Physics 125cCourse Notes

Approximate Methods040415 F. Porter

Contents

1 Introduction 2

2 Variational Method 22.1 Bound on Ground State Energy . . . . . . . . . . . . . . . . . 22.2 Example: Helium Atom . . . . . . . . . . . . . . . . . . . . . 32.3 Other Applications of Variational Method . . . . . . . . . . . 62.4 Variational Theorem . . . . . . . . . . . . . . . . . . . . . . . 72.5 The “Ritz Variational Method” . . . . . . . . . . . . . . . . . 8

3 The WKB Approximation 103.1 Example: Infinite Square Well . . . . . . . . . . . . . . . . . . 123.2 Example: Harmonic Oscillator . . . . . . . . . . . . . . . . . . 13

4 Method of Stationary Phase 144.1 Application: Asymptotic Bessel Function . . . . . . . . . . . . 154.2 Application: Quantum Mechanics of Free Particle Asymptotic

Wave Function . . . . . . . . . . . . . . . . . . . . . . . . . . 164.3 Application: A Scattering Problem . . . . . . . . . . . . . . . 18

5 Stationary State Perturbation Theory 195.1 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6 Degenerate State Perturbation Theory 25

7 Time-dependent Perturbation Theory 267.1 The Time-Ordered Product . . . . . . . . . . . . . . . . . . . 287.2 Transition Probability, Fermi’s Golden Rule . . . . . . . . . . 297.3 Coulomb Scattering . . . . . . . . . . . . . . . . . . . . . . . . 367.4 Decays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377.5 Adiabatically Increasing Potential . . . . . . . . . . . . . . . . 39

8 Eigenvalues – Comparison Theorems 41

1

9 Exercises 45

1 Introduction

Typically, problems in quantum mechanics are difficult to solve exactly withanalytic methods. We thus resort to approximate methods, or to numericalmethods. In this note, I review several approximate approaches.

2 Variational Method

There are many applications of the technique of varying quantities to find auseful extremum. This is the gist of the “variational method”. As a meansof finding approximate solutions to the Schrodinger equation, a commonapproach is to guess an approximate form for a solution, parameterized insome way. The parameters are varied until an extremum is found. Weillustrate this approach with examples.

2.1 Bound on Ground State Energy

Given a system with Hamiltonian H, and ground state energy E0, we maynote that for any state vector |ψ〉 we must have:

〈ψ|H|ψ〉〈ψ|ψ〉

≥ E0. (1)

This suggests that we may be able to use some set of functions ψ, parame-terized in some way, to obtain an upper limit on the ground state energy,even if we cannot solve the problem exactly. With careful choice of “trial”function, we may even be able to get a good approximation to the energylevel. Thus, the program is to find the minimum of the quantity in Eqn. 1over variations in the parameter space to get a “least” upper bound on E0

for our trial wave functions:

δ〈ψθθθ|H|ψθθθ〉〈ψθθθ|ψθθθ〉

δθθθ= 0. (2)

Here, the parameter set to be varied is denoted θθθ.

2

2.2 Example: Helium Atom

The Coulombic Hamiltonian for the helium atom is:

H =p2

1

2m+

p22

2m− 2α

|x1|− 2α

|x2|+

α

|x1 − x2|(3)

= − 1

2m(∇2

1 + ∇22) −

2α

|x1|− 2α

|x2|+

α

|x1 − x2|, (4)

where α = e2, m is the mass of the electron, and we are neglecting the motionof the nucleus. Let

r1 = |x1|, (5)

r2 = |x2|, (6)

r12 = |x1 − x2|. (7)

Toward guessing a “good” trial wave function, note that, if the interactionterm α/r12 were not present, the ground state wave function would be simplya product of two hydrogenic ground state wave functions in x1 and x2:

ψ(x1,x2) =Z3

πa30

e− Z

a0(r1+r2)

, a0 =1

mα. (8)

There is no reason to expect that the α/r12 term to be especially “small”compared with the other terms, so the perturbation theory approach (dis-cussed later) may not work especially well here. However, let us use theabove wave function here in the variational method, and let Z be a variableparameter, to get an upper bound on the helium ground state energy.

We need to evaluate the expectation value of H. The kinetic energy ofone electron is:

〈ψ| p21

2m|ψ〉 =

∫

(∞)d3(x1)

Z3

πa30

e−Zr1/a0p2

1

2me−Zr1/a0

∫

(∞)d3(x2)

Z3

πa30

e−2Zr2/a0 (9)

= Z2 × kinetic energy of hydrogen atom ground state (10)

= Z2 1

2mα2. (11)

Thus,

〈ψ| p21

2m+

p22

2m|ψ〉 = Z2mα2. (12)

3

Similarly,

〈ψ| − 2α

r1|ψ〉 = 2Z × potential energy of hydrogen atom ground state

= −2Zmα2. (13)

A remark is in order: The “2” on the left hand side is for the Coulombpotential felt by electron number one in the Z = 2 field of the helium nucleus.Note that this “2” is from the Hamiltonian for helium - it is not a variationalparameter. The factor of Z that appears on the right side is a variationalparameter, since it arises from the trial wave function (〈1/r1〉 = Z/a0). Thus,we so far have:

〈ψ| 1

2m(p2

1 + p22) − 2α

(1

|x1|+

1

|x2|

)|ψ〉 =

1

2mα2(2Z2 − 8Z). (14)

It remains to evaluate the “interaction energy” between the two electrons,for our trial wave function:

〈ψ| αr12

|ψ〉 = α∫

(∞)

∫

(∞)d3(x1)d

3(x2)

(Z3

πa30

)2e− 2Z

a0(r1+r2)

|x1 − x2|. (15)

Let us digress for a moment here to obtain a couple of handy integrals.

Theorem: Let u, v, and w be three positive real numbers (one of whichmay also be zero). Then

1.

I(u, v;x,x′) ≡∫

(∞)d3(y)

exp(−u|y − x| − v|y − x′|)|y − x||y − x′|

=4π(e−v∆ − e−u∆

)

∆(u2 − v2), (16)

where ∆ ≡ |x − x′|.2.

J(u, v, w) ≡∫

(∞)d3(x)

∫

(∞)d3(y)

exp(−u|x| − v|y| − w|x− y|)|x||y||x− y|

=(4π)2

(u+ v)(v + w)(w + u). (17)

4

Proof: (sketch)

1. Let z = y−x, and let |z| = r. Then we may make the replacement

d3(y) → d3(z) = r2drd cos θdφ. (18)

Pick the 3-axis to be along x− x′ for the integration over angles:

|y − x′| = |z + x − x′| =√r2 + ∆2 + 2r∆ cos θ. (19)

Thus,

I(u, v;x,x′) = 2π∫ ∞

0re−urdr

∫ 1

−1d cos θ

exp(−v√r2 + ∆2 + 2r∆ cos θ)√

r2 + ∆2 + 2r∆ cos θ.

(20)Integrating over cos θ yields

I(u, v;x,x′) =2π

v∆

∫ ∞

0dre−ur

[e−v|r−∆| − e−v(r+∆)

]. (21)

Finally, integrate over r to obtain

I(u, v;x,x′) =4π(e−v∆ − e−u∆

)

∆(u2 − v2). (22)

2. Let x = |x|, y = |y|, and write:

J(u, v, w) = 4π∫ ∞

0x2dx2π

∫ ∞

0y2dy

∫ 1

−1d cos θ (23)

exp(−ux− vy − w√x2 + y2 − 2xy cos θ)

xy√x2 + y2 − 2xy cos θ

.

The integration then proceeds similarly as above.

We apply this theorem now to our problem.

〈ψ| αr12

|ψ〉 = α

(Z3

πa30

)2 ∫

(∞)

∫

(∞)d3(x)d3(y)

e− 2Z

a0(|x|+|y|)

|x − y|

= α

(Z3

πa30

)2

∂u∂vJ(u = 2Z/a0, v = 2Z/a0, 0) (24)

=1

2mα2 5

4Z. (25)

5

Thus,

〈ψ|H|ψ〉 =1

2mα2

(2Z2 − 27

4Z). (26)

The minimum is at Z = 27/16. Hence

〈ψ|H|ψ〉min = −1

2mα2

[2(

27

16

)2]

(27)

= −77.0 eV. (28)

Experimentally, the ground state energy of helium, from the first and secondionization energies, is

E0 = −(24.59 + 54.41) = −79.00 eV. (29)

We have come within about 2.5% of the right value by our variationalmethod with the “hydrogen” trial function. More careful variational cal-culations give good agreement. Note that the best value was obtained forZ = 27/16 instead of Z = 2. This is suggestive of the “screening” of the nu-cleus from each electron by the other electron, reducing the effective chargeby 5

16e.

2.3 Other Applications of Variational Method

We may derive other potentially useful relations in connection with the vari-ational approach. For example, let ψn|n = 0, 1, 2, . . . be an orthonormalset of true eigenfunctions of H, Hψn = Enψn, with E0 < E1 ≤ . . .. Then wecan expand our trial wave function (assumed to be normalized) in this basis:

ψ =∞∑

n=0

cnψn. (30)

We then obtain the bound:

〈ψ|H|ψ〉 = |c0|2E0 +∞∑

n=1

|cn|2En (31)

≥ |c0|2E0 + E1

∞∑

n=1

|cn|2 (32)

≥ (1 − |c0|2)(E1 − E0) + E0. (33)

6

Thus,

1 − |c0|2 = 1 − |〈ψ|ψ0〉|2 ≤〈ψ|H|ψ〉 − E0

E1 − E0, (34)

giving us a bound on how close our trial wave function is to the true groundstate wave function. Of course, this is useful only if we have sufficient knowl-edge of the spectrum.

Likewise, we can obtain a lower bound on E0:

Theorem: If we have a normalized function |ψ〉 such that

E0 ≤ 〈ψ|Hψ〉 ≤ E1, (35)

then

E0 ≥ 〈ψ|H|ψ〉 − 〈Hψ|Hψ〉 − 〈ψ|H|ψ〉2

E1 − 〈ψ|H|ψ〉. (36)

The proof of this will be left to the reader. To use this theorem, a lowerbound on E1 − 〈ψ|H|ψ〉 may be inserted.

2.4 Variational Theorem

We may put our intuition on a firmer foundation with the following “Varia-tional Theorem”:

Theorem: Let ψ ∈ H (such that 0 < 〈ψ|ψ〉 <∞), and define the followingfunctional on H:

E(ψ) ≡ 〈ψ|H|ψ〉〈ψ|ψ〉 , (37)

where H is the Hamiltonian operator, with a discrete spectrum. Then,any vector ψ for which the variation of E is stationary [that is, δE(ψ) =0] is an eigenvector in the discrete spectrum of H, and the correspond-ing eigenvalue of H is E(ψ).

Proof: Write:E(ψ)〈ψ|ψ〉 = 〈ψ|H|ψ〉. (38)

Take the variation of both sides:

δE〈ψ|ψ〉 + E〈δψ|ψ〉 + E〈ψ|δψ〉 = 〈δψ|H|ψ〉+ 〈ψ|H|δψ〉. (39)

7

Thus,〈ψ|ψ〉δE = 〈δψ|H − E|ψ〉 + 〈ψ|H − E|δψ〉. (40)

If (H−E)|ψ〉 = 0, then δE = 0, that is, if ψ is an eigenstate of H witheigenvalue E(ψ), then E is stationary. Suppose, instead, that δE = 0.In this case,

〈δψ|H − E|ψ〉 + 〈ψ|H − E|δψ〉 = 0. (41)

Note that ψ is complex, and the variation of the real and imaginaryparts are independent (the normalization is not constrained). We maydeal with this by considering the variation of iψ. If δE = 0, we have:

〈δ(iψ)|H − E|ψ〉 + 〈ψ|H − E|δ(iψ)〉 = 0. (42)

Hence,−i〈δ(ψ)|H − E|ψ〉 + i〈ψ|H − E|δ(ψ)〉 = 0. (43)

Combining with Eq. 41, we obtain:

〈δψ|H − E|ψ〉 = 0 (44)

〈ψ|H − E|δψ〉 = 0. (45)

This must hold for all variations δψ, implying that:

H|ψ〉 = E|ψ〉, (46)

completing the proof.

2.5 The “Ritz Variational Method”

We can modify our notion of taking a single trial wave function, with parame-ters to be varied, to a set of orthonormal trial wave functions. In particular,we could try to use an orthonormal set of solutions to a simpler, but prefer-ably related, problem.

Consider a finite set of such functions:

|n〉 : n = 0, 1, 2, . . . , N. (47)

Our trial wave function is constructed according to:

|ψ〉 =N∑

n=0

an|n〉. (48)

8

The an are N + 1 complex parameters to be varied. Let us impose thenormalization canstraint

N∑

n=0

|an|2 = 1, (49)

so that 〈ψ|ψ〉 = 1.We wish to find the parameters such that the variation of the expectation

of H vanishes:δ(〈ψ|H|ψ〉) = 0, (50)

subject to the constraint 49. Thus,

0 = δ

N∑

n=0m=0

〈ψ|m〉〈m|H|n〉〈n|ψ〉

(51)

= δ

N∑

n=0m=0

a∗man〈m|H|n〉

, (52)

subject to

0 = δ(1) = δ

(N∑

n=0

|an|2)

= δ

N∑

n=0m=0

a∗manδmn

. (53)

We may impose the constraint with the method of Lagrange multipliers:

0 = δ

(∑

m,n

a∗man〈m|H|n〉)− δ

(λ∑

m,n

a∗manδmn

)(54)

=∑

m,n

[δa∗man (〈m|H|n〉 − λδmn) + a∗mδan (〈m|H|n〉 − λδmn)] (55)

=∑

m,n

[δa∗man (〈m|H|n〉 − λδmn) + δama∗n (〈m|H|n〉∗ − λδmn)] , (56)

where the last equation was obtain via a relabeling of the m,n indices. Asin the previous section, we note that the real and imaginary parts may bevaried separately, and hence:

∑

m,n

δa∗man (〈m|H|n〉 − λδmn) = 0, (57)

∑

m,n

δama∗n (〈m|H|n〉∗ − λδmn) = 0. (58)

9

We may also vary each of the individual an’s separately, setting all δan = 0except for one. This gives:

∑

n

an (〈m|H|n〉 − λδmn) = 0. (59)

This may be rewritten, in our finite basis:

〈m|[∑

n

an(H − λI)|n〉]

= 0, or (60)

〈m|(H − λI)∑

n

an|n〉 = 0, ∀m. (61)

That is,∑n a

(i)n |n〉, i = 0, 1, . . . , N are eigenvectors of H, in this reduced

basis, with eigenvalues λ(i).If we managed to pick “good” functions |n〉, then the first few E(i) = λ(i)

may be good approximations to the first few energy levels of the full problem.

3 The WKB Approximation

The WKB (for Wentzel-Kramers-Brillouin) method makes use of the wavenature of the solutions to the Schrodinger equation between classical turningpoints. We’ll give a simple-minded treatment here; refinements are possible.

We wish to consider the problem of stationary states in a one-dimensionalpotential well (or equivalent one-dimensional potential for three-dimensionalproblems with spherical symmetry). Label the states by n, where n =0, 1, . . ., and E0 < E1 < . . .. Consider En, with classical turning points

x1, x2. The energy En corresponds to the (n+1)th state, or the nth “excitedstate.” The wave function will be oscillatory, with roughly n + 1 half-wavesbetween the classical turning points. Fig. 1 illustrates this for the fourthexcited state.

Let’s see how we can use this idea to estimate the energy levels. Consideran oscillatory solution of the form:

ψ(x) = A(x) sinφ(x), (62)

where A(x) > 0, and φ(x) is a phase increasing monotonically with x. When-ever φ(x) = kπ, where k is an integer, there is a node in ψ(x). Consider thechange in φ between turning points:

∆φ = φ(x2) − φ(x1). (63)

10

x

E

2 3 44(5 )

4

1

4-~~

π π πππ ππ

V(x)

x x1 2

Figure 1: Illustration of the WKB method. The classical turning points x1

and x2 are shown for the fourth excited state with energy E4. The phase iscounted between the turning points.

The figure is suggestive of ∆φ ≈ (n+ 12)π. More rigorously, we may state:

nπ < ∆φ ≤ (n + 1)π. (64)

It is readily seen that the right bound is achieved for an infinite square wellpotential. Typically, we simply make the choice (n + 1

2)π, although it may

not be difficult to do better in some circumstances.Now let us make an approximate calculation for this change in phase

according to the Schrodinger equation. For a region of constant V < En thewave function is

ψ(x) = A sin(x− x0)p, (65)

where A and x0 are constants, and

p =√

2m(E − V ). (66)

Thus, in a region of constant V , the phase varies as φ(x) = (x − x0)p. Thechange in phase as we increase x slightly is

dφ = pdx =√

2m(E − V )dx. (67)

11

Adding up all such contributions between the turning points yields:

∆φ =∫ x2

x1

dφ

dxdx ≈

∫ x2

x1

√2m(E − V )dx. (68)

The approximation is better the more slowly V (x) varies with x, since we haveused an integrand based on the assumption of constant V regions. Therefore,we have: ∫ x2

x1

√2m [En − V (x)]dx ≈ (n+

1

2)π. (69)

To use this result to estimate the energy levels En, we must first solve forthe turning points as a function of E:

V (x1) = V (x2) = E, x2 > x1, (70)

yielding the functions x1(E) and x2(E). Next, we solve for the function ofE:

f(E) =∫ x2(E)

x1(E)

√2m [En − V (x)]dx. (71)

Finally, we determine En as the solution to

f(En) = (n+1

2)π. (72)

This is the “WKB method”.

3.1 Example: Infinite Square Well

Consider the infinite square well potential:

V (x) =

0 0 < x < ∆∞ x ≤ 0 or x ≥ ∆.

(73)

We have x1 = 0, and x2 = ∆, independent of E. Thus,

f(E) =∫ ∆

0

√2mEdx =

√2mE∆. (74)

Letting

∆√

2mEn = (n+1

2)π, (75)

12

we obtain the result

En =(n+ 1

2)2π2

2m∆2, n = 0, 1, 2, . . . (76)

Actually, we can do better than this. As we noted earlier, the change inphase for this potential is really (n+1)π, since the turning points are nodes.Thus, we expect

En =(n + 1)2π2

2m∆2, n = 0, 1, 2, . . . (77)

This is, in fact, the exact result, which is to be expected, because the as-sumption of constant V is valid in this case.

3.2 Example: Harmonic Oscillator

Consider the one-dimensional simple harmonic oscillator potential:

V (x) =1

2kx2. (78)

Let ω0 =√k/m be the classical angular frequency. Find the turning points:

E =1

2kx2

1 =1

2kx2

2, (79)

−x1 = x2 = x0 ≡√

2E

k. (80)

Now find f(E):

f(E) =∫ x0

−x0

dx√

2m [E − V (x)] (81)

= 2√mkx2

0

∫ 1

0dx

√1 − x2 (82)

= 2√mkx2

0

∫ π/4

0cos2 θdθ (83)

= π√mk

x20

2(84)

= πE

ω0. (85)

13

Setting f(En) = (n+ 12)π, we arrive at the result:

En = (n+1

2)ω0, n = 0, 1, 2, . . . (86)

This turns out to be the exact spectrum for the bound states of the simpleharmonic oscillator potential.

4 Method of Stationary Phase

Suppose that we wish to evaluate an integral of the form (known as thegeneralized Fourier integral):

I(ε) =∫ ∞

−∞f(x)eiθ(x)/εdx, (87)

where f and θ are real, and ε > 0. If ε is very small, the phase oscillationis very rapid, and we may anticipate little contribution to the integral fromsuch a region. The dominant contribution may be expected to arise wherethe phase variation is slow, that is, where dθ/dx = 0. This is the idea behindthe method of stationary phase. Figure 2 illustrates the idea.

Let us pursue this notion. Suppose θ(x) has a (single) stationary pointat x = x0: θ

′(x0) = 0. We do a Taylor series expansion about this point,letting x = x0 +

√εu:

I(ε) =∫ ∞

−∞

[f(x0) +

√εuf ′(x0) + 1

2εu2f ′′(x0) +O(ε3/2)

]

expi[θ(x0)/ε+ 1

2u2θ′′(x0) + 1

3!

√εu3θ′′′(x0) +O(ε)

]√εdu (88)

=√εf(x0)e

iθ(x0)/ε∫ ∞

−∞

[1 +

√εuf ′(x0)

f(x0)+ 1

2εu2f

′′(x0)

f(x0)+O(ε3/2)

]

ei2u2θ′′(x0) exp

[i3!

√εu3θ′′′(x0) +O(ε)

]du (89)

=√εf(x0)e

iθ(x0)/ε∫ ∞

−∞

1 +

√ε

[uf ′(x0)

f(x0)+ 1

3!u3θ′′′(x0) +O(ε)

]

ei2u2θ′′(x0)du. (90)

The terms odd in u vanish on integration, leaving:

I(ε) =√εf(x0)e

iθ(x0)/ε∫ ∞

−∞ei2u2θ′′(x0)du

14

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

Figure 2: Illustration for the method of stationary phase. The smooth con-cave down curve is f(x). The smooth concave up curve is θ(x). The oscil-lating curve of constant amplitude is the real part of eiθ(x)/ε. The remainingoscillatory curve is the real part of the desired integrand: f(x)eiθ(x)/ε. In this

illustration, ε = 0.01, f(x) = 0.8 exp[−1

2(x− 0.2)2

], and θ(x) = x2.

=√εf(x0)e

iθ(x0)/εeiπ4

sign θ′′(x0)

√2π

|θ′′(x0)|[1 +O(ε)] . (91)

For multiple stationary points, we simply sum over this expression evaluatedat each stationary point.

4.1 Application: Asymptotic Bessel Function

Suppose we wish to evaluate the Jn(z) Bessel function for large z. We startwith the integral representation:

Jn(z) =1

inπ

∫ π

0eiz cosφ cos(nφ)dφ. (92)

The role of ε is carried by 1/z. The quantity θ(φ) is just cosφ. This hasstationary points at φ = 0 and φ = π, within the region of integration. There

15

is a potential difficulty with the fact that these points are at the limits of theintegration. However, we may use the fact that the integrand is symmetricabout φ = 0, and periodic, to write:

Jn(z) =1

2inπ

∫ 3π/2

−π/2eiz cosφ cos(nφ)dφ. (93)

We need to sum Eqn. 91 over the two stationary points, with

f(0) = cos(n0) = 1, (94)

f(0) = cos(nπ) = (−1)n, (95)

θ(0) = −θ(π) = 1, (96)

θ′′(0) = −θ′′(π) = −1. (97)

The result is, for large z:

Jn(z) ∼ 1√2πz

1

2πin

[ei(z−π/4 +

(eiπ)ne−i(z−π/4

](98)

=1√2πz

[ei(z−nπ/2−π/4 + e−i(z−nπ/2−π/4

](99)

=

√2

πzcos

(z − nπ

2− π

4

). (100)

This is the familiar result.The reader may wish to go back and verify that the application of Eqn. 91

is really proper, as we have violated some of the assumptions made in ob-taining that result. Indeed, a more general treatment may be obtained basedon the “Reimann-Lebesgue Lemma”:

Lemma: Let |f(x)| be an integrable function of real variable x, and θ(x) bea continuosly differentiable real function on the interval [a, b], exceptthat θ is nowhere constant on any finite subinterval of [a, b]. Then

I(ε) ≡∫ b

af(x)eiθ(x)/εdx→ 0 as ε→ 0 + . (101)

4.2 Application: Quantum Mechanics of Free Particle

Asymptotic Wave Function

Let’s try an application of the method of stationary phase in quantum me-chanics. Consider the free particle (in one dimension), and the question:

16

What is the asymptotic behavior of the wave function as t → ∞? Supposein particular, we are interested in a wave function which in momentum spaceis localized around a momentum q at time t = 0. The position space wavefunction at t = 0 is:

ψ(x, 0) =1

2π

∫dpψ(p)eipx. (102)

Taking ψ to be real, we find that |ψ(x, 0)| is symmetric about 0. If we furthersuppose that ψ > 0, then |ψ| is maximal at x = 0 (alternatively, we couldchoose our coordinate system such that this is the case1).

The time evolution of the wave function is:

ψ(x, t) =1

2π

∫dpψ(p)eipx−itp

2/2m. (103)

The phase of interest here is

θ(p)/ε = px− tp2/2m. (104)

This is stationary at p = p0, where:

0 =1

ε

dθ

dp

∣∣∣p0

= x− pomt. (105)

Now we use equation 91 to determine the asymptotic form of ψ(x, t) ast→ ∞, with ε = 1/t. The stationary point is p0 = mx/t, and

θ(p0) =m

2

x2

t2, (106)

θ′′(p0) = −1/m, (107)

f(p0) =1

2πψ(mx/t). (108)

Thus,

ψ(x, t) → 1√tψ(mx/t)eimx

2/2te−iπ/4√m/2π, as t→ ∞, (109)

= ψ(mx/t)

√m

2πte−iπ/4eimx

2/2t. (110)

1You may wish to recall that a translation by x0 in position corresponds to multiplica-tion by a phase e−ipx0 in momentum space

17

The probability desnity, at large times, is thus:

|ψ(x, t)|2 = |ψ(mx/t)|2 m2πt

. (111)

Note that this is peaked around x = qt/m, if ψ(p) is peaked around q, givingthe appropriate classical correspondence for the motion. The probabilitydensity falls as 1/t, but also spreads out as 1/t (due to the 1/t in the argumentof ψ). Probability is conserved:

∫|ψ(x, t)|2dx =t→∞

∫|ψ(mx/t)|2 m

2πtdx =

1

2π

∫|ψ(p)|2dp = constant.

(112)

4.3 Application: A Scattering Problem

Let us try another example, a one-dimensional scattering problem. Supposethat there is a localized potential V (x):

V (x) = 0, |x| > L. (113)

For a given (positive) momentum component, and x > |L|, E = p2/2m, andthe wave function corresponding to this momentum is:

ψE(x) =

eipx +R(E)e−ipx, x < −L,T (E)eipx, x > L.

(114)

The solution to the time-dependent Schrodinger equation, for |x| > L, is

ψ(x, t) =1

2π

∫dpψ(p)ψE(x)e−iEt, (115)

where we have now summed over the plane wave components to obtain aphysical wave packet. Again, we assume that ψ(p) is localized around p = q.Note that, if V (x) = 0, this is just

ψ(x, t) =1

2π

∫dpψ(p)eipxe−ip

2t/2m, (116)

as in the previous section.Consider x > L: Let T (E) = |T (E)|eiδ(E). Then,

ψ(x, t) =1

2π

∫dpψ(p)|T (E)|eiδ(E)ei(px−Et). (117)

18

The phase factor is:

φ =1

εθ = δ(E) + px− Et. (118)

This phase is stationary at dφ/dp = 0. The solution is p = p0(x, t). Now,significant contributions to ψ(x, t) will only occur when p = p0(x, t) is nearq. Hence, we may approximate

δ(E) ≈ δ(E0) + (E − E0)τ(E0), (119)

where E0 ≡ q2/2m and τ(E) ≡ ∂δ(E)/∂(E).Thus, our problem is to evaluate:

ψ(x, t) ≈ eiδ(E0)−iE0τ(E0)∫dp

2πψ(p)|T (E)|ei[px−E(t−τ(E0)]. (120)

This is just like our free particle case, Eqn. 103, except t is replaced witht− τ(E0). Hence, by the method of stationary phase, as t→ ∞, for x > L:

ψ(x, t) = ψ

[m

x

t− τ(E0)

]√m

2π [t− τ(E0)]e−iπ/4eimx

2/2[t−τ(E0)]|T (E)|ei[δ(E0)−E0τ(E0)].

(121)The asymptotic probability density is:

|ψ(x, t)|2 =

∣∣∣∣∣ψ[m

x

t− τ(E0)

]∣∣∣∣∣

2m

2π [t− τ(E0)]

∣∣∣∣∣T

[mx/(t− τ(E0))]2

2m

∣∣∣∣∣

2

,

(122)

where we have substituted E = [mx/(t−τ(E0))]2

2m.

This result is of the form of the free particle probability density times thetransmission probability. However, the time t is replaced by t− τ(E0). Thatis, the outgoing wave is delayed relative to free propagation (V = 0) by atime delay given by τ(E) = dδ/dE. It is suggested that the reader considerthe classical correspondence for this effect. We will encounter this notionagain when we discuss phase shifts in scattering theory.

5 Stationary State Perturbation Theory

We are frequently concerned with the problem of determining the stationarystate eigenvalues and eigenfunctions of the Hamiltonian. There is a rela-tively systematic iterative approach to solving this problem, if a suitable

19

first approximation can be found. This is the method of stationary stateperturbation theory, which we develop with a simplified approach here.

Suppose we are given a Hamiltonian of the form H = H0 + V , where weknow the eigenfunctions and eigenvalues corresponding to H0:

H0|n〉 = εn|n〉. (123)

We are interested in solving, at least approximately, the problem:

H|N〉 = EN |N〉. (124)

If V is “small”, we expect that |n〉 and εn will be approximate eigenfunctionsand eigenvalues of H. We use this idea to form an iterative expansion for|N〉 and EN .

We introduce a “bookkeeping” parameter, λ, to count powers (of V ) inthe expansion. Let H(λ) ≡ H0 + λV . We expect the eigenstates of H(λ) tovary smoothly from eigenstates of H0 at λ = 0 to eigenstates of H at λ = 1.Thus, consider the following series:

|N〉 = |n〉 + λ|N1〉 + λ2|N2〉 + . . . , (125)

En = εn + λEn1 + λ2En2 + . . . (126)

Note, however, that such an expansion does not always make sense. Forexample, we cannot do perturbation theory on the free particle problem tosolve a bound state problem, no matter how “weak” the potential.

We need to develop an algorithm to solve for the terms in the series.Assume our unperturbed system of eigenstates is orthonormal:

〈n|m〉 = δnm. (127)

Normalize the eigenstates of H so that:

〈n|N〉 = 1. (128)

Assuming our perturbation is not too large, this should be possible. Ofcourse, we won’t have 〈N |N〉 = 1 in general, and will have to renormalizethese functions at the end. Then we have

1 = 〈n|N〉 = 〈n|n〉 + λ〈n|N1〉 + λ2〈n|N2〉 + . . . (129)

20

The coefficient of each power of λ must vanish, so that:

〈n|Nk〉 = 0, k = 1, 2, . . . (130)

Now consider the Schrodinger equation:

(H0 + λV )|N〉 = EN |N〉, (131)

or

(H0 + λV )(|n〉 + λ|N1〉 + λ2|N2〉 + . . .) (132)

= (εn + λEN1 + λ2EN2 + . . .)(|n〉 + λ|N1〉 + λ2|N2〉 + . . .).

Equating powers of λ on both sides, we obtain:

λ0 : H0|n〉 = εn|n〉 (133)

λ1 : H0|N1〉 + V |n〉 = ε|N1〉 + EN1|n〉 (134)

λk : H0|Nk〉 + V |Nk−1〉 =k∑

j=0

Enj|Nk−j〉, (135)

where

En0 ≡ εn (136)

|N0〉 ≡ |n〉. (137)

Consider the λ1 equation, and take the scalar product with 〈n|:

〈n|H0|N1〉 + 〈n|V |n〉 = 〈n|εn|N1〉 + 〈n|EN1|n〉. (138)

Thus, the first order correction to the nth energy level is

EN1 = 〈n|V |n〉. (139)

Or, to first order in the potential V (setting λ = 1):

EN = εn + 〈n|V |n〉 +O(V 2). (140)

This is the most commonly used equation in stationary state perturbationtheory.

21

In general, we find from Eqn. 135:

Enk = 〈n|V |Nk−1〉. (141)

If we know the (k − 1)th order correction to the wave function, we canobtain the kth order energy correction. To find the wave function corrections,expand in the eigenstates of H0:

|Nk〉 =∑

m 6=n|m〉〈m|Nk〉, k = 1, 2, . . . , (142)

where the m = n terms in the sum are excluded since 〈n|Nk〉 = 0, ∀k > 0.Now take the scalar product of Eqn. 135 with 〈m| to get the expansioncoefficients:

〈m|H0|Nk〉 + 〈m|V |Nk−1〉 =k∑

j=0

Enj〈m|Nk−j〉 (143)

εm〈m|Nk〉 + 〈m|V |Nk−1〉 = εn〈m|Nk〉 +k−1∑

j=1

Enj〈m|Nk−j〉, (144)

where the j = k term in the sum may be excluded, since 〈m|n〉 = 0. Thus,if εm 6= εn:

〈m|Nk〉 =1

εn − εm

〈m|V |Nk−1〉 −

k−1∑

j=1

Enj〈m|Nk−j〉

. (145)

We’ll try an example – estimating the helium ground state energy. LetH = H0 + V , with

H0 =p2

1

2m+

p22

2m− 2α

r1− 2α

r2(146)

V =α

r12. (147)

The ground state wave function for the unperturbed H0 is just

|n = 0〉 = |0〉0 =Z3

πa30

e− Z

a0(r1+r2), with Z = 2. (148)

The unperturbed ground state energy is:

ε0 = −1

2mα2(2Z2) = −108.9 eV. (149)

22

The first order correction to the energy is

E01 = 0〈0|V |0〉0 =∫

(∞)d3(x1)

∫

(∞)d3(x2)

(Z3

πa30

)2

e− 2Z

a0(r1+r2) α

r12. (150)

This is an integral we already evaluated, hence

E01 =1

2mα2

(5

4Z)

Z=2= 34.0 eV. (151)

The ground state energy, to first order in perturbation theory is thus

E0 = ε0 + 0〈0|V |0〉0 +O(V 2) (152)

= −1

2mα2

(2Z2 − 5

4Z)

+ O(V 2) (153)

= −74.9 eV +O(V 2). (154)

Our first order perturbation theory correction to the energy is 34 eV, to beapplied to the zeroth order level at -109 eV. This is a rather large correction,so it isn’t surprising that the result is still 4.1 eV away from the observedvalue of -79.0 eV. Note that the variational calculation we performed doesbetter – it really is the same calculation, except that there in addition weallowed Z to vary to accommodate the screening effect.

We might guess that the size of the second order effect could be estimatedby squaring the size of the first order correction:

O(V 2) ≈ 74.9(

34.0

108.9

)2

= 7.3 eV. (155)

This is certainly of the right order, but is only an order-of-magnitude esti-mate. To do better, we should perform the second order perturbation theorycalculation:

En2 = 〈n|V |N1〉 (156)

=∑

m 6=n〈n|V |m〉〈m|N1〉 (157)

=∑

m 6=n

|〈m|V |n〉|2

εn − εm. (158)

The second order correction depends on the “overlap” between 〈m| and V |n〉where 〈m| 6= 〈n|, and also on the energy level spacing in the unperturbedHamiltonian.

23

5.1 Normalization

We chose the normalization 〈n|N〉 = 1 in order to make the expansion conve-nient. The drawback is that |N〉 is not normalized to unit total probability.We may renormalize to obtain a normalized wave function as follows: Let|N〉 =

√A|N〉 such that 〈N |N〉 = 1. The constant A = 1/〈N |N〉 is called

the “wave function renormalization constant”. Take

√A =

√A〈n|N〉 = 〈n|N〉. (159)

Let’s compute A to second order in V :

1

A= 〈N |N〉 (160)

=(〈n| + λ〈N1| + λ2〈N2|

) (|n〉 + λ|N1〉 + λ2|N2〉

)+O(λ3) (161)

= 1 + λ2〈N1|N1〉 +O(λ3) (162)

= 1 + λ2∑

m 6=n〈N1|m〉〈m|N1〉 +O(λ3) (163)

= 1 + λ2∑

m 6=n

|〈m|V |n〉|2

(εn − εm)2+ . . . (164)

We see that |N〉 is normalized already to one to first order, with correctionsonly appearing at second order.

To second order,

A = 1 − λ2∑

m 6=n

|〈m|V |n〉|2

(εn − εm)2+O(λ3)

=∂

∂εn

εn + λ〈n|V |n〉 + λ2

∑

m 6=n

|〈m|V |n〉|2

εn − εm

+O(λ3)

=∂En∂εn

+O(λ3). (165)

This result is actually valid to all orders in perturbation theory: If a systemis in eigenstate |N〉 =

√A|N〉 of the perturbed Hamiltonian, the probability,

A = |〈n|N〉|2, to observe it in the unperturbed state, |n〉, is just the partialderivative, ∂En/∂εn, of the perturbed energy with respect to the unperturbedenergy. The partial derivative here means keeping εm (m 6= n) and 〈m|V |n〉fixed.

24

6 Degenerate State Perturbation Theory

If εn = εm and 〈n|V |M〉 6= 0, then the perturbation theory we developedabove breaks down. This is not an essential difficulty, however, and weaddress getting around it here.

Suppose that a set of states,

|n1〉, |n2〉, . . . , |n`〉, (166)

are degenerate with respect to H0:

H0|ni〉 = εn|ni〉, i = 1, 2, . . . , `. (167)

As we already noticed, if 〈ni|V |nj〉 6= 0 for i 6= j, the perturbation theory wehave developed breaks down. However, we may choose any linear combina-tions of these degenerate states and again obtain eigenstates of H0 with thesame eigenvalue. Thus, choose the set which diagonalizes V in this subspace:

|n′i〉 =

∑

j=1

bij|nj〉, (168)

such that〈n′

i|V |n′j〉 = 0, i 6= j (169)

The matrix B = bij which does this is just the matrix formed by thenormalized eigenvectors of the V matrix (in this `-dimensional subspace):

VBj = λjBj, (170)

where∑

i=1

|(Bj)i|2 = 1. (171)

Thus,

|n′i〉 =

∑

j=1

(Bi)j|nj〉. (172)

With this change in basis, we recover our original first order perturbationtheory result for the wave functions:

|N ′i〉 = |n′

i〉 +∑

m/∈nk

|m〉〈m|V |n′i〉

εn − εm+ . . . (173)

25

We also recover our original second order perturbation theory result for theenergies:

Eni= εn + 〈n′

i|V |n′i〉 +

∑

m/∈nk

|〈m|V |n′i〉|2

εn − εm+ . . . (174)

7 Time-dependent Perturbation Theory

We now add an element of time variation to our discussion of perturbations.We’ll develop the basic ideas here; they are especially useful in the applicationto scattering theory, which we’ll treat at more length in a later note.

Suppose that, at some time t < t0, the system is in a state |ψ0t 〉, satisfying

the Schrodinger equation:

i∂t|ψ0t 〉 = H0|ψ0

t 〉, t < t0. (175)

At time t = t0, we turn on a perturbing potential, Vt. For t > t0, we thusmust solve:

i∂t|ψt〉 = (H0 + Vt)|ψt〉, t > t0, (176)

with the boundary condition

|ψt〉 = |ψ0t 〉, t ≤ t0. (177)

Often, Vt is “small enough” so that we can find an approximate solution,even if an exact solution is too daunting a prospect. As for the stationarystate perturbation theory, we make an expansion in powers of Vt. Note that|ψt〉 contains the time dependence from H0, and if H0 Vt, we expect thisto be a large portion of the time dependence. As this isn’t usually the timedependence of interest in such problems, it is convenient to factor it out bywriting:

|ψt〉 = e−iH0t|ψ(t)〉. (178)

Then,i∂t|ψt〉 = H0|ψt〉 + e−iH0ti∂t|ψ(t)〉, (179)

and hence:

e−iH0ti∂t|ψ(t)〉 = (i∂t −H0)|ψt〉 (180)

= Vt|ψt〉 (181)

= Vte−iH0t|ψ(t)〉. (182)

26

If we defineV (t) ≡ eiH0tVte

−iH0t, (183)

we may writei∂t|ψ(t)〉 = V (t)|ψ(t)〉. (184)

We see that V (t) looks like a Hamiltonian for “state” |ψ(t)〉. We call |ψ(t)〉the state vector in the interaction representation, implying that the timedependence is due only to the interaction. The operator V (t) is the inter-action representation for operator Vt. We may note that, if Vt = 0, theinteraction representation is just the Heisenberg representation.

Now integrate with respect to time:

∫ t

t0∂t|ψ(t)〉dt = |ψ(t)〉 − |ψ(t0)〉, (185)

or,

|ψ(t)〉 = |ψ(t0)〉 +1

i

∫ t

t0V (t1)|ψ(t1)〉dt1. (186)

This suggests that we try an iterative solution, which we hope converges.Thus, to first order in V :

|ψ(t)〉 = |ψ(t0)〉 +1

i

∫ t

t0V (t1)|ψ(t0)〉dt1. (187)

To find the second order solution, we substitute the approximation of Eqn. 187into Eqn. 186:

|ψ(t)〉 = |ψ(t0)〉+1

i

∫ t

t0dt1V (t1)|ψ(t0)〉+

(1

i

)2 ∫ t

t0dt1V (t1)

∫ t1

t0dt2V (t2)|ψ(t0)〉.

(188)In general, we see that the nth order correction in this expansion is:

(1

i

)n ∫ t

t0dt1

∫ t1

t0dt2 . . .

∫ tn−1

t0dtnV (t1)V (t2) . . . V (tn)|ψ(t0)〉. (189)

If we define a kernel as K(t′, t) = V (t)θ(t′− t), we see that this series is reallyin the form of a Neumann series for the solution of an integral equation.

27

7.1 The Time-Ordered Product

There is another interesting way to express this series result. Let us define theconcept of a time-ordered product, denoted by ABt, of two operatorsA and B, according to:

A(t1)B(t2)t =A(t1)B(t2), if t1 ≥ t2,B(t2)A(t1) if t1 < t2.

(190)

With this idea, consider

[∫ t

t0V (t1)dt1

]2

t

=∫ t

t0dt1V (t1)

∫ t

t0dt2V (t2)

t

(191)

=∫ t

t0dt1

∫ t

t0dt2 V (t1)V (t2)t (192)

=∫ t

t0dt1

∫ t1

t0dt2 V (t1)V (t2)t +

∫ t

t0dt2

∫ t2

t0dt1 V (t2)V (t1)t

= 2∫ t

t0dt1

∫ t1

t0dt2 V (t1)V (t2)t . (193)

In general, we find:

[∫ t

t0V (t1)dt1

]n

t

=∫ t

t0dt1

∫ t

t0dt2 . . .

∫ t

t0dtn V (t1)V (t2) . . . V (tn)t

= n!∫ t

t0dt1

∫ t1

t0dt2 . . .

∫ tn−1

t0dtnV (t1)V (t2) . . . V (tn). (194)

Thus,1

inn!

[∫ t

t0V (t1)dt1

]n

t

|ψ(t0)〉 (195)

is the nth order term in our expansion for |ψ(t)〉. We might think of this interms of the picture in Fig. 3: Imagine that, in propagating from time t0 totime t, the wave “interacts” with the potential at discrete times t1, t2, . . . tn.To get the total evolution of the wave, we must integrate over all possibleinteraction times, and sum over all possible numbers of interactions. Whenwe sum over all terms in this expansion, we find:

|ψ(t)〉 =exp

[−i∫ t

t0V (t1)dt1

]

t

|ψ(t0)〉. (196)

28

t

t

t

t

t

0

n

n-1

1

V

V

V

Figure 3: Illustration of nth order interaction term in time evolution from t0to t.

7.2 Transition Probability, Fermi’s Golden Rule

Suppose a system is initially (at t = t0) in eigenstate |i〉 of H0:

|ψ(t)〉 = |i〉, where H0|i〉 = εi|i〉. (197)

Let |f〉 denote an arbitrary eigenstate of H0. For convenience, we choose |i〉,|f〉 to be in the interaction representation.

We wish to address the question: At time t > t0, what is the probabilitythat the system will be observed in state |f〉, i.e., what is the probabilitythat the interaction has caused the transition |i〉 → |f〉? In the interactionrepresentation, this amplitude is 〈f |ψ(t)〉. Using Eqn. 187, we have to firstorder in V :

〈f |ψ(t)〉 = 〈f |i〉 +1

i

∫ t

t0dt1〈f |V (t1)|i〉. (198)

We find for the matrix element in the integrand:

〈f |V (t1)|i〉 = 〈f |eiH0t1Vt1e−iH0t1 |i〉 (199)

= ei(εf−εi)t1〈f |Vt1|i〉. (200)

29

Hence, if 〈f |i〉 = 0, the transition amplitude is

〈f |ψ(t)〉 =1

i

∫ t

t0dt1e

i(εf−εi)t1〈f |Vt1|i〉. (201)

The transition probability is:

Pi→f(t) = |〈f |ψ(t)〉|2 =

∣∣∣∣∫ t

t0dt1e

i(εf−εi)t1〈f |Vt1 |i〉∣∣∣∣2

. (202)

For example, suppose the potential is “turned on” at time t = 0, and isconstant thereafter:

Vt =

0, t < t0 = 0,V = V (x), t > 0.

(203)

In this case,

Pi→f(t) =

∣∣∣∣∫ t

t0dt1e

i(εf−εi)t1〈f |V |i〉∣∣∣∣2

(204)

=

[sin

εf−εi

2t

εf−εi

2

]2

|〈f |V |i〉|2 . (205)

This probability is plotted as a function of the energy difference in Fig. 4.The zeros occur at εf = εi± 2πn

t, n = 1, 2, . . .. The magnitudes of the bumps

decrease as 1/(εf − εi)2.

For very small times,

Pi→f(t) ≈ t2 |〈f |V |i〉|2 , (206)

approximately independent of εf − εi, if |〈f |V |i〉|2 is not very dependent. Ast increases, the probability is largest for states with εf near εi – the heightof the central bump varies approximately as t2, and the width as 1

t, yielding

a total probability to be in the central bump that grows approximately ast. This may be thought of in terms of an “uncertainty relation”: If theperturbation turns on, or acts, in a very short time ∆t, transitions may beinduced in first order to a wide range of energy states,

∆ε∆t>∼2π

∆t∆t = 2π. (207)

30

0

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4

Figure 4: The i→ f transistion probability as a function of energy difference.The vertical axis is Pi→f(t)/ |〈f |V |i〉|2 in units of t2. The horizontal axis isεf − εi in units of 2π/t.

But as the interaction is turned on more slowly, or has acted for a longertime, the uncertainty in energy induced by the perturbation decreases, i.e.,energy is conserved to ∆ε ∼ 2π/t.

If the levels εf and εi are discrete, with εf 6= εi, the transition prob-ability simply oscillates with period T = 2π/|εf − εi|. If |f〉 and |i〉 aredegenerate, then the probability grows as t2. This cannot continue indefi-nitely, since probabilities are bounded by one. Eventually, higher orders inthe perturbation become important.

Consider the case where |f〉 is drawn from a continuum of energy states(or, perhaps a very closely spaced spectrum). For example, we could bedealing with a free particle in H0. In this case it makes more sense to ask forthe transition probability to some set of states in a neighborhood of |n〉. Forexample, for a free particle, we are interested in the transition probabilityto phase space volume element d3(p) about p. Since the area of the centralbump grows as t, we expect the transition probability to a set of such states

31

with εf ≈ εi to grow linearly with time. Hence, the transition rate to such aset of states is a constant. Let us calculate the transition rate for this case.

We must sum Eqn. 205 over the region of interest (call this region R):

∑

f∈RPi→f(t) =

∫

Rdεfρ(εf )

[sin

εf−εi

2t

εf−εi

2

]2

|〈f |V |i〉|2 , (208)

where ρ(εf) is the number of states per unit energy, known as the densityof states. Let’s suppose |〈f |V |i〉|2 doesn’t change much over the region ofinterest, and take this quantity outside the integration:

∑

f∈RPi→f(t) ≈ |〈f |V |i〉|2

∫

Rdεfρ(εf)

[sin

εf−εi

2t

εf−εi

2

]2

, (209)

As t becomes larger, the central bump falls entirely within R, and then thedensity of states can also be considered effectively constant over the sharplypeaked integrand. Hence, we take ρ(εf) outside the integral also, in thislimit:

∑

f∈RPi→f(t) ≈

[|〈f |V |i〉|2 ρ(εf )

]εf=εi

∫

Rdεf

[sin

εf−εi

2t

εf−εi

2

]2

, (210)

Finally, since the central peak is contained entirely within R, we may let thelimits of integration go to ±∞.

We wish to evaluate the integral:

∫ ∞

−∞dεf

[sin

εf−εi

2t

εf−εi

2

]2

= 2t∫ ∞

−∞dx

sin2 x

x2. (211)

One way to compute this integral is to notice that

sin2 x =1

2(1 − cos 2x) = <1 − e2ix

2, (212)

and consider the contour integral:

1

2

∮

C

1 − e2iz

z2dz = 0 (213)

around the contour in Fig. 5 The integral around the large semicircle is zero

32

x

y

Rε

C

Figure 5: Contour for the evaluation of the integral in Eqn. 213.

in the limit R → ∞. The desired integral is thus minus the integral aroundthe small semicircle, in the limit ε→ 0:

∫ ∞

−∞dx

sin2 x

x2= − lim

ε→0

∫ 0

πiε

1 − e2iεeiθ

2ε2e2iθdθ (214)

= − i

2(−2i)

∫ 0

πdθ (215)

= π. (216)

Hence, ∑

f∈RPi→f(t) ≈ Γt, (217)

where the transition rate, Γ, is:

Γ = 2π[|〈f |V |i〉|2 ρ(εf)

]εf=εi

. (218)

Equation 218 is an important result; it is known as Fermi’s Golden Rule.Our discussion is evidently not valid when:

1. The time t is too “short”. We must have the central bump within theregion of interest. That is, we must have (∆ε)R large compared with

33

2π/t, i.e.,

t >2π

(∆ε)R. (219)

2. The time t is too “long”. If t is too long, then there may be only afew states within the central bump (not a problem if the spectrum iscontinuous, of course). Suppose δε is the level spacing in the region ofinterest. This spacing must be small compared with 2π/t for the aboveanalysis. That is, we must have

t 2π

δε. (220)

Furthermore, if t becomes too long, the initial state becomes depleted,and the transition rate will no longer be constant.

Let us apply this framework to the case of a particle in a box of volumeL3.2 Turn on the potential V (x) inside the box. Start with a particle inmomentum state p, and ask for the rate at which it transitions to othermomentum states, p′. The matrix element of V (x) between momentumstates is

〈p′|V |p〉 =∫

L3d3(x)

e−ip′·x

L3/2V (x)

eip·x

L3/2(221)

= V (p′ − p)/L3, (222)

where V is the fourier transform (in the box) of V (x).To put this into the golden rule, we restate the golden rule somewhat:

We notice that as t grows, the function

[sin

εf−εi

2t

εf−εi

2

]2

→t→∞ 2πtδ(εf − εi). (223)

Then our transition rate is:

Γ = 2π |〈f |V |i〉|2 δ(εf − εi). (224)

This version of Fermi’s Golden Rule must be applied in the context of a sumover states |f〉, that is, there must be an integral over the delta function.

2We have in mind that we will eventually take the limit as the box size becomes infinite,and develop this into a theory for scattering.

34

Using Eqn. 224, the transition rate for p → p′ is:

Γp→p′ = 2π|V (p′ − p)|2

L6δ(ε′p − εp), (225)

where

εp =p2

2m. (226)

The rate Γ here is actually a differential decay rate. Let us apply it to obtainthe rate of scattering, dΓ, into an element of solid angle dΩ′:

dΓ =∑

p′∈dΩ′Γp→p′. (227)

To perform this summation, we need the density of states, i.e., we needthe number of states in phase space element d3(p′). On dimensional grounds,we must have a number of states dN ′:

dN ′ ∝ L3d3(p′), (228)

and it remains to determine the constant of proportionality. Let’s considerthe problem in one dimension. The free particle wave functions are

ψp(x) =1√Le±ipx. (229)

Imposing periodic boundary conditions ensures no net flux of particles outof the box3:

ψ(x) = ψ(x+ L) (230)

ψ′(x) = ψ′(x + L). (231)

Thus, we must have eipL = 1, or pL = 2πn, where n is an integer. Hence,

dN

dp=

L

2π. (232)

We generalize to three dimensions to obtain

dN =L3

(2π)3d3(p). (233)

3Note that we are really thinking in terms of eventually letting the box boundaries gooff to infinity, and the constraint we want is conservation of probability. We don’t carehere whether the wave function goes to zero at the box boundary.

35

Thus, we have

dΓ =∫

p′∈dΩ′

L3

(2π)3d3(p′)2π

|V (p′ − p)|2

L6δ(εp′ − εp) (234)

=dΩ′

L3

m

(2π)2

∫ ∞

0p′dεp′|V (p′ − p)|2δ(εp′ − εp) (235)

=dΩ′

L3

mp

(2π)2|V (p′ − p)|2, (236)

where p′ = |p|ΩΩΩ′, with ΩΩΩ

′a unit vector in dΩ′.

We may think of this example as a “scattering experiment” since thepotential is effectively “turned on” as the incident particle nears it. As long asthe potential falls off rapidly enough at large distances, the use of free particlewave functions for the incident wave at early times and for the scattered waveat late times is a plausible approximation to make. If we suppose that wehave a beam of incident particles, then Eqn. 236 tells us the rate at whichparticles scatter into dΩ′ per incident particle in volume L3. The flux ofparticles per incident particle of momentum p in volume L3 is just:

Number of beam/area/time =1

L3|v| =

p

mL3. (237)

This may be seen by interpreting the factor 1/L3 as the number of beamparticles per unit volume (i.e., we have normalized our wave to one parti-cle in the box of volume L3), and |v| (the speed of a beam particle) givesthe distance per unit time. Dividing the rate by this flux, we obtain thedifferential scattering cross section:

dσ

dΩ′ =m2

(2π)2|V (p′ − p)|2. (238)

This formula is referred to as the Born Approximation (or, as the “first”Born approximation) for the differential cross section. Notice that the sizeof the box has disappeared once we have divided out the incident flux; weexpect the formula to apply in the limit of infinite spatial extent (i.e., in thecontinuum limit).

7.3 Coulomb Scattering

In Coulomb scattering, we consider the potential

V (x) =q1q2r, (239)

36

where r = |x|. We need the Fourier transform:

V (p′ − p)/q1q2 =∫

(∞)d3(x)

1

re−i(p

′−p)·x (240)

= 2π∫ ∞

0rdr

∫ 1

−1d cos θ exp(−i|p′ − p|r cos θ) (241)

= 2π∫ ∞

0rdr

e−i|p′−p|r − ei|p

′−p|r

−i|p′ − p|r(242)

=2πi

(p′ − p)2(−2i)

∫ ∞

0sin xdx (243)

=4π

(p′ − p)2. (244)

Actually, the integral∫∞0 sin xdx = 1 may be a bit suspicious. However, we

get the same result if we consider scattering on a Yukawa potential and takethe limit as the Yukawa range goes to infinity. Writing

(p′ − p)2 = 2p2(1 − cos θ), (245)

we obtain the differential cross section:

dσ

dΩ=

m2

4π2

16π2

4p4(1 − cos θ)2(q1q2)

2 (246)

=(q1q2)

2m2

p4(1 − cos θ)2(247)

=(q1q2)

2

16E2 sin4 θ2

, (248)

where p2 = 2mE, and θ is the scattering angle. This result may be recog-nized as the Rutherford cross section. There should be some real concernwhether we had any business applying the Born approximation here, sincethe Coulomb potential falls off so slowly with distance. We’ll discuss thisissue further later.

7.4 Decays

We are sometimes faced with the problem of describing a decay process, suchas the radioactive decay of a nucleus, in which a particle of momentum p

37

is produced. To be more explicit, suppose that we have a nucleus in aninitial state |i〉 which decays to a final state |f〉 plus a particle in momentumstate |p〉. We’ll neglect the nuclear recoil here, i.e., we’ll assume that themomentum p is small compared with the nuclear masses involved.

Assume that we know the interaction matrix element, 〈f ;p|V |i〉 for thedecay. The differential rate to emit the particle into solid angle element dΩis

dΓ = dΩ∫ ∞

0

mpL3

(2π)3dεp2π|〈f ;p|V |i〉|2δ(Ei − Ef − εp) (249)

= dΩmpL3

(2π)2|〈f ;p|V |i〉|2, (250)

where we have used

dN =L3

(2π3)d3(p) = dΩ

L3

(2π)3pmdεp, (251)

Ei is the energy of the nucleus in the initial state, Ef is the energy of the

nucleus in the final state, εp = p2/2m, p = pΩ, and p =√

2m(Ei − Ef ).Note that the wave function of the emitted particle will have a normalizationproportional to 1/L3/2, so the rate will actually be independent of L.

Integrating over dΩ gives the total decay rate:

Γ =mpL3

4π2

∫

(4π)dΩ|〈f ;p|V |i〉|2. (252)

If at t = 0 we have N0 nuclei in state |i〉, then at later time t we will haveseen ∆N = N0Γt decays, according to first order perturbation theory. Forlarge times, this must break down, since the initial state becomes depleted.We may rectify this by writing

dN

dt= −N(t)Γ, (253)

i.e., the rate of observing nuclear decays is proportional the number of avail-able nuclei, as well as to the decay rate of a nucleus. Thus, we have thefamiliar exponential decay law:

N(t) = N0e−Γt. (254)

38

7.5 Adiabatically Increasing Potential

Earlier, we considered that our potential was turned on more-or-less suddenlyat some time t0. Let us consider the situation where the potential is turnedon very slowly, compared with some relevant time scale. We’ll see that thiscase is not fundamentally different from our previous discussion, in the spiritof the perturbative nature of the interaction.

Let us consider a situation where we imagine a slow turn-on of a potential,for example, suppose we have an atom in state |i〉 which we subject to anelectromagnetic field which is slowly increased from zero. A measure of“slow” here must mean that the rate of energy change associated with theexternal field must be small compared with the orbit frequency of the atomicelectrons, i.e., the time scale for significant variation must be longer than

∼ a0

v∼ 1

mα

1

α∼ 10−16 s. (255)

Formally, we may turn on a potential, V, slowly by writing

Vt = eγtV, (256)

where γ > 0 so that V−∞ → 0. We’ll assume here that V itself is independentof time. To first order in time-dependent perturbation theory:

|ψ(t)〉 = |ψ(t0)〉 +1

i

∫ t

t0dt1V (t1)|ψ(t0)〉. (257)

Consider the transition from |i〉 to |f〉, assuming |i〉 and |f〉 are orthogonal.The transition amplitude, in first order, is:

〈f |ψ(t)〉 =1

i

∫ t

t0→−∞dt1〈f |V (t1)|i〉 (258)

=1

i

∫ t

−∞dt1〈f |eiH0t1Vt1e

−iH0t1 |i〉 (259)

=1

i

∫ t

−∞dt1e

i(εf−εi)t1eγt1〈f |V |i〉 (260)

=eγt+i(εf−εi)t

εi − εf + iγ〈f |V |i〉. (261)

The resulting transition probability is:

|〈f |ψ(t)〉|2 =e2γt

(εf − εi)2 + γ2|〈f |V |i〉|2. (262)

39

The dependence of this probability on energy is in the form of a Breit-Wigner distribution, or of a Cauchy probability distribution. The energy

0

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4

Figure 6: The i→ f transistion probability as a function of energy difference.The smooth curve is γ2/[(εf − εi)

2 + γ2], plotted for γ = 2. The horizontalaxis is εf − εi in units of 2π/t. The wavy curve is reproduced from Fig. 4 forcomparison.

spread of this distribution is of order γ, which may be interpreted roughlyas the inverse of the length of time the potential has been “on”.4

If |f〉 is one of a continuum of states, we may calculate the transition rateto such states according to:

d

dt|〈f |ψ(t)〉|2 = e2γt

2γ

(εf − εi)2 + γ2|〈f |V |i〉|2. (263)

Consider the limit of arbitrarily slow turn-on: γ → 0, and

limγ→0

e2γt2γ

(εf − εi)2 + γ2→ Aδ(εf − εi), (264)

4Note that the standard deviation, or rms spread, of a Cauchy distribution is infinite.Hence, we use here something like the half width at half maximum as our measure ofenergy spread.

40

where A is a constant to be determined by matching the normalization inthe γ → 0 limit:

A =∫ ∞

−∞dεf

2γ

(εf − εi)2 + γ2(265)

= 2∫ ∞

−∞

dx

1 + x2(266)

= 2π. (267)

Thus, we have the transition rate

Γi→f = 2π|〈f |V |i〉|2δ(εf − εi), (268)

which we recognize as Fermi’s Golden rule once again! We observe that thisrule is robust with respect to the details of how the perturbing potential isturned on.

8 Eigenvalues – Comparison Theorems

We return to some further discussion of some techniques similar to our dis-cussion of the variational method. Consider the problem of a particle in a(time-independent) potential, and ask what we might say qualitatively aboutthe existence and number of bound states, and related questions.

We start with the following “comparison theorem”:

Theorem: Consider a self-adjoint Hamiltonian:

H = − 1

2m∇2 + V (x). (269)

Let θ > 0. Then

H(θ) = − 1

2m∇2 + θ2V (θx) (270)

is also self-adjoint, and if ψ(x) ∈ DH , then ψ(θx) ∈ DH(θ). If λ is aneigenvalue of H (λ ∈ Σ(H)), then θ2λ is an eigenvalue of H(θ), and wehave:

Σ [H(θ)] =θ2λ|λ ∈ Σ(H)

. (271)

In particular, if the negative spectrum of H is discrete (with only 0 asa possible point of accumulation), then the number of negative eigen-values of H(θ) is the same as of H.

41

0

0

L

L/

(x)

( x)

ψ

ψ θ

θFigure 7: Illustration of the scaling of the wave function corresponding tothe scaling of the potential.

Proof: Domain is a question of boundary conditions – the vectors must scalein the same way the potential is scaled. See Fig. 7 for an illustration.

Let Hψ(x) = λψ(x). Consider:

H(θ)ψ(θx) =[− 1

2m∂2x + θ2V (θx)

]ψ(θx) (272)

= θ2[− 1

2m∂2θx + V (θx)

]ψ(θx) (273)

= θ2λψ(θx). (274)

This “comparison theorem”, relating the spectra of two related operators,tells us, for example, that any valid formula which gives an upper/lower limiton the number of negative eigenvalues in terms of the potential must beinvariant under the substitution V (x) → θ2V (θx), for θ > 0. For example,consider

H =p2

2m− V0 +

1

2kx2. (275)

The energy levels are at −V0 + (n + 1/2)ω, where ω =√k/m. The number

42

of negative eigenvalues is n = [V0

ω− 1

2]. Now consider modifying the potential

from V (x) = −V0 + 12kx2 to

θ2V (θx) = −θ2V0 + θ4 1

2kx2, (276)

giving the spectrum:

Σ[H(θ)] =θ2[−V0 +

(n+

1

2

)]. (277)

The solution for the number of negative eigenvalues is the same as before.Now consider, for a finite dimensional Hilbert space the following “min-

max” theorem:

Theorem: Let Q be a Hermitain N ×N matrix, with eigenvalues:

λ1 ≤ λ2 ≤ . . . ≤ λN . (278)

Let Pk, 1 ≤ k ≤ N be the set of Hermitian projections onto a k-dimensional subspace. Then:

λn = minF∈Pn

(max

ψ∈F,‖ψ‖=1〈ψ|Q|ψ〉

), (279)

where we use the symbol F to mean both the projection operator ontoa subspace, and the corresponding subspace itself.

This theorem tells us that we can find the nth eigenvalue by first findingthe maximum of 〈ψ|Q|ψ〉 for all unit vectors ψ in a fixed n-dimensionalsubspace F , and then minimizing the result as a function of F . If we don’tminimize, then we obtain an upper bound on λn. The case n = 1 correspondsto the variational principle we have already discussed, since the subspaces arethen one-dimensional, hence there is only one unit vector in each subspace(i.e., there is no maximization step required), and our minimization step isonly to the extent of our trial function parameterization. The case n = N isalso trivial, since then F = H, and there is thus no minimization step.

Proof: We have already dealt with the trivial cases n = 1 and n = N , so wenow suppose that 1 < n < N . Since Q is Hermitian, it has the spectraldecomposition:

Q =N∑

k=1

λkEk, (280)

43

where EkE` = δk` and I =∑Nk=1Ek. That is, the Ek are Hermitian

projections into one-dimensional subspaces. Then we can write:

Q = λnI +N∑

k=n

(λk − λn)Ek −n∑

k=1

(λn − λk)Ek. (281)

Hence, for any ‖ψ‖ = 1:

〈ψ|Q|ψ〉 = λn + |N∑

k=n

(λk − λn)Ek| − |n∑

k=1

(λn − λk)Ek|, (282)

where taking the absolute values does not alter the validity, since eachterm in the two sums is non-negative. Now let F be any n-dimensionalsubspace, and let ψ ∈ F . Select ψ to be orthogonal to E1, E2, . . . , En−1,that is, orthogonal to these n−1 projections. This is certainly possible,since there are n independent directions in F . Then we have:

〈ψ|Q|ψ〉 = λn + |N∑

k=n

(λk − λn)Ek| (283)

≥ λn. (284)

Now select F orthogonal to En+1, En+2, . . . , EN . Again, this is certainlypossible, since there are still n directions left. In this case,

〈ψ|Q|ψ〉 = λn − |n∑

k=1

(λn − λk)Ek| (285)

≤ λn. (286)

For both statements 284 and 286 to be true, we must have the statementin the theorem.

From the minmax theorem, it follows that:

Theorem: Let Q and V be N ×N Hermitian matrices, and let:

Q ≡ Q + V. (287)

Also, let the spectra of these operators be denoted:

Σ(Q) = λ1 ≤ λ2 ≤ . . . ≤ λN, (288)

Σ(Q) = λ1 ≤ λ2 ≤ . . . ≤ λN, (289)

Σ(V ) = V1 ≤ V2 ≤ . . . ≤ VN. (290)

44

Thenλn + VN ≥ λn ≥ λn + V1. (291)

It also follows that:

Theorem: Let Q be an N ×N Hermitian matrix with spectrum

Σ(Q) = λ1 ≤ λ2 ≤ . . . ≤ λN. (292)

Let Pk be any k-dimensional Hermitian projection, 1 ≤ k ≤ N . Wemay regard the operator PkQPk as a Hermitian k × k matrix, whenrestricted to the subspace Pk. Let the eigenvalues of this restrictedmatrix be

λ1 ≤ λ2 ≤ . . . ≤ λk. (293)

Thenλn ≥ λn, for n = 1, 2, . . . , k. (294)

We may see a classical analog of these theorems in terms of a system ofcoupled oscillators: First, if we add more springs to a system, none of thenormal-mode frequencies decrease. Second, if we remove degrees of freedomby clamping, then the k remaining frequencies will be at least as large as thek lowest frequencies of the original unclamped system.

9 Exercises

1. Prove the theorem quoted in section 2.3:

Theorem: If we have a normalized function |ψ〉 such that

E0 ≤ 〈ψ|H|ψ〉 ≤ E1, (295)

then

E0 ≥ 〈ψ|H|ψ〉 − 〈Hψ|Hψ〉 − 〈ψ|H|ψ〉2

E1 − 〈ψ|H|ψ〉 . (296)

2. Let us pursue our variational approach to the estimation of ground stateenergy levels of atoms to the “general” case. We consider an atom with

45

nuclear charge Z, and N electrons. The Hamiltonian of interest is:

H(Z,N) = Hkin − ZVc + Ve (297)

Hkin =N∑

n=1

p2n

2m, (298)

where (299)

Vc = αN∑

n=1

1

|xn|(300)

Ve = α∑

N≥j>k≥1

1

|xk − xj|(301)

m = electron mass (302)

α = fine structure constant. (303)

Denote the ground state energy ofH(Z,N) by −B(Z,N), with B(Z, 0) =0.

(a) Generalize the variational calculation we performed for the groundstate of helium to the general Hamiltonian H(Z,N). Thus, se-lect your “trial function” to be a product of N identical “hydro-gen atom ground state” functions. Determine the resulting lowerbound B(Z,N) on B(Z,N) (i.e., an upper limit on the groundstate energies).

(b) Make a simple table comparing your variational bounds with theobserved ground state energies for lithium, beryllium, and nitro-gen. Note that a simple web search for “ionization potentials” willget you a multitude of tables of observed values, or you can lookat a reference such as the CRC Press’s Handbook of Chemistryand Physics. The table entries are typically of the form:

B(Z,N) − B(Z,N − 1).

(c) Do your results make sense? If not, can you figure out what iswrong, and whether the calculation we did for He is to be trusted?

3. We consider the quantum mechanics of a particle in the earth’s gravi-tational field:

V (r) = −GMm

r(304)

46

= −G Mm

R + z(305)

≈ −GMm

R+mgz (306)

where (307)

M = mass of earth (308)

m = mass of particle (309)

r = distance from center of earth (310)

G = Newton’s gravitational constant (311)

R = radius of earth (312)

z = height of particle above surface of earth (313)

g = GM/R2. (314)

We may drop the constant term in our discussion, and consider only themgz piece, with z R. We further assume that no angular momen-tum is involved, and treat this as a one dimensional problem. Finally,assume that the particle is unable to penetrate the earth’s surface.

(a) Make a WKB calculation for the energy spectrum of the particle.

(b) If the particle is an atom of atomic weight A ∼ 100, use the resultof part (a) to estimate the particle’s ground state energy (in eV).Is sunlight likely to move the particle into excited states?

(c) Now make a variational calculation for the ground state energy(i.e., an upper bound thereon). Pick a “sensible” trial wave func-tion, at least in the sense that it satisfies the right boundary con-ditions. Compare your result with the ground state level from theWKB approximation.

4. We discussed the method of stationary phase in section 4. Recall thatthe problem it addresses is to evaluate integrals of the form:

I(ε) =∫ ∞

−∞f(x)eiθ(x)/εdx, (315)

where f and θ are real, and ε > 0. We showed that, in the situationwhere ε is very small, and θ has a stationary point at x = x0, thisintegral is approximately:

I(ε) =√εf(x0)e

iθ(x0)/εeiπ4sign[θ′′(x0)]

√2π

|θ′′(x0)|[1 +O(ε)] . (316)

47

If there is more than one stationary point, then the contributions areto be summed.

To get a little practice applying this method, evaluate the followingintegral for large t:

J(t) =∫ 1

0cos

[t(x3 − x)

]dx. (317)

5. I suggested in section 4.3 that you consider the classical correspondencefor the time delay (or advance) of the asymptotic motion due to scatter-ing on a potential. Let us pursue this here. Consider one-dimensionalmotion. A particle of mass m is incident from the left on a potential:

V (x) =−K x ∈ (−∆/2,∆/2)0 otherwise.

(318)

We wish to solve for the motion for large x at large times.

(a) Let’s do the quantum mechanics calculation first. Suppose thatour momentum space wave function at early time is a gaussianwave packet:

ψ(p) =

[1√2πσ

e−12(

p−qσ )

2]1/2

. (319)

What is ψ(x, t) for large times and large x? Describe the motion,relative to what it would be if K = 0.

(b) Now do the same problem classically. That is, solve for the motionat large times and large x. Again, compare the result with whatit would be for K = 0. Contrast with the quantum result.

6. We have solved the Schrodinger equation for the Hydrogen atom withHamiltonian:

H0 =p2

2m− e2

r.

The kinetic energy term is non-relativistic – the actual kinetic energywill have relativistic corrections.

(a) Obtain an expression for the next order relativistic (kinetic en-ergy) correction to the energy spectrum of hydrogen. It is con-venient to avoid taking multiple derivatives by using the unper-turbed Schrodinger equation to eliminate them. Thus, write your

48

expression in terms of the unperturbed energies and expectationvalues of e2

rand ( e

2

r)2. Do not actually do the integration over

r here, but reduce the problem to such integrals. Make sure youunderstand all of your steps.

(b) Now apply your formula to obtain the first-order relativistic ki-netic energy correction to the ground state energy of hydrogen.Express your answer as a multiple of the unperturbed ground stateenergy, and also calculate the size of the correction in eV.

7. Let us consider an example of the use of degenerate stationary stateperturbation theory. Thus, let us take the hydrogen atom, with unper-turbed Hamiltonian H0 = P 2

2m− α

r, and consider the effect of putting this

atom in a uniform external electric field: E = Eez. We are interestedin calculating, to first order in perturbation theory, the shifts in then = 2 energy levels. Note that the n = 2 level is four-fold degenerate,corresponding to the eigenstates: |2S0〉, |2P1〉, |2P0〉, |2P−1〉, neglectingspins.

(a) Write down the perturbing potential, V . [Note that we need onlyconsider the electron’s coordinates, relative to the nucleus – why?]Calculate the commutator [V, Lz], and hence determine the matrixelements of V between states with different eigenvalues of Lz.

(b) You should have found a “selection rule” which simplifies the prob-lem. What is the degeneracy that needs to be addressed in theproblem now that you have made this calculation?

(c) Using the invariance of the hydrogen atom Hamiltonian underparity, write down the remaining matrix elements of V which needto be determined, and compute their values.

(d) Now complete your degenerate perturbation theory calculation todetermine the splitting of the states in the applied electric field.Calculate numerical splittings (in eV) for an applied field of 100kV/cm. Also, estimate the “typical” electric field felt by the elec-tron, due to the nucleus, in a hydrogen atom. Was the use ofperturbation theory reasonable for this problem?

8. It may happen that we encounter a situation where the eigenvalues ofH0, call them εn and εm, are nearly, but not quite equal. In this case,

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we cannot use degenerate perturbation theory, and ordinary perturba-tion theory looks unreliable. Let us try to deal with such a situation:Suppose the two eigenstates |n〉 and |m〉 of H0 have nearly the same en-ergy (and all other eigenstates don’t suffer this disease, for simplicity).Let H = H0 + V , and write

V =∑

i,j

|i〉〈i|V |j〉〈j| (320)

H0|i〉 = εi|i〉, (321)

where

〈i|j〉 = δij· (322)

Let

V = V1 + V2, (323)

with

V1 ≡ |m〉〈m|V |m〉〈m| + |n〉〈n|V |n〉〈n| + (324)

+|m〉〈m|V |n〉〈n| + |n〉〈n|V |m〉〈m| (325)

and V2 is everything else.

If we can solve exactly the problem with H1 = H0 + V1, then the trou-blesome 1/(εn− εm) terms are avoided by the exact treatment, and wemay treat V2 as a perturbation in ordinary perturbation theory (since〈i|V2|j〉 = 0 for i, j = n,m). All states |i〉, i 6= n,m, are eigenstatesof H1, since V1|i〉 = 0 in this case. However, |n〉 and |m〉 are not ingeneral eigenstates of H1.

(a) Solve exactly for the eigenstates and eigenvalues of H1, in thesubspace spanned by |n〉, |m〉. Express your answer in terms of

εn, εm, 〈m|V |n〉, 〈n|V |n〉, 〈m|V |m〉.

(You may also use the shorthand

E(1)n,m = εn,m + 〈n,m|V |n,m〉

if you find it convenient.)

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(b) As an application, consider an electron in a weak one-dimensionalperiodic potential (“lattice”) V (x) = V (x + d). Assume the lat-tice has a size L = Nd, and that we the have periodic boundarycondition on our wave functions: ψ(x) = ψ(x + L). With thisboundary condition, the unperturbed wave functions are planewaves, ψp(x) = 1√

Leipx, where p = 2πn/L, n=integer, and the

unperturbed eigenenergies are εn = p2

2m=(

2πnL

)21

2m. We expand

the potential in a Fourier series:

V (x) =∞∑

n=−∞ein2πx/dVn

If we label our eigenfunctions by |p〉 = 1√Le2πinpx/L, determine all

nonvanishing matrix elements of V :

〈q|V |p〉

Express your answer in terms of Vn.

(c) Suppose εnp and εnq are not close to each other ∀nq, given somenp. Calculate the perturbed wave function in ordinary first orderperturbation theory corresponding to unperturbed wave functionψp(x). Also, calculate the energy to 2nd order. Express youranswer in terms of Vn and the unperturbed energies.

(d) What is the condition on np (and hence on p) so that |p〉 will benearly degenerate in energy with another eigenstate of H0?

(e) Assume that the condition in (d) exists, and use part (a) to solvethis “almost degenerate” case for the eigenenergies. Complete thegraph in Fig. 8 for higher values of |p|.

9. When we calculated the density of states for a free particle, we used a“box” of length L (in one dimension), and imposed periodic boundaryconditions to ensure no net flux of particles into or out of the box.We have in mind that we can eventually let L → ∞, and are reallyinterested in quantities per unit length (or volume). Let us justifymore carefully the use of periodic boundary conditions, i.e., we wish toconvince ourselves that the intuitive rationale given above is correct.To do this, consider a free particle in a one-dimensional “box” from

51

E p

p

0|V |

ππ- /d /d0

Figure 8: Energy versus momentum for the one-dimensional lattice problem(6).

−L/2 to L/2. Remembering that the Hilbert space of allowed states isa linear space, show that the periodic boundary condition:

ψ(−L/2) = ψ(L/2), (326)

ψ′(−L/2) = ψ′(L/2) (327)

gives acceptable wave functions. “Acceptable” here means that theprobability to find a particle in the box must be constant. Are thereother acceptable choices?

10. See if you can generalize the result for the first Born approximation:

dσ

dΩ′ =m2

(2π)2|V (p′ − p)|2. (328)

to the case where the scattered particle (mass mf ) may have a differentmass than the incident particle (mass mi).

11. We consider the potential (called the “Yukawa potential”):

V (x) =Ke−µr

r, r = |x|,

52

with real parameters K and µ > 0. The parameter K can be regardedas the “strength” of the potential (“interaction”), and 1

µis effectively

the “range” of distance over which the potential is important. µ itselfhas units of mass – note that as µ→ 0 we obtain the Coulomb potential:µ can be thought of as the mass of an “exchanged particle” whichmediates the force. In electromagnetism, this is the photon, henceµ→ mγ = 0

(a) Find a condition on K and µ which guarantees that there are atleast n bound states in this potential. You will likely fashion anduse some kind of “comparison” theorem in arriving at your result.You should give at least a “heuristically convincing” argument, ifyou don’t actually prove it.

(b) Using the Born approximation for the differential cross sectionthat we developed in our discussion of time-dependent perturba-tion theory, calculate the differential cross section, dσ

dΩ, for scatter-

ing on this potential. Consider the limit µ→ 0 and compare withthe Coulomb differential cross section we obtained in the notes.

(c) Integrate your differential cross section over all solid angles toobtain the “total cross section”. Again, consider the limit µ→ 0.Hence, what is the total cross section for scattering on a Coulombpotential?

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