore we wish to understandterms which contain J+J− which we have written in terms of the operators J2 and Jz above.From the relations above we have the folowing,

〈j,m|J−J+|j,m〉 = 〈j,m|(J2 − Jz(Jz + h))|j,m〉 = (j −m)(j +m+ 1)h2〈j,m|j,m〉. (192)

and

〈j,m|J+J−|j,m〉 = 〈j,m|(J2 − Jz(Jz − h))|j,m〉 = (j +m)(j −m+ 1)h2〈j,m|j,m〉. (193)

Given that these cannot be negative, we get the following inequalities,

(j −m)(j +m+ 1) ≥ 0 (194)

and

40

(j +m)(j −m+ 1) ≥ 0 (195)

from which we can derive,

−j ≤ m ≤ j (196)

Also, since the vanishing of a norm of a vector is a necessary condition for the vector vanishing, wehave the following results,

J+|j,m〉 = 0 (197)

when (j −m)(j +m+ 1) = 0. Also,

J−|j,m〉 = 0 (198)

when (j +m)(j −m+ 1) = 0. Since m is in the interval ±j then these conditions amount to m = +jand m = −j respectively. For m 6= j, J+|j,m〉 is a vector of angular momentum and hence,

J2J+|j,m〉 = J+J2|j,m〉 = j(j + 1)h2J+|j,m〉 (199)

We can then use the commutator [Jz, J+] = hJ+, to write the following.

JzJ+|j,m〉 = J+(Jz + h)|j,m〉 = (m+ 1)hJ+|j,m〉 (200)

Therefore, J+|j,m〉 is an eigenvector of J2 and Jz with eigenvalues j(j+1)h2 and (m+1)h respectively.We can apply the same procedure to find J−|j,m〉 is an eigenvector as well with eigenvalues j(j+1)h2

and (m− 1)h respectively.We can put all of these results together to formulate the following important theorem,

If |j,m〉 is a vector of angular momentum, then(i) necessarily j ≤ m ≤ j

(ii) if m = j, J+|j,m〉 = 0if m 6= j, J+ is necessarily a vector of angular momentum (j,m+1) and a norm of [j(j+1)−m(m+1)]

(ii) if m = −j, J−|j,m〉 = 0ifm 6= −j, J− is necessarily a vector of angular momentum (j,m−1) and a norm of [j(j+1)−m(m−1)]

Therefore, for a fixed value of j, there exists an integer p such that Jp+|j,m〉 is a non-zero eigenvectorof angular momentum. Since m + p = j (with p and integer), we have that j −m is an integer. Wecan apply the same argument for Jq−|j,m〉 implying that p + q = 2j is also a non-negative integer.Since 2j is a positive integer, j = 0, 1

2 , 1, 32 , 2, ... The only only possible eigenvalues of Jz are integral

and half integral numbers.

In a similar manner to the ladder operators above, we can construct the following relations,

J+|j,m〉 =√j(j + 1)−m(m+ 1)h|j,m+ 1〉. (201)

and

J−|j,m〉 =√j(j + 1)−m(m− 1)h|j,m− 1〉. (202)

41

16 Spin Operators (Landau - Chapter 5), Cohen-Tannoudji-Chapter 4, Messiah-Chapter 13)

In both classical and quantum mechanics, conservation of angular momentum is the direct result ofof the isotropy of space. In the previous section, we discussed the matrix mechanics of the state |l,m〉where 〈θ, φ|l,m〉 are the spherical harmonics Y m

l discussed early on in the course with regards tothe quantum rigid rotor. We have been discussing angular momentum of an object (for example anelectron) circling around a fixed point. It turns out that there is also an internal angular momentumof a particle and this is demonstrated by the Stern-Gerlach experiment. Unfortunately, that the norelativistic level, discussed here, this concept of an internal angular momentum needs to be addedin “by hand” and does not come out (obviously) from theory. It does appear in a full relativistictreatment of the hydrogen atom and we will not attempt to pursue this here.

Experimentally, we are presented with the fact that there is an internal angular momentum withj = 1

2 . We call this new degree of freedom “spin” and outline some basic properties of this. We haveseen above that with every measurable physical quantity there must be associated an observable. Thisis a Hermitian operator whose eigenvectors form a basis in state space. We will construct this statespace and the properties of it.

16.0.1 Observable Sz

Given the doublet degeneracy associated with this internal angular momentum demonstrated by theStern-Gerlach experiment, we know that our basis must consist of two states that will be denoted as|+〉 and |−〉. We will therefore define the following eigenvalues of Sz.

Sz|+〉 =1

2|+〉 (203)

Sz|−〉 = −1

2|−〉

Invoking orthonormality four our new basis vectors,

〈+|+〉 = 〈−|−〉 = 1 (204)

〈+|−〉 = 0

Note that unlike above for the case of angular momentum where we required L2 and Lz to form aCSCO, Sz alone forms a CSCO here. For completeness, we note the following closure relation,

|+〉〈+| + |−〉〈−| = 1 (205)

We have therefore constructed a basis and properties of a two level system. Such formalism is important(particularly in quantum optics). We can construct a general vector in our new spin space as follows,

|ψ〉 = α| + 〉+ β|−〉 (206)

where

|α|2 + |β|2 = 1 (207)

In the {|+〉, |−〉} basis, the matrix representing Sz is diagonal,

Sz =1

2

(1 00 −1

)42

Like the case of angular momentum above, the three components of angular momentum do not com-mute. We can construct the operators Sx and Sy by first creating raising and lowering operators S+

and S−. These two operators can be created by inspection and also using the general theory createdabove.

S+ =1

2

(0 10 0

)and

S− =1

2

(0 01 0

)From these two matrices and noting Sx = 1

2(S+ + S−) and Sy = 12(S+ − S−), we get the following

Sx =1

2

(0 11 0

)and

Sy =1

2

(0 −ii 0

)These matrices following the commutator relations for angular momentum operators discussed above,namely [Sy, Sz] = iSx, [Sz, Sx] = iSy, [Sx, Sy] = iSz. As a matter of notation, the three 2× 2 matriceslisted above are known as the Pauli spin matrices σ. We state the definitions here for completeness,

σx =

(0 11 0

)

σy =

(0 −ii 0

)

σz =

(1 00 −1

)While this last step and definition might seem arbitrary (pointless) at this stage, these matrices havea number of helpful properties and become useful in field theory. For example, as outlined in Cohen-Tannoudji, they form a convenient basis of the 2× 2 matrix space. They also can be used in dealingwith two level atoms by assigning them an “fictitious” angular momentum (effectively just countingstates).

17 Time independent perturbation theory (Messiah-Chapter 16, Landau-Chapter6)

17.1 Non Degenerate case

It is assumed that the Hamiltonian can be written as the sum of an “unperturbed Hamiltonian” H0

and a perturbation which, for convenience, we will write as H1 = λV . We will take λ to be real andfor it to be time-independent. V is also time independent and like its counterpart H0, Hermitian. Wecan write the total Hamiltonian as,

H = H0 +H1 ≡ H0 + λV (208)

We also take the case that H0 is solvable with discrete eigenvalues E0, E1, E2, ... and eigenstatesH0|E0

i α〉 = E0i |E0

i α〉, where the additional parameter α distinguishes between eigenvectors belongingto a degenerate eigenvalue. We note that the eigenstates of H0 form a complete orthonormal basis.

43

The spectrum of H varies continuously with λ and we wish to solve for the problem such that whenλ→ 0, H and the eigenvalues and vectors returns to that of H0. We will first consider the case whenthe solutions of H0 are nondegenerate. We therefore want to solve the following problem,

H|ψ〉 = E|ψ〉 (209)

where the solution to H0 is known with ground state |0〉 and excited states |φm〉. These states forma complete orthonormal basis and hence any state can be written as a linear combination of |φm〉.If the perturbation λ is sufficiently small, we can assume that we can expand both E and |ψ〉 asfollows,

E = E0 + λε1 + λ2ε2 + ...+ λnεn + ... (210)

|ψ〉 = |0〉+ λ|1〉+ λ2|2〉+ ...+ λn|n〉+ ...

We have written the solutions such that limλ→0 the solutions return to |0〉 which is the ground statesolution of H0.The equation we wish to solve can then be written by putting this all together,

(H0 + λV )|ψ〉(|0〉+ λ|1〉+ λ2|2〉+ ...

)=(E0a + λε1 + λ2ε2 + ...

) (|0〉+ λ|1〉+ λ2|2〉+ ...

)(211)

For this equation to be true ∀ λ, we can equate equal orders of λ. Therefore, for zeroeth order in λ

H0|0〉 = E0|0〉 (212)

which is not very helpful as we knew this already. For 1st-order in λ terms,

(H0 − E0) |1〉+ (V − ε1) |0〉 = 0 (213)

For 2nd-order in λ terms,

(H0 − E0) |2〉+ (V − ε1) |1〉 − ε2|0〉 = 0 (214)

For general order in perturbation λ,

(H0 − E0) |n〉+ (V − ε1) |n− 1〉....− εn|0〉 = 0 (215)

Using the equation above for the first order correction, we can derive an expression for ε1 by acting〈0| on both sides of the equation to get,

〈0| (H0 − E0) |1〉+ 〈0| (V − ε1) |0〉 = 0. (216)

Given that H0 is Hermitian, the first term is zero and we are left,

ε1 = 〈0|V |0〉 (217)

.Therefore, the energy to first order is,

E = E0 + λ〈0|V |0〉+O(λ2) ≡ E0 + 〈0|H1|0〉+O(λ2) (218)

44

.We now want to find an expression for |1〉. We write this as a sum |1〉 =

∑n cn|φn〉 therefore reducing

the problem to solving for the individual coefficients cn. We can immediately solve for one of thesecoefficients by noting that the first order correction to |ψ〉must ensure normalisation toO(λ2). Anotherway of writing that is, 〈ψ|ψ〉 = 1 +O(λ2). This has to be true for the approximation to be consistentwith our postulates of quantum mechanics and our boundary conditions.

Writing the general expression for 〈ψ|ψ〉, we find that for this be true, that 〈0|1〉 must be zero andand therefore |1〉 =

∑n6=0 cn|φn〉. We therefore omit |0〉 by hand to satisfy the criterion that as |ψ〉

must be normalised. Substituting gives,

|ψ〉 = |0〉+ λ∑n 6=0

cn|φn〉+O(λ2) (219)

.The problem is therefore to solve for the coefficients cn. Substituting into,

(H0 − E0) |1〉+ (V − ε1) |0〉 = 0 (220)

gives

(H0 − E0)∑n6=0

cn|φn〉+ (V − ε1) |0〉 = 0 (221)

Acting 〈φm| on both sides and using the properties of the complete orthonormal basis φm, gives thefollowing equation for cm,

cm =〈φm|V |0〉E0 − Em

(222)

Therefore, the first order correction to |ψ〉

|1〉 =∑n6=0

〈φn|V |0〉E0 − En

|φn〉 (223)

Putting this all together gives,

|ψ〉 = |0〉+ λ∑n6=0

〈φn|V |0〉E0 − En

|φn〉+O(λ2) = |0〉+∑n6=0

〈φn|H1|0〉E0 − En

|φn〉+O(λ2) (224)

Note, that the perturbation has the effect of mixing in higher energy eigenstates with a weightingfactor given by the energy difference. Therefore, we expect that the bigger the energy difference,the less mixing. We can also see here why we need to make this distinction between degenerate andnon-degenerate cases. If there is a degeneracy, our current formalism would introduce a singularity forthe correction (which is supposed to be small!). We will deal with the case of degenerate perturbationtheory in the next section.

We can follow the same procedure to calculate the second order correction to the energy. The equationinvolving ε2 from above is,

(H0 − E0) |2〉+ (V − ε1) |1〉 − ε2|0〉 = 0 (225)

45

Acting 〈0| on both sides of this equation and substituting in the known first order corrections give thefollowing expression for the corrections to the energy,

E = E0 + 〈0|H1|0〉+∑n6=0

|〈φn|H1|0〉|2

E0 − En+O(λ3) (226)

.While we will stop here, the procedure for deriving higher order corrections for the both the eigenstatesand energy is very similar to outlined here, though somewhat tedious.

17.2 Degenerate levels and the secular equation

Note, from the expressions above it should be clear that this formalism does not work in the case ofdegenerate levels as if Ep = E0 then the equations become singular and the corrections are no longersmall. Therefore, for the case of degenerate eigenvalues, we must treat the perturbation differently.Lets step back to the following step above,

H|ψ〉 = (H0 +H1)|ψ〉 = E|ψ〉 (227)

Lets take the approximation that our basis of the unperturbed Hamiltonian {φp} is a good basis statefor |ψ〉

|ψ〉 =∑i

ai|φi〉 (228)

Where the eigenvalues of each member of the basis state is degenerate. Putting this together gives,

∑i

(E0 +H1)ai|φi〉 =∑i

Eai|φp〉 (229)

Acting by 〈φk| on both sides of this equation gives us the following,

(E − Ek)ak =∑i

〈φk|H1|φi〉ai (230)

Recall the fact that by definition Ek(λ) = Ek+E+..., (recall the unperturbed energies are degenerate)where E = λε1 is our first order correction. Substituting gives,

Eak =∑i

〈φk|H1|φi〉ai (231)

or

∑i

(〈φk|H1|φi〉 − Eδi,k

)ai = 0 (232)

This is a system of linear equations for quantities ai and has solutions which are not zero if thedeterminant of the coefficients vanishes. We therefore obtain the secular equation,

|〈φk|H1|φi〉 − Eδi,k| = 0 (233)

By solving the secular equations for the possible roots E, we can then substitute and find solutionsfor ai to get the eigenvectors.

46