102

Chapter 7. Approximation Methods – Part I Very few problems in quantum mechanics can be solved exactly (analytically), therefore approx-

imation methods have to be employed in almost all realistic situations. There are a small number

of situations where an exact solution of a simple equation gives a remarkably accurate account of

experimental data, but eventually refinement of experimental technique or previously overlooked

effects will require a more complicated theoretical description, which no longer yields to an ex-

actly analytic solution. Indeed the majority of physicists dedicate their lives to either measure or

evaluate such small effects by making good approximations.

There is an enormous arsenal of approximation methods in quantum mechanics. In this class

we’ll study some major analytic approximation methods, which are still widely used today and

are the foundation for numerical computational methods. With the advent powerful computers,

these approximation methods can be vastly more precise than was imaginable originally, provid-

ing indispensible qualitative and even quantitative understanding to complex and realistic quan-

tum systems.

In part I we will study the WKB method, the variational principle, and the time-independent

perturbation theory (Non-degenerate). In part II (Chapter 8), we will study and the degenerate

perturbation theory and the time-dependent perturbation theory.

7.1 WKB This method is also called semiclassical approximation, developed by Wentzel, Krammers and

Brillouin in 1926. The basic idea is that the Schrödinger Equation can be solved easily for con-

stant potential, so it can be solved approximately for a slowly-varying potential.

−2

2md 2

dx 2+V(x)

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥ψ(x) = Eψ(x) . (7.1)

If V(x) = 0 , ψ(x) = eikx = exp

i

2mE x

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟.

If V(x) =V0, ψ(x) = eikx = exp

i

2m(E −V

0) x

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟.

Note: k = 2m(E −V

0) / can be real or imaginary for a piecewise potential.

We thus guess that for an arbitrary V(x)

103

ψ(x) = eik(x )x = exp

i

σ(x)

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟, (7.2)

where σ(x) is a complex function in general.

Plugging Eq. (7.2) into Eq. (7.1) converts the equation for ψ(x) to σ(x) :

′ψ (x) =

i′σ (x)ψ(x), (7.3)

′′ψ (x) =i′′σ (x)−

′σ (x)

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

2⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ψ(x) , (7.4)

0 = ′σ( )2

− i ′′σ + 2m[V(x)−E ] . (7.5)

So far, no approximations are made, and Eq. (7.5) is mathematically identical to the Schrödinger

Equation [Eq. 6.1(7.1)].

The classical limit: Quantum mechanics reduces to classical mechanics when → 0 . WKB

makes good use of this to approximate solutions of Eq. (7.5), therefore it is called the semi-

classical approximation. In particular, WKB treats as a small parameter, expanding σ(x) in

terms of :

σ(x) = σ0(x)+ (−i)σ

1(x)+ (−i)2σ

2(x)+ (7.6)

0 = ′σ

0+ (−i) ′σ

1+ (−i)2 ′σ

2+⎡

⎣⎢⎤⎦⎥2− i ′′σ

0+ (−i) ′′σ

1+ (−i)2 ′′σ

2+⎡

⎣⎢⎤⎦⎥+ 2m[V(x)−E ] . (7.7)

Collecting the n terms to rewrite Eq. (7.7) into the form of 0 = nF

n(x)

n=0

∞

∑ , so that we can

require Fn(x) = 0 for n = 0,1,2; thus we obtain a set of equations:

0 = ( ′σ0)2 + 2m[V(x)−E ], (7.8)

0 = 2 ′σ0′σ1

+ ′′σ0 , (7.9)

0 = ( ′σ1)2 + 2 ′σ

0′σ2

+ ′′σ1 , (7.10)

and so on. Usually, if V(x) doesn’t vary extremely fast, σ2 and higher-order terms can be safely

neglected, leading to the first-order WKB approximation:

σ(x)≈ σ0(x)− iσ

1(x). (7.11)

Here σ0(x) and σ1

(x) can be easily found for an arbitrary potential V(x) by solving Eqs.

(7.8) and (7.9):

104

′σ0(x) = ±p(x), where p(x)≡ 2m[E −V(x)] . (7.12)

Here p(x) is the classical momentum if E >V ; when E <V , it is an imaginary function. Then

σ

0(x) = const. ± p( ′x )d ′x

a

x

∫ , (7.13)

where a depends on the boundary conditions.

Eq. (7.9) ⇒ ′σ1(x) =−

′′σ0(x)

2 ′σ0(x)

=−12

ddx

ln[ ′σ0(x)] =−

12

ddx

ln[±p(x)] , we obtain

σ

1(x) = const. −

12

ln[±p(x)]. (7.14)

Plugging Eqs. (7.14), (7.13) and (7.11) into Eq. (7.2), the (first-order) WKB approximation is

ψ(x)≈A

1

p(x)exp

i

p( ′x )d ′xa

x

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥+

A2

p(x)exp −

i

p( ′x )d ′xa

x

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥ . (7.15)

Remarks: (1) Classically, the momentum p(x) = 2m[E −V(x)] , and the probability of find the

particle at position x is proportional to 1/ p(x) . Quantum mechanically, the WKB approxima-

tion leads to ψ(x)

2∝ 1/ p(x) for E >V . (2) If V(x) =V

0, a constant potential, then Eq. (7.15)

reduces to ψ(x) = B1eipx/ + B

2e−ipx/ , with

p = 2m(E −V

0) , which are the solutions to the

Schrödinger equation for a constant potential. (3) A modern topic, the path integral method, has a

strong connection to the WKB approximation [Eq. (7.15)].

Turning Points

For an arbitrary potential V(x) ,

we will encounter the turning-

point problem when employing

WKB. Consider V(x) shown in

the figure on the right. Here

limx→−∞V(x)→∞ . The turning point is defined as x = b , where E =V(b) and the classical

momentum vanishes: p(b) = 0 .

In region I ( x >b ), E >V(x) , the phase integral starts from the turning point x = b :

105

σ

0(x) = p( ′x )d ′x = 2m E −V( ′x )( ) d ′x

b

x

∫b

x

∫ . (7.16)

Here p(x) is real, so σ0 is a pure quantum phase factor for wave function ψ(x) :

ψ(x) =1

p(x)A

1eiσ0(x )/ + A

2e−iσ0(x )/⎡

⎣⎢⎤⎦⎥. (7.17)

A very useful trick is to use another two parameters, A and φ , so that

ψ(x) =A

p(x)cosσ

0(x)

+φ⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥=

A

p(x)cos

1

p( ′x )d ′xb

x

∫ +φ⎡

⎣⎢⎢

⎤

⎦⎥⎥ . (7.18)

In region II ( x <b ), E <V(x) , p(x) is purely imaginary. We set p(x) = iq(x) , where q(x)

= 2m[V(x)−E ] , and it is real.

σ

0(x) = p( ′x )d ′x = i q( ′x )d ′x

x

b

∫x

b

∫ , (7.19)

ψ(x) =A

1

q(x)exp −

1

q( ′x )d ′xx

b

∫⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

+A

2

q(x)exp +

1

q( ′x )d ′xx

b

∫⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟. (7.20)

Here when x →−∞ , V(x)→∞ therefore ψ(x)→ 0 . But the second term in Eq. (7.20) will in-crease exponentially as x →−∞ , thus only the first term in Eq. (7.20) survives. Set B = A

1,

ψ(x) =B

q(x)exp −

1

q( ′x )d ′xx

b

∫⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟. (7.21)

Question: How to determine the relationship between A and B, and the phase shift φ ? Em-

ploying the boundary conditions at x = b :

ψ

I(x)

x=b= ψ

II(x)

x=b, (7.22)

ddxψ

I(x)

x=b

=ddxψ

II(x)

x=b

. (7.23)

However, at x → b both p(x) and q(x) vanish, so ψ(x) and

ddxψ(x) diverge, we can’t apply

the boundary conditions of Eqs. (7.22) and (7.23). Question: Why does it break down?

Every approximation has its limit beyond which the approximation method breaks down.

This is very important! Always keep in mind that you have to check if the approximation method

can be safely applied (i.e., the accuracy is acceptable); otherwise, your results are misleading. So

under what conditions can we apply the (first-order) WKB?

106

The central approximation we made in (first-order) WKB is Eq. (7.11), i.e.,

σ(x) = σ0(x)+ (−i)σ

1(x)+ (−i)2σ

2(x)+ ≈ σ0

(x)− iσ1(x). This approximation is well held

if the recursive expansion of Eq. (7.7) can be cut off at the second order:

0 = ( ′σ

0)2 + 2m(V −E)⎡

⎣⎢⎤⎦⎥ − i 2 ′σ

0′σ1

+ ′′σ0( )+O(2), (7.24)

which requires that | (′σ0)2 + 2m(V −E) | | 2 ′σ

0′σ1

+ ′′σ0| . Since in the WKB solution, we

require ′σ0

= p(x) , we choose (′σ0)2 as a measure of the first term, and for the similar reasons we

choose ′′σ0

as a measure of the second term. Then the validity condition for WKB is

| (′σ0)2 | | ′′σ

0|, (7.25)

leading to 1

′′σ0

( ′σ0)2

= ddx

1′σ0

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

=ddx

p(x)

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

=12π

dλ(x)dx

. We obtain the condition under

which the WKB approximation can be applied,

dλ(x)dx

1 . (7.26)

Physically, it means that the de Broglie wavelength varies slowly over space.

Close to the turning point, i.e., when x → b , p→ 0 , and because

dλ(x)dx

= 2πdp−1(x)

dx= 2π

1p2(x)

dp(x)dx

∝1

p2(x),

dλ(x)dx

x→b

→∞ , violating Eq. (7.26).

Question: then how to treat the region close to

x = b ? When an approximation method breaks

down, we can use (1) higher-order approximation

or (2) exact method if possible. Here we can solve

the Schrödinger equation near x = b , where

V(x) =V(b)+ ′V (b)(x −b), (7.27)

for | x −b |< δ→ 0 .

We have solved ψ(x) using WKB in region I ( x >b ) and in region II (x <b) :

ψI(x) =

A

p(x)cos

1

p( ′x )d ′xb

x

∫ +φ⎡

⎣⎢⎢

⎤

⎦⎥⎥ , where p(x) = 2m[E −V(x)] (7.28)

107

ψII(x) =

B

q(x)exp −

1

q( ′x )d ′xx

b

∫⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟, where q(x) = 2m[V(x)−E ] (7.29)

We need to find the relation between A and B, and the phase shift φ .

(1) Solving Schrödinger Eq. in | x −b |< δ for potential V(x) =V(b)+ ′V (b)(x −b) to get ψ0(x) .

(2) Now we can enforce the boundary conditions:

matching ψ0(x) and ψI

(x) at x = b + δ , and ψ0(x) and ψII

(x) at x = b− δ .

(3) Take the limit of δ→ 0 .

The procedure and math are quite involved. The final results are

A = 2B, φ =−π / 4 (7.30)

Applications

Example 1. Bound states in a potential well: V(x)→∞

when x →±∞ . The key is using the boundary condi-

tions at x = x1 and x = x

2, the two turning points.

Start the phase integral from the left,

ψI(x) =

A1

p(x)cos

1

p( ′x )d ′x −π4x1

x

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥ . (7.31)

Or we can start the phase integral from the right,

ψI(x) =

A2

p(x)cos

1

p( ′x )d ′x −π4x

x2

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥ =

A2

p(x)cos

1

p( ′x )d ′x +π4x2

x

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥ . (7.32)

But Eqs. (7.31) and (7.32) should be the same eigenfunction; therefore, A

1= A

2 and

1

p( ′x )d ′x −π4x1

x

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥ −

1

p( ′x )d ′x +π4x2

x

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥ = nπ , for n = 0,1,2 ⇒

p(x)dxx1

x2

∫ = (n +12)π (7.33)

This is the WKB quantization condition for a potential well, very similar to the semi-empirical

Bohr-Sommerfeld quantization condition:

p(x)dxx1

x2

∫ = (n +1)π, for n = 0,1,2 (7.34)

108

The difference rises from the treatment of boundary conditions:

Bohr-Sommefeld: ψ(x) vanishes at x = x1 and x

2

WKB: ψ(x) starts to exponentially decay at x = x1 and x

2

Now we consider a concrete example:

V(x) = α | x |, with α> 0 . First, we compute the posi-

tion of turning points for eigenenergy En: x2

= En/α

and x1=−x

2=−E

n/α .

Then, Eq. (7.33) ⇒

(n +12)π = 2m E

n−α | x |( ) dx

x1

x2

∫

= 2 2m(En−αx) dx

0

x2

∫ =4 2m

3αE

n3/2

(7.35)

We find the eigenenergies

E

n= n +

12

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

3απ

4 2m

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

2/3

. (7.36)

How accurate is the WKB? We can also solve this problem numerically using a computer. If

we define rn≡

En(WKB)

En(exact)

, then r0= 0.7162 , r1 = 0.9923 , r2

= 0.9942 , r3= 0.9985 …

For n ≥10 , the WKB energies are essentially the same as the exact values.

Example 2. Tunneling rate of an energy barrier.

We have studied a tunneling rate for a square bar-

rier; however, in reality, the energy barrier can

have a rather complicated profile, as shown here.

Then how can we evaluate the tunneling rate

T =ψ(x

2)

2

ψ(x1)

2 using the WKB approximation?

109

For x1< x < x

2, E <V(x) , and p(x) = 2m(E −V ) = iq(x) , where p(x) is imaginary

while q(x) is real. The WKB wavefunction:

ψ(x)≈A

q(x)exp −

1

q( ′x )d ′xx1

x

∫⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

. (7.37)

Here we omit the exponentially increasing component. The tunneling rate

T ≈q(x

1)

q(x2)exp −

1

q(x)dxx1

x2

∫⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2

. (7.38)

Normally the tunneling rate T 1 and q(x1)/q(x

2) 1 , then

T ≈ e−2γ, with γ =

1

q(x)dxx1

x2

∫ . (7.39)

Here γ is Gamow’s factor. An example is alpha decay, which is your homework problem.

7.2 Variational Method Perturbation theory requires exact solutions to H0

; however, it is often very difficult to solve the

eigenstate problem. The variational theory offers a powerful method to estimate the ground-state

energy E0 and wavefunctions ψ0

(r) .

The variational theorem: For an arbitrary wavefunction φ(r)

φ H φ

φ φ≥E

0. (7.40)

Proof: The eigenstates of H , ψ

n, form a basis to expand

φ :

φ = c

nψ

nn∑ , (7.41)

where c

n= ψ

nφ . Since

H ψ

n= E

nψ

n,

φ H φ = |c

n|2 E

n

n

∑ ≥E0

|cn|2

n

∑ , (7.42)

and

110

φ φ = |c

n|2

n∑ , (7.43)

therefore, the theorem (7.40) is proved.

The variational method is based on the variational theorem. The left hand side of (7.40),

E[φ(r)]≡φ H φ

φ φ=

φ*(r)Hφ(r)d 3r∫|φ(r) |2 d

3r∫, (7.44)

is a functional of φ(r) . Since E[φ(r)]≥E0, one can make E[φ] as small as possible by ‘vary-

ing’ φ to get close to E0 [and φ(r) also becomes close to ψ0

(r)]. φ(r) is called the trial wave-

function, which depends on m parameters λ1, λ2

, , λm

, and then E[φ] also depends on these

parameters. The minimum of E[φ] [constrained by the expression of φ(r) ] is found by solving

the following set of m equations,

∂E∂λ

1

= 0, ∂E∂λ

2

= 0, , ∂E∂λ

m

= 0. (7.45)

A good format of φ(r) will result in a good estimation of the ground-state energy. If φ(r)

happens to be exact, then the exact ground-state energy is obtained. But how to construct a trial

wavefunction? We appeal to physical intuition — the asymptotic behaviors at boundary or large

distance. Different forms of φ(r) will give different estimation of E0; in practice, we can use a

crude method to get crude φ(r) , then we use the variational method to refine φ(r) and E0.

Example: the ground-state energy of the one-dimensional infinite potential well

V(x) =0, for | x |<a

∞, for | x |≥a.

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

(7.46)

We have found the exact solution

ψ

0(x) =

1

acosπx2a

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

, (7.47)

E

0=2π2

8ma2. (7.48)

111

Because wavefunction must vanish at x = ±a , a trial wavefunction must satisfy φ(±a) = 0 , and

we have many choices. The simplest one is φ(x) = A(a2−x 2) , where A is the normalization pa-

rameter (NOT a variational parameter). Therefore, we don’t need to solve Eq. (7.45), instead, to-

tal energy is evaluated directly from Eq. (7.44):

E =(a2−x 2)

d 2

x 2(a2−x 2)dx

−a

a

∫(a2−x 2)2 dx

−a

a

∫=

10π2

π2!2

8ma2= 1.0132E

0. (7.49)

Next let’s try a bit more sophisticated one, φ(x) = A(|a |λ − | x |λ) , with a variational param-

eter λ , which is determined by evaluating E(λ) ,

E(λ) =

(λ+1)(2λ+1)2λ−1

2

4ma2, (7.50)

and then solving the equation

dE(λ)dλ

= 0 . We find that E(λ) has a minimum at

λ =

1+ 62≈ 1.7247 , with

Emin

=5 + 2 6π2

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟E

0≈ 1.00298 E

0.

7.3 Time-Independent Perturbation (Non-Degenerate)

Very few problems can be solved analytically; however, practically in all realistic cases the ef-

fects of different factors are not equipollent. Thus one can first single out the most significant

factors, which normally can be solved either analytically or approximately. The influence of ne-

glected factors can then be taken into account approximately as a perturbation.

Usually (not always), perturbations only slightly modify the wavefunctions and energy lev-

els, so that the qualitative picture of the perturbed system retains the main features of the unper-

turbed one, and the changes in wavefunctions and energy levels due to perturbation can be ex-

pressed in terms of the unperturbed wavefunctions and energy levels. In this class we will focus

on these well-behaved perturbations. However, in certain cases an infinitesimal perturbation can

generate significant effects on the unperturbed system, and the associated quantum chaos is cur-

rently among the frontiers in research on the foundation of quantum mechanics.

7.3.1 Formal Theory (Non-Degenerate)

112

Let’s start with Hamiltonian H = H0

+ ′H , where ′H is much weaker than H0.

What we know: (1) the eigenstates n(0) and eigenvalues En

(0) for H0: H

0n

(0) = En

(0)n

(0)

(2) ′Hnm

= n(0) ′H m

(0)

Goal: the eigenstates n and eigenvalues En

for H : H n = E

nn , from perturbation theory

in terms of En(0) ,

′Hnm

, and n(0) . We can write H = H

0+λ ′H , where 0≤λ≤1 , then

En= E

n(0) +λE

n(1) +λ2E

n(2) + (7.51)

n = n(0) +λ n(1) +λ2 n(2) + (7.52)

We want to express the j-th order corrections to energy, En( j ) , and to eigenstate,

n( j ) , in

terms of En(0) ,

′Hnm

, and n(0) . In principle, one can obtain the exact solutions to En

and n by

summing all orders (from 0 to infinity) of corrections; however, in practice, we normally do first-

and second-order corrections to eigenenergies, while only first-order corrections to eigenstates.

Assume all En(0) are significantly different, i.e.,

E

n

(0)−Em

(0) ′Hnm

(7.53)

which is the condition for applying time-independent non-degenerate perturbation theory.

H n = En

n ⇒ H0

+λ ′H( ) n(0) +λ n

(1) +λ2n

(2) +( ) = E

n

(0) +λEn

(1) +λ2E

n

(1) +( ) n(0) +λ n

(1) +λ2n

(2) +( )

From the above we derive one equation for each order. We plug the zeroth-order solutions to the

first-order equation to get the first-order corrections, and then plug the first-order solutions to the

second-order equation to get the second-order corrections, and so on.

Zeroth-order: H

0n

(0) = En

(0)n

(0)

First-order: H

0−E

n

(0)( ) n(1) = E

n

(1)− ′H( ) n(0) (7.54)

Second-order: H

0−E

n

(0)( ) n(2) = E

n

(1)− ′H( ) n(1) + E

n

(2)n

(0) (7.55)

Multiply n(0) on both sides of Eq. 6.43, using the zeroth-order solution

H

0n

(0) = En

(0)n

(0)

113

n

(0)H

0−E

n

(0)( ) n(1) = 0 = n

(0)E

n

(1)− ′H( ) n(0) = E

n

(1)− n(0) ′H n

(0)

⇒

E

n

(1) = n(0) ′H n

(0) = ′Hnn

(7.56)

Multiply m(0) on both sides of Eq. (7.54), where m ≠ n

m(0)

H0−E

n

(0)( ) n(1) = m

(0)E

n

(1)− ′H( ) n(0)

⇒ (Em

(0)−En

(0)) m(0)

n(1) = E

n

(1)δn,m− ′H

mn=− ′H

mn, since m ≠ n

We can express n(1) = C

nm(1) m(0)

m∑ , a linear combination of the basis

n(0){ } , and then

C

nm

(1) = m(0)

n(1) =

′Hmn

En

(0)−Em

(0), for m ≠ n ; while Cnn

(1) = 0 . Thus we obtain

n(1) = C

nm

(1)m

(0)

m

∑ =′Hmn

En

(0)−Em

(0)m

(0)

m≠n

∑ (7.57)

Here Eqs. (7.56) and (7.57) are the first-order non-degenerate time-independent perturbations

to energy levels and eigenstates, respectively. Notes: We can obtain this form of the first-order

correction to eigenstates only if E

n(0)−E

m(0) ≠ 0 , i.e., non-degeneracy. When

E

n

(0)−Em

(0) ′Hnm

, the first-order corrections are much smaller than their corresponding unper-

turbed results.

Now we derive the second-order energy corrections En(2) from Eq. (7.55). Using the similar

approach, multiply n(0) on both sides, we get

n(0)

H0−E

n

(0)( ) n(2) = n

(0)E

n

(1)− ′H( ) n(1) + n

(0)E

n

(2)n

(0)

⇒ 0 = En

(1)n

(0)n

(1) − n(0) ′H n

(1) + En

(2)

⇒ En

(2) = n(0) ′H n

(1) −En

(1)n

(0)n

(1)

Using the first-order results: n

(1) =′Hmn

En

(0)−Em

(0)m

(0)

m≠n

∑ ⇒ n(0) n(1) = 0 , therefore

E

n

(2) = n(0) ′H n

(1) = Cnm

(1)n

(0) ′H m(0)

m≠n

∑ =′Hnm

2

En

(0)−Em

(0)m≠n

∑ . (7.58)

114

You can derive n(2) , En

(3) , and so on; but in practice, they are rarely used. A summary:

En

(1) = ′Hnn

; n(1) =

′Hmn

En

(0)−Em

(0)m

(0)

m≠n

∑

En

(2) =′Hnm

2

En

(0)−Em

(0)m≠n

∑ ⇒

λ = 1, H = H0

+ ′H

En≈ E

n

(0) + ′Hn,n

+′Hnm

2

En

(0)−Em

(0)m≠n

∑

n ≈ n(0) +

′Hmn

En

(0)−Em

(0)m

(0)

m≠n

∑

7.3.2 Applications of the Non-Degenerate Time-Independent Perturbation Theory Example 1: Charged harmonic oscillator in a weak electric field. The Hamiltonian is

H =−

2

2md

2

dx2

+12mω2

x2 + Fx = H

0+ ′H , (7.59)

where ′H = Fx and F = qε , with ε the magnitude of electric field and q the charge.

We know that H

0n

(0) = En

(0)n

(0) with E

n(0) = (n +

12)ω .

We need to work out ′Hnk

= F n(0)

x k(0) , using the ladder operators

x = x0(a +a†) , where

x

0=

2mω

; then

′Hnk

=Fx

0n k = n−1

Fx0

n +1 k = n +1

0 Otherwise

⎧

⎨

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

(7.60)

En

(1) = ′Hnn

= 0 ,

En

(2) =| ′H

nk|2

En

(0)−Ek

(0)k≠n

∑ =| ′H

n,n−1|2

En

(0)−En−1(0)

+| ′H

n,n+1|2

En

(0)−En+1(0)

=(Fx

0)2

n

ω+

(Fx0)2(n +1)−ω

=−F

2x

02

ω=−

F2

2mω2

⇒ E

n≈ E

n(0) + E

n(1) + E

n(2) = (n +

12)ω−

F 2

2mω2.

115

It is interesting to compare the exact solution, which is straightforward to obtain, since a

charged simple harmonic oscillator in electric filed is still a SHO (Why? Consider the classical

SHO under a constant force). Set y = x +

Fmω2

, the Hamiltonian

H =−2

2md

2

dy2

+12mω2

y2−

F2

2mω2 ⇒ −

2

2md

2

dy2

+12mω2

y2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟ψ(y) = (E

n+

F2

2mω2)ψ(y)

⇒ En

+F

2

2mω2= (n +

12)ω ⇒ E

n= (n +

12)ω−

F2

2mω2.

Question: the exact solution is the same as that using the second-order perturbation theory, why?

Next, we employ the perturbation theory to find the wavefunctions:

ψ

n

(1)(x) = Cnk

(1)ψk

(0)(x)k≠n

∑ , where Cnk

(1) =′Hkn

En

(0)−Ek

(0). (7.61)

In this case, Cnk(1) ≠ 0 only for k = n ±1.

Eq. (7.60) ⇒ C

n,n−1(1) =

Fx0

n

ω and

C

n,n+1(1) =−

Fx0

n +1ω

, we obtain

ψ

n(x)≈ ψ

n(0)(x)+

Fx0

!ωn ψ

n−1(0) (x)− n +1 ψ

n+1(0) (x)⎡

⎣⎢⎤⎦⎥. (7.62)

Let’s compare with the exact results for the ground state.

From perturbation theory,

ψ

0(x)≈ ψ

0(0)(x)−

Fx0

!ωψ

1(0)(x) = 1−

Fx!ω

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ψ

0(0)(x) , (7.63)

since ψ

1(0)(x) =

xx

0

ψ0(0)(x) . (How would you prove that?)

The exact solution:

ψ

0(y) = A

0e−

y2x0

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

2

, with A0

=1

(2πx02)1/4

, y = x +F

mω2. (7.64)

The first-order Taylor expansion assuming small F :

ψ

0(x)≈A

0e−

x2x0

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

2

e−

1

2x02

xF

mω2

≈ ψ0(0)(x) 1−

12x

02

xFmω2

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟= 1−

Fxω

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟ψ

0(0)(x). (7.65)

116

Example 2: Fine structure of the hydrogen atom

The above figure shows the hydrogen optical absorption spectral lines on a logarithmic scale,

which can be accurately explained by solving the Schrödinger equation for Coulomb potential.

Specifically,

E

n=−

1n2

E0, where the Rydberg energy E

0=2

2mea

02

= 13.60 eV. (7.66)

117

The eigenfunctions ψnlm(r,θ,φ) can be written as

r,θ,φ n,lm = R

n(r)Y

lm(θ,φ). (7.67)

However, when resolution is increased, one finds small devia-

tions from theory. For example, for the 3s to 2p transition, using the

reduced mass, you’ll get 656.47 nm for hydrogen and 656.29 nm for

deuterium, respectively. But experimentally, physicists measured a

small shift of 0.15 nm (0.00042 eV) with a tiny splitting of 0.016 nm

(0.000045 eV). These effects are called the fine structure of H spec-

tral lines, and the physical origins are rather complicated.

If you look even closer, you will find more complicated

and even weaker effects, including the Lamb shifts and the

hyperfine structure. Lamb shifts are due to electron and

vacuum interaction, which are significant only for the s-

wave, while the hyperfine structure is due to nuclear spin.

The energy hierarchy is summarized on the right.

Here we will not discuss the Lamb shift and the hyperfine

structure. The Hamiltonian at the fine-structure level:

H = H0

+ HKE

+ HDW

+ Hso

. (7.68)

The perturbation ′H = H

KE+ H

DW+ H

so, accounting for the relativistic kinetic energy, the

Darwin and the spin-orbit coupling corrections, respectively.

(1) The relativistic kinetic energy corrections.

The kinetic energy: K.E. = p2c2 +m2c 4 −mc2 ≈

p2

2m−

p4

8m3c2, (7.69)

for low-energy electrons. Then

H

KE=−

p̂4

8m3c

2, (7.70)

E

KE,nlm(1) =−

18m3c2

n,lm p̂4 n,lm . (7.71)

118

You can imagine that evaluating Eq. (7.71) in real-space integrals is very difficult. Here is

the trick. We note that

p

4 = 4m2 p2

2m

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

= 4m2H

0+

e2

r

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

(7.72)

so that

n,lm p̂4 n,lm = 4m2 (E

n(0))2 + 2E

n(0)e2 1

r+e 4 1

r 2

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥, (7.73)

where

1r

= n,lm1r

n,lm and

1r 2

= n,lm1r 2

n,lm . We find that (Homework problem)

e2 1

r=−2E

n(0) , (7.74)

e 4 1

r 2=

4nE02

l +1/ 2. (7.75)

The first-order relativistic kinetic energy corrections are

E

KE,nlm(1) =−

α4

2mc2 −

34n 4

+1

n3(l +1/ 2)

⎡

⎣⎢⎢

⎤

⎦⎥⎥ (7.76)

(2) The Darwin correction.

This is the correction due to the relativistic Dirac equation. Physically, when an electron ap-

proaches the nuclear, it will be “trembling”, a phenomenon called “zitterbewegung”. It is a con-

sequence of the interference of the electron with its antiparticle, which appears as a virtual parti-

cle when the energy of the electron comes close to its rest energy. The consequence of the zitter-

bewegung is that the electron sees a smeared-out Coulomb potential of the nucleus. Mathemati-

cally, the Darwin correction is found by acting with the Laplacian on the potential.

In the case of the Coulomb potential, this yields

∇2V(r) ∇2(1/ r) δ(r) , (7.77)

so that

H

DW= gδ(r), where g =

2π2e

2

2m2c

2. (7.78)

E

DW,nlm

(1) = nlm HDW

nlm = g | ψnlm

(r = 0) |2 . (7.79)

119

Because u

nl(r) = rR

nl(r) = rA

nl

ra

0

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

l

exp −Zrna

0

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟L

n−l−1(2l+1) 2Zr

na0

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟ [See Eq. (4.66) in Chapter 4],

we obtain

ψnlm

(r = 0) =

0 (l > 0)

1

π

1na

0

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

3/2

(l = 0),

⎧

⎨

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

(7.80)

and

EDW,nlm(1) =

0 (l > 0)

α4

2n3mc2 (l = 0),

⎧

⎨

⎪⎪⎪⎪

⎩⎪⎪⎪⎪

(7.81)

which indicates that the Darwin correction is positive and only applies to the s-wave.

(3) The spin-orbit interaction.

The spin-orbit coupling Hamiltonian is

H

so=

e2

2m2c

2r

3S ⋅L (7.82)

The total angular momentum is defined as

J = S + L , (7.83)

therefore

S ⋅L =

12(J 2−L2−S 2) . (7.84)

We need to construct the common eigenstates for J2 , L2 and S2 , designated as j,m;l,s ,

where j = | l −s |, | l −s | +1, , j + s−1, j + s and m = m

j=−j, − j +1, , j −1, j .

In this basis,

′j , ′m ; ′l ,s S ⋅L j,m;l,s = δ

j , ′jδ

m, ′mδ

l , ′l

2

2j(j +1)− l(l +1)−s(s +1)⎡⎣⎢

⎤⎦⎥ . (7.85)

Here s = 1/ 2 , so that j = l ±1/ 2 ( l ≠ 0 ) and j = s = 1/ 2 ( l = 0 ). For l = 0 , there is no

spin-orbit coupling and Eso(1) vanishes. For l ≠ 0 , we obtain

120

Eso,nlm(1) =

2e2

4m2c2

1r 3

l (j = l +1/ 2)

−(l +1) (j = l −1/ 2)

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

(7.86)

Using the result of (Again your homework problem)

1r 3

=1a

03

1n3l(l +1/ 2)(l +1)

, (7.87)

one finds the first-order spin-orbit interaction corrections

Eso,nlm(1) =

α4

4mc2 1

n3l(l +1/ 2)(l +1)

l (j = l +1/ 2)

−(l +1) (j = l −1/ 2)

⎧⎨⎪⎪⎪

⎩⎪⎪⎪

(7.88)

All these three corrections are of the α4 effects, and we combine them to write down the to-

tal fine-structure energy shifts

E

f.s.,nj(1) =−

α4

2n3mc2 1

j +1/ 2−

34n

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

. (7.89)

Some remarks:

1) Here j = l ±1/ 2 for l > 0 , while j = 1/ 2 for l = 0 .

2) For l > 0 , ΔEf.s.(1) comes from the spin-orbit interaction and relativistic kinetic energy;

however, when l = 0 , i.e., for the s-wave, ΔEf.s.(1) comes from the Darwin correction and

relativistic kinetic energy.

3) The exact fine-structure energy levels obtained from the Dirac Equation are

Enj

= mc2 1+α

n−(j +1/ 2)+ (j +1/ 2)2−α2

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

−12

−mc2 . (7.90)

Expand Eq. 6.79 to order α4 , we recover the perturbative results (Eq. 6.78), amazing!

4) Evaluating the correction due to the spin-orbit interaction requires the degenerate pertur-

bation theory, which will be discussed in next Chapter.