7 Perturbation Theory I: Time Independent Case

We’ve now come about as far as we can (in this course) relying purely on symmetry

principles. The dynamics of systems of genuine physical interest is rarely simple enough to

hope that we can solve it exactly, just using the general constraints of symmetry. In these

circumstances we need to make some sort of approximation, treating our system as being

‘close to’ some other system whose dynamics is su�ciently simple to be controllable. That

is, we treat the di↵erence between our actual system, in whose experimental properties

we are interested, and our model system, whose description is simple enough that we can

handle it, as a perturbation. The art of a theoretical physicist lies in striking this balance;

if one’s model system is too simplistic, it might not provide a reasonable guide to the

behaviour of the real case40, while if the model is overly complicated it may itself prove

impossible to understand.

There’s nothing inherently quantum mechanical about the need to approximate a com-

plicated system with a simpler one, but, fortunately, perturbation theory in quantum me-

chanics often turns out to be considerably easier than in classical dynamics, largely because

of the vector space nature of H. In this chapter and the next, we’ll study various techniques

to handle quantum mechanical perturbation theory, beginning with the simplest cases.

7.1 An Analytic Expansion

Let H be the Hamiltonian of the experimental system we wish to understand, and H0

be the Hamiltonian of our model system whose eigenstates and eigenvalues we already

know. We hope that � := H �H0 is in some sense ‘small’, so that it may be treated as a

perturbation. More specifically, we look for a parameter — let’s call it � — that our true

Hamiltonian depends on such that at � = 0, H = H0. For � 2 [0, 1], define

H� := H0 + �� . (7.1)

We can think of H� as the Hamiltonian of an apparatus that is equipped with a dial that

allows us to vary �. At � = 0 the system is our model case, and at � = 1 it is the case of

genuine interest.

We now seek the eigenstates |E�i of H�. This may look as though we’ve made the

problem even harder — we now need to find the eigenstates not just of our model and

experimental systems, but of a 1-parameter family of interpolating systems. Our key

assumption is that since H� depends analytically on �, so too do its eigenstates. In essence,

this amounts to the assumption that small changes in the system will lead to only small

changes in the outcome. Every mountain climber knows that this assumption can be

dreadfully false and we’ll see that it can easily fail in QM too, but for now let’s see where

it takes us.40Milk production at a dairy farm was low, so the farmer wrote to the local university, asking for help

from academia. A multidisciplinary team of professors was assembled, headed by a theoretical physicist,

and two weeks of intensive on-site investigation took place. The scholars then returned to the university,

notebooks crammed with data, where the task of writing the report was left to the team leader. Shortly

thereafter the physicist returned to the farm, saying to the farmer,“I have the solution, but it works only

in the case of spherical cows in a vacuum”.

– 82 –

If indeed |E�i depends analytically in �, then we can expand it as

|E�i = |ai+ �|bi+ �2|ci+ · · · (7.2)

and similarly expand the eigenvalues

E(�) = Ea + �Eb + �2Ec + · · · . (7.3)

Plugging these expansions into the defining equation H�|E�i = E(�)|E�i we obtain

(H0 + ��)�|ai+ �|bi+ �2|ci+ · · ·

�=�Ea + �Eb + �2Ec + · · ·

� �|ai+ �|bi+ �2|ci+ · · ·

�.

(7.4)

Since we require this to hold as � varies, it must hold for each power of � separately. Thus

we find an infinite system of equations

H0|ai = Ea|ai ,H0|bi+�|ai = Ea|bi+ Eb|ai ,H0|ci+�|bi = Ea|ci+ Eb|bi+ Ec|ci ,

...

(7.5)

and so on.

The first of these equations simply states that |ai is an eigenstate of the model Hamil-

tonian H0 with eigenvalue Ea. This is not surprising; under our analyticity assumptions

the terms of O(�0) are all that would survive when the dial is set to � = 0, which is

indeed the model system. Henceforth, we relabel |ai ! |n(0)i and Ea ! E(0)n to reflect this

understanding. (The notation |n(0)i is intended to imply the nth energy eigenstate of the

model Hamiltonian H0, with eigenvalue E(0)n .)

To determine Eb we contract the second of equations (7.5) with ha| ⌘ hn(0)| to find

hn(0)|H0|bi+ hn(0)|�|n(0)i = E(0)n hn(0)|bi+ Eb (7.6)

Since the model Hamiltonian is self–adjoint (as for any Hamiltonian),

hn(0)|H0| b i = h b |H0|n(0)i = E(0)n hn(0)| b i , (7.7)

so (7.6) becomes

Eb = hn(0)|�|n(0)i . (7.8)

In other words, to first order in �, the change in the energy of our system as we move

away from the model system is given by the expectation value h�in(0) of the change in the

Hamiltonian when the system is in its original state |n(0)i.To find the perturbed state, to first order in � we must understand |bi. We can expand

|bi in the complete set {|n(0)i} of eigenstates of the original system as

|bi =X

n

bn|n(0)i . (7.9)

– 83 –

Using this expression in the second of equations (7.5) and contracting with hm(0)| (where|m(0)i 6= |n(0)i, the initial state we are perturbing) gives

(E(0)n � E(0)

m ) bm = hm(0)|�|n(0)i (7.10)

and so, provided E(0)m 6= E(0)

n ,

bm =hm(0)|�|n(0)iE(0)

n � E(0)m

. (7.11)

This will hold provided the energy levels of our model system are non–degenerate; we’ll

examine how to handle the more general case including degeneracy in section 7.2. In the

non–degenerate case, equation (7.11) determines all the expansion coe�cients bm in |biexcept bn. Fortunately, one can argue that bn = 0 from the requirement that |E�i remains

correctly normalised — I’ll leave this as an exercise, but the essential idea is that if we

move a point on a unit sphere, then to first order r · �r = 0 so that the variation is only

non–zero in directions orthogonal to the original vector. With bn = 0 we have

|bi =X

m 6=0

hm(0)|�|n(0)iE(0)

n � E(0)m

|m(0)i (7.12)

as the first–order perturbation of the state when the unperturbed state is |n(0)i.

We can also examine the second–order perturbation of the energy levels, Ec. To do

so, contract the third of equations (7.5) with hn(0)|. Using the facts that hn(0)|H0|ci =

E(0)n hn(0)|ci and hn(0)|bi = 0 we have

Ec = hn(0)|�|bi =X

m 6=n

hn(0)|�|m(0)ihm(0)|�|n(0)iE(0)

n � E(0)m

=X

m 6=n

|hn(0)|�|m(0)i|2

E(0)n � E(0)

m

,

(7.13)

using our expression (7.12) for |bi. We could go on to higher order in �, next finding |ciin terms of the original states {|n(0)i} and then finding the third–order energy shift Ed

etc., but in practice the summations become increasingly messy and we hope (!) that the

first few terms already provide a good guide to the behaviour of the system near � = 1.

(In high–energy quantum field theory, modern experiments typically cost many millions

of dollars so it’s especially important to have extremely accurate theoretical predictions

to compare to. Consequently, there’s a whole industry of people whose life’s work is to

compute higher and higher order terms in perturbation series such as these.) Fortunately,

in the Tripos you’ll never be asked to do anything beyond 2nd order.

Combining our results, we have that the energy levels of the perturbed system are

En(�) = E(0)n + �hn(0)|�|n(0)i+ �2

X

m 6=n

|hn(0)|�|m(0)i|2

E(0)n � E(0)

m

+O(�3) (7.14)

accurate to second order in �. Recall that this expression is derived under the assumptions:

1) that the new energies E(�) and new states |E�i are analytic at � = 0 and 2) that the

model system is non–degenerate so E(0)m = E(0)

n i↵ |m(0)i ⌘ |n(0)i. Let’s now take a look

at the use of this formula in a number of examples.

– 84 –

7.1.1 Fine structure of hydrogen

Our treatment of the hydrogen atom in the last chapter assumed the electron was moving

non–relativistically in the Coulomb field of the proton. Since |E/µc2| = ↵2/2n2 ⌧ 1,

non–relativistic quantum mechanics should indeed be a good approximation. Nonetheless,

better agreement with experiment is obtained by describing the electron using the rela-

tivistic Dirac equation41. The Dirac equation itself is beyond the scope of this course, but

one consequence is that the Hamiltonian of the hydrogen atom is modified to

H = H0 +HSO (7.15)

where H0 includes the pure Coulomb potential as before and

HSO =↵~32µ2c

1

|X|3 S · L (7.16)

is known as the spin–orbit Hamiltonian. The energy levels found from H0 are often known

as the gross structure of hydrogen, whereas the corrections induced by the spin–orbit Hamil-

tonian are called fine structure42

Since the coe�cient of the operator S · L is positive, spin–orbit coupling lowers the

total energy when the spin and orbital angular momentum are antiparallel. We write

S · L =1

2

�(L+ S)2 � L2 � S2

�=

1

2

�J2 � L2 � S2

�(7.17)

soHSO is diagonal in a basis of simultaneous eigenstates of J2, L2 and S2. In particular, s·Lannihilates any state of hydrogen in which ` = 0, since then j = s = 1

2 . As an important

special case, there is no spin–orbit coupling in the ground state. However, excited states

with ` 6= 0 do feel the spin–orbit coupling, and the eigenvalues of S · L are

1

2

✓j(j + 1)� `(`+ 1)� 3

4

◆=

(12` when j = `+ 1

2

�12(`+ 1) when j = `� 1

2 .(7.18)

Using our general expression (7.8), the first order shift in the energy of the |n, `,mi hydro-genic state is

�E = hn, `,m|HSO|n, `,mi = Z4↵4

4n3µc2

�((`+ 1)(`+ 1

2) when j = `+ 12

�`(`+ 12) when j = `� 1

2 ,(7.19)

for ` > 0. (To obtain this result, we’ve used (7.18) to evaluate the result of the S · Loperator, then used the fact that

Z1

r3| n`m(x)|2 d3x =

Z3

a0n3`(`+ 12)(`+ 1)

(7.20)

41A better approximation still — in precise agreement with the most accurate measurements ever per-

formed in any branch of science — comes from quantum field theory.42The Dirac equation predicts further corrections, including a term coupling the electron’s spin to the

magnetic field it experiences as a consequence of moving through the electrostatic field of the proton.

– 85 –

to evaluate hn, `,m|1/|X|3|n, `,mi; the details of how to perform this integral are not

important, though you can reproduce it as an exercise if you wish.) Thus, when j = `+ 12

we have energy levels

Ej=`+1

2n,`,m

= �1

2Z2↵2µc2

1

n2� ↵2Z2

2n3(`+ 12)(`+ 1)

!(7.21a)

to first order in the spin–orbit coupling, whereas when j = `� 12 we have

Ej=`�1

2n,`,m

= �1

2Z2↵2µc2

1

n2+

↵2Z2

2n3`(`+ 12)

!. (7.21b)

Note that these energy levels now depend on ` as well as the principal quantum number n.

Thus states with di↵erent total orbital angular momentum (and also di↵erent alignment

of their spin and orbital angular momentum) have di↵erent energies, so the degeneracy of

the hydrogenic gross structure spectrum is partially lifted. We also note that, as expected,

the state has lower energy, so is more tightly bound, when the spin and orbital angular

momentum are anti–aligned with eachother (j = ` � 12) and higher energy when they are

aligned (j = `+ 12).

Equations (7.21a)-(7.21b) show that the correction to the energy from the fine structure

is suppressed by Z2↵2 compared to the gross structure, where ↵ ⇡ 1/137. In hydrogen, Z =

1 so in the n = 1 ground state the fine structure contribution is ⇠ 105 times smaller than

the gross structure, suggesting our treatment of the spin–orbit coupling using perturbation

theory was fully justified. However, the importance of the fine structure contribution

increases as one proceeds to heavier elements with higher Z. In particular, from (7.19) we

see that the energy di↵erence between the states with principal quantum number n but

j = `± 12 is

Ej=`+1

2n,`,m

� Ej=`�1

2n,`,m

=Z4↵4

2n3

µc2

`(`+ 1)(7.22)

This gives the splitting between the two 2p states of Hydrogen to be 4.53 ⇥ 10�5 eV, in

excellent agreement with the measured value of 4.54⇥ 10�5 eV. However, these splittings

reach around 10% of the gross energy as one reaches the middle of the periodic table.

7.1.2 The ground state of helium

After hydrogen, helium is the next most abundant element, making up about a quarter of

the ordinary matter in the Universe. Much of this helium was created during the period

of nucleosynthesis, a tiny fraction of a second after the Big Bang.

Let’s use perturbation theory to find an estimate of the ground state energy of helium.

Working in the centre of mass frame and treating the nucleus as stationary at the origin,

the (gross structure) Hamiltonian is

H = HH(P1,X1) +HH(P2,X2) +e2

|X1 �X2|(7.23)

– 86 –

where (X1,X2) are the position operators and (P1,P2) the momentum operators for the

two electrons, and

HH(P,X) :=P2

2m� Ze2

|X| (7.24)

is the hydrogenic Hamiltonian describing the kinetic term of an electron, together with its

Coulomb attraction to the nucleus of charge +Ze (with Z = 2 for helium). The remaining

term in (7.23) is the electrostatic repulsion between the two electrons. We’ll take the model

Hamiltonian to be H0 = HH(P1,X1)+HH(P2,X2) and hope to treat the electron–electron

interaction as a perturbation.

Before proceeding, we should ask what is the small, dimensionless parameter in which

we’re performing our perturbative expansion. One might hope that, as in the case of the

fine structure of hydrogen, this is the fine structure constant ↵ = e2/~c ⇡ 1/137. However,

it’s clear that this cannot be the case here43. The electron–electron interaction is of order

e2 ⇠ ↵, but so too is the hydrogenic potential term. Furthermore, the virial theorem says

that in any state | i, for the 1/r hydrogenic Coulomb potential, 2hT i = �hV i so the

kinetic terms in the hydrogenic term are likewise of order ↵. In fact, we will perform

a perturbation in the inverse atomic number 1/Z, treating this as a continuous small

parameter. For helium we have 1/Z = 1/2 which, whilst certainly less than 1, is not very

small. One should not expect low–order perturbative results to be especially accurate.

In the unperturbed Hamiltonian, single–electron states are just the hydrogenic states

|n, `,mi with wavefunction n,`,m(x) := hx|n, `,mi and energy

En = �4↵2

2n2mc2 . (7.25)

Since helium has atomic number 2, this is four times the corresponding energy in hydrogen.

However, note that we do not write this as a function of 1/Z; the model Hamiltonian and

its eigenstates should be considered as independent of our expansion parameter. Since

electrons are fermions, the total state of the neutral helium atom must be antisymmetric

under their exchange. In particular, the ground state is

| 0i = |1, 0, 0i ⌦ |1, 0, 0i ⌦ 1p2(|" i|# i � |# i|" i) (7.26)

where the final factor is the antisymmetric spin state of the two spin-12 electrons. (Note

that, at the level of the gross structure of helium, the Hamiltonian (7.23) is independent

of the spins of the two electrons, so we are free to seek simultaneous energy and spin

eigenstates.) This state is an eigenstate of the model Hamiltonian H0 with energy

E(0)0 = �4↵2

2mc2 � 4↵2

2mc2 = �4↵2mc2 ⇡ �108.8 eV , (7.27)

43↵ does turn out to be the expansion parameter one typically uses when studying the perturbative

scattering of free electrons o↵ a Coulomb potential, as you’ll see if you take Applications of Quantum

Mechanics next term. There, one treats the entire electrostatic term in HH as a perturbation of the kinetic

term, describing a free electron. You might think that even this is a strange thing to do: given that we’ve

solved the bound state spectrum of hydrogen exactly, shouldn’t we also be able to do something similar for

the scattering states? I might agree with you...

– 87 –

being the sum of the energies of the two electrons individually.

We now take account of the electron–electron repulsion. To use perturbation theory,

we set

� =2

Zand � =

e2

|X1 �X2|=

2↵~c|X1 �X2|

. (7.28)

Our � is chosen so that i) it is independent of the expansion parameter and ii) �� is

precisely the electron–electron potential when � = 1, appropriate for helium. Following

the general results above, the first order shift in the ground state energy is

h 0|�| 0i =Zh 0|x0

1,x02ihx0

1,x02|�(X1,X2)|x1,x2ihx1,x2| 0i d3x01 d3x02 d3x1 d3x2

=

Z 0(x1,x2)

↵~c|x1 � x2|

0(x1,x2) d3x1 d

3x2

=

Z↵~c

|x1 � x2|| 100(x1)|2 | 100(x2)|2 d3x1 d3x2 ,

(7.29)

Here 100(x) are the single–electron n = 1 wavefunctions

100(x) =1p⇡

✓2

a2

◆ 3

2

e�|x|/a2 where a2 :=~

2↵mc. (7.30)

The length scale a2 is half the Bohr radius in hydrogen. In going to the last line of (7.29),

we’ve used the fact that the interaction � is independent of spin, and the spin states

in (7.26) have norm 1.

Performing the integral is straightforward but somewhat tedious44, and isn’t the sort

of thing I’m going to ask you to reproduce in an exam. One finds

h 0|�| 0i =5↵~c8a2

=5

4↵2mc2 (7.31)

44In case you’re curious: Choose the z-axis of x2 to be aligned with whatever direction x1 is. Then

|x1 � |bfx2| =

p|r

2

1+ r

2

2� 2r1r2 cos ✓2|, independent of �2. So

h 0|�| 0i =4↵~ca3

2

Z| 100(x1)|

2

"Zr2

2 sin ✓2 e�2r2/a2

p|r

2

1+ r

2

2� 2r1r2 cos ✓2|

dr2 d✓2

#d3x1 .

The integral in square brackets can be done using

Z ⇡

0

sin ✓2 d✓2p|r

2

1+ r

2

2� 2r1r2 cos ✓2|

=1

r1r2

Z ⇡

0

d

d✓2

q|r

2

1+ r

2

2� 2r1r2 cos ✓2| d✓2 =

(2/r1 when r1 > r2

2/r2 when r1 < r2

.

Thus the radial integral dr2 must be broken into two regions. Letting ⇢1 := 2r1/a2 be a rescaled radial

coordinate, we have

h 0|�| 0i =8↵~ca3

2

Z| 100(x1)|

2

Z r1

0

r2

2

r1e�2r2/a2

dr2 +

Z 1

r1

r2 e�2r2/a2

dr2

�d3x1

=4↵~ca2

Z| 100(x1)|

2 2� e�⇢1(2 + ⇢1)⇢1

d3x1 =

2↵~ca2

Z 1

0

⇢1 e�⇢1

�2� e�⇢1(2 + ⇢1)

�d⇢1 =

5↵~c4a2

.

This is the value that was used in the text. And no, I’m not going to derive the second–order result for

you.

– 88 –

Figure 12: First ionisation energies of the ground states of the first few elements. Figure

by Agung Karjono at Wikipedia.

As we suspected, this perturbation is also of order ↵2 just like the leading–order term.

Including the factor of � = 2/Z, to first–order in perturbation theory the ground state of

helium is

E0(1/Z)Z=2 = �4↵2mc2✓1� 5

8

1

Z+ · · ·

◆

Z=2

⇡ �74.8 eV . (7.32)

This is in significantly better agreement with the experimental value E0 ⇡ �79.0 eV than

our zeroth–order estimate (7.27). Including also the second–order term leads to

E0(1/Z) = �4↵2mc2✓1� 5

8

1

Z+

25

256

1

Z2+O(1/Z3)

◆(7.33)

giving E0(1/2) ⇡ �77.5 eV. Given the crudity of our expansion this is in remarkably close

agreement with the experimental value.

In heavy atoms, the ground state energy itself is di�cult to measure experimentally

since it involves computing the energy required to strip o↵ all the electrons from the

nucleus. Fortunately, for the same reason it’s also rarely the directly relevant quantity.

Instead, chemists and spectroscopists are usually more interested in the first ionisation

energy, defined to be the energy required to liberate a single electron from the atom.

Hydrogen only has one electron, so its first ionization energy is just the 13.6 eV which we

must supply to eject this electron from the ground state of the atom. In the case of helium,

once one electron is stripped away the remaining electron sees the full force of the Coulomb

potential from the Z = 2 nucleus, so will have binding energy �54.4 eV. The first ionisation

energy is thus the di↵erence ⇡ (�54.4 + 79.0) eV = 24.6 eV. This is significantly greater

than the ionisation energy of hydrogen, and in fact helium has the greatest first ionisation

energy of any of the elements, reflecting its status as the first noble gas (see figure 12).

7.1.3 The quadratic Stark E↵ect

Suppose a hydrogen atom is placed in an external electric field E = �r�. Most electric

fields we can generate in the laboratory are small compared to the strength of the Coulomb

– 89 –

field felt by the electron due to the nucleus45, so perturbation theory should provide a good

estimate of the energy shifts such an external field induces. By definition of the electrostatic

potential �, the field changes the energy of the atom by �E = e(�(xp)� �(xe)) where xp

and xe are the locations of the proton and electron. We assume the field is approximately

constant on scales of the typical atomic radius, so

�E ⇡ �e r ·r� = �e r ·E (7.34)

where r := xe � xp. Let’s choose the direction of the electric field to define the z-axis,

and set E := |E| so as to avoid confusion with the energy. Thus E = E z and the e↵ect of

the external electric field is to add a new term to the Hamiltonian H0 of the unperturbed

atom:

H = H0 � e EX3 (7.35)

where X3 is the position operator for the z-coordinate of the electron relative to the proton.

Let’s consider the e↵ect of this electric field on the ground state |n, `,mi = |0, 0, 0i ofthe hydrogen atom. From (7.8) the first order change in the energy is

Eb = �e E h1, 0, 0|X3|1, 0, 0i (7.36)

This ground state has even parity, P|1, 0, 0i = +|1, 0, 0i (it’s wavefunction involves the

‘trivial’ spherical harmonic Y00), so applying the parity operator gives

h1, 0, 0|X|1, 0, 0i = h1, 0, 0|P�1XP|1, 0, 0i = �h1, 0, 0|X|1, 0, 0i , (7.37)

where the last equality follows from the definition of the parity operator acting on the

position operator. In particular, h1, 0, 0|X3|1, 0, 0i = 0 for the z-component, so in this

example the ground state energy is una↵ected to first order in the external electric field.

At second order, the shift in energy is given by (7.13) to be

Ec = e2E21X

n=2

X

`<n

`X

m=�`

|h1, 0, 0|X3|n, `,mi|2E1 � En

(7.38)

in terms of a sum over a complete set of states other than the ground state. In the second

problem set you proved that

hn0, `0,m0|X|n, `,mi (7.39)

vanishes unless |`0�`| = 1 and |m0�m| 1. Applying this to our case, h1, 0, 0|X3|n, `,mi =0 unless ` = 1 andm = �1, 0, 1, which greatly simplifies the sums in (7.38). We can simplify

still further by noting that [Lz, X3] = 0 so X3|n, `,mi is an eigenstate of Lz with eigenvalue

m~ and hence orthogonal to |1, 0, 0i unless m = 0. In conclusion, the only non–vanishing

terms in (7.38) are

Ec = e2E21X

n=2

|h1, 0, 0|X3|n, 1, 0i|2E1 � En

(7.40)

45With modern high intensity lasers, electric fields comparable to the nuclear Coulomb potential can be

generated. The behaviour of a hydrogen atom in such a laser must be studied by other methods.

– 90 –

involving just a single sum. This is the second–order change in the ground state energy

due to the presence of the external electric field.

The fact that the ground state energy changes only at second order, and hence ⇠ E2,

is easy to understand physically. In response to the applied electric field, the probability

density of the atom’s charge changes by an amount that is proportional to the coe�cients

bm, which are themselves proportional to E . In other words, the applied electric field E

polarizes the atom, generating a dipole moment D / E. In this field, the energy of such a

dipole is �D ·E ⇠ E2. The original ground state wavefunction was spherically symmetric,

so in this state the atom had no dipole moment before the field was applied.

7.2 Degenerate Perturbation Theory

Our derivation of the coe�cients

bm =hm(0)|�|n(0)iE(0)

n � E(0)m

(7.41)

of the first–order shift in the state breaks down if there are states |m(0)i that have the same

energy as the state |n(0)i that we’re attempting to perturb. Indeed, naively applying (7.41)

in this case appears to show that the small perturbation can cause an extremely dramatic

shift in the state of the system.

There’s nothing particularly surprising about this. Consider a marble lying in the

bottom of a bowl at the minimum of the gravitational potential. If we tilt the bowl

slightly, the marble will move a little, resettling in the bottom of the tilted bowl as it

adjusts to the new minimum of the potential. However, if the marble initially lies at rest

on a smooth table, so that it initially has the same energy no matter where on the table

it lies, then tilting the table even very slightly will lead to a large change in the marble’s

location.

To handle this degenerate situation, we first observe that the only states that will

acquire a significant amplitude as a result of the perturbation are those that are initially

degenerate with the original state. In other words, to good approximation, the state to

which the system changes will be a linear combination of those having the same zeroth–

order energy as our initial state. In many situations, there are only a finite number of these.

We then diagonalise the matrix hr|�|si of the perturbing Hamiltonian in the subspace

spanned by these degenerate states. Since this subspace was completely degenerate wrt

H0, in this subspace the eigenstates of the perturbation � are in fact eigenstates of the full

Hamiltonian H0 +�.

Let’s see how this works in more detail. Suppose V ⇢ H is an N–dimensional subspace

with H0| i = EV | i for all | i 2 V . For r = 1, . . . , N , we let {|ri} be an orthonormal

basis of V and define PV to be the projection operator PV : H ! V . We can write this

projection operator as

PV =NX

r=1

|rihr| (7.42)

– 91 –

in terms of our orthonormal basis. Similarly, we let P? := 1 � PV denote the projection

onto H?, where

H? := { |�i 2 H : h |�i = 0 8 | i 2 V } . (7.43)

Note that, as projection operators, P 2V= PV and P 2

? = P? and that PV P? = P?PV = 0.

Also, we have

[H0, PV ] = 0 and [H0, P?] = 0 (7.44)

since V was defined in terms of a degenerate subspace of H0.

Now let’s consider a perturbed Hamiltonian H = H0+��. Any eigenstate | �i of thefull Hamiltonian obeys

0 = (H0 � E(�) + ��)| �i= (EV � E(�) + ��)PV | �i+ (H0 � E(�) + ��)P?| �i

(7.45)

where in the second line we’ve separated out the terms in V from those in H?. Acting on

the left with either PV or P? and using (7.44), we obtain the two equations

0 = (EV � E(�) + �PV�)PV | �i+ �PV�P?| �i (7.46a)

and

0 = �P?�PV | �i+ (H0 � E(�) + �P?�)P?| �i , (7.46b)

in which we have separated out the e↵ects of the perturbation within V and its complement.

Suppose now that we’re perturbing a state | 0i that, in the absence of the perturbation,

lies in V . We write| �i = | 0i+ �| 1i+ �2| 2i+ · · ·

E(�) = E(0) + �E(1) + �2E(2) + · · ·(7.47)

just as before, noting that P?| �i is necessarily at least first–order in �. To zeroth order,

equation (7.46a) tells us that E(0) = EV as expected. At first order, this equation becomes

(�E(1) + PV�)| 0i = 0 or equivalently PV�PV | 0i = E(1)| 0i . (7.48)

This is a remarkable equation! It tells us that our starting point | 0i 2 V had to be

an eigenstate of PV�PV — the perturbation, restricted to V . In other words, we weren’t

allowed to set up the perturbation series assuming we were beginning with any old | 0i 2 V ;

as soon as the perturbation is turned on it forces our degenerate states to line up with the

eigenstates of the perturbation. This is the quantum analogue of the fact that the location

of a marble on a perfectly flat table is extremely sensitive to any tilt.

The unperturbed Hamiltonian H0 gave us no preferred basis {|ri} of its degenerate

subspace V , so let’s now choose the |ris to be eigenstates of the perturbation PV�PV .

Choosing our initial state | 0i = |ri to be one of these, the first order shift in its energy is

E(1)r = hr|PV�PV |ri = hr|�|ri (7.49)

This is exactly what we would have found before, but here it’s very important that we’re

considering the e↵ects of perturbations when we start with an eigenstate of �. Had we

– 92 –

started with any other state in V our assumption that the perturbed state was analytic in

�, and in particular that it continuously deforms the unperturbed state, would have been

invalid. In fact, since |ri is an eigenstate of both H0 and �, we have

H|ri = (H0 + ��)|ri = (EV + �E(1)r )|ri (7.50)

so that in the subspace V we have solved the perturbed spectrum exactly. This is easier

than completely solving the spectrum ofH because we are only concerned with the subspace

V , which in problems of interest is typically of low dimension.

As we have seen, degeneracy in the energy spectrum often arises as a result of some

symmetry. If this symmetry is dynamical, such as the U(d) symmetry of the d–dimensional

harmonic oscillator, or the su(2)⇥su(2) symmetry of the gross structure of hydrogen, it will

almost inevitably by broken by a perturbation. Even symmetries such as rotations that are

inherited from transformations of space may be broken in the presence of external fields

— even if one’s atom is spherically symmetric, the laboratory in which one conducts an

experiment likely won’t be. For this reason, perturbations typically lift degeneracy. As we

have seen, even very small perturbations can have a dramatic e↵ect in immediately singling

out a preferred set of states in an otherwise degenerate system; we’ll return to this point in

section 9.4 when trying to understand ‘collapse of the wavefunction’ in quantum mechanics.

If you take the Applications of Quantum Mechanics course next term, you’ll see that the

lifting of degeneracy is also important in giving crystals their electronic properties: when

a large number of identical metal atoms are brought together, ignoring the inter–atomic

interactions, all the valence electrons in the di↵erent atoms will be degenerate. The small

e↵ect of the potential from nearby atoms breaks this degeneracy and makes these valence

electrons prefer to delocalise throughout the crystal.

7.2.1 The linear Stark E↵ect

As an application of degenerate perturbation theory, let’s again consider the Stark e↵ect but

now for the 2s state of our atom. This state is degenerate with the three46 2p states, so we

must diagonalise the perturbation47 eEX3 in the basis {|2, 0, 0i, |2, 1, 1i, |2, 1, 0i, |2, 1,�1i}spanning this degenerate subspace. We find

�|V= eE

0

BBB@

0 0 a 0

0 0 0 0

a⇤ 0 0 0

0 0 0 0

1

CCCA(7.51)

in this subspace, where

a = h2, 0, 0|X3|2, 1, 0i = 2⇡

Z 2,0,0(r) 2,1,0(r, ✓) r cos ✓ r

2d(cos ✓) = �3a0 , (7.52)

46Since the applied electric field does not coupled to spin, we ignore the e↵ect of spin in this discussion

– at this order, the Stark e↵ect will not lift the degeneracy of the gross structure of hydrogen wrt the

electron’s spin.47A reminder: we define the z-direction to be the direction of the electric field, andX3 is the corresponding

operator.

– 93 –

with a0 the Bohr radius. The eigenvalues of� in this subspace are thus {3eEa0, 0, 0,�3eEa0}with corresponding eigenstates

1p2(|2, 0, 0i � |2, 1, 0i) , |2, 1, 1i , |2, 1,�1i and

1p2(|2, 0, 0i+ |2, 1, 0i) . (7.53)

In the presence of the electric field, the n = 2 state of lowest energy is (|2, 0, 0i+|2, 1, 0i)/p2

and we conclude that even for a tiny electric field, the n = 2 states will jump to this

preferred state. Note that the degeneracy between |2, 1, 1i and |2, 1,�1i is not lifted by

this perturbation. These states have their orbital angular momentum maximally aligned

or anti-aligned with the direction of the electric field; classically, they corresponds to orbits

confined to a plane perpendicular to E.

From our discussion of the quadratic Stark e↵ect above, we know that a change in

energy / E requires the dipole moment D of an atom to be independent of E . Since

this n = 2 state has Eb = �3eEa0, we conclude that the n = 2 states of hydrogen do

indeed have a permanent electric dipole moment of magnitude 3ea0. Classically, for an

electron in an elliptical orbit, this result is expected because the electron would spend

more time at the apocentre (the point of its orbit furthest from the atom’s centre of mass)

than at the pericentre (the point nearest to the centre of mass). If electron’s orbit was a

perfect Keplerian ellipse, the atom would have a permanent electric dipole moment aligned

parallel to the orbit’s major axis. However, any small deviation from the 1/r potential will

cause the major axis of the ellipse to precess, and the time–averaged dipole moment would

vanish. For hydrogen, the potential does di↵er from 1/r, though only very slightly due to

the e↵ects of fine structure. Hence even a weak external field can prevent precession and

give rise to a steady electric dipole.

Quantum mechanically, in the presence of the electric field the lowest energy n = 2

state (|2, 0, 0i + |2, 1, 0i)/p2 is no longer an eigenstate of L2, because the field applies

torque to the atom, changing it’s total orbital angular momentum. However, this lowest

energy state is still an eigenstate of Lz with eigenvalue zero. Recalling that we took the

z-direction to be the direction of the applied field, we see that the angular momentum is

perpendicular to E, as expected from the classical picture of an ellipse with major axis

aligned along E.

7.3 Validity of Perturbation Theory

We began our study of perturbation theory by assuming that the states and energy eigen-

values of the full Hamiltonian depended analytically on a dimensionless parameter � con-

trolling the perturbation. Even when the individual coe�cients of powers of � are finite,

this is often not the case because the infinite perturbative series itself may fail to converge,

or may converge only for some range of �. The issue is that we really have an expansion in������

hm(0)|�|n(0)iE(0)

m � E(0)m

����� where m 6= n ,

and the coe�cients of � may grow too rapidly. Heuristically, the condition for convergence

is thus that the typical energy splitting |hm(0)|�|n(0)i| induced by the perturbation should

– 94 –

be much smaller than the initial energy di↵erence E(0)m �E(0)

m . However, a detailed criterion

is often hard to come by since the higher terms in the perturbation expansion involve

complicated sums (or integrals) over many di↵erent intermediate states.

To illustrate this in a simple context, let’s consider in turn the following three pertur-

bations of a 1d harmonic oscillator potential

H =P 2

2m+

1

2m!2X2 +

8>><

>>:

��m!2x0X

+12�m!2X2

+�✏X4

(7.54)

where x0 and ✏ are constants. Of course, the first two can be solved exactly — we’d never

really use perturbation theory to study them.

In the first case, we have

H =P 2

2m+

1

2m!2(X � �x0)

2 � �2

2m!2x20 (7.55)

from which we easily see that the exact energies are

En =

✓n+

1

2

◆~! � �2

2m!2x20 , (7.56)

and that the exact position space wavefunctions are hx|n�i = hx� �x0|n(0)i, just a trans-

lation of the unperturbed wavefunctions. If we instead tackled this problem using pertur-

bation theory, we’d find

En(�) = E(0)n � �m!2x0hn(0)|X|n(0)i+ �2m2!4x20

X

k 6=n

|hk|X|ni|2(n� k)~!

=

✓n+

1

2

◆~! � �2

2m!2x20

(7.57)

to second order in �. To obtain this result, we note that the first–order term vanishes (e.g.

by parity), while since X is a linear combination of creation and annihilation operators

A† and A, the only |ki = |n + 1i and |ki = |n � 1i terms can contribute to the sum in

the second–order term. Going further, we’d find that there are no higher corrections in �,

though this is not easy to see directly. Thus, in this case, the perturbative result converges

to the exact result, and the radius of convergence is infinite. This reflects the fact that the

perturbation ��m!2x0X did not really change the character of the original Hamiltonian.

No matter how large � is, for large enough x the perturbation remains negligible.

Turning to the second case, it’s again immediate that the exact energy levels are

En(�) =

✓n+

1

2

◆~!(�) where !(�) := !

p1 + � . (7.58)

This has a branch cut starting at � = �1, so the energy is only analytic in � inside the

disc |�| < 1. Again using perturbation theory, we find

En(�) = E(0)n +

�

2m!2hn(0)|X2|n(0)i+O(�3)

=

✓n+

1

2

◆~!✓1 +

�

2� �2

8+O(�3)

◆ (7.59)

– 95 –

agreeing to this order with the Taylor expansion of the exact answer. Continuing further,

we’d find that this Taylor series does indeed converge provided |�| < 1, and that it then

converges to the exact answer. The physical reason why the perturbation series diverges

when |�| � 1 is simply that if � = �1, the ‘perturbation’ has completely cancelled the

original harmonic oscillator potential, so we’re no longer studying a system that can be

treated as a harmonic oscillator in the first instance. Once � < �1 the harmonic oscillator

potential is turned upside down, and we do not expect our system to possess any stable

bound states.

Finally, consider the case

H = HSHO + �✏X4 . (7.60)

I do not know whether this model has been solved exactly, but it can be treated perturba-

tively. After a fair amount of non-trivial calculation48 one obtains the series

E0(�) = ~! +1X

n=1

(�✏)nan (7.61)

for the ground state energy including the quartic interaction, where the coe�cients behave

as

an =(�1)n+1

p6

⇡3/23n �(n+

1

2)

✓1� 95

72

1

n+O(n�2)

◆. (7.62)

On account of the �-function, these grow faster than factorially with n, so the series (7.61)

has zero radius of convergence in �! Once again, this is easy to see from the form of the

perturbed Hamiltonian: even though we may only care about � > 0, our assumption that

the perturbation expansion is analytic in � at � = 0 means that, if it converges, it will

do so for a disc � 2 D ⇢ C. For any � < 0, the Hamiltonian of the quartic oscillator is

unbounded below, so there cannot be any stable bound states that are analytic in � at

� = 0.

Let me comment that even when perturbative series do not converge, they may still

provide very useful information as an asymptotic series. You’ll learn far more about these

if you take the Part II course on Asymptotic Methods next term, but briefly, we say a

series SN (�) =P

N

n=0 an�n is asymptotic to an exact function S(�) as � ! 0+ (written

SN (�) ⇠ S(�) as �! 0+) if

lim�!0+

1

�N

�����S(�)�NX

n=0

an�n

����� = 0 . (7.63)

In other words, if we just include a fixed number N of terms in our series, then for small

enough � � 0 these first N terms di↵er from the exact answer by less that ✏�N for any ✏ > 0

(so the di↵erence is o(N)). However, if we instead try to fix � and improve our accuracy

by including more and more terms in the series, then an asymptotic series will eventually

48I don’t expect you to reproduce it – if you’re curious you can find the details in Bender, C. and Wu,

T.T., Anharmonic Oscillator II: A Study of Perturbation Theory in Large Order, Phys. Rev. D7, 1620-1636

(1973).

– 96 –

diverge. Most of the perturbative series one meets in the quantum world (including most

Feynman diagram expansions of Quantum Field Theory) are only asymptotic series. Just

as in our toy examples above, the radius of convergence of such series is often associated

with interesting physics.

– 97 –