41

CHAPTER OUTLINE

41.1 An Interpretation of Quantum Mechanics41.2 A Particle in a Box41.3 The Particle Under Boundary Conditions41.4 The Schrödinger Equation41.5 A Particle in a Well of Finite Height41.6 Tunneling Through a Potential Energy Barrier41.7 The Scanning Tunneling Microscope

Oscillator

41.8 The Simple Harmonic

Quantum Mechanics

ANSWERS TO QUESTIONS

Q41.1 A particle’s wave function represents its state, containing all theinformation there is about its location and motion. The squaredabsolute value of its wave function tells where we wouldclassically think of the particle as a spending most its time. Ψ 2

is the probability distribution function for the position of theparticle.

Q41.2 The motion of the quantum particle does not consist of movingthrough successive points. The particle has no definite position.It can sometimes be found on one side of a node andsometimes on the other side, but never at the node itself. Thereis no contradiction here, for the quantum particle is moving asa wave. It is not a classical particle. In particular, the particledoes not speed up to infinite speed to cross the node.

Q41.3 Consider a particle bound to a restricted region of space. If its minimum energy were zero, then theparticle could have zero momentum and zero uncertainty in its momentum. At the same time, theuncertainty in its position would not be infinite, but equal to the width of the region. In such a case,the uncertainty product ∆ ∆x px would be zero, violating the uncertainty principle. This contradictionproves that the minimum energy of the particle is not zero.

Q41.4 The reflected amplitude decreases as U decreases. The amplitude of the reflected wave isproportional to the reflection coefficient, R, which is 1−T , where T is the transmission coefficient asgiven in equation 41.20. As U decreases, C decreases as predicted by equation 41.21, T increases, andR decreases.

Q41.5 Consider the Heisenberg uncertainty principle. It implies that electrons initially moving at the samespeed and accelerated by an electric field through the same distance need not all have the samemeasured speed after being accelerated. Perhaps the philosopher could have said “it is necessary forthe very existence of science that the same conditions always produce the same results within theuncertainty of the measurements.”

Q41.6 In quantum mechanics, particles are treated as wave functions, not classical particles. In classicalmechanics, the kinetic energy is never negative. That implies that E U≥ . Treating the particle as awave, the Schrödinger equation predicts that there is a nonzero probability that a particle can tunnelthrough a barrier—a region in which E U< .

491

492 Quantum Mechanics

Q41.7 Consider Figure 41.8, (a) and (b) in the text. In the square well with infinitely high walls, theparticle’s simplest wave function has strict nodes separated by the length L of the well. The particle’s

wavelength is 2L, its momentum hL2

, and its energy pm

hmL

2 2

22 8= . Now in the well with walls of only

finite height, the wave function has nonzero amplitude at the walls. The wavelength is longer. Theparticle’s momentum in its ground state is smaller, and its energy is less.

Q41.8 Quantum mechanically, the lowest kinetic energy possible for any bound particle is greater thanzero. The following is a proof: If its minimum energy were zero, then the particle could have zeromomentum and zero uncertainty in its momentum. At the same time, the uncertainty in its positionwould not be infinite, but equal to the width of the region in which it is restricted to stay. In such acase, the uncertainty product ∆ ∆x px would be zero, violating the uncertainty principle. Thiscontradiction proves that the minimum energy of the particle is not zero. Any harmonic oscillatorcan be modeled as a particle or collection of particles in motion; thus it cannot have zero energy.

Q41.9 As Newton’s laws are the rules which a particle of large mass follows in its motion, so theSchrödinger equation describes the motion of a quantum particle, a particle of small or large mass. Inparticular, the states of atomic electrons are confined-wave states with wave functions that aresolutions to the Schrödinger equation.

SOLUTIONS TO PROBLEMS

Section 41.1 An Interpretation of Quantum Mechanics

P41.1 (a) ψ x Ae A x Ai x A kx Ai kxi xa f e j e j a f a fe j= = × + × = +

×5 00 10 10 1010

5 10 5 10.

cos sin cos sin goes through

a full cycle when x changes by λ and when kx changes by 2π . Then kλ π= 2 where

k = × =−5 00 10210 1. mπλ

. Then λπ

=×

= × −2

5 00 101 26 10

1010 m

m.

.e j

.

(b) ph

= =× ⋅×

= × ⋅−

−−

λ6 626 101 26 10

5 27 1034

1024.

..

J s m

kg m s

(c) me = × −9 11 10 31. kg

Km v

mpm

e

e= = =

× ⋅

× ×= × =

××

=−

−−

−

−

2 2 2 24 2

3117

17

192 2

5 27 10

2 9 11 101 52 10

1 52 101 60 10

95 5.

..

..

. kg m s

kg J

J J eV

eVe je j

P41.2 Probability P xa

x adx

aa

xaa

a

a

a

a

a

= =+

= FHGIKJFHGIKJFHGIKJ

− −

−

−z zψ

π πa f

e j2

2 211

tan

P = − − = − −FHGIKJ

LNM

OQP =

− −11 1

14 4

12

1 1

π ππ π

tan tan a f

Chapter 41 493

Section 41.2 A Particle in a Box

P41.3 E1192 00 3 20 10= = × −. . eV J

For the ground-state, Eh

m Le1

2

28= .

(a) Lh

m Ee

= = × =−

84 34 10 0 434

1

10. . m nm

(b) ∆E E Eh

m Lh

m Le e

= − =FHG

IKJ −FHG

IKJ =2 1

2

2

2

248 8

6 00. eV

P41.4 For an electron wave to “fit” into an infinitely deep potential well, an integralnumber of half-wavelengths must equal the width of the well.

nλ2

1 00 10 9= × −. m so λ =×

=−2 00 10 9.

nhp

(a) Since Kpm

h

mhm

nn

e e e= = =

×=

−

2 2 2 2 2

9 22

2 2 2 2 100 377

λe je j

e j. eV

For K ≈ 6 eV n = 4

(b) With n = 4, K = 6 03. eV

FIG. P41.4

P41.5 (a) We can draw a diagram that parallels our treatment of standingmechanical waves. In each state, we measure the distance dfrom one node to another (N to N), and base our solution uponthat:

Since dN to N =λ2

and λ =hp

ph h

d= =λ 2

.

Next, Kpm

hm d de e

= = =× ⋅

×

L

NMMM

O

QPPP

−

−

2 2

2 2

34 2

312 81 6 626 10

8 9 11 10

.

.

J s

kg

e je j

.

Evaluating, Kd

=× ⋅−6 02 10 38

2. J m2

Kd

=× ⋅−3 77 10 19

2. eV m2

.

In state 1, d = × −1 00 10 10. m K1 37 7= . eV .

In state 2, d = × −5 00 10 11. m K 2 151= eV.

In state 3, d = × −3 33 10 11. m K3 339= eV .

In state 4, d = × −2 50 10 11. m K4 603= eV .FIG. P41.5

continued on next page

494 Quantum Mechanics

(b) When the electron falls from state 2 to state 1, it puts out energy

E hfhc

= − = = =151 37 7 113 eV eV eV.λ

into emitting a photon of wavelength

λ = =× ⋅ ×

×=

−

−

hcE

6 626 10 10

113 1 60 1011 0

34 8

19

.

..

J s 3.00 m s

eV J eV nm

e je ja fe j

.

The wavelengths of the other spectral lines we find similarly:

Transition 4 3→ 4 2→ 4 1→ 3 2→ 3 1→ 2 1→ E eVa f 264 452 565 188 302 113 λ nma f 4.71 2.75 2.20 6.60 4.12 11.0

P41.6 λ = 2D for the lowest energy state

Kpm

hm

hmD

ph h

D

= = = =× ⋅

× ×= × =

= = =× ⋅

×= × ⋅

−

− −

−

−

−−

2 2

2

2

2

34 2

27 14 214

34

1420

2 2 8

6 626 10

8 4 1 66 10 1 00 108 27 10 0 517

26 626 10

3 31 10

λ

λ

.

. .. .

..

J s

kg m J MeV

J s

2 1.00 10 m kg m s

e je j e j

e j

P41.7 ∆Ehc h

m Lh

m Le e

= =FHG

IKJ − =

λ

2

22 2

2

282 1

38

Lh

m ce= = × =−3

87 93 10 0 79310λ

. . m nm

P41.8 ∆Ehc h

m Lh

m Le e

= =FHG

IKJ − =

λ

2

22 2

2

282 1

38

so Lh

m ce=

38

λ

P41.9 The confined proton can be described in the same way as a standingwave on a string. At level 1, the node-to-node distance of the standingwave is 1 00 10 14. × − m , so the wavelength is twice this distance:hp= × −2 00 10 14. m.

The proton’s kinetic energy is

K mvpm

hm

= = = =× ⋅

× ×

=×

×=

−

− −

−

−

12 2 2

6 626 10

2 1 67 10 2 00 10

3 29 102 05

22 2

2

34 2

27 14 2

13

19

λ

.

. .

..

J s

kg m

J1.60 10 J eV

MeV

e je je j

FIG. P41.9

continued on next page

Chapter 41 495

In the first excited state, level 2, the node-to-node distance is half as long as in state 1. Themomentum is two times larger and the energy is four times larger: K = 8 22. MeV .

The proton has mass, has charge, moves slowly compared to light in a standing wave state, andstays inside the nucleus. When it falls from level 2 to level 1, its energy change is

2 05 8 22 6 16. . . MeV MeV MeV− = − .

Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at thespeed of light, and that it has an energy of +6 16. MeV .

Its frequency is fEh

= =× ×

× ⋅= ×

−

−

6 16 10 1 60 10

6 626 101 49 10

6 19

3421

. .

..

eV J eV

J s Hz

e je j.

And its wavelength is λ = =×

×= ×−

−cf

3 00 101 49 10

2 02 108

21 113.

..

m s s

m .

This is a gamma ray, according to the electromagnetic spectrum chart in Chapter 34.

P41.10 The ground state energy of a particle (mass m) in a 1-dimensional box of width L is EhmL1

2

28= .

(a) For a proton m = × −1 67 10 27. kge j in a 0.200-nm wide box:

E1

34 2

27 10 222 3

6 626 10

8 1 67 10 2 00 108 22 10 5 13 10=

× ⋅

× ×= × = ×

−

− −

− −.

. .. .

J s

kg m J eV

e je je j

.

(b) For an electron m = × −9 11 10 31. kge j in the same size box:

E1

34 2

31 10 218

6 626 10

8 9 11 10 2 00 101 51 10 9 41=

× ⋅

× ×= × =

−

− −

−.

. .. .

J s

kg m J eV

e je je j

.

(c) The electron has a much higher energy because it is much less massive.

P41.11 EhmL

nn =FHGIKJ

2

22

8

E1

34 2

27 14 214

6 626 10

8 1 67 10 2 00 108 21 10=

× ⋅

× ×= ×

−

− −

−.

. ..

J s

kg m J

e je je j

E1 0 513= . MeV E E2 14 2 05= = . MeV E E3 19 4 62= = . MeV

Yes , the energy differences are ~1 MeV , which is a typical energy for a γ-ray photon.

496 Quantum Mechanics

*P41.12 (a) The energies of the confined electron are Eh

m Lnn

e

=2

22

8. Its energy gain in the quantum

jump from state 1 to state 4 is h

m Le

2

22 2

84 1−e j and this is the photon energy:

hm L

hfhc

e

2

215

8= =

λ. Then 8 152m cL he = λ and L

hm ce

=FHGIKJ

158

1 2λ

.

(b) Let ′λ represent the wavelength of the photon emitted: hc h

m Lh

m Lh

m Le e e′= − =

λ

2

22

2

22

2

284

82

128

.

Then hc

hc

h m L

m L he

eλλ′

= =2 2

2 2

15 8

8 1254

e j and ′ =λ λ1 25. .

Section 41.3 The Particle Under Boundary Conditions

Section 41.4 The Schrödinger Equation

P41.13 We have ψ ω= −Aei kx tb g and∂∂

= −2

22ψψ

xk .

Schrödinger’s equation:∂∂

= − = − −2

22

22ψ

ψ ψx

km

E Ua f .

Since kp

hp2

2

2

2

2

2

2

2 2= = =

πλ

πa f b gand E U

pm

− =2

2.

Thus this equation balances.

P41.14 ψ x A kx B kxa f = +cos sin∂∂

= − +ψx

kA kx kB kxsin cos

∂∂

= − −2

22 2ψ

xk A kx k B kxcos sin − − = − +

2 22

mE U

mEA kx B kxa f a fψ cos sin

Therefore the Schrödinger equation is satisfied if

∂∂

= −FHGIKJ −

2

2 22ψ

ψx

mE Ua f or − + = −FHG

IKJ +k A kx B kx

mEA kx B kx2

22

cos sin cos sina f a f.

This is true as an identity (functional equality) for all x if Ekm

=2 2

2.

*P41.15 (a) With ψ x A kxa f a f= sin

ddx

A kx Ak kxsin cos= andddx

Ak kx2

22ψ = − sin .

Then − = + = = = = =2 2

2

2 2 2 2

2 2

2 2 22

2 2

4

4 2 2 212m

ddx

km

A kxh

m

pm

m vm

mv Kψ π

π λψ ψ ψ ψ ψsin

e je ja f

.

(b) With ψπλ

x Ax

A kxa f = FHGIKJ =sin sin

2, the proof given in part (a) applies again.

Chapter 41 497

P41.16 (a) x xL

xL

dxL

xx

Ldx

L L

= FHGIKJ = −FHG

IKJz z2 2 2 1

212

42

0 0

sin cosπ π

xL

xL

L xL

xL

xL

LL L

= − +LNM

OQP =

12

116

4 4 42

2

0

2

20π

π π πsin cos

(b) Probability= FHGIKJ = −LNM

OQPz 2 2 1 1

442

0 490

0 510

0 490

0 510

Lx

Ldx

Lx

LL x

LL

L

L

L

sin sin.

.

.

.ππ

π

Probability= − − = × −0 0201

42 04 1 96 5 26 10 5. sin . sin . .

ππ πa f

(c) ProbabilityxL

xL L

L

−LNMOQP = × −1

44

3 99 100 240

0 2602

ππ

sin ..

.

(d) In the n = 2 graph in Figure 41.4 (b), it is more probable to find the particle either near xL

=4

or xL

=34

than at the center, where the probability density is zero.

Nevertheless, the symmetry of the distribution means that the average position is L2

.

P41.17 Normalization requires

ψ 2 1dxall spacez = or A

n xL

dxL

2 2

0

1sinπFHGIKJ =z

An x

Ldx A

LL2 2

0

2

21sin

πFHGIKJ = FHG

IKJ =z or A

L=

2.

P41.18 The desired probability is P dxL

xL

dxL L

= = FHGIKJz zψ

π2

0

42

0

42 2sin

where sincos2 1 22

θθ

=−

.

Thus, PxL

xL

L

= −FHGIKJ = − − +FHG

IKJ =

14

4 14

0 0 0 0 2500

4

ππ

sin . .

P41.19 In 0 ≤ ≤x L , the argument 2π x

L of the sine function ranges from 0 to 2π . The probability density

2 22

Lx

LFHGIKJFHGIKJsin

π reaches maxima at sinθ = 1 and sinθ = −1 at

22

π πxL

= and 2 3

2π πxL

= .

∴ The most probable positions of the particle are at at and xL

xL

= =4

34

.

498 Quantum Mechanics

*P41.20 (a) Probability = = FHGIKJ = − FHG

IKJ

LNM

OQPz z zψ

π π1

2

0

2

0 0

2 11

2dx

Lx

Ldx

Lx

Ldxsin cos

= − FHGIKJ

LNM

OQP = − F

HGIKJ

12

2 12

2

0Lx

L xL L Lππ

ππ

sin sin

(b)

1.2

1

0.8

0.6

0.4

0.2

00 0.5 1 1.5

Probability Curve for an InfinitePotential Well

L

FIG. P41.20(b)

(c) The probability of finding the particle between x = 0 and x = is 23

, and between x = and

x L= is 13

.

Thus, ψ 12

0

23

dxz =

∴ − FHGIKJ =L L

12

2 23π

πsin , or u u− =

12

223π

πsin .

This equation for L

can be solved by homing in on the solution with a calculator, the result

being L= 0 585. , or = 0 585. L to three digits.

Chapter 41 499

P41.21 (a) The probability is P dxL

xL

dxL

xL

dxL L L

= = FHGIKJ = −FHG

IKJz z zψ

π π2

0

32

0

3

0

32 2 12

12

2sin cos

PxL

xL

L

= −FHGIKJ = −FHG

IKJ = −FHG

IKJ =

12

2 13

12

23

13

34

0 1960

3

ππ

ππ

πsin sin . .

(b) The probability density is symmetric about xL

=2

.

Thus, the probability of finding the particle between

xL

=23

and x L= is the same 0.196. Therefore, the

probability of finding it in the range L

xL

323

≤ ≤ is

P = − =1 00 2 0 196 0 609. . .a f .

FIG. P41.21(b)

(c) Classically, the electron moves back and forth with constant speed between the walls, andthe probability of finding the electron is the same for all points between the walls. Thus, theclassical probability of finding the electron in any range equal to one-third of the available

space is Pclassical =13

.

P41.22 (a) ψπ

12

xL

xL

a f = FHGIKJcos ; P x x

Lx

L1 12 22a f a f= = F

HGIKJψ

πcos

ψπ

22 2

xL

xL

a f = FHGIKJsin ; P x x

Lx

L2 22 22 2a f a f= = F

HGIKJψ

πsin

ψπ

32 3

xL

xL

a f = FHGIKJcos ; P x x

Lx

L3 32 22 3a f a f= = F

HGIKJψ

πcos

(b)

x x

0 0

n = 1

n = 2

n = 3

ψ 2ψ

− L2

− L2

L2

L2

∞ ∞ ∞ ∞

FIG. P41.22(b)

500 Quantum Mechanics

P41.23 Problem 43 in Chapter 16 helps students to understand how to draw conclusions from an identity

(a) ψ x AxL

a f = −FHGIKJ1

2

2ddx

AxL

ψ= −

22

ddx

AL

2

22ψ

= −

Schrödinger’s equationddx

mE U

2

2 22ψ

ψ= − −a f

becomes − = − −FHGIKJ +

− −

−

2 21

2 12 2

2

2 2

2 2 2 2

2 2 2

AL

mEA

xL

m x A x L

mL L x

e j e je j

− = − + −12 2

2

2 2

2

4LmE mEx

LxL

.

This will be true for all x if both12 2L

mE=

andmE

L L2 2 41

0− =

both these conditions are satisfied for a particle of energy EL m

=2

2 .

(b) For normalization, 1 1 122

2

2

22

2

2

4

4= −FHGIKJ = − +

FHG

IKJ− −

z zAxL

dx Ax

LxL

dxL

L

L

L

123 5

23 5

23 5

1615

1516

23

2

5

42 2= − +

LNM

OQP

= − + + − +LNM

OQP =FHGIKJ =

−

A xxL

xL

A L LL

L LL

AL

AL

L

L

.

(c) P dxL

xL

xL

dxL

xxL

xL L

L L L

L

L

L

L

L

L

= = − +FHG

IKJ = − +

LNM

OQP

= − +LNM

OQP− − −

z zψ 2

3

3 2

2

4

43

3 3

2

5

53

315

161

2 1516

23 5

3016 3

281 1 215

P = =4781

0 580.

P41.24 (a) Setting the total energy E equal to zero and rearranging the Schrödinger equation to isolatethe potential energy function gives

U xm

ddx

a f = FHGIKJ

2 2

221ψ

ψ.

If ψ x Axe x La f = − 2 2.

Thenddx

Ax AxLe

L

x L2

23 2

44 62 2

ψ= −

−

e j

orddx

x L

Lx

2

2

2 2

4

4 6ψψ=

−e j a f

and U xmL

xL

a f = −FHG

IKJ

2

2

2

224

6 .

(b) See the figure to the right.

FIG. P41.24(b)

Chapter 41 501

Section 41.5 A Particle in a Well of Finite Height

P41.25 (a) See figure to the right.

(b) The wavelength of the transmitted wavetraveling to the left is the same as the originalwavelength, which equals 2L .

FIG. P41.25(a)

P41.26

FIG. P41.26

Section 41.6 Tunneling Through a Potential Energy Barrier

P41.27 T e CL= −2 where Cm U E

=−2 a f

22 2 9 11 10 8 00 10

1 055 102 00 10 4 58

31 19

3410CL =

× ×

×× =

− −

−−

. .

.. .

e je je j

(a) T e= =−4.58 0 010 3. , a 1% chance of transmission.

(b) R T= − =1 0 990. , a 99% chance of reflection.FIG. P41.27

P41.28 C =× − × ⋅

× ⋅= ×

− −

−−

2 9 11 10 5 00 4 50 1 60 10

1 055 103 62 10

31 19

349 1

. . . .

..

e ja fe j kg m s

J s m

T e

T

CL= = − × × = −

= ×

− − −

−

2 9 1 12

3

2 3 62 10 950 10 6 88

1 03 10

exp . exp .

.

m me je j a f

FIG. P41.28

P41.29 From problem 28, C = × −3 62 109 1. m

10 2 3 62 106 9 1− −= − ×exp . me jL .

Taking logarithms, − = − × −13 816 2 3 62 109 1. . me jL .

New L = 1 91. nm

∆L = − =1 91 0 950 0 959. . . nm nm nm .

502 Quantum Mechanics

*P41.30 The original tunneling probability is T e CL= −2 where

Cm U E

=−

=× × − ×

× ⋅= ×

− −

−−2 2 2 9 11 10 12 1 6 10

6 626 101 448 1 10

1 2 31 19 1 2

3410 1a fc h a fe jπ . .

..

kg 20 J

J s m .

The photon energy is hfhc

= =⋅

=λ

1 2402 27

eV nm546 nm

eV. , to make the electron’s new kinetic energy

12 2 27 14 27+ =. . eV and its decay coefficient inside the barrier

′ =× × − ×

× ⋅= ×

− −

−−C

2 2 9 11 10 20 14 27 1 6 10

6 626 101 225 5 10

31 19 1 2

3410 1

π . . .

..

kg J

J s m

a fe j.

Now the factor of increase in transmission probability isee

e e eC L

CLL C C

− ′

−− ′ × × ×= = = =

− −2

22 2 10 0 223 10 4.459 10 1

85 9a f m m. . .

Section 41.7 The Scanning Tunneling Microscope

P41.31 With the wave function proportional to e CL− , the transmission coefficient and the tunneling currentare proportional to ψ 2 , to e CL−2 .

Then,II

e

ee

0 5000 515

1 352 10 0 0 500

2 10 0 0 51520 0 0 015.

..

. .

. .. . nm

nm

nm nm

nm nm

a fa f

b ga fb ga f

a f= = =−

−.

P41.32 With transmission coefficient e CL−2 , the fractional change in transmission is

e e

ee

L L

L

− − +

−−−

= − = =2 10 0 2 10 0 0 002 00

2 10 020 0 0 002 001 0 0392 3 92%

. . .

.. . . .

nm nm nm

nm

b g b gb gb g

b g .

Section 41.8 The Simple Harmonic Oscillator

P41.33 ψ ω= −Be m x2 2b g so ddx

mx

ψ ωψ= −FHGIKJ and

ddx

mx

m2

2

22ψ ωψ

ωψ= FHG

IKJ + −FHG

IKJ .

Substituting into Equation 41.13 gives m

xm mE m

xω

ψω

ψ ψω

ψFHGIKJ + −FHG

IKJ = −FHG

IKJ + FHG

IKJ

22

2

222

which is satisfied provided that E =ω2

.

Chapter 41 503

P41.34 Problem 43 in Chapter 16 helps students to understand how to draw conclusions from an identity.

ψ = −Axe bx2 so

ddx

Ae bx Aebx bxψ= −− −2 2

2 2

andddx

bxAe bxAe b x e b b xbx bx bx2

22 3 2 22 4 4 6 4

2 2 2ψψ ψ= − − + = − +− − − .

Substituting into Equation 41.13, − + = −FHGIKJ + FHG

IKJ6 4

22 22

2b b xmE m

xψ ψ ψω

ψ .

For this to be true as an identity, it must be true for all values of x.

So we must have both − = −62

2bmE

and 4 22

bm

= FHGIKJ

ω.

(a) Therefore bm

=ω

2

(b) and Ebm

= =3 3

2

2

ω .

(c) The wave function is that of the first excited state .

P41.35 The longest wavelength corresponds to minimum photon energy, which must be equal to thespacing between energy levels of the oscillator:

hc kmλ

ω= = so λ π π= = ××F

HGIKJ =

−

2 2 3 00 109 11 10

600831 1 2

cmk

..

m s kg

8.99 N m nme j .

P41.36 (a) With ψ ω= −Be m x2 2b g , the normalization condition ψ 2 1dxxall z =

becomes 1 2 212

2 2 2 2

0

22 2

= = =−

−∞

∞−

∞z zB e dx B e dx Bm

m x m xω ω πω

b g b g

where Table B.6 in Appendix B was used to evaluate the integral.

Thus, 1 2= Bmπω

and Bm

=FHGIKJ

ωπ

1 4

.

(b) For small δ, the probability of finding the particle in the range − < <δ δ2 2

x is

ψ δ ψ δ δω

πδ

δ2

2

22 2 0

1 2

0dx B em

−

−z = = =FHGIKJa f .

504 Quantum Mechanics

*P41.37 (a) For the center of mass to be fixed, m v m v1 1 2 2 0+ = . Then

v v v vmm

vm m

mv= + = + =

+1 2 1

1

21

2 1

21 and v

m vm m1

2

1 2=

+. Similarly, v

mm

v v= +2

12 2 and

vm v

m m21

1 2=

+. Then

12

12

12

12

12

12

12

12

12

12

1 12

2 22 2 1 2

2 2

1 22

2 12 2

1 22

2

1 2 1 2

1 22

2 2 2 2

m v m v kxm m v

m m

m m v

m mkx

m m m m

m mv kx v kx

+ + =+

++

+

=+

++ = +

b g b gb gb g

µ

(b)ddx

v kx12

12

02 2µ +FHG

IKJ = because energy is constant

012

212

2= + = + = +µ µ µvdvdx

k xdxdt

dvdx

kxdvdt

kx .

Then µ a kx= − , akx

= −µ

. This is the condition for simple harmonic motion, that the

acceleration of the equivalent particle be a negative constant times the excursion from

equilibrium. By identification with a x= −ω 2 , ωµ

π= =k

f2 and fk

=1

2π µ.

P41.38 (a) With x = 0 and px = 0, the average value of x2 is ∆xa f2 and the average value of px2 is

∆pxb g2 . Then ∆∆

xpx

≥2

requires

Epm

kp

pm

kp

x

x

x

x

≥ + = +2 2

2

2 2

22 2 4 2 8.

(b) To minimize this as a function of px2 , we require

dEdp m

kpx x

2

2

401

2 81

1= = + −a f .

Thenkp mx

2

481

2= so p

mk mkx2

2 1 22

8 2=FHG

IKJ =

and Emkm

kmk

km

km

≥ + = +2 2

28 4 4

2

a f

Ekmmin = =

2 2ω

.

Chapter 41 505

Additional Problems

P41.39 Suppose the marble has mass 20 g. Suppose the wall of the box is 12 cm high and 2 mm thick. Whileit is inside the wall,

U mgy= = =0 02 9 8 0 12 0 023 5. . . . kg m s m J2b ge ja f

and E K mv= = = =12

12

0 02 0 8 0 006 42 2. . . kg m s Jb gb g .

Then Cm U E

=−

=× ⋅

= ×−−2 2 0 02 0 017 1

1 055 102 5 1034

32 1a f b gb g. .

..

kg J

J s m

and the transmission coefficient is

e e e eCL− − × × − × − × − × −= = = = =−

2 2 2 5 10 2 10 10 10 2 30 4.3 10 4.3 10 1032 3 29 29 29 30

10 10. .

~e je j e j .

P41.40 (a) λ = = × −2 2 00 10 10L . m

(b) ph

= =× ⋅×

= × ⋅−

−−

λ6 626 10

3 31 1034

1024.

. J s

2.00 10 m kg m s

(c) Epm

= =2

20 172. eV

P41.41 (a) See the figure.

FIG. P41.41(a)

(b) See the figure.

FIG. P41.41(b)

(c) ψ is continuous and ψ → 0 as x→±∞ . The function can be normalized. It describes aparticle bound near x = 0 .

(d) Since ψ is symmetric,

ψ ψ2 2

0

2 1dx dx−∞

∞ ∞z z= =

or 22

212 2

0

20A e dx

Ae ex−

∞−∞z =

−

FHGIKJ − =α

α e j .

This gives A = α .

(e) P a e dx e ex

x− →

−

=

− −= =−FHGIKJ − = − =z1 2

2 2

0

1 22 2 12

22

1 1 0 632α αα

αα αα

αb g b g e j e j e j1 2 .

506 Quantum Mechanics

P41.42 (a) Use Schrödinger’s equation

∂∂

= − −2

2 22ψ

ψx

mE Ua f

with solutions

ψ 11 1= + −Ae Beik x ik x [region I]

ψ 22= Ceik x [region II]. FIG. P41.42(a)

Where kmE

12

=

and km E U

22

=−a f

.

Then, matching functions and derivatives at x = 0

ψ ψ1 0 2 0b g b g= gives A B C+ =

and ddx

ddx

ψ ψ1

0

2

0

FHGIKJ = FHG

IKJ gives k A B k C1 2− =a f .

Then Bk kk k

A=−+

11

2 1

2 1

and Ck k

A=+

21 2 1

.

Incident wave Aeikx reflects Be ikx− , with probability RBA

k k

k k

k k

k k= =

−

+=

−

+

2

22 1

2

2 12

1 22

1 22

1

1

b gb g

b gb g

.

(b) With E = 7 00. eV

and U = 5 00. eV

kk

E UE

2

1

2 007 00

0 535=−

= =..

. .

The reflection probability is R =−

+=

1 0 535

1 0 5350 092 0

2

2

.

..

a fa f .

The probability of transmission is T R= − =1 0 908. .

Chapter 41 507

P41.43 Rk k

k k

k k

k k=

−

+=

−

+1 2

2

1 22

2 12

2 12

1

1

b gb g

b gb g

2 2

2km

E U= − for constant U

212

2km

E= since U = 0 (1)

222

2km

E U= − (2)

Dividing (2) by (1),kk

UE

22

12 1 1

12

12

= − = − = so kk

2

1

12

=

and therefore, R =−

+=

−

+=

1 1 2

1 1 2

2 1

2 10 029 4

2

2

2

2

e je j

e je j

. .

FIG. P41.43

P41.44 (a) The wave functions and probability densities are the same as those shown in the two lowercurves in Figure 41.4 of the textbook.

(b) P dxx

dx

x x

1 12

0 150

2

0 150

0 350

0 150

21 00 1 00

2 002

1 00 21 00

= = FHGIKJ

FHG

IKJ

= − FHG

IKJ

LNM

OQP

z zψπ

ππ

. .

.

.

.sin

.

..

sin.

nm

0.350 nm

nm

0.350 nm

nm nm

nm nm

4 nmb g

In the above result we used sin sin2

214

2axdxx

aaxz = FHG

IKJ −FHGIKJ a f .

Therefore, P xx

10 150

1 001 00 2

1 00= − F

HGIKJ

LNM

OQP.

.sin

. .

nm nm

2 nm nm

0.350 nm

b gπ

π

P1 1 00 0 350 0 1501 00

0 700 0 300 0 200= − − −RSTUVW =. . .

.sin . sin . .nm nm nm

nm2

b g a f a fπ

π π .

(c) Px

dxx x

22

0 150

0 350

0 150

0 3502

1 0021 00

2 002

1 008

41 00

= FHGIKJ = − F

HGIKJ

LNM

OQPz.

sin.

..

sin..

.

.

.ππ

π

P xx

20 150

0 350

1 001 004

41 00

1 00 0 350 0 1501 004

1 40 0 600

0 351

= − FHGIKJ

LNM

OQP = − − −RST

UVW=

..

sin.

. . ..

sin . sin .

..

.

ππ

ππ πa f a f a f

(d) Using En hmLn =2 2

28, we find that E1 0 377= . eV and E2 1 51= . eV .

508 Quantum Mechanics

P41.45 (a) fEh

= =× ⋅

×FHG

IKJ = ×

−

−1 80

6 626 10

1 60 104 34 10

34

1914.

.

..

eV

J s

J1.00 eV

Hza f

e j

(b) λ = =×

×= × =−c

f3 00 104 34 10

6 91 10 6918

147.

..

m s Hz

m nm

(c) ∆ ∆E t ≥2

so ∆∆ ∆

Et

ht

≥ = =× ⋅

×= × = ×

−

−− −

2 46 626 10

4 2 00 102 64 10 1 65 10

34

629 10

π πa f e j.

.. .

J s

s J eV

*P41.46 (a) Taking L L Lx y= = , we see that the expression for E becomes

Eh

m Ln n

ex y= +

2

22 2

8e j .

For a normalizable wave function describing a particle, neither nx nor ny can be zero. The

ground state, corresponding to n nx y= = 1, has an energy of

Eh

m Lh

m Le e1 1

2

22 2

2

281 1

4, = + =e j .

The first excited state, corresponding to either nx = 2 , ny = 1 or nx = 1 , ny = 2 , has an energy

E Eh

m Lh

m Le e2 1 1 2

2

22 2

2

282 1

58, ,= = + =e j .

The second excited state, corresponding to nx = 2 , ny = 2 has an energy of

Eh

m Lh

m Le e2 2

2

22 2

2

282 2, = + =e j .

Finally, the third excited state, corresponding to either nx = 1 , ny = 3 or nx = 3, nx = 1 , has an

energy

E Eh

m Lh

m Le e1 3 3 1

2

22 2

2

281 3

54, ,= = + =e j .

(b) The energy difference between the secondexcited state and the ground state is givenby

∆E E Eh

m Lh

m L

hm L

e e

e

= − = −

=

2 2 1 1

2

2

2

2

2

2

4

34

, ,

.

E1, 3 , E 3, 1

E2, 2

E1, 2 , E 2, 1

E1, 1

energy

hm Le

2

2

0Energy level diagram

FIG. P41.46(b)

Chapter 41 509

P41.47 x x dx2 2 2=−∞

∞z ψ

For a one-dimensional box of width L, ψπ

n Ln x

L= F

HGIKJ

2sin .

Thus, xL

xn x

Ldx

L Ln

L2 2 2

0

2 2

2 22

3 2= F

HGIKJ = −z sin

ππ

(from integral tables).

P41.48 (a) ψ 2 1dx−∞

∞z = becomes

Ax

Ldx A

L xL

xL

AL

L

L

L

L2 2

4

42

4

422

214

42 2

1cos sinπ

ππ π

ππF

HGIKJ = FHG

IKJ + F

HGIKJ

LNM

OQP = FHG

IKJLNMOQP =− −

zor A

L2 4= and A

L=

2.

(b) The probability of finding the particle between 0 and L8

is

ψπ

π2

0

82 2

0

8 2 14

12

0 409dx Ax

Ldx

L L

z z= FHGIKJ = + =cos . .

P41.49 For a particle with wave function

ψ xa

e x aa f = −2for x > 0

and 0 for x < 0 .

(a) ψ xa f 2 0= , x < 0 and ψ 2 22x

ae x aa f = − , x > 0

(b) Prob x x dx dx< = = =−∞ −∞z z0 0 0

20 0

a f a f a fψ

FIG. P41.49

(c) Normalization ψ ψ ψx dx dx dxa f 2 20

2

0

1−∞

∞

−∞

∞z z z= + =

02

0 1 1

02

1 0 865

02

0

20

2

0

2

0

20

2

dxa

e dx e e

x a dxa

e dx e e

x a x a

ax a

ax a a

−∞

−∞

− ∞ −∞

− − −

z zz z

+ FHGIKJ = − = − − =

< < = = FHGIKJ = − = − =

e j

a fProb ψ .

510 Quantum Mechanics

P41.50 (a) The requirement that n

Lλ2

= so ph nh

L= =λ 2

is still valid.

E pc mc Enhc

Lmc

K E mcnhc

Lmc mc

n

n n

= + ⇒ = FHGIKJ +

= − = FHGIKJ + −

b g e j e j

e j

2 2 2 22 2

22

2 2 2

2

2

(b) Taking L = × −1 00 10 12. m, m = × −9 11 10 31. kg , and n = 1, we find K1144 69 10= × −. J .

Nonrelativistic, EhmL1

2

2

34 2

31 12 214

8

6 626 10

8 9 11 10 1 00 106 02 10= =

× ⋅

× ×= ×

−

− −

−.

. ..

J s

kg m J

e je je j

.

Comparing this to K1, we see that this value is too large by 28 6%. .

P41.51 (a) Ue

d

e

dk e

de=

∈− + − + − +FHG

IKJ + −

LNM

OQP =

−

∈= −

2

0

2

0

2

41

12

13

112

17 3

47

3π πa f b g

(b) From Equation 41.12, K Eh

m d

hm de e

= = =22

8 9 361

2

2

2

2e j.

(c) E U K= + and dEdd

= 0 for a minimum:73 18

02

2

2

3k ed

hm d

e

e

− =

dh

k e m

hm k ee e e e

= = =×

× × ×=

−

− −

3

7 18 42

6 626 10

42 9 11 10 8 99 10 1 60 100 049 9

2

2

2

2

34 2

31 9 19 2a fe je j

a fe je je j.

. . ..

C nm .

(d) Since the lithium spacing is a, where Na V3 = , and the density is NmV

, where m is the mass

of one atom, we get:

aVmNm

m= FHGIKJ =FHG

IKJ =

× ×FHG

IKJ = × =

−−

1 3 1 3 27 1 3101 66 10 7

5302 80 10 0 280

density kg

kg m m nm

.. .

(5.62 times larger than c).

Chapter 41 511

P41.52 (a) ψ ω= −Bxe m x2 2b g

ddx

Be Bxm

xe Be Bm

x e

ddx

Bxm

xe Bm

xe Bm

xm

xe

ddx

Bm

xe

m x m x m x m x

m x m x m x

m

ψ ω ω

ψ ω ω ω ω

ψ ω

ω ω ω ω

ω ω ω

ω

= + −FHGIKJ = − FHG

IKJ

= −FHGIKJ − FHG

IKJ − FHG

IKJ −FHG

IKJ

= − FHGIKJ

− − − −

− − −

−

2 2 2 2 2

2

22 2 2 2

2

22

2 2 2 2

2 2 2

22

2

3

b g b g b g b g

b g b g b g

b gx m xBm

x e2 2

23 2+ FHG

IKJ

−ω ωb g

Substituting into the Schrödinger Equation (41.13), we have

− FHGIKJ + FHG

IKJ = − + FHG

IKJ

− − − −322

23 2

22

22 22 2 2 2

Bm

xe Bm

x emE

Bxem

x Bxem x m x m x m xω ω ωω ω ω ωb g b g b g b g .

This is true if − = −32

ωE

; it is true if E =3

2ω

.

(b) We never find the particle at x = 0 because ψ = 0 there.

(c) ψ is maximized if ddx

xmψ ω

= = − FHGIKJ0 1 2 , which is true at x

m= ±

ω.

(d) We require ψ 2 1dx−∞

∞z = :

1 2 214 2

2 2 2 2 23

2 1 2 3 2

3 2

2 2

= = = =−

−∞

∞−z zB x e dx B x e dx B

m

B

mm x m xω ω π

ω

πω

b g b gb g a f .

Then Bm m

= FHGIKJ =FHG

IKJ

2 41 2

1 4

3 4 3 3

3

1 4

πω ω

π.

(e) At xm

= 2ω

, the potential energy is 12

12

422 2 2m x m

mω ω

ωω= F

HGIKJ = . This is larger than the

total energy 3

2ω

, so there is zero classical probability of finding the particle here.

(f) Probability = = FH IK =− −ψ δ δω ω2 22

2 22 2

dx Bxe B x em x m xb g b g

Probability = FHGIKJFHGIKJ = F

HGIKJ

− −δπ

ωω

δωπ

ω ω2 481 2

3 24

1 24m

me

mem mb g b g

512 Quantum Mechanics

P41.53 (a) ψ 2

0

1dxLz = : A

xL

xL

xL

xL

dxL

2 2 2

0

162

82

1sin sin sin sinπ π π πFHGIKJ +

FHGIKJ +

FHGIKJFHGIKJ

LNM

OQP =z

AL L x

Lx

Ldx

AL x

Lx

Ldx A

L L xL

L

L

x

x L

2

0

2 2

0

2 3

0

216

28

21

172

1617

2163

1

FHGIKJ +FHGIKJ +

FHGIKJFHGIKJ

LNMM

OQPP =

+ FHGIKJFHGIKJ

LNMM

OQPP = + F

HGIKJ

LNMM

OQPP =

z

z=

=

sin sin

sin cos sin

π π

π ππ

π

AL

2 217

= , so the normalization constant is AL

=2

17.

(b) ψ 2 1dxa

a

−z = : A

xa

Bx

aA B

xa

xa

dxa

a2 2 2 2

22

21cos sin cos sin

π π π πFHGIKJ +

FHGIKJ +

FHGIKJFHGIKJ

LNM

OQP =

−z

The first two terms are A a2 and B a2 . The third term is:

22

22 2

42 2

83 2

0

2

3

A Bxa

xa

xa

dx A Bxa

xa

dx

a A B xa

a

a

a

a

a

a

cos sin cos cos sin

cos

π π π π π

ππ

FHGIKJFHGIKJFHGIKJ

LNM

OQP = F

HGIKJFHGIKJ

= FHGIKJ =

− −

−

z z

so that a A B2 2 1+ =e j , giving A Ba

2 2 1+ = .

*P41.54 (a) x xa

e dxax0

1 22

0= FHGIKJ =−

−∞

∞z π, since the integrand is an odd function of x.

(b) x xa

x e dxax1

3 1 224

02

=FHGIKJ =−

−∞

∞

z π, since the integrand is an odd function of x.

(c) x x dx x x x x x dx01 0 12

0 1 0 112

12

12

= + = + +−∞

∞

−∞

∞z zψ ψ ψ ψb g a f a f

The first two terms are zero, from (a) and (b). Thus:

x xa

ea

xe dxa

x e dx

aa

a

ax ax ax01

1 42

3 1 42

2 1 22

0

2 1 2

3

1 2

2 2 242

2

22 1

4

12

= FHGIKJ

FHGIKJ =

FHGIKJ

=FHGIKJFHGIKJ

=

− −

−∞

∞−

∞

z zπ π π

ππ

, from Table B.6

Chapter 41 513

P41.55 With one slit open P1 12= ψ or P2 2

2= ψ .

With both slits open, P = +ψ ψ1 22 .

At a maximum, the wave functions are in phase Pmax = +ψ ψ1 22c h .

At a minimum, the wave functions are out of phase Pmin = −ψ ψ1 22c h .

Now PP

1

2

12

22 25 0= =

ψ

ψ. , so

ψψ

1

25 00= .

and PP

max

min

.

.

.

.

.

..=

+

−=

+

−= = =

ψ ψ

ψ ψ

ψ ψ

ψ ψ

1 22

1 22

2 22

2 22

2

2

5 00

5 00

6 00

4 00

36 016 0

2 25c hc h

c hc h

a fa f .

ANSWERS TO EVEN PROBLEMS

P41.212 P41.22 (a) ψ

π1

2x

Lx

La f = F

HGIKJcos ;

P xL

xL1

22a f = FHGIKJcos

π;

ψπ

22 2

xL

xL

a f = FHGIKJsin ;

P xL

xL2

22 2a f = FHGIKJsin

π;

ψπ

32 3

xL

xL

a f = FHGIKJcos ;

P xL

xL3

22 3a f = FHGIKJcos

π;

P41.4 (a) 4; (b) 6 03. eV

P41.6 0 517. MeV, 3 31 10 20. × ⋅− kg m s

P41.838

1 2h

m ce

λFHGIKJ

P41.10 (a) 5 13. meV ; (b) 9 41. eV ; (c) The muchsmaller mass of the electron requires it tohave much more energy to have the samemomentum. (b) see the solution

P41.12 (a) 158

1 2h

m ce

λFHGIKJ ; (b) 1 25. λ P41.24 (a)

2

2

2

224

6mL

xL

−FHG

IKJ ; (b) see the solution

P41.26 see the solutionP41.14 see the solution;

2 2

2km

P41.28 1 03 10 3. × −

P41.16 (a) L2

; (b) 5 26 10 5. × − ; (c) 3 99 10 2. × − ; P41.30 85.9(d) see the solution

P41.32 3 92%.P41.18 0 250.

P41.34 (a) see the solution; bm

=ω

2; (b) E =

32

ω ;

P41.20 (a) L L− F

HGIKJ

12

2π

πsin ; (b) see the solution; (c) first excited state

(c) 0 585. L

514 Quantum Mechanics

P41.36 (a) Bm

=FHGIKJ

ωπ

1 4

; (b) δω

πmFHGIKJ

1 2P41.48 (a)

2L

; (b) 0 409.

P41.50 (a) nhc

Lm c mc

2

22 4 2F

HGIKJ + − ;P41.38 see the solution

P41.40 (a) 2 00 10 10. × − m; (b) 3 31 10 24. × ⋅− kg m s ; (b) 46 9. fJ ; 28 6%.(c) 0 172. eV

P41.52 (a) 3

2ω

; (b) x = 0 ; (c) ±mω

;P41.42 (a) see the solution; (b) 0 092 0. , 0 908.

(d) 4 3 3

3

1 4m ωπ

FHG

IKJ ; (e) 0; (f) 8

1 24δ

ωπ

meF

HGIKJ

−P41.44 (a) see the solution; (b) 0 200. ; (c) 0 351. ;(d) 0 377. eV , 1 51. eV

P41.54 (a) 0; (b) 0; (c) 2 1 2aa f−P41.46 (a) h

m Le

2

24,

58

2

2h

m Le

, h

m Le

2

2 , 5

4

2

2h

m Le

;

(b) see the solution, 3

4

2

2h

m Le