Physics 43 Chapter 41 Homework #11 Key

1. A particle in an infinitely deep square well has a wave function given by ( )

=

Lx

Lx πψ 2sin2

2

for 0 ≤ x ≤ L and zero otherwise. (a) Determine the expectation value of x. (b) Determine the probability of finding the particle near L/2, by calculating the probability that the particle lies in the range 0.490L ≤ x ≤ 0.510L. (c) What If? Determine the probability of finding the particle near L/4, by calculating the probability that the particle lies in the range 0.240L ≤ x ≤ 0.260L. (d) Argue that the result of part (a) does not contradict the results of parts (b) and (c).

(a) π π = = − ∫ ∫2

0 0

2 42 2 1 1sin cos2 2

L Lx x

x x dx x dxL L L L

π π π

π = − + =

2 2

200

4 4 41 1 sin cos2 16 2

L Lx x xx L LxL L L L L

(b) Probability π ππ

= = − ∫0.510 0.510

2

0.4900.490

2 42 1 1sin sin4

L L

LL

x xLdx xL L L L L

Probability ( )π ππ

−= − − = × 510.020 sin 2.04 sin 1.96 5.26 104

(c) Probability ππ

− − = ×

0.2602

0.240

41 sin 3.99 104

L

L

xxL L

(d) In the = 2n , it is more probable to find the particle either

near =4Lx or =

34Lx than at the center, where the

probability density is zero.

Nevertheless, the symmetry of the distribution means that

the average position is 2L .

2. Show that the wave function ψ = Aei(kx – ωt) is a solution to the Schrödinger equation where U = 0.

3. Consider a particle moving in a one-dimensional box for which the walls are at x = –L/2 and x = L/2. (a) Write the wave functions and probability densities for n = 1, n = 2, and n = 3. (b) Sketch the wave functions and probability densities.

Model: A particle is confined in a rigid one-dimensional box of length L. Visualize:

Solve: (a) ψ (x) is zero because it is physically impossible for the particle to be there because the box is rigid. (b) The potential energy within the region −L/2 ≤ x ≤ L/2 is U(x) = 0 J. The Schrödinger equation in this region is

( ) ( ) ( )2

22 2

2d x m E x xdxψ

ψ β ψ= − = −

where 22mEβ = . (c) Two functions ψ (x) that satisfy the above equation are sinβx and cosβx . A general solution to the Schrödinger equation in this region is

ψ (x) = Asinβx + Bcosβx where A and B are constants to be determined by the boundary conditions and normalization. (d) The wave function must be continuous at all points. ψ = 0 just outside the edges of the box. Continuity requires that ψ also be zero at the edges. The boundary conditions are ψ (x = −L/2) = 0 and ψ (x = L/2) = 0. (e) The two boundary conditions are

( )2 sin cos sin cos 02 2 2 2L L L LL A B A Bβ β β βψ − = − + − = − + =

( )2 sin cos 02 2L LL A Bβ βψ = + =

These are two simultaneous equations. Unlike the boundary conditions in the particle in a box problem of Section 41.4, there are two distinct ways to satisfy these equations. The first way is to add the equations. This gives

2 cos 02LB β =

⇒ B = 0 ⇒ψ (x) = Asinβx

To finish satisfying the boundary conditions,

sin 02Lβ =

⇒ βL = 2π , 4π , 6π , … = 2nπ with n = 1, 2, 3, …

With this restriction on the values of β , the wave function becomes ( ) ( )sin 2x A n x Lψ π= . Using the definition of β from part (b), the energy is

( ) ( )2 2 2

22 2

22

2 8nn hE nmL mLπ

= = n = 1, 2, 3, …

The second way is to subtract the second equation from the first. This gives

2 cos 02LA β − =

⇒ A = 0 ⇒ψ (x) = Bcosβx

To finish satisfying the boundary conditions,

cos 02Lβ =

⇒ βL = π , 3π , 5π , … = (2n – 1)π n = 1, 2, 3, …

With this restriction on the values of β , the wave function becomes ( ) ( )( )cos 2 1x B n x Lψ π= − . Using the definition of β , the energy is

( ) ( )2 2 2 2

22 2

2 12 1

2 8nn hE n

mL mLπ−

= = − n = 1, 2, 3, …

Summarizing this information, the allowed energies and the corresponding wave functions are

( )

( ) ( )

( ) ( )

22

1 1 12

22

1 1 12

2 1cos 2 1 , 9 , 25 ,

8

2sin 2 4 , 16 , 36 ,

8

n

n

n x hB E n E E EL mL

xn x hA E n E E EL mL

π

ψπ

− = − =

= = =

where E1 = h2/8mL2. (f) The results are actually the same as the results for a particle located at 0 ≤ x ≤ L. That is, the energy levels are the same and the shapes of the wave functions are the same. This has to be, because neither the particle nor the potential well have changed. All that is different is our choice of coordinate system, and physically meaningful results can’t depend on the choice of a coordinate system. The new coordinate system forces us to use both sines and cosines, whereas before we could use just sines, but the shapes of the wave functions in the box haven’t changed.

4 A one-dimensional harmonic oscillator wave function is 2bxAxe−=ψ

(a) Find b and the total energy E. (b) Is this a ground state or a first excited state? Problem 41 in Chapter 16 helps students to understand how to draw conclusions from an identity.

ψ −=2bxAxe so ψ − −= −

2 222bx bxd Ae bx Aedx

and

ψ ψ ψ− − −= − − + = − +

2 2 22

2 3 2 22 2 4 4 6 4bx bx bxd bxAe bxAe b x e b b x

dx

Substituting into the Schrödinger equation,

ωψ ψ ψ ψ − + = − +

22 2 226 4 mE mb b x x

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

So we must have both − = −

2

26 mEb and ω =

224 mb

(a) Therefore ω=

2mb

(b) and ω= =

23 32

bEm

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

5. An electron has a kinetic energy of 12.0 eV. The electron is incident upon a rectangular barrier of height 20.0 eV and thickness 1.00 nm. By what factor would the electron’s probability of tunneling through the barrier increase assuming that the electron absorbs all the energy of a photon with wavelength 546 nm (green light)?

The original tunneling probability is −= 2CLT e where

( )( ) ( )( )π − −

−−

× × − ×−= = = ×

× ⋅

1 21 2 31 1910 1

34

2 2 9.11 10 kg 20 12 1.6 10 J21.448 1 10 m

6.626 10 J sm U E

C

The photon energy isλ

⋅= = =

1 240 eV nm2.27 eV

546 nmhchf , to make the electron’s new kinetic energy

+ =12 2.27 14.27 eV and its decay coefficient inside the barrier

( )( )π − −

−−

× × − ×′ = = ×

× ⋅

1 231 1910 1

34

2 2 9.11 10 kg 20 14.27 1.6 10 J1.225 5 10 m

6.626 10 J sC

Now the factor of increase in transmission probability is ( ) − −

′−′− × × ×

− = = = =9 10 1

22 2 10 m 0.223 10 m 4.45

2 85.9C L

L C CCL

e e e ee

6. An electron having total energy E = 4.50 eV approaches a rectangular energy barrier with U = 5.00 eV and L = 950 pm as. Classically, the electron cannot pass through the barrier because E < U. However, quantum-mechanically the probability of tunneling is not zero. Calculate this probability, which is the transmission coefficient.

( )( )( )− −

−−

× − × ⋅= = ×

× ⋅

31 199 1

34

2 9.11 10 5.00 4.50 1.60 10 kg m s3.62 10 m

1.055 10 J sC

( )( ) ( )− − −

−

= = − × × = −

= ×

2 9 1 12

3

exp 2 3.62 10 m 950 10 m exp 6.88

1.03 10

CLT e

T

7. A particle is described by the wave function

(a) Determine the normalization constant A. (b) What is the probability that the particle will be found between x = 0 and x = L/8 if its position is measured? (c) What is the most probable position in the box? (d) What is the expectation value in the box?

(a) ψ∞

−∞

=∫ 2 1dx becomes

π π π ππ π−−

= + = = ∫4 4

2 2 2 2

44

2 41cos sin 12 4 2 2

L L

LL

x x xL LA dx A AL L L

or =2 4AL

and =2AL

.

(b) The probability of finding the particle between 0 and 8L is

πψ

π = = + = ∫ ∫

8 82 2 2

0 0

2 1 1cos 0.4094 2

L Lx

dx A dxL

8. What is the probability that an electron will tunnel through a 0.45nm gap from a metal to a STM probe if the work function is 4.0eV? The work function E0 = 4.0 eV is the energy barrier U0 – E that an electron must either go over (photoelectric effect) or tunnel through. From Equation 41.40, the electron’s penetration distance is

( ) ( )( )34

11

31 190

1.05 10 J s 9.72 10 m 0.0972 nm2 2 9.11 10 kg 4.0 eV 1.6 10 J/eVm U E

η−

−

− −

×= = = × =

− × × ×

The tunneling probability through the barrier is ( ) ( )2 0.50 nm 0.0972 nm2 5

tunnel 3.40 10wP e eη −− −= = = ×

9. Find the expectation value for the for the first two states of a harmonic oscillator. 10. Find the most probable position for the the first two states of a harmonic oscillator.

(a) π

∞

−

−∞

= = ∫

21 2

0 0axax x e dx , since the integrand is an odd function of x.

(b) π

∞

−

−∞

= =

∫2

1 232

1

4 0axax x x e dx , since the integrand is an odd function of x.

(c) ( ) ( ) ( )ψ ψ ψ ψ∞ ∞

−∞ −∞

= + = + +∫ ∫20 1 0 101 0 1

1 1 12 2 2

x x dx x x x x x dx

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