Download Chapter 6 Quantum Mechanics in One Dimension. Solutions

Chapter 6
Quantum Mechanics in One
Dimension. Solutions of Selected
Problems
6.1
Problem 6.13 (In the text book)
A proton is confined to moving in a one-dimensional box of width 0.200 nm.
(a) Find the lowest possible energy of the proton.
(b) What is the lowest possible energy of an electron confined to the same box?
(c) How do you account for the large difference in your results for (a) and (b)?
Solution
(a) The energy of a particle of mass min a on dimensional box of length L is:
En =
n 2 π 2 ~2
n2 π 2 (h/2π)2
n2 h2
=
=
2mL2
2mL2
8mL2
The lowest energy (n = 1) of a proton with mass mp = 1.67 × 10−27 kg in a box with
L = 2 × 10−10 m w get:
2
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF
SELECTED PROBLEMS
E1 =
h2
8mp L2
(6.626 × 10−34 J · s)2
8 × 1.67 × 10−27 kg × (2 × 10−10 m)2
= 8.22 × 10−22 J
8.22 × 10−22 J
=
1.60 × 10−19 J/eV
= 5.13 × 10−3 eV
=
(b) The lowest energy of an electron in a similar box is:
E1 =
h2
8me L2
(6.626 × 10−34 )2
=
8 × 9.11 × 10−31 × (2 × 10−10 )2
= 1.506 × 10−18 J
= 9.40 eV
(c) The electron energy is much larger than the proton energy because the proton mass
mp ≈ 2000me .
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
6.2. PROBLEM 6.20 (IN THE TEXT BOOK)
6.2
3
Problem 6.20 (In the text book)
Consider a particle with energy E bound to a finite square well of height U and width 2L
situated on − L ≤ x ≤ +L. Because the potential energy is symmetric about the midpoint
of the well, the stationary state waves will be either symmetric or antisymmetric about this
point.
(a) Show that for E < U , the conditions for smooth joining of the interior and exterior
waves lead to the following equation for the allowed energies of the symmetric waves:
k tan kL = α
(symmetric case)
p
p
where α = (2m/~2 )(U − E) and k = 2mE/~2 is the wavenumber of oscillation in
the interior.
(b) Show that the energy condition found in (a) can be rewritten as
√
k sec kL =
2mU
~
Apply the result in this form to an electron trapped at a defect site in a crystal, modeling the defect as a square well of height 5 eV and width 0.2 nm. Solve the equation
numerically to find the ground-state energy for the electron, accurate to ± 0.001eV .
Solution
The Schrödinger equation for such a problem can written as:
d2 ψ
2m
=
{U (x) − E} ψ(x)
dx2
~2
Inside the well U (x) = 0 and outside the well U (x) = U independent of x, so Schrödinger
equation becomes:

2m


{U − E} ψ(x)
for x > +L and x < −L (exterior)
d2 ψ  ~2
= 
dx2
2m

 −
Eψ(x)
for − L < x < +L (interior)
~2
Using k 2 = 2mE/~2 and α2 = (2m/~2 )(U − E), Schrödinger, then becomes:
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
4
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF
SELECTED PROBLEMS
2
dψ
=
dx2
(
α2 ψ(x)
− k 2 ψ(x)
for x > +L and x < −L (exterior)
for − L < x < +L (interior)
Solution of the interior equation are cos kx and sin kx. The cos kx solution is symmetric
around x = 0, while the sin kx is antisymmetric around x = 0. Since we are considering the
symmetric case, then the sin kx solution must be discarded. So we then have:
− L < x < +L
ψ(x) = A sin kx
The solutions of the exterior equation are of the form e±αx . Since the wavefunction must be
zero ant ∞, we then must discard the e+αx and the exterior solution becomes:
ψ(x) = Ce−α|x|
x > L or x < −L
where A and C are constants.
(a) At x = L, the the interior and exterior wavefunctions and their derivatives must match,
i.e.
A cos kL = Ce−αL
−Ak sin kL = −Cαe−αL
(6.1)
(6.2)
Dividing Equation (6.2) by Equation (6.1) we get:
k tan kL = α
(6.3)
(b) Note that:
k 2 + α2 =
2mE 2m
2mU
+ 2 (U − E) =
2
~
~
~2
from which we get:
α2 =
2mU
− k2
2
~
Using Equation (6.3) we get:
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
6.2. PROBLEM 6.20 (IN THE TEXT BOOK)
5
r
2mU
− k2
2
~
r
2mU L2
− k 2 L2
kL tan kL =
~2
2mU L2
(kL)2 tan2 kL =
− k 2 L2
~2
2mU L2
(kL)2 (tan2 kL + 1) =
~2
k tan kL =
It can be shown that (tan2 kL + 1) sec2 kL, the last equation than becomes:
2mU L2
2
√~
2mU
kL sec kL =
L
~
√
2mU
k sec kL =
~
(kL)2 sec2 kL =
(6.4)
(6.5)
For an electron in a a well of height U = 5 eV and width 2L = 0.2 nm, using Equation
(6.4) we get:
√
2me U L
√ ~
kL
2me c2 U L
=
cos kL
p ~c
2 × 511 × 103 × 5 × (0.1)2
=
197.3
kL
= 1.1457
cos kL
kL sec kL =
(6.6)
Equation (6.6) can be solved numerically, using sophisticated methods. However, it is
not very difficult to solve it by try and error. Trying kL = 1 rad makes the right
had side of Equation (6.6) larger than 1.1457, trying kL = 1.5 rad gives even larger
value. So, obviously kL < 1 rad trying values less than one quickly gives a solution with
kL = 0.799, which gives k = 7.99 nm−1 , the energy is then given by:
2
~2 k 2
~2 c k 2
(197.3 × 7.99)2
E=
=
= 2.432 eV
=
2me
2me c2
2 × 511 × 103
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
6
6.3
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF
SELECTED PROBLEMS
Problem 6.29 (In the text book)
An electron is described by the wavefunction
(
0
ψ(x) =
Ce−x (1 − e−x )
for x < 0
for x > 0
where x is in nanometers and C is a constant
(a) Find the value of C that normalizes ψ.
(b) Where is the electron most likely to be found; that is, for what value of x is the probability
for finding the electron largest?
(c) Calculate < x > for this electron and compare your result with its most likely position.
Comment on any differences you find.
Solution
(a) Normalization requires:
+∞
Z
|ψ|2 dx
1 =
−∞
+∞
Z
C 2 e−2x (1 − e−x )2 dx
=
0
= C
2
Z
+∞
e−2x (1 − 2e−x + e−2x ) dx
Z0 +∞
(e−2x − 2e−3x + e−4x ) dx
0
1
1 1
2
= C
−2 +
2
3 4
2
C
=
12
√
C =
12
= C2
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
6.3. PROBLEM 6.29 (IN THE TEXT BOOK)
7
(b) The most likely location for the electron is where the probability |ψ|2 of finding the
electron is largest, and that is where the wavefunction ψ is largest,i.e.
dψ
dx
d
=
C(e−x − e−2x )
dx = C 2 −e−x + 2e−2x
= C 2 e−x 2e−x − 1
0 =
The right hand side of the last equation vanishes when x = ∞ and when 2e−x = 1, or
x = ln 2 = 0.693 nm. The most likely location of the the electron is xmp = 0.693 nm.
(c) The average location of the electron is given by:
Z
+∞
x|ψ|2 dx
<x> =
−∞
= C
2
Z
+∞
xe−2x (1 − 2e−x + e−2x ) dx
Zo +∞
x(e−2x − 2e−3x + e−4x ) dx
1
1
1
2
−2 +
C
4
9 16
13
C2
144
12 × 13
144
1.083 nm
= C2
o
=
=
=
=
< x > is larger than xmp due to the fact that values of x larger that xmp have higher
probability than those lower than xmp .
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
8
6.4
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF
SELECTED PROBLEMS
Problem 6.32 (In the text book)
Calculate < x >, < x2 >, and ∆x for a quantum oscillator in its ground state. Hint: Use
the integral formula
r
Z ∞
1 π
2 −ax2
a>0
xe
dx =
4a a
0
Solution
2
The quantum oscillator wavefunction is given by ψ(x) = Ce−ax , where C =
a = mω/2~. The expectation value of x is then:
Z +∞
Z +∞
2
2
2
< x >=
x|ψ| dx = C
xe−2ax dx
−∞
p
4
a/π and
−∞
The integrand in the above equation is an odd function of x so integration from − ∞ to
+ ∞ gives zero, so < x >= 0. Now, < x2 > is:
Z
2
+∞
x2 |ψ|2 dx
<x > =
−∞
= 2C
=
=
=
=
2
Z
+∞
2
x2 e−2ax dx
0
r
1
π
2
2C
8a 2a
r
r
a 1
π
2
π 8a 2a
1
√
4 2a
~
√
2 2mω
Now ∆x is given by:
p
√
∆x = < x2 > −(< x >)2 = < x2 > =
Physics 205:Modern Physics I, Chapter 6
Fall 2004
s
~
√
2 2mω
Ahmed H. Hussein
6.5. PROBLEM 6.37 (IN THE TEXT BOOK)
6.5
9
Problem 6.37 (In the text book)
Nonstationary states. Consider a particle in an infinite square well described initially by a
wave that is a superposition of the ground and first excited states of the well:
Ψ(x, 0) = C [ψ1 (x) + ψ2 (x)]
√
(a) Show that the value C = 1/ 2 normalizes this wave, assuming ψ1 and ψ2 are themselves
normalized.
(b) Find ψ(x, t) at any later time t.
(c) Show that the superposition is not a stationary state, but that the average energy in this
state is the arithmetic mean (E1 + E2 )/2 of the ground- and first excited-state energies
E1 and E2 .
Solution
(a) The normalized wavefunctions and energies of a particle in a an infinite square well, with
width of L, are given by:
r
nπx 2
sin
L
L
2 2 2
nπ ~
=
2mL2
ψn (x) =
En
for o < x < L and n = 1, 2 3, · · ·
The normalization of the superpositioned wave function requires:
Z
+∞
|Ψ|2 dx
1 =
−∞
Z
+∞
{ψ1∗ + ψ2∗ } {ψ1 + ψ2 } dx
−∞
Z
Z
Z
Z
2
2
∗
∗
2
|ψ1 | dx + |ψ2 | dx + ψ1 ψ2 dx + ψ2 ψ1 dx
= C
= C
2
The first two integrals on the right hand side are unity each since each wave function is
normalized, while the last two integral are actually the same since both ψ1 and ψ2 are
real. Using the actual form of the each wavefunction we get:
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
10
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. SOLUTIONS OF
SELECTED PROBLEMS
Z
2
ψ1 ψ2 dx =
L
Z
L
sin
πx 0
L
sin
2πx
L
Using sin θ1 sin θ2 = 12 {sin(θ1 − θ2 ) − cos(θ1 + θ2 )}, the last equation becomes:
Z
L
3πx
cos
− cos
dx
L
L
0
(
L )
1 L h πx iL
3πx
L
sin
sin
−
L π
L o
3π
L
0
1 L
πL
L
3πL
sin
−
sin
L π
L
3π
L
1 L
L
sin π −
sin 3π
L π
3π
0
1
ψ1 ψ2 dx =
L
=
=
=
=
Z
πx The normalization condition becomes:
1 = C
2
Z
Z
2
|ψ1 | dx +
2
|ψ2 | dx +
Z
ψ1∗ ψ2
Z
dx +
ψ2∗ ψ1
dx
= C 2 (1 + 1 + 0 + 0)
= 2C 2
1
C = √
2
(b) The time dependent wavefunction is:
Ψ(x, t) = C ψ1 (x)e−iω1 t + ψ2 (x)e−iω2 t
since ω = E/~,
Ψ(x, t) = C ψ1 (x)e−iE1 t/~ + ψ2 (x)e−iE2 t/~
(c) The time dependent wavefunction Ψ(x, t) is a stationary state only if it is an eigenfunc∂
tion of the enrgy operator, [E] = i~ ∂t
, i.e. if:
[E]Ψ(x, t) = EΨ(x, t)
Physics 205:Modern Physics I, Chapter 6
Fall 2004
(6.7)
Ahmed H. Hussein
6.5. PROBLEM 6.37 (IN THE TEXT BOOK)
11
Let us see if the above condition in Equation (6.7) can be satisfied:
−iE1
−iE2
−iE1 t/~
−iE2 t/~
[E]ψ = C i~
ψ1 e
+ i~
ψ2 e
~
~
= C E1 ψ1 e−iE1 t/~ + E2 ψ2 e−iE2 t/~
Since E1 6= E2 , then obviously, the stationary state condition in Equation (6.7) is not
satisfied, and Ψ(x, t) is not a stationary state.
the average energy < E > of the state is:
Z
Ψ∗ [E]Ψ dx
= C 2 ψ1∗ (x)e+iE1 t/~ + ψ2∗ (x)e+iE2 t/~ [E] ψ1 (x)e−iE1 t/~ + ψ2 (x)e−iE2 t/~ dx
= C 2 ψ1∗ (x)e+iE1 t/~ + ψ2∗ (x)e+iE2 t/~ E1 ψ1 (x)e−iE1 t/~ + E2 ψ2 (x)e−iE2 t/~ dx
Z
Z
2
2
2
= C E1 |ψ1 | dx + E2 |ψ2 | dx
<E> =
The cross product terms vanish as in part (a). Since ψ1 and ψ2 are both normalized and
C 2 = 12 , then:
1
< E >= (E1 + E2 )
2
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein