Quantum Mechanics of Atoms & Molecules · Quantum Mechanics of Atoms & Molecules J.Pearson July 24,...

Quantum Mechanics of Atoms & Molecules

J.Pearson

July 24, 2008

Abstract

These are a set of notes I have made, based on lectures given by Y.Xian at the University ofManchester Jan-June ’08. Please e-mail me with any comments/corrections: [email protected].

CONTENTS 3

Contents

1 Review of Quantum Mechanics 1

1.1 State and Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Transposed Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.2 Hermitian Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.3 Correspondence Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Example: Orthogonality of Eiegenstates . . . . . . . . . . . . . . . . . . . . . 2

1.2.2 Hamiltonian Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Harmonic Oscillator & Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3.1 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3.3 Creation & Destruction Operators . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3.4 The Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Matrix Representation of QM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4.1 Example: Electron Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Angular Momentum, Hydrogen & Helium Atoms 12

2.1 Angular Momenta & Their Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1.1 Angular Momentum Operators in Spherical Polars . . . . . . . . . . . . . . . 14

2.1.2 Eigenstates & Eigenvalues of Angular Momentum . . . . . . . . . . . . . . . 15

2.1.3 Angular Momentum Commutators . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1.4 Addition of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2 Hydrogen-like Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Introducing Many-Body Quantum Physics 21

3.1 Symmetry of Exchanged Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Independent Particle Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.2.1 The Slater Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.3 Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 CONTENTS

3.4 The Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4.1 Atomic Spectral Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.5 Ground State Energy of Helium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 Approximation Methods in Quantum Mechanics 28

4.1 Time Independant Perturbation Theory (non-degenerate) . . . . . . . . . . . . . . . 28

4.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.1.2 Perturbation Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.1.3 Solution for First Order Perturbation Equation . . . . . . . . . . . . . . . . . 29

4.1.4 Solution for Second Order Perturbation Equation . . . . . . . . . . . . . . . . 30

4.1.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.2 Degenerate Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.3 Time Dependent Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.3.1 Newton’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.3.2 Time Dependant Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.3.3 Perturbation Expansion Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.3.4 Example: Hydrogen hit by Electric Field . . . . . . . . . . . . . . . . . . . . 44

4.3.5 Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.4 Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.4.1 Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.4.2 Example: Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.4.3 Example: Helium Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5 Introduction to Molecular Quantum Mechanics 52

5.1 Born-Oppenheimer Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5.2 Molecular Orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5.3 Hydrogen Molecular Ion H+2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.4 Structures of Homonuclear Diatomic Molecules . . . . . . . . . . . . . . . . . . . . . 56

A Worked Examples 58

A.1 Number of Symmetric or Anti-symmetric States . . . . . . . . . . . . . . . . . . . . . 58

1

1 Review of Quantum Mechanics

This section covers basics, and re-does some concepts in a more formal language.

1.1 State and Operator

In classical mechanics we use r(t), p(t) to describe the state of a classical particle: its position andmomentum. Which is a set of 6 numbers.In quantum mechanics, we use a wavefunction ψk(r), k = 1, 2, . . .. We will sometimes denote thisthe ‘state function’. ψ covers a given space.We see that physical observables become operators in QM:

Aψk(r) = ψ′k(r)

Where the action of an operator on some state changes the state to something else. The physicalmeasurement is an expectation value: ⟨

A⟩

=∫d3r ψ∗kAψk

1.1.1 Transposed Operator

For any two wavefunctions Ψ and Φ, we can write the integral, which is the same as the transpose:∫d3rΦ(AΨ) =

∫d3rΨAΦ

Where A is the transpose of A. The transpose of a matrix is when you swop rows for columns. Inthis context, instead of operating to the right, you operate to the left, with the transpose operator,which we then rearrange.

1.1.2 Hermitian Operator

Now, if we conjugate the transposed operator; i.e. (A)∗, we define this as A†. That is, we havefound the Hermitian conjugate of the operator A.e.g. Let A ≡ ∂

∂x , then:

A = −A = − ∂

∂x⇒ A† = (A)∗ = − ∂

∂x

Now, to find the transpose of the above differential operator, we do the following:∫dxφ1(x)Aφ2(x) =

∫dxφ1(x)

∂φ2(x)∂x

Now, we can integrate this by parts, so that:∫ ∞−∞

dxφ1(x)∂φ2(x)∂x

= [φ1(x)φ2(x)]∞−∞ −∫ ∞−∞

dx∂φ1(x)∂x

φ2(x)

2 1 REVIEW OF QUANTUM MECHANICS

Now, as wavefunctions go to zero at ±∞, the middle term is zero. Also, if we recall that aninterpretation of the transpose of an operator is when it acts to the left (rather than right), thenwe have our result. ∫ ∞

−∞dxφ1(x)

∂φ2(x)∂x

= −∫ ∞−∞

dx∂φ1(x)∂x

φ2(x)

⇒∫ ∞−∞

dxφ1(x)Aφ2(x) = −∫ ∞−∞

dx Aφ1(x)φ2(x)

Hence, we see that ∂∂x is not a Hermitian operator.

e.g. Let A ≡ −i ∂∂x , then:

A = −A = i∂

∂x⇒ A† = (A)∗ = −i ∂

∂x

So, we see that an operator is Hermitian, if it satisfies:

A = A† (1.1)

All physical operators are Hermitian operators.

1.1.3 Correspondence Principle

Here, we just say that the QM operators can be found from the classical version:

r → r = r

p → p = −ih∇

And both are Hermitian operators.

1.2 Schrodinger Equation

An eigenequation:Aφ = aφ

Where φ is an eigenstate of the operator A, with eigenvalue a. Generally, this is:

Aφn = anφn n = 1, 2, . . .

1.2.1 Example: Orthogonality of Eiegenstates

Here, we show that two eigenstates of an Hermitian operator, with different eigenvalues are alwaysorthogonal. We start with:

Aφn = anφn

Let us then multiply this by some other eigenstate, and integrate over all space:∫d3r φ∗1Aφ2 = a2

∫d3r φ∗1φ2

1.3 Harmonic Oscillator & Dirac Notation 3

Now, we use the Hermiticity of A to write:∫d3r φ∗1Aφ2 =

∫d3r (Aφ1)∗φ2 = a1

∫d3r φ∗1φ2

Where the middle step has applied the operator from the left, conjugated it (Hermitian conjugate)and rearranged the terms. We then evaluate the eigenvalue equation, noting that Hermitian oper-ators have real eigenvalues (which is not proved here, but has been in previous courses). We noticethat the LHS of the two above equations are the same, hence their RHS are:

a2

∫d3r φ∗1φ2 = a1

∫d3r φ∗1φ2

Which is the same as:(a2 − a1)

∫d3r φ∗1φ2 = 0

Hence, if a2 6= a1, then: ∫d3r φ∗1φ2 = 0

1.2.2 Hamiltonian Operator

The energy, classically, is the sum of kinetic and potential. Quantum mechanically it is hence just:

H =p2

2m+ V (r)

Where p2 = p · p = −h2∇2.

The Schrodinger equation provides us with the dynamics of the system:

ih∂

∂tΨ(r, t) = HΨ(r, t) (1.2)

Letting H be time independant results in the solution being:

Ψ(r, t) = Φ(r)e−iEt/h

And the Schrodinger equation reduces to:

HΦ(r) = EΦ(r)

Which is an eigen equation, with E being the eigenvalue, and Φ the eigenstate of H.

1.3 Harmonic Oscillator & Dirac Notation

Suppose (classicaly) we consider the potential V = 12kx

2 = 12mω

2x2, we find we have the equationof motion:

md2x

dt2= −kx


Which gives us an oscillation about the equilibrium point, and solutions:

x(t) = x0 sin(ωt+ θ0)p(t) = ωx0 cos(ωt+ θ0)

Where x0, θ0 are from initial conditions.

In quantum mechanics, we find that the ground state wavefunction is:

ψ0(x) = Ae−αx2

α ≡ mω

2h

Where A comes from normalisation:∫ ∞−∞

ψ20 dx = 1 ⇒ A =

(mω2h

)1/4

With ground state energy E0 = 12 hω, and general energy:

En =(n+ 1

2

)hω n = 0, 1, 2, . . .

1.3.1 Dirac Notation

We write the wavefunction φ(r) as a state vector, or just state, |φ〉. And its conjugate φ∗ → 〈φ| .We need the state and its ‘pair’ in order to find expectation values.

We know that a new state can be found from a linear combination of other states; hence, in Diracnotation this is just:

|ψ〉 = c1 |φ1〉+ c2 |φ2〉

The inner product between two state vectors is just:

〈φ1|φ2〉 =∫d3r φ∗1φ2 (1.3)

For any state ψ, we can form it from a linear combination of orthonormal eigenfunctions:

|ψ〉 =∑n

cn |φn〉 (1.4)

So, we say that φn;n = 1, 2, . . . is a complete orthonormal set. Remember, orthonormality isdefined as:

〈φn|φm〉 = δnm (1.5)

It is the combined statement that a single state is normalised (when n = m), and that two differentstates are orthogonal.


1.3.2 Operators

We have, generally, that AΦ = Ψ. That is, in Dirac notation:

A |Φ〉 = |Ψ〉

That is:|AΦ〉 = |Ψ〉

To find the Hermitian conjugate, start by taking the complex conjugate Ψ∗ = (AΦ)∗; which, inDirac notation is:

〈Ψ| = A∗〈Φ| = 〈Φ| A†

So: (〈φ1| A |φ2〉

)∗= 〈φ2| A† |φ1〉

Product of Operators Suppose that we have:

C = A · B

What is C†? To find this, consider:〈ψ| C |φ〉

So:

〈ψ| C |φ〉 = 〈ψ| A · B |φ〉= 〈ψ| A |Bφ〉= 〈A†ψ|Bφ〉= 〈B†A†ψ|φ〉= 〈C†ψ|φ〉

That is: C† = B†A†.

Consider the eigenequation A |ψ〉 = a |ψ〉. Then, we say that a is the eigenvalue of A, havingeigenstate ψ.

Expectation Values These are given by (for the unnormalised case):⟨A⟩

=〈ψ| A |ψ〉〈ψ|ψ〉

(1.6)

1.3.3 Creation & Destruction Operators

We define the destruction operator thus:

a ≡√mω

2hx+ i

p√2mhω

(1.7)


Taking the Hermitian conjugate of this (as x, p are unnafected by this operation - they representphysical observables!) has the effect of just swapping the i for −i. We call this the creation operator :

a† =√mω

2hx− i p√

2mhω(1.8)

Just to reiterate the point, (ip)† = i†p† = (−i)p. In example sheets, we have derived the commutator(infact, its not very hard to actually do!): [

a, a†]

= 1 (1.9)

1.3.4 The Harmonic Oscillator

Now, the Hamiltonian H = p2

2m + 12mω

2x2 may be written in terms of the creation and destructionoperators. To see this, consider:

a+ a† = 2√mω

2hx

⇒ x =12

√2hmω

(a+ a†)

a− a† = 2ip√

2mhω

⇒ p =i

2

√2mhω(a† − a)

So that, computing the elements of the Hamiltonian:

p2 = −2mhω4

((a†)2 + a2 − a†a− aa†

)x2 =

14

2hmω

((a†)2 + a2 + a†a+ aa†

)⇒ H =

p2

2m+

12mω2x2

= hω(a†a+ 12)

Where we have made use of the above commutator.

So then, lets figure out how to use the operators.

Suppose that:a |0〉 = 0

Then let us find some state |0〉 for which the above is satisfied, then the above statment is proven.This is of course equivalent to:

a†a |0〉 = 0

Now, we notice the above is pretty much the same as the Hamiltonian. So, let us operate with theHamiltonian on the state:

H |0〉 = hω(a†a+ 12) |0〉

= hωa†a |0〉+ hω 12 |0〉

= 12 hω |0〉


Hence, we see that |0〉 is an eigenstate of the Hamiltonian, with eigenvalue E0 ≡ 12 hω.

Now, let us find what |0〉 would be, in ‘real space’. Let us denote it ψ0(x). So, we have alreadystated that a |0〉 = 0; so: (√

mω

2hx+ i

p√2mhω

)ψ0(x) = 0

Which, after putting p = −ih ∂∂x , we get:√

mω

2hxψ0(x) +

h√2mωh

dψ0(x)dx

= 0

Rearranging results in:

−mωxh

ψ0(x) =d

dxψ0(x)

Separating out:

−mωhx dx =

dψ0(x)ψ0(x)

Which is trivially integrated to give:

ψ0(x) = Ae−mω2hx2

Where the A will be the normalisation constant, and is found to be:

A =(mωπh

)1/4

Hence, we have found a state |0〉.

Now, let us consider the action of the creation operator on the state |0〉. We will pre-suppose thatit gives:

a† |0〉 = |1〉

We would like to find out if the following is true:

H |1〉 = E1 |1〉

We have already seen that H can be written in terms of a†a, so that H |1〉 = a†a |1〉. We have alsostated that |1〉 = a† |0〉. So:

H |1〉 = a†a |1〉 = a†aa† |0〉

Now, the commutation relation is just:

aa† − a†a = 1 (1.10)

Hence:aa† = 1 + a†a

Therefore, we have:a†a |1〉 = a†aa† |0〉 = a†(1 + a†a) |0〉


And orthogonality:

〈1|2〉 = (1, 0, . . .)

01...

= 0

Thus, the orthonormality relation:

〈φn|φm〉 = δnm (1.15)

Now, we know that any state can be expressed as a linear combination of eigenstates: |ψ〉 =∑n cn |φn〉. Hence:

|ψ〉 =

c1

c2...

Hence:

〈ψ|ψ〉 = |c1|2 + |c2|2 + . . . =∑k

|ck|2

Where the ci may be complex numbers.The matrix representation for an operator, in the basis of its eigenstates, is a diagonal matrix whoseelements are its eigenvalues:

A =

a1 . . .... a2

. . .

So, just to be clear, the operator A has eigenstates |k〉, and eigenvalues ak; which are the diagonalentires in its matrix-representation.

Now, suppose we have some arbitrary operator B; then its matrix representation will be:

B =

B11 B12 . . .B21 B22 . . .

.... . .

Where the element may be computed thus:

Bk′k = 〈k′| B |k〉

Which, in real space is:

Bk′k =∫φ∗k′Bφk d

3r

Where the φk are eigenstates of the operator A; so, we say that B uses a basis which are theeigenstates of A. Again, to be clear where this comes from:Let us have the eigenequation B |λ〉 = λ |λ〉. Then, we may express the states |λ〉 in terms ofeigenstates of A; that is, states satisfying A |k〉 = k |k〉. That is, we may write:

|λ〉 =∑k

ck |k〉

1.4 Matrix Representation of QM 11

Then, we have that:B |λ〉 = λ |λ〉 ⇐⇒

∑k

ckB |k〉 = λ∑k

ck |k〉

Now, let us multiply by the conjugate: 〈k′| , so:∑k

ck〈k′| B |k〉 = λ∑k

ck〈k′|k〉

However, the states |k〉 satisfy the standard orthonormality condition; i.e. that 〈k′|k〉 = δk′k; hence:∑k

ck〈k′| B |k〉 = λck′

Then, if we define:Bk′k ≡ 〈k′| B |k〉

The above just becomes matrix multiplication:∑k

Bk′kck = λck′

Thus, supposing we find the eigenvectors of the matrix B, then we would be finding them in thebasis of the eigenstates of A.Thus, to find the eigenvalues of B, we have the determinant:∣∣∣∣∣∣∣

B11 − λ B12 . . .B12 B22 − λ . . .

.... . .

∣∣∣∣∣∣∣ = 0

We we have the same number of eigenvalues as the dimension of the matrix.

1.4.1 Example: Electron Spin

We have, for an electron with s = 12 , 2 spin states (2 ways of projecting onto the magnetic axis).

We write:

Sz = 12 h

(1 00 −1

)Where we have written the matrix representation of the Sz operator in a basis which is its owneigenstates. That is, its eigenvectors are:

|1〉 =(

10

)≡ | ↑〉

|2〉 =(

01

)≡ | ↓〉

So that:

Sz | ↑〉 = +12 h | ↑〉

Sz | ↓〉 = −12 h | ↓〉

12 2 ANGULAR MOMENTUM, HYDROGEN & HELIUM ATOMS

So, we can write the ladder operators:

S± ≡ Sx ± iSy

In the basis of eigenstates of the Sz operator. That is:

S+ =(〈1| S+ |1〉〈1| S+ |2〉〈2| S+ |1〉〈2| S+ |2〉

)So, to evaluate the elements, we have the relations that:

S+ |1〉 = S+ | ↑〉 = 0 S+ |2〉 = S+ | ↓〉 = h | ↑〉

So that, using the relations, and orthogonality 〈i|j〉 = δij , we have that:

S+ = h

(0 10 0

)Similarly, we have that:

S− = 12 h

(0 11 0

)So, we can find Sx = 1

2(S+ + S−), and Sy = 12i(S

+ − S−); giving:

Sx = 12 h

(1 01 0

)Sy = 1

2i h

(0 1−1 0

)So, we can also easily find:

S2 = S2x + S2

y + S2z = 3

4 h2

(1 00 1

)Infact, we have that for a spin s system, the representation of S2 is always just:

S2 = h2s(s+ 1)1

Where 1ij = δij ; the identity matrix.

2 Angular Momentum, Hydrogen & Helium Atoms

2.1 Angular Momenta & Their Addition

Let us consider the 3D Hamiltonian, with a time-independant potential term. Thus:

H = − h2

2m∇2 + V (r)

Now, ∇2 is easy to write down in Cartesian coordinates:

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2

2.1 Angular Momenta & Their Addition 13

In spherical coordinates (r, θ, ϕ), its a little harder:

∇2 =1r2

∂

∂rr2 ∂

∂r+

1r2 sin θ

∂

∂θsin θ

∂

∂θ+

1r2 sin2 θ

∂2

∂ϕ2

Where this can be derived by considering the transformation expressions between the two coordinatesystems. Now, it can be shown that L2 is just the angular part of ∇2. That is:

L2 = −h2

[1

r2 sin θ∂

∂θsin θ

∂

∂θ+

1r2 sin2 θ

∂2

∂ϕ2

]And, we have that Lz can be written:

Lz = −ih ∂

∂ϕ(2.1)

So that we have:

∇2 =1r2

∂

∂rr2 ∂

∂r− L2

h2r2

Hence, the Hamiltonian can be written:

H = − h2

2m1r2

∂

∂rr2 ∂

∂r+ V (r) +

12m

L2

r2

Now, suppose our eigenstate for the Hamiltonian is:

Hψ(r, θ, ϕ) = Eψ(r, θ, ϕ)

And, also, suppose that the eigenstate may be separated thus:

ψ(r, θ, ϕ) = R(r)Y (θ, ϕ) (2.2)

And also, that Y (θ, ϕ) is an eigenstate of L2; that is:

L2Y (θ, ϕ) = λY (θ, ϕ)

Hence, we have that Hψ = Eψ reduces to:[− h2

2m1r2

∂

∂rr2 ∂

∂r+ V (r) +

λ

2mr2

]R(r) = ER(r)

Where we have that λ = h2`(`+ 1), where ` = 0, 1, . . ..

We can easily show that[L2, Lz

]= 0; and that:

LzY (θ, ϕ) = mhY (θ, ϕ)


2.1.1 Angular Momentum Operators in Spherical Polars

We have already touched upon these, but here we state them all specifically.

We wish to convert the Laplacian into spherical polars, from Cartesian. That is, a coordinatetransformation (x, y, z)→ (r, θ, ϕ). So, we have:

x = r sin θ cosϕ r =√x2 + y2 + z2

y = r sin θ sinϕ tan2 θ =x2 + y2

z2

z = r cos θ tanϕ =y

x

Now, the Laplacian, in cartesian coordinates is:

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2

So, we must transform each component. The x transforms like:

∂

∂x=∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ

Each component becomes:

∂

∂x= sin θ cosϕ

∂

∂r+

1r

cos θ cosϕ∂

∂θ− 1r

sinϕsin θ

∂

∂ϕ

∂

∂y= sin θ sinϕ

∂

∂r+

1r

cos θ sinϕ∂

∂θ− 1r

cosϕsin θ

∂

∂ϕ

∂

∂z= cos θ

∂

∂r− 1r

sin θ∂

∂θ

Everything must now be sqaured; which is a little tedious, and wont be repeated here! We will endup with:

∇2 =1r2

∂

∂rr2 ∂

∂r+

1r2 sin θ

∂

∂θsin θ

∂

∂θ+

1r2 sin2 θ

∂2

∂ϕ2

So, now consider Lz:

Lz = −ih(x∂

∂y− y ∂

∂x

)= −ih ∂

∂ϕ

After substitution of relevant terms. Similalry, we can compute the other terms; giving us a full setof angular momentum operators in spherical polar coordinates:

Lx = −ih(− sinϕ

∂

∂θ− cot θ cosϕ

∂

∂ϕ

)(2.3)

Ly = −ih(

cosϕ∂

∂θ− cot θ sinϕ

∂

∂ϕ

)(2.4)

Lz = −ih ∂

∂ϕ(2.5)


Also, from the defintion of the ladder operators, L± = Lx ± iLy, we can also derive:

L± = −ihe±iϕ(±i ∂∂θ− cot θ

∂

∂ϕ

)(2.6)

If we go through the algebra, we also find that:

− 1h2 L

2 =1

sin θ∂

∂θsin θ

∂

∂θ+

1sin2 θ

∂2

∂ϕ2

So, comparing terms, we are able to get to the form of the Schrodinger equation:[(− h2

2m1r2

∂

∂rr2 ∂

∂r+ V (r)

)+

L2

2mr2

]Ψ(r, θ, ϕ) = EΨ(r, θ, ϕ)

2.1.2 Eigenstates & Eigenvalues of Angular Momentum

Suppose that we further separate:Y (θ, ϕ) = Θ(θ)Φ(ϕ)

Now, as Lz = −ih ∂∂ϕ , we have that:

LzΦ(ϕ) = λmΦ(ϕ)

That is:−ih ∂

∂ϕΦ = λmΦ

The solution is easy, and is just:Φ(ϕ) = 1√

2πeiλmϕ/h

Where the factor of 1√2π

comes from normalisation.We now employ the boundary conditions that Φ(ϕ) = Φ(ϕ+ 2π); hence:

eiλmϕ/h = eiλm(ϕ+2π)/h

That is:eiλm2π/h = 1

That is to say, λmh is an integer, m, say. Hence:

λm = mh m = 0,±1,±2, . . .

Thus, we have that:|m〉 = Φm(ϕ) = 1√

2πeimϕ

So:Y`m(θ, ϕ) = Θ`m(θ)Φm(ϕ)

Where we label states as ‘forethought’. So, we have:

Lz |m〉 = hm |m〉 (2.7)

So, how about:L2Y`m(θ, ϕ) = λY`m(θ, ϕ)

That is:L2 |`,m〉 = λ |`,m〉


2.1.3 Angular Momentum Commutators

We can fairly easily show that: [Lz, L±

]= ±hL± (2.8)[

L+, L−]

= 2hLz (2.9)

Compare the last one with the commutator for the creation and destruction operators:[a, a†

]= 1.

Ladder Operators Let us first consider the action of the operator L± on a state |`,m〉. Let usfirst consider:

Lz |`,m〉 = hm |`,m〉

Now, suppose that the state |`,m〉 above is actually the state L± |`,m〉. So, we have:

Lz

(L± |`,m〉

)= LzL

± |`,m〉

Let us now use the commutator: LzL± − L±Lz = ±hL±; so that the above just reads:

LzL± |`,m〉 =

(±hL± + L±Lz

)|`,m〉

We can evaluate the far-right term, giving:

LzL± |`,m〉 =

(±hL± +mhL±

)|`,m〉

= h(m± 1)L± |`,m〉

That is, just:LzL

± |`,m〉 = h(m± 1)L± |`,m〉

That is, L± |`,m〉 is an eigenstate of Lz, with eigenvalue h(m± 1). That is, the eigenstate:

L± |`,m〉 = c± |`,m± 1〉

Just to clarify; if the above is an eiegenstate of Lz, then:

LzL± |`,m〉 = Lzc

± |`,m± 1〉 = h(m± 1)c± |`,m± 1〉

Where c± is a normalisation constant, which shall be determined later.

L2 Degeneracy Here, we show that states with the same `, but different m are degenerate.Consider the commutator

[L2, L±

]= 0. That is just:

L2L± = L±L2

Thus, putting a state |`,m〉 on the left:

L2L± |`,m〉 = L±L2 |`,m〉


We can evaluate the action of the operator on the immediate left of the state:

L2c± |`,m± 1〉 = L±λ`,m |`,m〉

And again:c±λ`,m±1 |`,m± 1〉 = λ`,mc

± |`,m± 1〉Hence, we see that:

λ`,m±1 = λ`,m

That is, the eigenvalues of L2 are independant of m. So, we denote them λ`.

Limit on m Here, we show that |m| ≤ `.Now,: ⟨

L2 + L2⟩> 0

Hence, as:L2 = L2

x + L2y +m2h2

Then |m| must be finite, with maximum value `. This is just saying that, the contributions fromall directions must be taken into consideration; then we have a minimum |m| = 0, and maximum|m| = `. Hence, we have that:

m = 0,±1,±2, . . . ,±` (2.10)

So that there are 2`+ 1 values of m for a given `; where ` = 0, 1, 2, . . ..

Eigenvalue of L2 Let us start with the equation:

L2 |`,m〉 = λ` |`,m〉

So, what is λ`? We have previously shown that it is independant of m. Now:

L2 = L2x + L2

y + L2z

Now, using ladder operators, and the commutator[L+, L−

]= 2hLz, we are able to fairly easily

derive:L2 = L−L+ + hLz + L2

z

Now, we have the ladder operator relations:

L+ |`, `〉 = 0 L− |`,−`〉 = 0

That is, operating on the state with max/min values of m. So, let us then write:

L2 |`, `〉 = (L−L+ + hLz + L2z) |`, `〉

= (0 + hh`+ (hm)2) |`, `〉= h2`(`+ 1) |`, `〉

However, we have seen that the eigenvalue of L2 is actually independant of the value of m in thestate |`,m〉. Hence:

L2 |`, `〉 = L2 |`,m〉Hence, we have that:

L2 |`,m〉 = h2`(`+ 1) |`,m〉 (2.11)


What is Constant in Ladder Operations? Here, we shall consider what the constant in thefollowing, is:

L± |`,m〉 = c± |`,m± 1〉Now, consider:

(L+ |`,m〉)∗ = 〈`,m| (L+)† = 〈`,m| L−

Now the, suppose we consider:

(c+ |`,m+ 1〉)2 = (c+ |`,m+ 1〉)∗(c+ |`,m+ 1〉)= (c+)2〈`,m+ 1|`,m+ 1〉= (c+)2

Thus, doing the same to the other side:

(c+)2 = 〈`,m| L−L+ |`,m〉

We use a previously derived relation:

(c+)2 = 〈`,m| L2 − hLz − L2z |`,m〉

That is just:(c+)2 = (h2`(`+ 1)− h2m− h2m2)〈`,m|`,m〉

Thus, as the state is orthonormal:

c+ = h√`(`+ 1)−m(m+ 1) (2.12)

Similarly:c− = h

√`(`+ 1)−m(m− 1)

Finding Highest Order Spherical Harmonic Now, we have:

L+ |`, `〉 = 0

We also have:

L+ = −iheiϕ(∂

∂θ− cot θ

∂

∂ϕ

)And:

|`, `〉 → Y`,`(θ, ϕ) = Θ`(θ)Φ`(ϕ)

And, we have already solved for:

Φ`(ϕ) =1√2πei`ϕ

So, we can derive:Y`,`(θ, ϕ) = eci`ϕ sin` θ

Where the normalisation constant:

c =(−1)`√

4π

√(2`+ 1)!!

(2`)!!

Hence, we may find a general spherical harmonic by applying L−.


General Angular Momentum Theorem Almost everything we have done is general, exceptfor the direct specification of operators (which isnt really necessary), and when we wrote:

m = 0,±1,±2, . . . ,±`

The general relations we can use are:

L2 |`,m〉 = h2`(`+ 1) |`,m〉 (2.13)Lz |`,m〉 = mh |`,m〉 (2.14)

m = −`, (−`+ 1), . . . , 0, . . . , (`− 1), ` (2.15)

2.1.4 Addition of Angular Momenta

Consider:L = L1 + L2

Where each has its own states:|`1,m1〉 |`2,m2〉

So, what are the eigenvalues of L2, in terms of eigenvalues of L21, L

22? Now, the states of each

operator are known thus:

L2i |ì,mi〉 = h2ì(ì + 1) |ì,mi〉 i = 1, 2

Lzi |ì,mi〉 = hmi |ì,mi〉 mi = −ì, (−ì + 1), . . . , (ì − 1), ì

Good quantum numbers are those for which we can find common sets of eigenstates for their oper-ators. So, for example, we have, as always:[

L2, Lz

]= 0

However, we also have that: [L2, Lz1

]6= 0

And similar for Lz2.Hence, we see that `, `z1 are not good, as `, `z are. So, we use the states:

|`1, `2, `,m〉

So that:

L2i |`1, `2, `,m〉 = h2ì(ì + 1) |`1, `2, `,m〉

L2 |`1, `2, `,m〉 = h2`(`+ 1) |`1, `2, `,m〉Lz |`1, `2, `,m〉 = hm |`1, `2, `,m〉

We also see that the possible values of `, given `1, `2 are:

` = (`1 + `2), (`1 + `2)− 1, . . . , |`1 − `2|

And m = −`, (−`+ 1), . . . , 0, . . . , (`− 1), `.


Example Two particle, spin 12−system.

Consider the Hamiltonian:H = αS1 · S2

Then, via:S = S1 + S2

We see that:S2 = S2

1 + S22 + 2S1 · S2

So that:H =

α

2[S2 − S2

1 − S22 ]

Now, as s1 = s2 = 12 , we have that s2

1 = s22 = 3

4 h2. Hence:

H =α

2[S2 − 3

2 h2]

Thus:

H |s,m〉 = α2 (s(s+ 1)h2 − 3

2 h2) |s,m〉

= Es |s,m〉

Now, by the addition of angular momentum, we see that s = 0, 1. Hence, we have:

E1 =αh2

8s = 1

andE0 = −3

4αh2 s = 0

Thus, we see that E0 is the ground state, and is a singlet (m = 0 is only possible value). E1 is atriplet state, as m = −1, 0, 1.

Now, we can guess at the states. Using the following notation:

| ↑〉i = |si = 12 ,m = +1

2〉| ↓〉i = |si = 1

2 ,m = −12〉

So, we have that the ground state looks like:

|0, 0〉 =1√2

( | ↑〉1 | ↓〉2 − | ↓〉1 | ↑〉2)

The excited states are:

|1,−1〉 = | ↓〉1 | ↓〉2|1, 1〉 = | ↑〉1 | ↑〉2

|1, 0〉 =1√2

( | ↑〉1 | ↓〉2 + | ↓〉1 | ↑〉2)

We see that:S+i | ↓〉i |〉j = |〉jc+

i | ↑〉iThat is, only operates on its ‘specific’ state.

2.2 Hydrogen-like Atoms 21

2.2 Hydrogen-like Atoms

We have seen that, for a time-independant radial-only potential:

H = − h2

2m1r2

∂

∂rr2 ∂

∂r+ V (r) +

12mr2

L2

We have states:H |n, `,m,ms〉 = En |n, `,m,ms〉

Where:

En = −13.6Z2

n2eV

In real space, the states are:

|n, `,m,ms〉 → Rn`(r)Y`m(θ, ϕ)χms

3 Introducing Many-Body Quantum Physics

3.1 Symmetry of Exchanged Particles

Consider a two-identical-particle system. Suppose that their wavefunction is ψ(x1, x2), where xi isthe ‘coordinate’ of particle i. An example of such a system is the two eletrons in the helium atom,where we will have that the coordinate is given by x = (r, σ), where r is the spatial coordinate,and σ the spin of the particle.

Now, consider the definition of the exchange operator, such that:

P1↔2ψ(x1, x2) = ψ(x2, x1) (3.1)

Now, as a slight aside, it will share a common set of eigenstates with the Hamiltonian, where:

H =p2

1

2m+

p22

2m+

e2

4πε0

1r12

So that: [P1↔2, H(1, 2)

]= 0

So, let ψ be an eigenstate of the exchange operator P1↔2. That is:

P1↔2ψ(x1, x2) = pψ(x1, x2)

However, by definition, we have that the following is true:

pψ(x1, x2) = ψ(x2, x1)

That is, we equate the eigenvalue equation with the action of the operator. Now, let us write downthe eigenvalue equation, after we have operated twice with the exchange operator:

P 21↔2ψ(x1, x2) = p2ψ(x1, x2)

22 3 INTRODUCING MANY-BODY QUANTUM PHYSICS

This is, however, also equivalent to:

P1↔2ψ(x2, x1) = ψ(x1, x2)

Hence, by inspection, we see that:

p2 = 1 ⇒ p = ±1

So, we have two classifications of particles:Those for whom p = +1, we classify as bosons, and those for whom p = −1, we call fermions. So,from the above considerations, we see that bosons are symmetric, and fermions anti-symmetric.A more general analysis (which is beyond us) reveals that bosons have integer-spin, and fermionshalf-integer.The general version of the exchange operation, for a system with N particles, is just:

Pm↔nψ(x1, . . . , xm, . . . , xn, . . . , xN ) = (±1)ψ(x1, . . . , xn, . . . , xm, . . . , xN )

3.2 Independent Particle Approximation

Now, consider the general Hamiltonian, for an N -particle system:

H =N∑i=1

Hi + V

Where the ‘interaction potential’ is given by:

V =12

N∑i 6=2

V (rij)

Where the factor of one-half comes from double counting, and the symmetry inherent in the potentialterm. So, for example, we have that the interaction potentials for helium (2 electrons interactingvia the Coulomb force), or lithim (3 interacting electrons; Coulomb again) are given by:

V =e2

4πε0

1r12

V =e2

4πε0

(1r12

+1r13

+1r23

)rij ≡ ri − rj

Now, consider the simplest approximation: ignore any interaction between the particles:

H =N∑i=1

Hi

This is known as the independent particle approximation. The reason we do this, is to keep thequantum nature of the particles, but simplify things by ignoring any coupling they may have. Now,suppose we have solved the single particle Schrodinger equation:

H1Φk(x1) = EkΦk(x1)

3.2 Independent Particle Approximation 23

That is, we find that one particle is in energy level k. Then, by the above approximation, the totalN -body wavefunction is just the product of single particle wavefunctions. Namely:

Ψ(x1, x2, . . . , xN ) ∝ Φk1(x1)Φk2(x2) . . .ΦkN (xN )

And that the total energy of the system is just the sum of the individual components:

E =N∑i=1

Eki

We have done things like this before, but usually done wit ‘extra dimensions’, such as going from the1D to 3D boxes, multiplying the different wavefunctions together, and adding the energies. Now,suppose we want to write a specific 2-body wavefunction, that will by symmetric; i.e., that willrepresent a boson. Then, we may choose:

ψB(x1, x2) = 1√2[φk1(x1)φk2(x2) + φk2(x1)φk1(x2)]

So, we see that particle i is in an energy state ki. We see that it satisfies the symmetry requirementof a bosonic particle, by just interchanging the coordinate labeling, and noticing that ψB(x1, x2) =ψB(x2, x1). Now, for a fermionic system:

ψF (x1, x2) = 1√2[φk1(x1)φk2(x2)− φk2(x1)φk1(x2)]

Again, we see that it satisfies the anti-symmetry requirement. That is, ψF (x1, x2) = −ψF (x2, x1).Notice, we can also construct the following states, where the particles are in the same state:

ψB(x1, x2) = φk1(x1)φk1(x2)

This construction is not possible for fermions.

3.2.1 The Slater Determinant

Let us reconsider the fermionic wavefunction above:

ψF (x1, x2) = 1√2[φk1(x1)φk2(x2)− φk2(x1)φk1(x2)]

Now, notice that the wavefunction combination is actually a 2x2 determinant:∣∣∣∣ φk1(x1) φk1(x2)φk2(x1) φk2(x2)

∣∣∣∣ = φk1(x1)φk2(x2)− φk2(x1)φk1(x2)

So, we generalise this, to being able to find the N -particle fermionic wavefunction, via the Slaterdeterminant :

φ(x1, x2, . . . , xN ) =1√N !

∣∣∣∣∣∣∣∣∣φk1(x1) φk1(x2) . . . φk1(xN )φk2(x1) φk2(x2) . . . φk2(xN )

.... . .

...φkN (x1) φkN (x2) . . . φkN (xN )

∣∣∣∣∣∣∣∣∣Now, from determinant theory (which we havnt done properly!), is we swop two columns in thedeterminant, then the sign of the determinant swops. This will manifest itself as the anti-symmetryof fermions; as we will be swopping the coordinates. Also, from determinant theory, if any two ofthe energy states, or coordinates, are the same, then the determinant vanishes. (ASIDE: this sortof determinant is known as a alternant.) So, we arrive at the the Pauli Exclusion Principle.


3.3 Pauli Exclusion Principle

So, putting everything above together, we are able to write:

A state can only be occupied by, at most, a single fermion; but may be occupied by any numberof bosons.

This is a very important result, and is the reason things like Bose-Einstein condensation is allowed,or neutron star retain their size.

3.4 The Periodic Table

Let us now consider fermions; as they are the ‘class’ of particle of which most things are made!Now, let us write a total wavefunction as the product of its spatial components, and spin:

ψ(x1, x2, . . . , xN ) = φ(r1, r2, . . . , rN )χ(σ1, σ2, . . . , σN ) (3.2)

Hence, if we require ψ to be anti-symmetric, we require one (and only one) of φ or χ to be symmetric.That is, if φ is symmetric, then the spin wavefunction χ must be anti-symmetric, in order that thetotal wavefunction ψ be anti-symmetric. The reverse is also obviously true.

Now, as we have previously seen, the single particle wavefunction may be characterised by 4quantum numbers, in the state |n, `,m,ms〉; so that:

φn`m(r1)χms(σ1)→ |n, `,m,ms〉

Now, let us use some notation to label the orbital angular momentum states:

` = 0⇒ s ` = 1⇒ p ` = 2⇒ d

So, when we say the ‘p’ state, we are referring to the state with ` = 1. Now, we must begin tocount how many electrons we can put in a particular state, taking the Pauli exclusion principle intoaccount: no two electrons may be in the same state.So, for the p state, we have possible m = −1, 0, 1; within each of which can have spin up or down.Hence, in the p state, we can have a maximum of 6 electrons. In general, for a state `, we can fita maximum of 2(2` + 1) electrons into that state. The outer-factor of ‘2’ comes from the 2 spinstates, and the inner ‘2`+ 1’ from the m degeneracy.Thus, the maximum number of electrons in an s ‘shell’ is 2; in p is 6; in d is 10.In its ground state, an atom will have electrons which occupy the lowest possible states, in order ofenergy. These levels are ordered thus:

Orbital: 1s 2s 2p 3s 3p 4s 3d 4p 5s . . .Electron number: 2 4 10 12 18 20 30 36 38 . . .

Where the ‘electron number’ is a running total of the number of electrons that is allowed in theparticular ground state configuration. Note, each state adds a certain number of extra allowedelectrons; as previously discussed. The number infront of the shell label refers to the principlequantum number n. That is, for a particular n (= 2, say), only ` = 0, . . . , n − 1 shells is allowed

3.4 The Periodic Table 25

(` = 0, 1; in our example for n = 2). So, the electronic configuration of the ground state can bewritten:

(1s)2(2s)2(2p)6(3s)2(3p)6(4s)2(3d)10(4p)6 . . .

Notice, the 4s state is in the middle of a 3p and 3d state; this is due to the energy levels overlappingsomewhat. The superscript shows the maximum number of electrons allowed in that particularstate. So, for example, carbon, with 6 electrons, its ground state has electronic configuration:

C → (1s)2(2s)2(2p)2

This does not, however, give any information about the total angular momentum of the system.We use the atomic spectral term to show such information.

3.4.1 Atomic Spectral Term

Here, we use the symbol:

2S+1LJ (3.3)

To denote a particular wavefunction of an atomic state; where S is the total spin, L its total orbitalangular momentum, and J its total angular momentum. We use the same ‘letter’ notation for L,but use captiol letters instead (so, S means L = 0; P means L = 1; etc). So, for example:

2P3/2

Denotes a state where L = 1, S = 12 and J = 3

2 . We use Hund’s rules to determine the atomicspectral term for the ground state of an atom:

• For a given shell, the term with the greatest possible value of S gives the lowest energy;

• The greatest possible value of L (for the S) has the lowest energy;

• For half-or-less filling, J = |L−S| gives the lowest energy. For more-than-half filling, J = L+Sgives the lowest energy.

This comes about by finding the state for which the spins are parallel (first), that is, symmetric;and antisymmetric orbital wavefunction. Let us look at some examples to see how this works:

• Helium (Z = 2): has electron configuration (1s)2. Hence, S = 0, L = 0. Hence, its groundstate term is 1S0; where J = 0.

• Carbon (Z = 6): has electron configuration (1s)2(2s)2(2p)2. The greatest value of S is 1; fromhaving two parallel spins in the p state. Thus, L = 1. Hence, also, as less-than-half filling ofthis state, we have that J = |L− S| = 0. Hence, its term is 3P0.


3.5 Ground State Energy of Helium

Consider the Hamiltonian of a 2 electron problem:

H = H1 + H2 + V

Where the single-particle Hamiltonians just consider the Coulomb interaction between the electron,and nucleus; and the interaction potential looks at the interaction between two electrons. That is:

Hi = − h2

2m∇2i −

Ze2

4πε0riV =

e2

4πε0

1r12

r12 = |r1 − r2|

Notice, this Hamiltonian is spin-independant. Now, the ground state of helium will be when bothelectrons are in the 1s state, with anti-parallel spin. That is, the electron configuration will be(1s)2, and atomic term 1S0. Now, the total (ground state) wavefunction will be:

ψ0(1, 2) = φ1s(r1)φ1s(r2)χ(σ1, σ2)

Now, as we said, the spin wavefunction is anti-symmetric. Therefore, the spatial part must besymmetric. So, a two-particle anti-symmetric spin wavefunction may be written:

χ(σ1, σ1) = 1√2( | ↑〉1 | ↓〉2 − | ↓〉1 | ↑〉2)

Now we can look up the ground state wavefunction for a single electron:

φ1s(r) =

√Z3

πa30

e−Zr/a0

And its energy:

En = − e2

8πε0a0

Z2

n2

So, let us consider the state where n = 1: the ground state. Now, the energy of the two-particleground state will be given by the expectation of it Hamiltonian, with its total wavefunction; thus:

Eg = 〈ψ0| H |ψ0〉

Which is just:

Eg = 〈ψ0|(H1 + H2 +

e2

4πε0

1r12

)|ψ0〉

Now, because of the symmetry in the single particle hamiltonians, we are able to write the aboveas:

Eg = 2〈ψ0| H1 |ψ0〉+e2

4πε0〈ψ0|

1r12|ψ0〉

The first expression we know:〈ψ0| H1 |ψ0〉 = E1

Where the ‘lack of superscript g’ denotes the single particle energy state, with n = 1. Hence:

Eg = 2E1 +e2

4πε0〈ψ0|

1r12|ψ0〉

3.5 Ground State Energy of Helium 27

Now, to do the integral:

〈ψ0|1r12|ψ0〉 =

∫d3r1d

3r2φ∗1s(r1)φ∗1s(r2)

1|r1 − r2|

φ1s(r1)φ1s(r2)

Where we have noted that the spin part will normalise to unity. Now, upon insertion of the formof the ground state wavefunction, this becomes:

1π2

(Z

a0

)6 ∫d3r1d

3r21

|r1 − r2|e−2Z(r1+r2)/a0

This integral is hard to do; and involves Fourier transforms. So, we quote the answer to be:

58Z

a0

Hence, we have that the ground state energy of helium is given by:

Eg = −2e2

8πε0a0Z2 +

e2

4πε0Z2 5

8Z

a0

So, cleaning up; using the Rydberg energy:

Eg =(−2Z2 +

54Z

)R∞

Where R∞ = 13.6eV. Hence, for Z = 2, in the helium atom, we have:

Eg = −5.50R∞

Thus, we have computed that the ground state energy of helium is -74.8eV. Experimentally, it isfound to be −5.81R∞; so, an agreement to a few percent using the independant particle methodused.

28 4 APPROXIMATION METHODS IN QUANTUM MECHANICS

4 Approximation Methods in Quantum Mechanics

4.1 Time Independant Perturbation Theory (non-degenerate)

4.1.1 Introduction

In general, we have to solve:Hψp = Epψp

Which, in general is very hard to do. Now, suppose we can write the Hamiltonian in terms of someperturbation, and an unperturbed Hamiltonian thus:

H = H0 + V (4.1)

So, suppose the unperturbed Schrodinger equation has been solved:

H0φk = ekφk

(note, we may have different sets of quantum numbers for the full and unperturbed system). Thatis, φk, ek are known. Then, we use our knowledge of H0 to solve for the perturbed system. Now,by linear combination, we know that we can write:

ψp =∑k

cpkφk

Then, we need to find cpk.

4.1.2 Perturbation Expansion

Now, obviously, if there is no perturbation potential; i.e. V = 0; then the ‘perturbed’ solutions arejust the same as the ‘unperturbed’. This is referred to as the zeroth order perturbation. That is, wesee that Ek = ek and ψk = φk. Now, suppose we write:

H(λ) = H0 + λV

Where 0 ≤ λ ≤ 1. So, the case λ = 0 is the zeroth order case; and λ = 1 is the perturbed case. Letus assume that p = k. Now, let us expand the eigenvalues and eigenstates of H(λ):

Ek = E(0)k + λE

(1)k + λ2E

(2)k + λ3E

(3)k + . . . (4.2)

ψk = ψ(0)k + λψ

(1)k + λ2ψ

(2)k + λ3ψ

(3)k + . . . (4.3)

So, we have that the actual eigenvalues and eigenstates can be found by adding successive correc-tions. To find these corrections, let us write the Schrodinger equation:

Hψk = Ekψk

Then, substituting our expansions in:

(H0 + λV )(ψ(0)k + λψ

(1)k + λ2ψ

(2)k + . . .) = (E(0)

k + λE(1)k + λ2E

(2)k + . . .)(ψ(0)

k + λψ(1)k + λ2ψ

(2)k + . . .)(4.4)

4.1 Time Independant Perturbation Theory (non-degenerate) 29

Now, let us expand this, and equate the coefficients of the same power of λ:

H0ψ(0)k = E

(0)k ψ

(0)k (4.5)

H0ψ(1)k + V ψ

(0)k = E

(0)k ψ

(1)k + E

(1)k ψ

(0)k (4.6)

H0ψ(2)k + V ψ

(1)k = E

(0)k ψ

(2)k + E

(1)k ψ

(1)k + E

(2)k ψ

(0)k (4.7)

Where we have written the zeroth, first and second orders. Note, the zeroth order equation is justour unperturbed Schrodinger equation: ψ(0)

k = φk, and E(0)k = ek. Let us consider the first and

second order.

4.1.3 Solution for First Order Perturbation Equation

Now, let us write the first order equation again:

H0ψ(1)k + V ψ

(0)k = E

(0)k ψ

(1)k + E

(1)k ψ

(0)k

And, using our zeroth-order results:

ψ(0)k = φk E

(0)k = ek

We can write it as: (V − E(1)

k

)φk =

(ek − H0

)ψ

(1)k (4.8)

Now, we can write the first-corrections to the eigenstates in terms of a linear combination of theunperturbed states; by the linear completeness principle:

ψ(1)k =

∑j

c(1)kj φj

Hence, we write (4.8) as: (V − E(1)

k

)φk =

∑j

(ek − H0

)c

(1)kj φj

But, we can act with the Hamiltonian, giving:(V − E(1)

k

)φk =

∑j

(ek − ej) c(1)kj φj

Now, we multiply everything by φ∗n, on the left, and integrate:∫φ∗n

(V − E(1)

k

)φk d

3r =∫φ∗n∑j

(ek − ej) c(1)kj φjd

3r

That is: ∫φ∗nV φkd

3r − E(1)k

∫φ∗nφk d

3r =∑j

(ek − ej) c(1)kj

∫φ∗nφjd

3r


Let us use the orthonormality of the eigenstates:∫φ∗nV φkd

3r − E(1)k δnk =

∑j

(ek − ej) c(1)kj δnj

That is:Vnk − E(1)

n δnk = (ek − en)c(1)kn

Where:Vnk ≡

∫φ∗nV φkd

3r

So, relabelling the indices:

Vij − E(1)i δij = (ej − ei)c(1)

ji (4.9)

Looking at the case i = j, then the above is obviously just:

Vii = E(1)i

That is:

E(1)i =

∫φ∗i V φid

3r (4.10)

And, the case i 6= j, we have, from rearranging (4.9):

c(1)ij =

Vjiei − ej

(4.11)

Note, the diagonal elements, cii are taken to be zero. Thus, we have that the energy and state, upto first order corrections, are given by:

Ek ≈ ek + E(1)k = ek + Vkk (4.12)

ψk ≈ φk +∑j 6=k

Vjkek − ej

φj (4.13)

Hence, we see that to determine the first-order energy correction, we only need to know the zeroth-order wavefunction, ψ(0)

k = φk; and not the first-order wavefunction correction. This is generallytrue: to find the n-order energy correction, one needs the n− 1 wavefunction correction.

4.1.4 Solution for Second Order Perturbation Equation

The second order equation is just:

H0ψ(2)k + V ψ

(1)k = E

(0)k ψ

(2)k + E

(1)k ψ

(1)k + E

(2)k ψ

(0)k

Which we may rewrite as:

E(2)k ψ

(0)k = (V − E(1)

k )ψ(1)k + (H0 − E(0)

k )ψ(2)k


Where we now have that E(0)k , ψ

(0)k , E

(1)k , ψ

(1)k are known. Again, we write our second-order correction

to the wavefunction as a linear combination of the unperturbed states:

ψ(2)k =

∑j

c(2)kj φj ψ

(1)k =

∑j

c(1)kj φj

Substituting this in, as well as the unperturbed terms::

E(2)k φk =

∑j

c(1)kj (V − E(1)

k )φj +∑j

c(2)kj (H0 − ek)φj

Actioning our Hamiltonian:

E(2)k φk =

∑j

c(1)kj (V − E(1)

k )φj +∑j

c(2)kj (ej − ek)φj

As before, let us multiply by φ∗i , and integrate:∫d3rφ∗iE

(2)k φk =

∑j

c(1)kj

[∫d3rφ∗i V φj − E

(1)k

∫d3rφ∗iφj

]+∑j

c(2)kj (ej − ek)

∫d3rφ∗iφj

Again, orthonormality leave us with:

E(2)k δik =

∑j

c(1)kj

[Vij − E(1)

k δij

]+∑j

c(2)kj (ej − ek)δij

Using the RHS kronecker:

E(2)k δik =

∑j

c(1)kj

[Vij − E(1)

k δij

]+ c

(2)ki (ei − ek)

Now, let us look at the i = k case (i.e. non-zero LHS). Then, we initially note that ei − ei = 0. So:

E(2)k =

∑j

c(1)kj

[Vkj − E

(1)k δkj

]=

∑j

[c

(1)kj Vkj − c

(1)kj E

(1)k δkj

]=

∑j

[c

(1)kj Vkj − c

(1)jj E

(1)j

]=

∑j

c(1)kj Vkj

We noted that cii = 0; which is a result we used before. Now, we have an expression for c(1)kj , in

(4.11). So, using that, and being careful of the indices, we write the above as:

E(2)k =

∑j 6=k

Vjkek − ej

Vkj


Now, we note that VjkVkj = |Vkj |2. Then:

E(2)k =

∑j 6=k

|Vkj |2

ek − ej

Therefore, the energy, up to second order corrections; is given by:

Ek ≈ E(0)k + E

(1)k + E

(2)k

That is:

Ek ≈ ek + Vkk +∑j 6=k

|Vkj |2

ek − ej

Where:Vkn ≡

∫d3rφ∗kV φn

We have not discussed the second-order correction to the wavefunction though1.

So, to summarise what we have so far, in second order non-degenerate perturbation theory.The total energy is given by:

Ek ≈ ek + Vkk +∑k 6=k′

|Vk′k|2

ek − ek′(4.14)

The wavefunction is given by (only up to first order correction terms):

ψk ≈ φk +∑k 6=k′

Vk′k

ek − ek′φk′ (4.15)

Where:Vk′k = 〈k′|V |k〉 H = H0 + V H0φk = ekφk

4.1.5 Examples

Helium Ground State Energy Here, we have the Hamiltonian H0 = H1+H2, with perturbationpotential:

V =e2

4πε0r12r12 ≡ |r1 − r2|

So, the ground state energy, up to first order corrections, is:

Eg = eg + Vgg

Now, let the wavefunction be that of the unperturbed state:

ψg = ψ(0)g = φ1s(r1)φ1s(r2) 1√

2[ | ↑〉1 | ↓〉2 − | ↓〉1 | ↑〉2]

1This may be found on http://myweb.tiscali.co.uk/jonathanp, under ‘Physics Stuff’, then the document entitled‘Index Notation”.


We have previously discussed how to get to this wavefunction. So, the zeroth-order energy is just:

eg = 〈ψ(0g | H0 |ψ(0)

g 〉

And that is just twice (as 2 particles) the standard hydrogenic result:

eg = 2Z2R∞ = −108.8eV

So, to computethe first order correction, we find:

Vgg = 〈ψ(0g | V |ψ(0)

g 〉

This result is merely quoted to be Vgg = 34eV. The integral is, in general, too hard to do! Hence,the total ground state energy of helium, up to first order corrections is:

Eg = −108.8 + 34eV = −74.8eV

Compare this with the experimentally determined −79eV. So, a pretty good agreement.

Anharmonic Oscillator Consider an anharmonic oscillator; with the following Hamiltonain:

H = H0 + V

H0 = − h2

2m∂2

∂x2+

12mω2x2 V = αx3

Where α is ‘small’. Hence, we can use perturbation theory.

Now, our unperturbed eigenequation has been solved, and looks like:

H0 |n〉 = hω(n+ 12) |n〉 n = 0, 1, 2, . . .

Thus, the zeroth order energy is:en = hω(n+ 1

2)

Now, recall the following relationships, from our previous creation/destruction operators discussion:

a†a |n〉 = n |n〉 a† |n〉 =√n+ 1 |n+ 1〉 a |n〉 =

√n |n− 1〉

From now on, I shall supress the hat on the operator. It should be obvious that it is an operator.Hence, our unperturbed Hamiltonian can be written:

H0 = hω(a†a+ 12)

Also recall our definitions of the operators:

a ≡√mω

2hx+

ip√2mhω

a† =√mω

2hx− ip√

2mhω

Hence, we can combine these, to form:

x = x =

√h

2mω(a+ a†) p =

1i

√mωh

2(a− a†)


Therefore, we can write our perturbation potential in terms of the creation/destruction operators:

V = αx3 = α

(h

2mω

)3/2

(a+ a†)3

Now, let us compute the first order correction to the energy. That is:

Vnn = 〈n| V |n〉

To do this, we express the potential as:

V = β(a+ a†)3 = β(a+ a†)(a+ a†)2 β ≡ α(

h

2mω

)3/2

= β(a+ a†)(a2 + aa† + a†a+ a†2)= β(a+ a†)(a2 + 2a†a+ 1 + a†2)

After we have used the commutator [a, a†] = 1. Thus, in this way, we write the energy correctionas:

Vgg = β〈n| (a+ a†) (a2 + 2a†a+ 1 + a†2) |n〉

Now, we ‘split this up’, into the bit next to the bra-state, and next to the ket-state. So, the ket-statebit:

(a2 + 2a†a+ 1 + a†2) |n〉 = (a2 + 2n+ 1 + a†2) |n〉

As we have used the previously stated relation that a†a |n〉 = n |n〉. Thus, carrying on evaluatingthe above relation, we see that the ‘ket-state bit’ looks like:

(a2 + 2n+ 1 + a†2) |n〉 → ζ1 |n− 2〉+ η1 |n〉+ ξ1 |n+ 2〉

Where ζ1, η1, ξ1 are constants, determined from the creation/destruction relations. Infact, theirvalue is not important in the current discussion, but they will be left in for completeness. Now,consider the ‘bra-state bit’ of the potential:

〈n| (a+ a†)

Now, consider: (a† |n〉

)†=(√n+ 1 |n+ 1〉

)†That is, doing the hermitian conjugating:

〈n| a = 〈n+ 1|√n+ 1

Similarly:〈n| a† = 〈n− 1|

√n

Hence, we see that:〈n| (a+ a†)→ ζ2〈n+ 1| + η2〈n− 1|

Therefore, in total, we see that the first-order correction to the energy is given by:

Vgg = (ζ2〈n+ 1| + η2〈n− 1| )(ζ1 |n− 2〉+ η1 |n〉+ ξ1 |n+ 2〉) = 0


As all states are orthogonal. Thus, the first order correction to the energy is zero Vgg = 0.

Let us determine the second order correction term; and to do so, we must compute:

Vn′n = 〈n′| V |n〉

To do this, we consider:β(a+ a†)(a2 + 2a†a+ 1 + a†2) |n〉

This gives:β(a+ a†)(c1 |n− 2〉+ c2 |n〉+ c3 |n+ 2〉)

This gives:

β(c4 |n− 3〉+ c5 |n− 1〉+ c6 |n+ 1〉+ c7 |n− 1〉+ c8 |n+ 1〉+ c9 |n+ 3〉)

Coefficents of similar states we can group, leaving four states:

V |n〉 → β(c4 |n− 3〉+ c10 |n− 1〉+ c11 |n+ 1〉+ c9 |n+ 3〉)

Thus, the first order correction energy term Vn′n = 〈n′| V |n〉 only has 4 non-zero components:

n′ = n− 3, n− 1, n+ 1, n+ 3

As:〈n′| V |n〉 = β〈n′| [(c4 |n− 3〉+ c10 |n− 1〉+ c11 |n+ 1〉+ c9 |n+ 3〉)]

If the coefficients are calculated, carefully, one finds:

c4 =√n(n− 1)(n− 2)

c10 = 3n3/2

c11 = 3(n+ 1)3/2

c9 =√

(n+ 1)(n+ 2)(n+ 3)

Hence, the second order energy term is given by:∑n′ 6=n

V 2n′n

en − en′=

V 2n−3,n

en − en−3+

V 2n−1,n

en − en−1+

V 2n+1,n

en − en−1+

V 2n+3,n

en − en+3

Where:en = (n+ 1

2)hω

These terms come out to be:

=1

3hωα2

(h

2mω

)3

n(n− 1)(n− 2)

+1hω

α2

(h

2mω

)3

9n3

− 1hω

α2

(h

2mω

)3

9(n+ 1)3

− 13hω

α2

(h

2mω

)3

(n+ 1)(n+ 2)(n+ 3)


Which turns out to be:

− 1hω

α2

(h

2mω

)3

(30n2 + 30n+ 11)

Thus, up to second order corrections, the energy is given by:

En = hω(n+ 12)− 1

hωα2

(h

2mω

)3

(30n2 + 30n+ 11)

With energy intervals, between adjacent states:

∆ ≡ En − En−1 = hω − 60hω

α2

(h

2mω

)3

n

Notice then, we see that the energy levels for the anharmonic oscillator are lower, and get closer-spaced as n increases; whereas the energy levels were all equally spaced for the harmonic oscillator.

Figure 1: The energy levels of the anharmonic oscillator H, relative to those of the harmonicoscillator H0. Notice that they are all lower & get closer, as n increased; whereas for the harmonicoscillator they all maintain the same spacing.

4.2 Degenerate Perturbation Theory

In the previous subsection, we used that the unperturbed states φk were non-degenerate. That is:

en 6= ek ∀n 6= k

Now, if we have an unperturbed Schrodinger equation of the form:

H0φnd = enφnd d = 1, 2, . . . ,M

So that now we see that the energies are independant of one of the quantum numbers of the system.Of course, either n or d can be sets of numbers themselves. Where the full Hamiltonian is the sumof the unperturbed, with the perturbation potential H = H0 + V . We say that d is the degenerateindex.

4.2 Degenerate Perturbation Theory 37

For an example of such a system, consider the Hydrogen atom. Its energies are given purely bythe principle (radial) quantum number n, but the states have angular dependancies:

en = −13.61n2eV ψn`m

So that here, d = (`,m).

So, let us start, using a similar technique as before. Suppose that the total Schrodinger equationis:

Hψnd = Endψnd

Where the perturbation terms lifts the degeneracy. We make the standard expansions of the energy& state:

End = E(0)nd + E

(1)nd + . . .

ψnd = ψ(0)nd + ψ

(1)nd + . . .

So that, to zeroth order, we have:H0ψ

(0)nd = E

(0)nd ψ

(0)nd

Leading us to identify:ψ

(0)nd = φnd E

(0)nd = en

And to first order:V ψ

(0)nd + H0ψ

(1)nd = E

(1)nd ψ

(0)nd + E

(0)nd ψ

(1)nd

So, we expand the first correction terms to the state, in terms of known unperturbed states:

ψ(1)nd =

∑αβ

c(1)nd,αβφαβ

Then, if we take the case n = α, the above may be written as a sum of those terms, and those inn 6= α:

ψ(1)nd =

∑β 6=d

c(1)nd,nβφnβ +O(n 6= α)

Notice that we exclude from the sum the case β = d, as this produces a 1/0 in the evaluation of thec(1); as, from before2:

c(1)nd,αβ =

Vnd,αβ

E(0)nd − E

(0)αβ

Vnd,αβ ≡ 〈nd| V |αβ〉

Now, we have already identified E(0)nd = en, hence, in the above, with n = α, we have:

c(1)nd,nβ =

Vnd,nβ

E(0)nd − E

(0)nβ

=Vnd,nβen − en

=∞

Notice that the energy is degenerate: it dosent care about the ‘second’ subscript. Hence, noticethat the coefficients diverge!

2I reproduce the previous result, but we take n = α here.


So, we have just seen that if we do with degenerate perturbation theory, what we did for non-degenerate, we get a problem. The source of the problem is in the assumption that quantumnumbers before and after perturbations are the same.Hence, let us remove this assumption. Let us say that the quantum numbers before are the set(n, d); and after (n, k).

Now, this is equivalent to a new basis set ; and instead of identifying ψ(0)nd with φnd, we change to

some new basis. So that:

Ωnk ≡M∑d=1

bkdφnd k = 1, . . . ,M (4.16)

Where the coefficients bkd are chosen to avoid divergences.

We have that the unperturbed Schrodinger equation is:

H0φnd = enφnd

So, the perturbation expansion:

Enk = E(0)nk + E

(1)nk + . . .

ψnk = ψ(0)nk + ψ

(1)nk + . . .

To zeroth order:H0ψ

(0)nk = E

(0)nk ψ

(0)nk

So that:E

(0)nk = en ψ

(0)nk = Ωnk =

∑d

bkdφnd

And to first order:(V − E(1)

nk )ψ(0)nk = (E(0)

nk − H0)ψ(1)nk

Where we now say that the first correction term to the wavefunction can be found in a linearcombination of the new basis:

ψ(1)nk =

∑α

ckαΩnα

So, using this, and that ψ(0)nk = Ωnk as well as E(0)

nk = en , we get:

(V − E(1)nk )Ωnk = (en − H0)

∑α

ckαΩnα

Let us multiply this whole expression by Ω∗nβ, and integrate:∫d3rΩ∗nβ(V − E(1)

nk )Ωnk =∫d3rΩ∗nβ(en − H0)

∑α

ckαΩnα

=∫d3rΩ∗nβ(en − en)

∑α

ckαΩnα

= 0

4.2 Degenerate Perturbation Theory 39

After we have used that H0Ωnα = enΩnα. Hence:∫d3rΩ∗nβ(V − E(1)

nk )Ωnk = 0

Now, noting orthogonality, and using some standard notation:∫d3rΩ∗nβ(V − E(1)

nk )Ωnk = Vnβ,nk − E(1)nk δβk

Which is obviously:

Vnβ,nk − E(1)nk δβk = 0 (4.17)

Now, let us consider two cases in (4.17):

• β = k. Then, this results in:

E(1)nk = Vnk,nk (4.18)

• β 6= k. Then, this gives:

Vnβ,nk =∫d3rΩ∗nβV Ωnk = 0 ∀β 6= k (4.19)

Thus, from these two cases, we see that we choose the new basis Ωnk so that V is diagonalised. Thediagonal elements are first order corrections to the energy (i.e. E(1)

nk ).

Example: Spin-Orbit Coupling This is responsible for the fine structure of atomic spectra.Let we consider the (unperturbed) Hamiltonian, with Coulomb potential, of a Hydrogen-like atom:

H0 = − h2

2m∇2 −

Ze2

4πε0

Then, we we know, the states and energies are given by:

φn`m = Rn`(r)Y`m(θ, φ)χ(σ)→ |n`m`sms〉 en = −13.6Z2 1n2eV

Thus, this is the unperturbed problem, which has been solved. Notice: the energy is only dependentupon n, and thus the degeneracy indices are (`,m).Now, one of the relativistic corrections, is to add on a perturbation term:

VLS = A(r)L · S

So that the total Hamiltonian is H = H0 + VLS . Rewriting the perturbation term, in the standardway:

VLS = 12A(r)(J2 − L2 − S2) J ≡ L+ S


Now, as we previously said, the original basis is |n`m`sms〉, and we want to choose some new basisthat will diagonalise the VLS term. Now, by inspection, we shall choose |n`sjmj〉. Hence, the firstorder energy correction term is:

E(1)LS = 〈n`sjmj | VLS |n`sjmj〉

=h2

2(j(j + 1)− `(`+ 1)− 3

4) 〈A(r)〉n`

Where we have used that s = 1/2. Also:

〈A(r)〉n` =∫r2drA(r)R2

n`(r) j = `+ 12 , `−

12

4.3 Time Dependent Perturbation Theory

Quantum systems respond to external fields. So, we want to determine probabilities for the systemmaking transitions to other states. By way of setting up the situation, we shall consider Newton’sequation, and how it comes about from quantum mechanics.

4.3.1 Newton’s Equation

Consider the time dependent Schrodinger equation:

ih∂

∂tψ(r, t) = Hψ(r, t) (4.20)

Let us take the Hermitian conjugate of this (remembering that H† = H):

− ih ∂∂tψ∗(r, t) = Hψ∗(r, t) (4.21)

Now, for some general operator, we have its expectation value:⟨A⟩

= 〈ψ| A |ψ〉 ≡∫d3r ψ∗Aψ

Where we have assumed that the states are normalised. Now, the rate of change of this expectationvalue is obviously:

∂

∂t

⟨A⟩

=∫d3r

∂

∂t

[ψ∗(r, t)Aψ(r, t)

]Which we can differentiate by parts:

∂

∂t

⟨A⟩

=∫d3r

[ψ∗(r, t)A

∂ψ(r, t)∂t

+ ψ∗(r, t)∂A

∂tψ(r, t) +

∂ψ∗(r, t)∂t

Aψ(r, t)

]Now, we have expressions for the differentials of the wavefunctions, namely (4.20) and (4.21) forthe function and its conjugate, respectively. Hence, being careful, we end up with:

∂

∂t

⟨A⟩

=∫d3r ψ∗(r, t)

∂A

∂tψ(r, t) +

1ih

∫d3r ψ∗(r, t)(AH − HA)ψ(r, t)

4.3 Time Dependent Perturbation Theory 41

Which is just:

∂

∂t

⟨A⟩

=

⟨∂A

∂t

⟩+

1ih

⟨[A, H

]⟩(4.22)

Hence, we have a very general expression for the rate of change of the expectation value (which,remember, this bit we are able to measure) of an operator.

So, for Newton, we have:dp

dt= F

Where we have that p = −ih∇, as being independent of t. Hence:

d

dt〈p〉 =

1ih

⟨[p, H

]⟩And, taking:

H =p2

2m+ V (r)

We see the commutator is just the term:[p, H

]=[p, V

]= −ih

[∇, V

]And: [

∇, V]ξ = ∇(V ξ)− V (∇ξ) = V∇ξ + ξ∇V − V∇ξ = ξ∇V

Hence:d

dt〈p〉 = −

⟨∇V

⟩Which is something we knew already!

4.3.2 Time Dependant Expansion

Suppose we have:H = H0 + V (t)

Where the unperturbed Hamiltonian is time independent. We shall also assume that the followinghas been solved:

H0φn(r) = Enφn(r)

And that we need to solve:

ih∂

∂tψ(r, t) = Hψ(r, t) (4.23)

Now:ψ(r, t) =

∑n

dn(t)φn(r)


Which of course has time dependence:

ψ(r, t) =∑n

cn(t)e−iEnt/hφn(r)

Let us substitute this into the Schrodinger equation (4.23):

ih∂

∂t

(∑n

cne−iEnt/hφn(r)

)=∑n

(H0 + V )cne−iEnt/hφn(r)

Where we can differentiate the LHS by parts, and action the unperturbed Hamiltonian on the RHSunperturbed state φ (which is time independent):

ih∑n

φn(r)(dcndt− iEn

hcn

)e−iEnt/h =

∑n

(En + V )cne−iEnt/hφn(r)

Cleaning up the LHS:∑n

(Encn + ih

dcndt

)φn(r)e−iEnt/h =

∑n

(En + V )cne−iEnt/hφn(r)

Notice that the Encn term cancels from both sides:

ih∑n

dcndtφn(r)e−iEnt/h =

∑n

V cne−iEnt/hφn(r)

Now, multiply both sides by φ∗m, and integrate:

ih∑n

dcndte−iEnt/h

∫φ∗mφn d

3r =∑n

cne−iEnt/h

∫φ∗mV φn d

3r

Using orthonormality, and standard notation:

ih∑n

dcndte−iEnt/hδnm =

∑n

cnVmne−iEnt/h

Where the sum on the LHS obviously will filter out only one non-zero term; thus:

ihdcmdt

e−iEmt/h =∑n

cnVmne−iEnt/h

Hence, using one more piece of notation, we have:

ihdcndt

=∑m

Vnmcmeiωnmt (4.24)

ωnm ≡ En − Emh

(4.25)

Vnm ≡∫φ∗nV φm d

3r (4.26)

Where, just to be explicit, φ(r) only, and V (t), c(t); functions of time.


4.3.3 Perturbation Expansion Theory

So, writing our Hamiltonian:H(t) = H0 + λV (t)

And a series expansion of our coefficents:

cn(t) = c(0)n + λc(1)

n + λ2c(2)n + . . .

Althogh not explicitly stated, all c(j)i are functions of time. Hence, we inserting this into our

expression for the rate of change of the coefficients:

ihdcndt

=∑m

Vnmcmeiωnmt

⇒ ihd

dt

[c(0n + λc(1)

n + . . .]

=∑m

λVnm(c(0m + λc(1)

m + . . .)eiωnmt

Notice that we have inserted our previously ignored book-keeping factors.

Then, to zeroth order, we have:

ihd

dtc(0)n = 0

Hence, c(0)n is independent of t.

To first order, we have:

ihd

dtc(1)n =

∑m

Vnmc(0)m eiωnmt (4.27)

To second order:ihd

dtc(2)n =

∑m

Vnmc(1)m eiωnmt

We will only consider the expansion up to first order. That is, we will only determine c(0)n , c

(1)n .

We proceed by using initial conditions:

• Let us apply the perturbation only after some time t > 0:

V (t) =

0 t ≤ 06= 0 t > 0

• Also, at t ≤ 0, suppose the system is in an eigenstate of H0, say |1〉. So that ψ(r, t = 0) =φ1(r). Hence, in the series expansion of coefficients, for t ≤ 0, we will only have non zerovalues for n = 1. We can write this as:

cn(t) = δn1 t ≤ 0

Therefore, as it is time-independant, we have:

c(0)n = δn1


Hence, using this, let us rewrite the first order expansion expression (4.27):

ihd

dtc(1)n =

∑m

Vnmδ1meiωnmt

= Vn1eiωn1t

Let us consider the case n = 1. Thus:

ihd

dtc

(1)1 = V11e

iω11t

Let us integrate this:

ih

∫ t

0dc

(1)1 =

∫ t

0V11(t′)eiω11t′dt′

Integrating the LHS:

ih

∫ t

0dc

(1)1 = ih[c(1)

1 (t)− c(1)1 (t = 0)]

Notice, on the far right, we have a c(1)n being evaluated at t ≤ 0. We have an expression for this,

from our initial conditions: c(1)1 (t = 0) = 1. Hence, using this:

ih[c(1)1 (t)− 1] =

∫ t

0V11(t′)eiω11t′dt′

Also, recall the definition of ωnm: it will be zero for n = m. Hence:

ih[c(1)1 (t)− 1] =

∫ t

0V11(t′)dt′

Thus:

c(1)1 = 1 +

1ih

∫ t

0V11(t′)dt′ (4.28)

And, for n 6= 1, we see that we have:

c(1)n =

1ih

∫ t

0Vn1(t′)eiωn1t′dt′ n 6= 1 (4.29)

Hence, we have that cn is a probability amplitude, and |cn|2 the actual probability. For example,|c(1)

1 |2 is the probability for the system to remain in its initial state (up to a first order approximation,of course). For n > 1, we have that |c(1)

n |2 is the probability for the system to make the transitionfrom the initial state |1〉 to the state |n〉.

4.3.4 Example: Hydrogen hit by Electric Field

Let us consider a hydrogen atom, being hit by the pulse of an electric field. Then, let the field be:

ε =

0 t ≤ 0ε0e−t/τ t > 0


Where τ > 0, and can be thought of as the severity of the field: big τ means that the field wont stay‘big’ for too long. Now, assume that initially the atom was in its ground state 1s. Let us determinewhat the probability is for the atom to be in the 2p state after a sufficiently long time. Now, wemust start with expressions for the ground state, first excited state wavefunctions of hydrogen, aswell as their energies:

φ1s(r) =1√πa3

0

e−r/a0 φ2p(r) =1√

32πa50

r cos θe−r/2a0

E1s = − e2

8πε0a0E2p =

E1s

4a0 =

4πε0h2

me2

Hence, we have that the probability for a transition from the initial state (ground state) to someother state (the 2p state) as being:

c(1)1s→2p =

1ih

∫ t

0V2p,1se

iω2p,1st′dt′

So:ω2p,1s =

E2p − E1s

h= − 3

4hE1s

Now, we must be careful: the atom is hit by an electric field. Hence, the associated potential is justcharge multiplied by the distance away:

V (t) = eε · r = eεz = eε0e−t/τr cos θ

Now to find the perturbation matrix-element expression:

V2p,1s(t) = 〈2p| V |1s〉 = eε0e−t/τ 〈2p| r cos θ |1s〉

The expression of which is pretty lengthy:

V2p,1s(t) = eε0e−t/τ

∫ ∫ ∫1√

32πa50

r cos θe−r/2a0 · r cos θ · 1√πa3

0

e−r/a0 · r2 sin θdrdθdφ

Let us bring together all the constants, and separate different integrals:

V2p,1s(t) = eε0e−t/τ 1

4πa40

√2

(∫ ∞0

r4e3r/a0dr

)(∫ π

0cos2 θ sin θdθ

)(∫ 2π

0dφ

)We can do the first integral by comparing with a standard integral, the solution to which we shallquote. We can do the other two intgegrals easily enough:

V2p,1s(t) = eε0e−t/τ 1

4πa40

√2

4!25a50

35· 3

5· 2π = ξe−t/τ

Where we have put everything obvious into a single constant ξ. Therefore:

c(1)1s→2p =

ξ

ih

∫ t

0e−t

′/τeiω2p,1st′dt′


The integral is fairly easy to evaluate:

c(1)1s→2p =

ξ

ih

et(iω2p,1s− 1τ

) − 1iω2p,1s − 1

τ

Therefore, to compute the transition probability, we must compute the square of the above expres-sion |c|2 = c∗c. Doing so, carefully (and supressing the subscripts on the ω):

|c(1)1s→2p|

2 =(

ξ

−ih

)(e−iωte−

1τ − 1

−iω − 1τ

)·(ξ

ih

)(eiωte−

1τ − 1

iω − 1τ

)

=ξ2

h2

(e−2t/τ − e−itωe−t/τ − eitωe−t/τ + 1

ω2 + 1/τ2

)

=ξ2

h2

(1 + e−2t/τ − 2e−t/τ cos(ωt)

ω2 + 1/τ2

)

Now, in the initial specification of the problem, we had the line after a sufficiently long time. Hence,taking t→∞ in the above expression, we simply get:

|c(1)1s→2p|

2 =ξ2

h2

1ω2 + 1/τ2

4.3.5 Fermi’s Golden Rule

Consider the following perturbation potential:

V (t) =

0 t ≤ 0V t > 0

That is, a step function, where V is independent of time. Let us consider the transition probabilityper unit time, for the system initially in state |1〉 at t ≤ 0, to end up in the state |2〉 at t > 0.

So, the coefficient we want is:

c(1)2←1(t) =

1ih

∫ t

0V21(t′)eiω2,1t′dt′

And, as the perturbation is independent of time, this easily becomes:

c(1)2←1(t) =

V21

ih

∫ t

0eiω2,1t′dt′

=V21

ih

1iω

(eiωt − 1)

=V21

hω(1− eiωt)

4.4 Variational Method 47

Hence, the probability for such a transition:

P2←1(t) = |c(1)2←1(t)|2

=|V21|2

h2ω2(1− e−iωt)(1− eiωt)

=|V21|2

h2ω22(1− cosωt)

=4|V21|2

h2ω2sin2(ωt/2)

Putting the subscript back on:

P2←1(t) =4|V21|2

h2ω221

sin2(ω21t/2) V21 = 〈2| V |1〉 (4.30)

Now, let us write:

D(ω, t) ≡ 2tπ

sin2(ωt/2)ω2

(4.31)

Then we see that we can write the probability as:

P2←1(t) =2πt|V21|2

h2 D(ω, t)

Now: ∫ ∞∞

D(x, t)dx =2tπ

∫ ∞−∞

sin2(xt/2)x2

dx = 1

That is:limt→∞

D(x, t) = δ(x)

Hence, remembering that:x ≡ ω = ω21 = E2 − E1

Therefore:

P2←1(t) =2πt|V21|2

hδ(E2 − E1)

Then, the probability, per unit time, is:

P2←1(t) =2π|V21|2

hδ(E2 − E1) (4.32)

4.4 Variational Method

The above perturbation theory was purely mechanical: there is a set method, and if we stick to it,we get an answer out. We cannot say the same thing of the variational method: we may call it an‘art’.


4.4.1 Variational Principle

The expectation value of a given Hamiltonian in any state ψ is always greater or equal to the exactground state energy E0: ⟨

H⟩

=〈ψ| H |ψ〉〈ψ|ψ〉

≥ E0

It is equal if the exact state ψ has been found. Let us proove it:

Suppose:H |φn〉 = En |φn〉 n = 0, 1, 2, . . .

Where we will have that the ground state corresponds to n = 0. Therefore E0 < En, if n > 0. Now,any state can be written as a linear combination of eigenstates:

|ψ〉 =∞∑n=0

cn |φn〉

Let us extract the n = 0 from the sum:

|ψ〉 = c0 |φ0〉+∞∑n=1

cn |φn〉

And acting our Hamiltonian on it:

H |ψ〉 = c0H |φ0〉+∞∑n=1

cnH |φn〉

= c0E0 |φ0〉+∞∑n=1

cnEn |φn〉

Then let us compute the expectation value of the energy⟨H⟩

= 〈ψ| H |ψ〉:

〈ψ| H |ψ〉 =

(c0〈φ0| +

∞∑n=1

cn〈φn|

)(c∗0E0 |φ0〉+

∞∑n=1

c∗nEn |φn〉

)

Which, by orthogonality 〈φn|φm〉 = δnm results in:

〈ψ| H |ψ〉 = |c0|2E0 +∞∑n=1

|cn|2En

Now, as En > E0, n > 0, we therefore have:

〈ψ| H |ψ〉 = |c0|2E0 +∞∑n=1

|cn|2En > |c0|2E0 + E0

∞∑n=1

|cn|2

That is:

〈ψ| H |ψ〉 > E0

(|c0|2 +

∞∑n=1

|cn|2)


It is evident that the lone |c0|2 may be absorbed into then summation, changing the lower limitaccordingly. Now, notice:

|ψ〉 =∞∑n=0

cn |φn〉 ⇒ 〈ψ|ψ〉 =∞∑n=0

|cn|2

And therefore:〈ψ| H |ψ〉 > E0〈ψ|ψ〉

And that gives us the pre-stated variational precept:⟨H⟩

=〈ψ| H |ψ〉〈ψ|ψ〉

≥ E0

Now, in general, we dont know E0. So, in the variational method, using a trial wavefunction willgive a value of

⟨H⟩

which will always be bigger than the ground state energy. However, as the

trial wavefunction gets closer to the actual wavefunction,⟨H⟩

will decrease untill the actual value

is reached. Hence, the lower the value of⟨H⟩

gives a better result, which is ‘truer’ than a highervalue.

The procedure:

• Choose a trial wavefunction ψ, with some variational parameters:

ψ = ψ(α1, α2, . . .)

• The evaluate the expectation value:

E(α1, α2, . . .) =〈ψ| H |ψ〉〈ψ|ψ〉

• Then to minimise the energy, find the variational derivatives:

∂E

∂α1= 0

∂E

∂α2= 0 . . .

And solve them for the variational parameters αi. These are then put back into E(α1, α2, . . .).

We shall proceed by examples.

4.4.2 Example: Hydrogen Atom

Let us find the ground state of hydrogen, with a trial wavefunction of the form:

ψ = Ce−αr C =

√α3

π


Where the normalisation constant has been found & written. So here, our variational parameter isa single α. We have the standard Hamiltonian:

H = − h2

2m∇2 − e2

4πε0r∇2 =

1r2

∂

∂r

(r2 ∂

∂r

)+ Θ

Where Θ is the angular part. It will not play a role here, due to the form of the wavefunction. Weget the expectation value of the Hamiltonian:

Eα =⟨H⟩

=h2α2

2m− e2α

4πε0

Differentiating this, to minimise the variational parameter:

∂

∂αEα =

h2α

m− e2

4πε0= 0

Solving trivially results in:

α =me2

4πε0h2 =

1a0

Where we have noticed the Bohr radius expression fall out. And therefore, using this α in Eα:

Eα = − 18π

e2

4πε0a0

Which is indeed the ground state of hydrogen, exactly. It is only exact as we ‘guessed’ the exactform of the ground state wavefunction.

4.4.3 Example: Helium Atom

We have the Hamiltonian:

H = − h2

2m∇2

1 −h2

2m∇2

2 −Ze2

4πε0

(1r1

+1r2

)+

e2

4πε0r12

And wavefunction:

ψ(1, 2) = φ1s(r1)φ1s(r2)χ12 φ1s(r) =

√Z3

πa30

e−Zr/a0

We have previously worked all this out:

χ12 =1√2

( | ↑〉1 | ↓〉2 − | ↓〉1 | ↑〉2)

And: ⟨H⟩

= 2⟨− h2

2m∇2 − Ze2

4πε0

1r

⟩+

e2

4πε0

⟨1r12

⟩


Which we have seen gives: ⟨H⟩

= (−2Z2 + 54Z)R∞ R∞ ≡

e2

8πε0a0

Plugging numbers in, Z = 2: ⟨H⟩

= Eg = −5.5R∞

Compare this with the experimental Eg = −5.81R∞: it is too high. So, we shall use the variationalmethod to improve this ground state energy, which should be lower. In the method above we usedwavefunctions with no free parameters. We shall now ‘vary’ the charge part of the wavefunction,as our trial:

φ1s(r) =

√Z ′3

πa30

e−Z′r/a0

So that our variational parameter is Z ′. Note, we shall still use the charge Z, for the nuclear charge,in the Hamiltonian. This time we get:⟨

H⟩Z′

= 2(Z ′2e2

8πε0a0− ZZ ′

4πε0a0

)+

54Z ′e2

8πε0a0

= 2(Z ′2 − 2ZZ ′ + 5

8Z′)R∞

To find the value of Z ′ that will minimise this, we must differentiate, and put equal to zero:

∂

∂Z ′

⟨H⟩Z′

= 0 ⇒ 2Z ′ − 2Z + 58 = 0

That is:Z ′ = Z − 5

16So the ground state energy:

Eg = 2(Z ′2 − 2ZZ ′ + 5

8Z′)R∞ = −2

(Z − 5

16

)2

R∞

Thus, using Z = 2, this results in: ⟨H⟩Z′

= −5.7R∞

Now, compare this with both⟨H⟩Z

= −5.5R∞ and the experimental -5.81R∞. Hence, our varia-tional parameter has given us a value closer to experimental data.

Now, the physical meaning of the variational Z ′ is that of a screening effect of the nuclear charge,due to the other electrons around.

52 5 INTRODUCTION TO MOLECULAR QUANTUM MECHANICS

5 Introduction to Molecular Quantum Mechanics

5.1 Born-Oppenheimer Approximation

Consider a general Hamiltonian:H = He + Hn + Hne

Where He is the Hamiltonian of the electrons in the molecule, Hn for the nuclei, and Hne theinteraction between the electron & nuclei.

We shall use x = (x1, x2, . . .) as the coordinates of electrons (where xi = (r, σ), both spatial &spin); and X = (X1, X2, . . .) for nuclei. The total wavefunction of the molecule is separable, thus:

Ψ(x,X) = ψe(x,X)ψn(X)

So that:HΨ(x,X) = EtΨ(x,X)

Now, in general: [Hn, ψe

]6= 0

Because Hn contains a KE term for the nucleus. However, the non-commuting part is ∝ 1/m; andthe mass of the nucleus is 2000times that of an electron. Thus, we ignore the non-commutivity:

Hnψe(x,X) = ψe(x,X)Hn

We have separable Schrodinger equations:(He + Hne

)ψe(x,X) = E(X)ψe(x,X)

Where E = E(X) is the eigenvalue for electrons motion at fixed nuclear configuration. We alsohave: (

Hn + E(X))ψn(X) = Etψn(X)

Where E(X) is the potential for the nuclear motion. This is called the Born-Oppenheimer approx-imation (BO approx).

Recall that for atomic spectral states are classified by angular momentum in 2S+1LJ . Where allterms have their usual meaning; total spin, total orbital angular momentum & total angular mo-mentum reading from left to right. We classify states like this because in atoms, angular momentumis conserved.

But, to diatomic molecules, angular momentum is not conserved. Only the component about theaxis connecting the two nuclei.

We shall choose the symbol Λ for as the orbital angular momentum quantum number componentalong the axis. So that we have molecular spectral term:

2S+1Λ

5.2 Molecular Orbitals 53

Where S is still the total spin. Just as we used s, p, d to denote L = 0, 1, 2; we use Λ = 0, 1, 2 asΣ,Π,∆. We also use u, g to denote parity being odd, even.Parity symmetry, for the Hamiltonian, is where everything remains the same as the electrons coor-dinate change sign. So, we have a more general molecular term:

2S+1Λu,g

5.2 Molecular Orbitals

The Born-Oppenheimer approximation of the previous section let us solve the Schrodinger equationfor electrons in a fixed nuclear configuration.

Let us approximate the nuclear wavefunction then; as a linear combination of atomic groundstates,associated with each nucleus:

ψ =∑i

ciφi (5.1)

Where φi is the atomic ground state at nucleus ‘i’. The coefficients ci are a variational parameter.

We let the Hamiltonian H be the electron part of the total molecular Hamiltonian (as the nuclearpart is fixed by the BO). We define a couple of matrix elements:

Hij ≡ 〈φi| H |φj〉 Sij ≡ 〈φi|φj〉 (5.2)

That is, the Hamiltonian matrix, and an overlap matrix. So, the energy expectation value of themolecule is:

E =⟨H⟩

=〈ψ| H |ψ〉〈ψ|ψ〉

That is, using our approximated wavefunction as a linear combination of atomic ground states:

〈ψ| H |ψ〉〈ψ|ψ〉

=〈ciφi| H |cjφj〉〈ciφi|cjφj〉

=cicj〈φi| H |φj〉cicj〈φi|φj〉

=cicjHij

cicjSij

Where, putting the summation signs back in, this is:

E =⟨H⟩

=

∑i.j cicjHij∑i,j cicjSij

So, our variational equations require:∂E

∂ck= 0


So, let us differentiate, carefully, our expression:

∂E

∂ck=

∂

∂ck

(cicjHij

cicjSij

)=

1cicjSij

∂

∂ck(cicjHij) + cicjHij

∂

∂ck

(1

cicjSij

)This far, we have done nothing except use the product rule, to separate out the numerator fromthe denominator. Let us differentiate the first term, ignoring the dividing factor for now; by theproduct rule:

∂

∂ck(cicjHij) = Hijci

∂cj∂ck

+Hijcj∂ci∂ck

= Hijciδjk +Hijcjδik

= Hikci +Hkjcj

And differentiating the second term:

∂

∂ck

(1

cicjSij

)= − 1

(cicjSij)2

∂

∂ck(cicjSij)

= − 1(cicjSij)2

(Sijciδjk + Sijcjδik)

= − 1(cicjSij)2

(Sikci + Skjcj)

And therefore, our variational equation:

∂E

∂ck=Hikci +Hkjcj

cicjSij−cicjHij(Sikci + Skjcj)

(cicjSij)2

If we recall:E =

cicjHij

cicjSij

And we notice this term as present in our variational equation, in the last term. Thus:

∂E

∂ck=Hikci +Hkjcj

cicjSij−E(Sikci + Skjcj)

(cicjSij)

And we can tidy this expression up, bringing it under a common denominator:

∂E

∂ck=

Hikci +Hkjcj − ESikci − ESkjcjcicjSij

=cj(Hkj − ESkj) + ci(Hik − ESik)

cicjSij

Note that there is a ‘k’ on the LHS, and so the summations on the RHS are over i & j only. Andof course, this is equal to zero, from the variational principle:

cj(Hkj − ESkj) + ci(Hik − ESik)cicjSij

= 0

5.3 Hydrogen Molecular Ion H+2 55

Therefore, each term in the numerator vanishes. Picking the second, and putting the summationsign back in: ∑

i

ci(Hik − ESik) = 0 (5.3)

As this is a linearly independant equation, we must have that the secular determinant is zero:

det |Hik − ESik| = 0 (5.4)

This forms a way of finding the coefficient & wavefunctions & the energies of molecular state.

5.3 Hydrogen Molecular Ion H+2

Consider a molecule of ionised hydrogen (i.e. with two protons, but one electron). The two protonsare fixed (by the BO approximation), with the electron free to move. Let the protons be at ±R/2from the origin, and the electron at r. Then, the Hamiltonian is the following:

H = − h2

2m∇2r −

e2

4πε0

1|r −R/2|

− e2

4πε0

1|r +R/2|

+e2

4πε0R

Reading from left to right: the kinetic energy of the electron the attraction between the electron& one of the protons, attraction between the electron & the other proton, and finally the repulsionbetween the two protons.

Now, these terms give rise to a total energy, with a minimum position, at R0. Let us compute theground state energy.

The total wavefunction of the molecule is the sum over the two atomic hydrogen ground statewavefunctions:

ψ = c1φ1s1 + c2φ

1s2

Where:φ1s

1 =1√πa3

0

e−|r−R/2|/a0 φ1s2 =

1√πa3

0

e−|r+R/2|/a0

So, the overlap integral:S ≡ S12 = S21 = 〈φ1|φ2〉

Notice that the individual ground state wavefunctions are normalised, so that S11 = S22 = 1. Alsonotice that due to the symmetry of the atomic wavefunctions (i.e. that the two atoms are the same),the off-diagonal elements are the same, for the overlap integral. We shall define the following:

α ≡ H11 = H22 = 〈φ1| H |φ1〉

Again, it is the case that the diagonal components of the Hamiltonian matrix are the same, due tothe fact that the two atoms are idential. The off-diagonal elements are denoted:

β ≡ H12 = H21 = 〈φ1| H |φ2〉


So, the secular equation: ∣∣∣∣ H11 − ES11 H12 − ES12

H21 − ES21 H22 − ES22

∣∣∣∣ = 0

That is, using our definitions: ∣∣∣∣ α− E β − ESβ − ES α− E

∣∣∣∣ = 0

We are able to find:E± =

α± β1± S

And the corresponding coefficients, for both E+, E−, respectively:

c+ ≡ c1 = c2 =1√

2(1 + S)

c− ≡ c1 = −c2 =1√

2(1− S)

Giving corresponding wavefunctions:

1σg ≡ ψ+ =1√

2(1 + S)(φ1 + φ2) E+ =

α+ β

1 + S

1σu ≡ ψ− =1√

2(1− S)(φ1 − φ2) E+ =

α− β1− S

We have also written the symbols for each state. The ψ+ is of even parity, and ψ− odd parity.

We can calculate the minimum position R0 = 1.3A, and dissociation energy ED ≡ E1s − E+ as170kJ/mol. Compare this with experimental data of Rexpt0 = 10.6Aand EexptD = 251kJ/mol.

5.4 Structures of Homonuclear Diatomic Molecules

Consider now a full H2 molecule: two protons & two electrons. Here, we will have two electrons atr1, r2 and two protons at fixed ±R/2. Hence, Hamiltonian:

H = − h2

2m∇2i −

e2

4πε0

1|ri −R/2|

− e2

4πε0

1|ri +R/2|

+e2

4πε0r12+

e2

4πε0R

Where i must run over both 1 & 2, to pick up both electrons. Again, reading left to right: thekinetic energy & attraction terms of the electrons with the first and second proton. With the finaltwo terms being the repulsion of the electrons & repulsion between the two protons.

A molecular wavefunction will be:

ψ(r1, r2, R) = ψ+(r1)ψ+(r2)χ12

Where we must have that ψ is anti-symmetric; so as ψ+ are symmetric, we must have the spin partbeing anti-symmetric:

χ12 = 1√2( | ↑〉1 | ↓〉2 − | ↓〉1 | ↑〉2)

5.4 Structures of Homonuclear Diatomic Molecules 57

We merely quote the results: R0 = 0.74A& ED = 350kJ/mol. The experimental values are R0 =0.742A& ED = 432kJ/mol.

Now, thus far, we have quantitatively discussed the ‘mixing’ of the 1s orbitals in molecular Hydro-gen. We had that:

ψ+ = c+(φa + φB) ψ− = c−(φa − φb)

Where φi are the 1s atomic orbitals, at nucleus i. We have that ψ+ is the bound σ orbital, and ψ−the unbound σ∗ orbital.

Now, we can also construct other molecular orbitals from other atomic orbitals. We can alsoconstruct 2s & 2pz bound & unbound orbitals (being denoted 2sσ & 2pσ for bound, and 2sσ∗ &2pσ∗ for unbound). Now, the symmetry of the z-components mean that the overlap has cylindricalsymmetry; so that the overlap of the wavefunctions is just at the ends.We can also construct π orbitals, which correspond to both px & py states overlapping.

So, combining these, the electronic configuration, for diatomic molecules:

(1sσg)2(1sσ∗u)2(2sσg)2(2sσ∗u)2(2pπu)4(2pσg)2(2pπ∗g)4(2pσ∗u)2

So, we see that the 2pσ/π seem to mix in order; and noting that a π orbital can contain 4 electrons(as it contains both x, y directions), whereas a σ only 2. In effect, each direction can contain only2 electrons.We shall talk about bonding order, where each direction has capability of 1 ‘bond’. Thus, a σ bondhas 1 bonds, and a π state 2 bonds. Conversly, a σ∗ has -1 bonds; and π -2 bonds. The bondingorder is the sum of these bond numbers. If the bonding number is greater than zero, then themolecule is bound; and if zero or less, unbound.

Let us consider an example.

Molecular Nitrogen N2 has 14 electrons, so, filling these up; its ground state electronic configura-tion:

(1sσg)2(1sσ∗u)2(2sσg)2(2sσ∗u)2(2pπu)4(2pσg)2

Its molecular ground state term is:1Σg

This has been decided by noting that the spin (the upper-left 1) is zero (as 2 electrons in the valentshell, combining to give ↑↓= 0 ⇒ 2s + 1 = 1). Its bonding order is 1 − 1 + 1 − 1 + 2 + 1 = 3.Hence, the Nitrogen molecule is bound, with triple bonds.

As another example, consider O2, which has 16 electrons. Hence, its electronic configuration:

(1sσg)2(1sσ∗u)2(2sσg)2(2sσ∗u)2(2pπu)4(2pσg)2(2pπ∗g)2

Its bonding order is 1 − 1 + 1 − 1 + 2 + 2 − 2 = 2. Hence, molecular oxygen is bound, with twobonds.

As a final example, consider H−2 ; molecular hydrogen, with an extra electron. Hence, its configu-ration:

(1sσg)2(1sσ∗u)1

Therefore, its bonding order is 1− 1/2 = 1/2. Therefore, bound, with half-bonding.

58 A WORKED EXAMPLES

A Worked Examples

A.1 Number of Symmetric or Anti-symmetric States

Consider a quantum system of two identical particles. In the independent particle approximation,given n single particle states, prove that there are 1

2n(n − 1) possible anti-symmetric states (i.e.fermionic) and 1

2n(n+ 1) possible symmetric (i.e. bosonic) states.

We shall prove this by induction; but we shall start by constructing a few states.

Suppose we have n = 2. Then, the (only) possible anti-symmetric states are:

ψa(r1, r2) = 1√2(φ1(r1)φ2(r2)− φ2(r1)φ1(r2))

The possible symmetric states are:

ψs(r1, r2) = 1√2(φ1(r1)φ2(r2)− φ2(r1)φ1(r2))

= φ1(r1)φ1(r2)= φ2(r1)φ2(r2)

These are just computed by inspection. Now, let us do this for n = 3. To do so, we shall use someinteresting representation/notation.

Figure 2: The possible states for a system of 3 particles can be seen by the number of ‘lines’ on therelevant diagram. (a) is for anti-symmetric, and (b) for symmetric. Note, the symmetric case has‘nodes’, whereby the lines can have the same start & end point.

Figure 3: The possible number of states for a system of 4 particles. Again, (a) is for anti-symmetric,and (b) symmetric. Notice that the number of states/lines is consistent with the theorem we aretrying to prove.

A.1 Number of Symmetric or Anti-symmetric States 59

So, we see that for both n = 3, 4, the theorem is consistent.

Suppose the theorem is true for n. That is (denoting the number of states by N):

Nna = 1

2n(n− 1)Nns = 1

2n(n+ 1)

So, if we show that it works for (n+ 1), then we have proved the theorem. That is, we must have:

Nn+1a = 1

2(n+ 1)(n+ 1− 1)Nn+1s = 1

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

So, for an additional state, we can construct n additional anti-symmetric states; and n+ 1 (1 fromnode) symmetric states. Thus, inserting these into Nn

i shows that Nn+1i are the consistent.

Thus proven.

Quantum Mechanics of Atoms & Molecules · Quantum Mechanics of Atoms & Molecules J.Pearson July 24,...

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Transcript of Quantum Mechanics of Atoms & Molecules · Quantum Mechanics of Atoms & Molecules J.Pearson July 24,...