2 Solutions

a constant. In this manner, given a Boltzmann distribution of energies atequilibrium, any mode having hf much greater than kBT is not excited.2. Describe the photoelectric effect experiments and Einstein’s explana-

tion for both the number and energy of the emitted electrons as a functionof the frequency of the incident light.Solution: Shine light on a metal. Visible light, no electrons given off.

Increase intensity, still no electrons given off. With uv (ultraviolet light)electrons given off. Increase intensity of uv light, more electrons given off,but the maximum kinetic energy of the electrons is unchanged. Einsteinexplanation was that light consists of photons. Radiation at frequency f ismade up of photons having energy hf . They can give up their energy tothe electrons in an all or nothing fashion. Therefore electrons are given offonly if light having frequency f > W/h(W =work function of the metal)is incident. Maximum kinetic energy = hf −W.3. What were Bohr’s postulates in his theory of the hydrogen atom?

How do these postulates explain the spectrum and stability of the hydrogenatom?Solution: The postulates are (1) The electron orbits the nucleus in

circular orbits having discrete values of angular momentum given by

L = mvr = n~,

where n = 1, 2, 3, 4,. . . .and (2) The electrons radiate energy only when they“jump” from a larger to a smaller orbit. The frequency of the radiationemitted is given by

fn2,n1 =En2 − En1

h,

Stability is explained since the electron only radiates in changing orbits. Ifit is in its lowest energy state, there is no lower energy state for it to go to.The first postulate of quantized angular momentum plus Newton’s SecondLaw led to quantized energy levels. The second postulate then explains thediscrete spectrum.4. Draw an energy level diagram for the hydrogen atom and indicate

on it the energy of the four lowest energy states. Also indicate the radius,electron velocity and angular momentum of these states as given by theBohr theory.Solution: rn = n2a0 = 5.29× 10−11n2 (m) vn = αc/n = 2.19× 10−6/n

(m/s) En = −(1/2)α2(mc2)/n2 = −13.6 eV/n2. For n = 1, L = ~, r =a0, v = αc,E = −13.6 eV; for n = 2, L = 2~, r = 4a0, v = αc/2, E = −3.4eV; for n = 3, L = 3~, r = 9a0, v = αc/3, E = −1.51 eV; for n = 4, L =4~, r = 16a0, v = αc/4, E = −0.85 eV;5. Calculate the wavelength of the n = 7 to n = 2 and of the n = 3 to

n = 2 transitions in hydrogen. What frequency of radiation is needed toexcite the hydrogen atom from its n = 1 to n = 3 state? How much energyis required to ionize a hydrogen atom from its n = 2 state?

0.1 Chapter 1. Introduction 3

Solution: E7 = −13.6/49 = −.278 eV E2 = −13.6/4 = −3.4eV E3 =−13.6/9 = −1.51eV.f72 = [−.278− (−3.4)]eV/4.14×10−15eV ·s = 7.54×1014Hz; λ = c/f =

3.97× 10−7m = 397nm.f32 = [−1.51− (−3.4)]eV/4.14×10−15eV ·s = 4.56×1014Hz; λ = c/f =

6.57× 10−7m = 657nm.E3 − E1 = 12.1eV . This is the energy needed to excite the transition.f = (E3 − E1)/h = 2.92 × 1015Hz. To ionize the hydrogen atom, we

must provide enough energy to give the electron a total energy of zero, inwhich case it is no longer bound to the proton. If it starts in n = 2 withE3 = −3.4eV , 3.4 eV of energy is needed.6. Radiation from the n = 2 to n = 1 state of hydrogen is incident on a

metal having a work function of 2.4 eV. What is the maximum energy ofthe emitted electrons from the metal?Solution: E2 − E1 = 10.2eV =photon energy= hf . Maximum K.E.

= hf −W (work function) = 10.2− 2.4 = 7.8eV.7. What is the significance of the equation λdB = h/p? What is meant

by the wave particle duality of matter? What determines when matter actsas a particle and when it acts as a wave?Solution: λdB = h/p implies that a wavelength can be associated with

matter. The wave-particle duality refers to the fact that sometimes matteracts as a wave and sometimes as a particle. It will act as a wave if its deBroglie wavelength is larger than or of the order of a characteristic lengthin the problem (such as beam size, obstacle size, slit width, orbit radius,etc.)8. Calculate the de Broglie wavelength for a particle of mass 1.0 gm

moving with a speed of 1.0 cm/year. Calculate the de Broglie wavelengthfor the electron in the n = 1 state of the Bohr atom.Solution: 1cm/yr= 3.2× 10−10m/s.λdB = h/p = h/mv = 6.63× 10−34J·s/[10−3kg·3.2× 10−10m/s] = 2.1×

10−11m.In the n = 1 state v = αc = c/137 = 2.2× 106m/s.λdB = h/p = h/mv = 6.63 × 10−34J·s/[9.1 × 10−31kg·2.2 × 106m/s] =

3.3× 10−10m.Note that the de Broglie wavelength is larger than the size of the orbit

which is 5.3× 10−11m. The electron acts as a wave.9. In general terms, discuss the measurement process in quantum me-

chanics. Why is it necessary to make measurements on a large number ofidentically prepared systems to obtain |ψ(r, t)|2? How does this differ fromNewtonian mechanics? Why is a single particle in a superposition state anintrinsically quantum object?Solution: Since the Schrödinger equation describes probability waves,

measuring one particle cannot give you the probability distribution |ψ(r, t)|2.By measuring the position of the particle on a large number of identi-cally prepared systems, you can deduce what |ψ(r, t)|2 is. The measurement

4 Solutions

FIGURE 1. Problem 1.10

process in each case irreversibly changes the wave function of the particle.In Newtonian mechanics, it is assumed that measurements can be madeon the position and velocity of a particle without significantly affecting itsmotion. There is no classical analogue of a single particle in a superposi-tion state. It simultaneously has the possibility of yielding a measurementhaving different values of a dynamic variable.10. In a two-slit experiment, particles are sent into the apparatus one

particle at a time (see figure). How many detectors does a single particletrigger? If the experiment is repeated many times, what will a graph of thenumber of counts in a detector vs detector position look like? Explain.Solution: One particle, one detector. |ψ(r, t)|2 just before the detectors

will look like the two-slit interference pattern. After many trials this patternwill be recorded by the detectors.11. When light is incident on a glass slab, some of the light is reflected.

This is a wave-like phenomenon (if a classical particle encounters a changein potential, it simply slows down or speeds up with no reflection), eventhough we are in a geometrical optics limit (neglect of diffraction). Whydoes a wave-like effect occur in this case? Is there any connection of thiswith the rainbow?Solution: Reflection is a wave-like phenomenon since a classical particle

that sees a step potential will either speed up or slow down as it enters thepotential region. A wave-like effect occurs because the index of refractionchanges in a distance that is small compared with a wavelength. If we wereto change the index continuously over a distance large compared to a wave-length, the reflection would be negligible. Similarly, although described bygeometrical optics, the rainbow is a wave effect. It occurs at an angle wherethe first internal reflection gives rise to a maximum scattered signal. Theangle differs slightly for different wavelengths, giving rise to the rainbow.

0.1 Chapter 1. Introduction 5

FIGURE 2. Problem 1.12.

12. The blackbody spectrum as a function of frequency uf (f) can beobtained using uf (f)df = u(ω)dω = 2πu(2πf)df . Plot [cuf (f)/4π] × 1018

as a function of frequency for T = 2.73 ◦K and find the maximum frequency[cuf (f)/4π is the power per unit area per unit frequency per unit solid angle- this corresponds to the flux incident on a detector per unit frequencyper steradian (sr)]. This Planck distribution corresponds to the cosmicmicrowave background. What is the energy per unit volume of the cosmicmicrowave background?Solution: A plot of

cuf (f)

4π=

1

2

~ (2πf)3

π2c21(

e2π~f/kBT − 1)

is shown. The maximum occurs at f ≈ 160 GHz, in the radio-frequencypart of the spectrum. The energy per unit volume is equal to 4σT 4/c =4.20× 10−14 J/m3.

6 Solutions

0.2 Chapter 2. Mathematical Preliminaries

1. Evaluate eiπ, eiπ/2, and e2.3i. If

ψ(x) =eiaxe−gx

2/2

x+ ib= u+ iv = reiθ,

find u, v, r, θ, assuming that x, a, b, g, r, θ are real. Evaluate |ψ(x)|2.Solution: eiπ = cosπ + i sinπ = −1; eiπ/2 = cos (π/2) + i sin (π/2) = i;

e2.3i = cos (2.3) + i sin (2.3) = −0.67 + 0.75i

ψ(x) =eiaxe−gx

2/2

x+ ib

x− ibx− ib =

eiaxe−gx2/2 (x− ib)

x2 + b2

=[cos (ax) + i sin (ax)] e−gx

2/2 (x− ib)x2 + b2

={[x cos (ax) + b sin (ax)] + i [−b cos (ax) + x sin (ax)]} e−gx2/2

x2 + b2.

Therefore

u =[x cos (ax) + b sin (ax)] e−gx

2/2

x2 + b2;

v =[−b cos (ax) + x sin (ax)] e−gx

2/2

x2 + b2;

|ψ(x)|2 =e−gx

2

x2 + b2;

r =√u2 + v2 =

√|ψ(x)|2 =

e−gx2/2

√x2 + b2

;

θ = tan−1(v/u) = tan−1

[−b cos (ax) + x sin (ax)

x cos (ax) + b sin (ax)

].

In general, for any complex function f = u+ iv, where u and v are real,

f∗ = u− iv;

Re[f ] =f + f∗

2;

Im[f ] =f − f∗

2i.

2. Given the function

f(x) =N

b2 + x2,

find N such that f2(x) is normalized; that is, find N such that∫ ∞−∞

f2(x)dx = 1,

0.2 Chapter 2. Mathematical Preliminaries 7

assuming that b is real. Find the Fourier transform a(k) of f(x). Evaluate

∆x2 =

∫ ∞−∞

(x− x̄)2f2(x)dx

∆k2 =

∫ ∞−∞

(k − k̄

)2 |a(k)|2 dk

and the product ∆x∆k. Does ∆x∆k = 1/2 for these functions? To evaluatethe integrals, you can use integral tables or Mathematica, Maple, or Matlab.Solution: Assume that b > 0. If b < 0, replace b by |b| in the equations

below. ∫ ∞−∞

f2(x)dx = N2

∫ ∞−∞

dx

(b2 + x2)2 = N2 π

2b3= 1;

N =

√2

πb3/2;

a(k) =1√2π

∫ ∞−∞

f(x)e−ikxdx =1√2π

√2

πb3/2

∫ ∞−∞

e−ikx

b2 + x2dx

= b1/2e−|k|b.

Both x̄ and k̄ equal zero. Thus

∆x2 =

∫ ∞−∞

x2f2(x)dx =2b3

π

∫ ∞−∞

x2dx

(b2 + x2)2 =

2b3

π

π

2b= b2;

∆k2 =

∫ ∞−∞

k2 |a(k)|2 dk = b

∫ ∞−∞

k2e−2|k|bdk =b

2b3=

1

2b2;

∆x∆k =1√2> 1/2.

This is not a minimum uncertainty packet.3-4. The k space amplitude for a free particle is given by

a(k) =

{a0 = 1/

√2k0 − k0 ≤ k ≤ k0

0 other wise.

The wave function ψ(x, 0) is the Fourier transform of a(k). Plot bothk0 |a(k)|2 as a function of k/k0 and |ψ(x, 0)|2 /k0 as a function of k0x.By “eyeballing”the graphs, estimate ∆x, ∆k, and their product. Now cal-culate ∆k and show that ∆x is infinite.Solution:You are given

a(k) =

{a0 = 1/

√2k0 − k0 ≤ k ≤ k0

0 other wise.

8 Solutions

FIGURE 3. Problem 2.3-4.

The wave function at t = 0 is

ψ(x, 0) =1√2π

∫ ∞−∞

dkΦ(k) exp[ikx]

=1√

2k0

√2π

∫ k0

−k0dk exp[ikx] =

i sin (k0x)√πk0x

.

Graphs of k0 |a(k)|2 and |ψ(x, 0)|2 /k0 are shown. The widths can be readfrom the graphs and ∆k ≈ k0 while ∆x ≈ 3/k0 such that ∆k∆x ≈ 3. Toevaluate them analytically,

∆x2 =

∫ ∞−∞

dxsin2 (k0x)

πk0=∞;

∆k2 =

∫ k0

−k0dk

k2

2k0=k2

0

3.

Thus, although the width of the spatial distribution appears to be of order2π/k0 (width of first lobe), the actual variance is infinite since the wavefunction falls off only as 1/x.

0.2 Chapter 2. Mathematical Preliminaries 9

5. If the functions f(x) and a(k) are Fourier transforms of one another,prove Parseval’s Theorem,∫ ∞

−∞|f(x)|2 dx =

∫ ∞−∞|a(k)|2 dk

Solution: ∫ ∞−∞|f(x)|2 dx =

∫ ∞−∞

f(x)f∗(x)dx

=1

2π

∫ ∞−∞

dk

∫ ∞−∞

dk′∫ ∞−∞

a(k)eikxa∗(k′)e−ik′xdx

=

∫ ∞−∞

dk

∫ ∞−∞

dk′a(k)a∗(k′)δ (k − k′)

=

∫ ∞−∞|a(k)|2 dk.

6. Prove

δ(ax) =δ(x)

|a|

δ(x2 − a2

)=

1

2 |a| [δ(x− a) + δ(x+ a)]∫ ∞−∞

dx′f(x′)dδ(x− x′)

dx′= −f ′(x).

Solution: Consider ∫ b

−bδ(ax)dx,

where b > 0. Let y = ax to arrive at

1

a

∫ b/a

−b/aδ(y)dy =

1

a×{

1 a > 0−1 a < 0

=1

|a| .

Therefore, as an integral operator δ(ax) = δ(x)/ |a|. The delta function isan even function of its argument.Next consider

δ(x2 − a2

)= δ [(x− a)(x+ a)] .

This function contributes only near x = ±a. Near x = −a

δ(x2 − a2

)≈ δ [(−2a)(x+ a)] =

1

2 |a|δ(x− a)

and near x = a

δ(x2 − a2

)≈ δ [(2a)(x− a)] =

1

2 |a|δ(x− a).

10 Solutions

Therefore

δ(x2 − a2

)=

1

2 |a| [δ(x− a) + δ(x+ a)] .

To prove ∫ ∞−∞

dx′f(x′)dδ(x− x′)

dx′= −f ′(x),

start with∫ ∞−∞

dx′d

dx′[f(x′)δ(x− x′)] =

∫ ∞−∞

dx′f ′(x′)δ(x− x′)

+

∫ ∞−∞

dx′f(x′)dδ(x− x′)

dx′= f(x′)δ(x− x′)|∞−∞ = 0.

Therefore ∫ ∞−∞

dx′f(x′)dδ(x− x′)

dx′= −f ′(x).

7. Given the probability distribution

P (x) =

{1 0 ≤ x ≤ 1

0 otherwise.

Calculate x̄ and∆x. If the probability distribution is shifted so it is centeredat x = 0, do x̄ and ∆x change?Solution:

x̄ =

∫ 1

0

xdx = 1/2;

x2 =

∫ 1

0

x2dx = 1/3;

∆x =

√x2 − x̄2 = 1/

√12.

If you shift the probability distribution so it is centered at x = 0, x̄ changes,but ∆x does not change?8. Use Taylor’s theorem to estimate 291/3 correct to order 0.001. Compare

your answer with the exact value.Solution: Let f(x) = x1/3 and expand about x = 27 with a = 2.

f(29) = f(27 + 2) ≈ f(27) +df

dx

∣∣∣∣x=27

2 +1

2!

d2f

dx2

∣∣∣∣x=27

4

= 3 +2

3 (27)2/3

+1

2

1

3

(−2

3

)4

(27)5/3

= 3.0722451.

0.2 Chapter 2. Mathematical Preliminaries 11

The error is of order

1

3!

1

3

(2

3

)(5

3

)8

(27)8/3

= 0.0000752.

The exact value is f(29) = 3.0723168 so the error is 0.0000752 which isconsistent with the error estimate.9. Show that

∫∞0dxdδ(x)

dx is not well defined. To do this use the definitionof the delta function as the limit of a Gaussian

δ(x) = lima→0

1√πa2

e−x2/a2 .

and show that the integral diverges as a→ 0.Solution:

∫ ∞0

dxdδ(x)

dx= lim

a→0

∫ ∞0

dx1√πa2

d

dxe−x

2/a2 = lima→0

1√πa2

e−x2/a2

∣∣∣∞0

= lima→0

−1√πa

which diverges. You can also "prove" this using

∫ ∞0

dxdδ(x)

dx= δ(x)|∞0 ∼ −∞.

12 Solutions

0.3 Chapter 3. Free-Particle Schrödinger Equation:Wave Packets

1. Suppose that a smooth wave packet has width σ at time t = 0. Explainwhy

σ(t) =

√σ2 +

(~tmσ

)2

is not a bad guess for the width of the wave packet at any time. Usingthis guess, find the time it takes for a wave packet that is confined to itsde Broglie wavelength to spread to twice its initial width in terms of theparticle’s energy and Planck’s constant.Solution: For a smooth packet we expect ∆p ∼ ~/σ or ∆v ∼ ~/mσ and

we can add the contributions in quadrature in analogy with a Gaussianpacket. At the very least we know this gives the correct answer at t = 0and for very large t. Given σ = λdB = h/mv. Then

σ(t) =

√σ2 +

(~tmσ

)2

= 2σ

σ2 +

(~tmσ

)2

= 4σ2

(~tmσ

)2

= 3σ2(~tmσ

)=√

3σ

t =√

3

(mσ2

~

)=√

3

(mh2

m2v2~

)=

2πh√

3

mv2=πh√

3

E.

2. Using the result of problem 3.1 and assuming that you are a pointparticle having mass 50 kg and are localized to 1.0 × 10−11 m, calculatehow long would it take for you to spread by this initial localization distance.Solution:We can use the result of the last problem:

t =√

3

(mσ2

~

).

is the time for a packet to spread to twice its initial width. Thus

t =√

3

(50× 10−22

10−34

)= 8.7× 1013 s ≈ 2× 106 yr.

3-4. Start with a one dimensional wave packet in coordinate space. Takea wave function such as ψ(x) = Ne−x

4

or ψ(x) = Ne−x6

, for which no