Physics 43 Chapter 41 Homework #11 Key
1. A particle in an infinitely deep square well has a wave function given by
ψ 2 (x ) =
2
 2πx 
sin 

L  L 
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).
L
L
4π x 
2
2
 2π x 
1 1
(a) =
x
x sin 2  =
x  − cos
 dx
 dx
L
L
L 
 L 
2 2
0
0
∫
1 x2
x =
L 2
∫
L
0
4π x
4π x 
1 L2  4π x
L
sin
+ cos
=
2 

L 16π  L
L
L 0
2
L
−
0.510L
(b)
Probability
=
4π x 
2
1 L
 2π x 
1
dx  x −
sin 2 
sin
=
L
L 4π
L  0.490L
 L 
L
0.490L
0.510L
∫
0.020 −
Probability =
1
5.26 × 10−5
( sin 2.04π − sin 1.96π ) =
4π
4π x 
x 1
Probability  −
=
sin
3.99 × 10−2

π
4
L
L

 0.240L
0.260L
(c)
(d)
In the n = 2 , it is more probable to find the particle either
L
3L
than at the center, where the
near x = or x =
4
4
probability density is zero.
Nevertheless, the symmetry of the distribution means that
L
the average position is .
2
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
d 2ψ ( x )
2m
=
− 2 Eψ ( x ) =
− β 2ψ ( x )
dx 2

where β = 2mE  2 .
(c) Two functions ψ (x) that satisfy the above equation are sinβ x and cosβ x . A general solution to the Schrödinger
equation in this region is
ψ (x) = Asinβ x + Bcosβ x
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
 βL 
 βL 
 βL 
 βL 
− A sin 
0
 + B cos  −
=
 + B cos 
=
2
2
2






 2 
ψ ( −L 2) =
A sin  −
 βL 
 βL 
 + B cos 
 =0
 2 
 2 
ψ ( L 2 ) = A sin 
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
 βL 
2 B cos 
 = 0 ⇒ B = 0 ⇒ ψ (x) = Asinβ x
 2 
To finish satisfying the boundary conditions,
 βL 
sin 
 = 0 ⇒ β L = 2π , 4π , 6π , … = 2nπ with n = 1, 2, 3, …
 2 
With this restriction on the values of β , the wave function becomes ψ ( x ) = A sin ( 2nπ x L ) . Using the definition of β
from part (b), the energy is
2nπ )  2
(=
2
En
=
2
2mL
( 2n )
2
h2
8mL2
n = 1, 2, 3, …
The second way is to subtract the second equation from the first. This gives
 βL 
−2 A cos 
0 ⇒ A = 0 ⇒ ψ (x) = Bcosβ x
=
 2 
To finish satisfying the boundary conditions,
 βL 
cos 
 = 0 ⇒ β L = π , 3π , 5π , … = (2n – 1)π
 2 
n = 1, 2, 3, …
With this restriction on the values of β , the wave function becomes
=
ψ ( x ) B cos ( 2 ( n − 1) π x L ) . Using the definition of
β , the energy is
=
En
2
2
( 2n − 1) π 2=
2
2mL
( 2n − 1)
2
h2
8mL2
n = 1, 2, 3, …
Summarizing this information, the allowed energies and the corresponding wave functions are
2

 ( 2n − 1) π x 
2 h
E1 , 9 E1 , 25E1 , 
=
( 2n 1)
 B cos 
 En =−
L
8mL2



ψ ( x) = 
2
x
2 h
 sin  ( 2n ) π=
2
4 E1 , 16 E1 , 36 E1 , 
=
A
E
n
(
)


n

8mL2
 L 

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.
ψ = Axe −bx
4 A one-dimensional harmonic oscillator wave function is
2
(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.
dψ
ψ = Axe− bx so= Ae− bx − 2bx 2 Ae− bx
dx
2
2
2
and
2
2
2
d2ψ
=
−2bxAe− bx − 4bxAe− bx + 4b2 x3 e− bx =
−6bψ + 4b2 x 2ψ
2
dx
Substituting into the Schrödinger equation,
 2mE 
 mω  2
−6bψ + 4b2 x 2ψ =
−
ψ + 
 xψ
  
  
2
For this to be true as an identity, it must be true for all values
of x.
So we must have both
(a)
Therefore
(b)
and
(c)
The wave function is that of the
b=
E
=
−6b =
−
mω
2
3b 2
=
m
3
ω
2
first excited state .
mω 
2mE
and 4b2 = 

2
  
2
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 T = e−2CL where
E) )
( 2m(U − =
12
C
=

2π ( 2 × 9.11× 10−31 kg ( 20 − 12) 1.6 × 10−19 J)
= 1.448 1× 1010 m −1
6.626 × 10−34 J⋅ s
12
hc 1 240 eV ⋅ nm
=
= 2.27 eV , to make the electron’s new kinetic energy
546 nm
λ
12 + 2.27 =
14.27 eV and its decay coefficient inside the barrier
The photon energy is hf
=
2π ( 2 × 9.11× 10−31 kg ( 20 − 14.27 ) 1.6 × 10−19 J)
=
C′
= 1.225 5 × 1010 m −1
6.626 × 10−34 J⋅ s
12
Now the factor of increase in transmission probability is
−9
10
e−2C′L
− C′ )
m −1
4.45
e2×10 m ×0.223×10 =
e=
= e2L ( C=
85.9
−2CL
e
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.
C
=
2 ( 9.11× 10−31 ) ( 5.00 − 4.50) ( 1.60 × 10−19 ) kg ⋅ m s
= 3.62 × 109 m −1
1.055 × 10−34 J⋅ s
T =e−2CL =exp  −2 ( 3.62 × 109 m −1 )( 950 × 10−12 m )  =exp ( −6.88)
1.03 × 10−3
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
dx = 1 becomes
L4
A
2
 2π x 
 L
cos2 
A2 
 dx =
L 
2π


−L 4
∫
or A 2 =
(b)
 4π x  
 π x 1
 L  π 
A2 
1
+ sin 
=
=



 L 4
 2π   2 
 L  −L 4
L4
2
4
and A =
.
L
L
The probability of finding the particle between 0 and
L8
∫
L8
ψ dx =A 2
2
0
L
is
8
1 1
 2π x 
= 0.409
 dx = +
L 
4 2π
∫ cos 
2
0
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
1.05 ×10−34 J s
9.72 ×10−11 m =
0.0972 nm
=
=
−31
−19
2m (U 0 − E )
2 ( 9.11×10 kg )( 4.0 eV ×1.6 ×10 J/eV )
η=

The tunneling probability through the barrier is
nm )
wη
Ptunnel
= e −2=
e −2( 0.50 nm ) ( 0.0972=
3.40 ×10−5
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)
=
x0
12
2
 a
=
x   e− ax dx
π
 
−∞
∫
∞
0 ,
since the integrand is an odd function of x.
12
 4a3 
2 − ax 2
(b)
x1 =
x
=
 x e dx
 π 
−∞
∫
∞
(c)
x
01
=
∫
−∞
x
0 , since the integrand is an odd function of x.
1
1
1
2
(ψ 0 +ψ 1 ) dx = x 0 + x 1 +
2
2
2
∞
∫ xψ
0
( x )ψ 1 ( x ) dx
−∞
The first two terms are zero, from (a) and (b). Thus: