ALL 10 CHAPTERS COVERED
SOLUTIONS MANUAL
,1 Problems Chapter 1: Introduction to Nanoelectronics
2 Problems Chapter 2: Classical Particles, Classical Waves, and
Quantum Particles
2.1. What is the energy (in Joules and eV) of a photon having wavelength 650 nm? Repeat for an electron
having the same wavelength and only kinetic energy.
Solution: For the photon,
hc hc hc 19
p = ; E= = 9
= 3:058 10 J (1)
E 650 10
EJ
EeV = = 1:91 eV.
jqe j
For the electron,
h2 h2 25
E= 2 = = 5:704 10 J (2)
2me jqe j 2me (650 10 9 )2
e
6
= 3:56 10 eV.
2.2. For light (photons), in classical physics the relation
c= f (3)
is often used, where c is the speed of light, f is the frequency, and is the wavelength. For photons,
is the de Broglie wavelength the same as the wavelength in (3)? Explain your reasoning. Hint: use
Einstein’s formula p
E = mc2 = p2 c2 + m0 c4 ; (4)
where m0 is the particle’s rest mass (which, for a photon, is zero).
Solution: Yes, these wavelengths are the same. From Einstein’s formula, E = pc for photons, and
using E = hf we have
E pc
c= f = = ; (5)
h h
so that we must have = h=p.
2.3. Common household electricity in the United States is 60 Hz, a typical microwave oven operates at
2:4 109 Hz, and ultraviolet light occurs at 30 1015 Hz. In each case, determine the energy of the
associated photons in joules and eV.
Solution:
E = hf; (6)
32 13
Eelec = h60 = 3:98 10 J = 2:48 10 eV
9 24 6
Eoven = h 2:4 10 = 1:59 10 J = 9:92 10 eV
15 17
Euv = h 30 10 = 1:99 10 J = 124:2 eV.
2.4. Assume that a HeNe laser pointer outputs 1 mW of power at 632 nm.
(a) Determine the energy per photon
Solution: Each photon carries
2 c (2 ) 3 108 19
Ep = }! = } =} = 3:145 10 J = 1:963 eV. (7)
632 10 9
1
, (b) Determine the number of photons per second, N .
Solution: The sum of all N photons has power
3
P = N Ep (1/s) J = 10 J/s (8)
3
10
!N = 19
= 3:179 7 1015 photons/s.
3:145 10
2.5. Repeat 2.4 if the laser outputs 10 mW of power. How does the number of photons per second scale
with power?
Solution: The sum of all N photons has power
3
P = N Ep (1/s) J = 10 10 J/s (9)
3
10 10
!N = 19
= 3:179 7 1016 photons/s.
3:145 10
The number of photons scales linearly with power.
2.6. Calculate the de Broglie wavelength of
(a) a proton moving at 437; 000 m/s,
(b) a proton with kinetic energy 1; 100 eV,
(c) an electron travelling at 10; 000 m/s.
(d) a 800 kg car moving at 60 km/h.
Solution: (a)
h h h 13
= = = 27 ) (437000)
= 9:065 10 m (10)
p mp v (1:673 10
(b)
r
1 (2) (1100 jqe j)
E= mp v 2 = 1100 jqe j ! v = = 4:59 105 m/s (11)
2 1:673 10 27
h h h 13
= = = = 8:631 10 m
p mp v (1:673 10 ) (4:59 105 )
27
(c)
h h h 8
= = = 31 ) (10000)
= 7:274 10 m (12)
p me v (9:1095 10
(d)
km 1m 1 hour 1 min 60
60 = = 16:67 m/s (13)
hour 10 3 km 60 min 60 s 10 3 (602 )
h h h
= = = = 4:969 10 38 m
p mv (800) (16:67)
2.7. Determine the wavelength of a 150 gram baseball traveling 90 miles/hour. Use this result to explain
why baseballs do not seem to di¤ract around baseball bats.
Solution:
90 miles 1 km 1m 1 hour 1 min
(14)
hour 0:6214 miles 10 km 60 min
3 60 s
90
= = 40:23 m/s (15)
0:6214 (10 3 ) (602 )
h h h
= = = = 1:098 10 34 m
p mv (150 10 3 ) (40:23)
The de Broglie wavelength is too small to observe di¤raction, since one would observe di¤raction on
size scales of the order of . The size scale of the bat is far to large.
2
, 2.8. How much would the mass of a ball need to be in order for it to have a de Broglie wavelength of 1 m
(at which point its wave properties would be clearly observable)? Assume that the ball is travelling 90
miles/hour.
Solution:
h h h h 35
= = = =1m!m= = 1:647 10 kg (16)
p mv m (40:23) 40:23
2.9. Determine the momentum carried by a 640 nm photon. Since a photon is massless, does this momentum
have the same meaning as the momentum carried by a particle with mass?
Solution:
h 9 h 27
= = 640 10 !p= 9
= 1:035 10 Js/m=kg m/s (17)
p 640 10
The momentum has essentially the same meaning as for a particle having mass: the photon momentum
exerts a force on objects (in general, force multiplied by time equals momentum) that can be used to,
for example, move objects.
2.10. Consider a 4 eV electron, a 4 eV proton, and a 4 eV photon. For each, compute the de Broglie
wavelength, the frequency, and the momentum.
Solution: For the photon,
h hc hc
= = = = 310:17 nm, (18)
p E 4 jqe j
E 4 jqe j
E = hf ! f = = = 9:672 Hz,
h h
E 4 jqe j
p= = = 2:136 10 27 kg m/s.
c c
For the electron,
h h
=p =p = 0:613 nm, (19)
2me E 2me 4 jqe j
E 4 jqe j
E = hf ! f = = = 9:672 Hz,
h h
p = me v = (me ) 1:186 106 = 1:080 10 24 kg m/s, since
s
1 (2) (4 jqe j)
E = me v 2 = 4 jqe j ! v = = 1:186 106 m/s
2 me
For the proton,
h h
=p =p = 0:0143 nm, (20)
2mp E 2mp 4 jqe j
E 4 jqe j
E = hf ! f = = = 9:672 Hz,
h h
p = mp v = mp (27683) = 4:630 10 23 kg m/s, since
s
1 2 (2) (4 jqe j)
E = mv = 4 jqe j ! v = = 27; 683 m/s
2 mp
Obviously, f is the same for all particles since E = hf . The momentum values are very small, but
smallest for the photon. The wavelength is far larger for the photon than for the electron, which itself
has a far larger wavelength than for the proton (the proton has far greater mass than the electron).
2.11. Determine the de Broglie wavelength of an electron that has been accelerated from rest through a
potential di¤erence of 1:5 volts.
3
, Solution:
h h
E = 1:5 eV, and thus =p =p = 1:001 nm. (21)
2me E 2me (1:5) jqe j
2.12. Calculate the uncertainty in velocity of a 1 kg ball con…ned to
(a) a length of 20 m,
(b) a length of 20 cm,
(c) a length of 20 m.
(d) What can you conclude about observing “quantum e¤ects” using 1 kg balls? What kind of objects
would you need to use to see quantum e¤ects on these length scales?
Solution: From
}
p x ; (22)
2
(a)
} }
v = = 2:637 10 30 m/s (23)
2m x 2 (1) (20 10 6 )
(b)
} } 34
v = 2)
= 2:637 10 m/s (24)
2m x 2 (1) (20 10
(c)
} } 36
v = = 2:637 10 m/s (25)
2m x 2 (1) (20)
(d) 1 kg balls are far too heavy to observe quantum e¤ects - one would need to have masses on
the order of an atomic particle to observe quantum e¤ects, since the value of h is so small.
2.13. If we know that the velocity of an electron is 40:23 0:01 m/s, what is the minimum uncertainty in
its position? Repeat for a 150 gram baseball travelling at the same velocity.
Solution: From
}
m v x ; (26)
2
for the electron,
} 3
x = 5:789 10 m. (27)
2me (0:01)
For the baseball,
} 32
x = 3:52 10 m. (28)
2 (:15) (0:01)
2.14. If a molecule having mass 2:3 10 26 kg is con…ned to a region 200 nm in length, what is the minimum
uncertainty in the molecule’s velocity?
Solution: From
}
m v x ; (29)
2
} }
v = 26 ) (200 9)
= 0:0115 m/s. (30)
2m x 2 (2:3 10 10
2.15. Determine the minimum uncertainty in the velocity of an electron that has its position speci…ed to
within 10 nm.
Solution: From
}
m v x ; (31)
2
} }
v = 9)
= 5788:5 m/s. (32)
2me x 2me (10 10
4
,2.16. Explain the di¤erence between a fermion and a boson, and give two examples of each.
Solution: Particles with integral (in units of }) spin are bosons. Examples are photons and phonons.
Particles with half-integral spin are fermions. Examples are electrons, protons, and neutrons.
3 Problems Chapter 3: Quantum Mechanics of Electrons
5 0
3.1. For the matrix operator L = , show that eigenvalues and eigenvectors are
1 2
0
= 2, x= ; (33)
7
= 5, x= ;
where ; 6= 0. That is, show that the preceding quantities satisfy the eigenvalue problem Lx = x.
Solution:
Lx = x (34)
5 0 0 0
=2
1 2
0 0 p
=
2 2
Lx = x (35)
5 0 7 7
= 5
1 2
35 35 p
=
5 5
n o
3.2. Consider the set of functions p1 einx ; n = 0; 1; 2; ::: .
2
(a) Show that this is an orthonormal set on the interval ( ; ).
Solution:
Z
1 1 1 ei(n m) e i(n m)
p einx p e imx dx = =0 if n 6= m (36)
2 2 2 i (n m)
Z
1 inx 1
p e p e inx dx = 1 if n = m
2 2
(b) On the interval ( =2; =2), is the set an orthogonal set, an orthonormal set, or neither?
Solution: Evaluating the integral one sees that the set is not orthogonal (and, hence, can’t be
orthonormal).
nq o
2
3.3. Consider the set of functions sin(nx); n = 1; 2; ::: on the interval (0; ).
(a) Show that this is an orthonormal set.
5
, Solution:
Z r r Z
2 2 1
sin(nx) sin(mx)dx = (cos (mx nx) cos (mx + nx)) dx
0 0
1 sin ((n m) ) sin ((n + m) )
= = 0 if n 6= m
n m n+m
Z Z
2 2 1 1
= sin2 (nx)dx = cos 2nx dx
0 0 2 2
2 1
= = 1 if n = m
2
(b) Determine an operator (including boundary conditions) for which the preceding set are eigenfunc-
tions. What are the eigenvalues?
Solutions: The operator is
d2
ob = 2 ; (37)
dx
the second derivative operator ( d2 =dx2 also works), acting on functions de…ned over 0 x .
Every function (x) = A sin (kx) + B cos (kx) is an eigenfunction with eigenvalue = k 2 , since
d2
(A sin (kx) + B cos (kx)) = k 2 (A sin (kx) + B cos (kx)) : (38)
dx2
If k is to be an integer, k = n, and B = 0, then the boundary condition is (0) = ( ) = 0. The
eigenvalues are simply n = 0; 1; 2; :::.
3.4. For the di¤erential operator L = d2 =dx2 , u (0) = u (a) = 0, determine eigenvalues and eigenfunc-
tions u. That is, solve
Lu = u; (39)
where u (x) is a nonzero function subject to the given boundary conditions. Normalize the eigenfunc-
tions, and show that the eigenfunctions are orthonormal.
Solution:
d2
u u=0 (40)
dx2 p p
u = A sin x + B cos x
d 2 p p p p
check: 2
A sin x + B cos x A sin x + B cos x =0
dx
u (0) = B = 0
p p
u (a) = A sin a=0! a = n , n = 0; 1; 2; :::
n 2
= :
a
Therefore
n
u = A sin x (41)
a
To normalize the eigenfunctions,
Z a
n
A2 sin2 ( x)dx = 1 (42)
0 a
Z a
1 1
A2 cos 2 nx dx =1
0 2 2 a
r
2 1 2
A a =1!A=
2 a
6
, Finally,
Z a
2 n m
sin( x) sin( x)dx = 0 if n 6= m (43)
a 0 a a
= 1 if n = m:
3.5. Repeat problem 3.4, but for boundary conditions u0 (0) = u0 (a) = 0, where u0 = du=dx.
Solution:
d2
u u=0 (44)
dx2 p p
u = A sin x + B cos x
0
u (0) = A = 0
p p p
u0 (a) = B sin a=0! a = n , n = 0; 1; 2; :::
n 2
= :
a
Therefore
n
u = B cos x: (45)
a
To normalize the eigenfunctions, set
Z a
n
B 2 cos2 x dx = 1 (46)
0 a
leading to r
"n
B= ; (47)
a
where
1; n=0
"n : (48)
2; n 6= 0
3.6. Assume that some observable of a certain system is measured and found to be n for some integer n.
By postulate 2, we know that immediately after the measurement the system is in state n , which is
an eigenstate of the measurement operator ob (i.e., where ob n = n n ).
(a) What can we conclude about the system’s state immediately before the measurement?
Solution: Nothing, other than that it was in a superposition with some content in n .
(b) Assume that the identical measurement is then performed on 100; 000 identical systems, and
each time the measurement result is the same, n . What can we infer about the system’s state
immediately before the measurement?
Solution: We can reasonably assume that before the measurement the system was in state n .
8
3.7. Assume that an electronic state has a lifetime of 10 s. What is the minimum uncertainty in the
energy of an electron in this state?
Solution: From
}
E t ; (49)
2
then
27
} } 5:273 10 8
E = 8)
= J=3:291 10 eV. (50)
2 t 2 (10 jqe j
7
,3.8. In the example of solving the one-dimensional Schrödinger equation on p. 65, we obtained the state
functions
(x; t) = (x) e iEn t=}
where
1=2
2 n
(x) = sin x ; n even, (51)
L L
1=2
2 n
= cos x ; n odd,
L L
are eigenfunctions of the second derivative operator d2 =dx2 , and where energy eigenvalues were found
to be
}2 n 2
En = : (52)
2m L
(a) Show that the odd eigenfunction (sine) can be written as
1 1 pn 1 i p}n x
(x) = p p ei } x p e (53)
2 i L i L
= + ;
and determine a similar expression for the even eigenfunction. The term + ( ) represents a
wave propagating with positive (negative) momentum. Thus, any state described by sine and
cosine can be thought of as representing a superposition of positive and negative momentum
states.
Solution:
1=2 1=2 n n
2 n 2 ei L x e i L x
(x) = sin x = ; (54)
L L L 2i
and using
n
k= ; (55)
L
it was shown in the example that
h hk n }
p= = = = pn ; (56)
2 L
so that
1=2
1 1 pn
i p}n x
(x) = ei } x
e : (57)
i 2L
(b) Although the decomposition of a standing wave into two counterpropagating waves, as in part (a),
is useful, it can be misinterpreted. Since the probability density (x; t) (x; t) is independent of
time, the expectation value of position, hxi, is independent of time, and so, really, we should not
think of the particle as “bouncing” back and forth in the con…ned space (otherwise, hxi would be
a function of t). Determine the expectation value of momentum, using either (3.217) or (3.219),
and discuss your answer in light of the above comment.
Solution: The expectation value for momentum is,
Z L=2
@
hpx i = i} dx (58)
L=2 @x
Z L=2
1 1 pn 1 pn
= p p e i}x p ei } x
L=2 2 i L i L
@ 1 1 pn 1 pn
i} p p ei } x p e i } x dx
@x 2 i L i L
Z L=2
i pn
= pn sin 2 x dx = 0:
L L=2 }
8
, Therefore, the average momentum is zero, meaning that there are an equal number of positive
and negative momentum states, and no net movement.
(c) Assume that the particle is in a state composed of the …rst two eigenfunctions,
!
1=2 1=2
1 2 i
E1 t 2 2 i
E2 t
(x; t) = p cos x e } + sin x e } : (59)
2 L L L L
Show that the expectation value of position as a function of time is
16 L 3 } 2
hxi = cos t : (60)
9 2 2 m L2
Interpret this solution, compared with the expectation value of position for a single stationary
state n , which is time-independent.
Solution:
Z L=2
hxi = x dx (61)
L=2
Z L=2
12 E1 t 2 E2 t
= x cos x ei } + sin x ei }
2L L=2 L L
iE1 t 2 i
E2 t
cos x e } + sin x e } dx
L L
Z L=2
1 2 (E1 E2 )t
= x cos2 x + cos x sin x ei }
L L=2 L L L
2 (E2 E1 )t 2
+ cos x sin x ei } + sin2 x dx
L L L
Z L=2
2 (E1 E2 ) t 2
= cos x cos x sin x dx
L } L=2 L L
16L (E1 E2 ) t
= cos ;
9 2 }
where we used the fact that the integral of an odd function over symmetric limits is zero. Since
!
2
(E1 E2 ) 1 }2 2 }2 2
= (62)
} } 2m L 2m L
!
2
} 2 2 3} 2
= = ;
2m L L 2m L2
then
16L 3} 2
hxi = cos t : (63)
9 2 2m L2
3.9. Since Schrödinger’s equation is a homogeneous equation, the most general solution for the state function
is a sum of homogeneous solutions (3.142),
X
(r; t) = an n (r) e iEn t=} : (64)
n
Show that if (r; 0) is known then an expression for the weighting amplitudes an can be determined.
Assume that the eigenfunction n form an orthonormal set. Hint: multiply
X
(r; 0) = an n (r) (65)
n
9
SOLUTIONS MANUAL
,1 Problems Chapter 1: Introduction to Nanoelectronics
2 Problems Chapter 2: Classical Particles, Classical Waves, and
Quantum Particles
2.1. What is the energy (in Joules and eV) of a photon having wavelength 650 nm? Repeat for an electron
having the same wavelength and only kinetic energy.
Solution: For the photon,
hc hc hc 19
p = ; E= = 9
= 3:058 10 J (1)
E 650 10
EJ
EeV = = 1:91 eV.
jqe j
For the electron,
h2 h2 25
E= 2 = = 5:704 10 J (2)
2me jqe j 2me (650 10 9 )2
e
6
= 3:56 10 eV.
2.2. For light (photons), in classical physics the relation
c= f (3)
is often used, where c is the speed of light, f is the frequency, and is the wavelength. For photons,
is the de Broglie wavelength the same as the wavelength in (3)? Explain your reasoning. Hint: use
Einstein’s formula p
E = mc2 = p2 c2 + m0 c4 ; (4)
where m0 is the particle’s rest mass (which, for a photon, is zero).
Solution: Yes, these wavelengths are the same. From Einstein’s formula, E = pc for photons, and
using E = hf we have
E pc
c= f = = ; (5)
h h
so that we must have = h=p.
2.3. Common household electricity in the United States is 60 Hz, a typical microwave oven operates at
2:4 109 Hz, and ultraviolet light occurs at 30 1015 Hz. In each case, determine the energy of the
associated photons in joules and eV.
Solution:
E = hf; (6)
32 13
Eelec = h60 = 3:98 10 J = 2:48 10 eV
9 24 6
Eoven = h 2:4 10 = 1:59 10 J = 9:92 10 eV
15 17
Euv = h 30 10 = 1:99 10 J = 124:2 eV.
2.4. Assume that a HeNe laser pointer outputs 1 mW of power at 632 nm.
(a) Determine the energy per photon
Solution: Each photon carries
2 c (2 ) 3 108 19
Ep = }! = } =} = 3:145 10 J = 1:963 eV. (7)
632 10 9
1
, (b) Determine the number of photons per second, N .
Solution: The sum of all N photons has power
3
P = N Ep (1/s) J = 10 J/s (8)
3
10
!N = 19
= 3:179 7 1015 photons/s.
3:145 10
2.5. Repeat 2.4 if the laser outputs 10 mW of power. How does the number of photons per second scale
with power?
Solution: The sum of all N photons has power
3
P = N Ep (1/s) J = 10 10 J/s (9)
3
10 10
!N = 19
= 3:179 7 1016 photons/s.
3:145 10
The number of photons scales linearly with power.
2.6. Calculate the de Broglie wavelength of
(a) a proton moving at 437; 000 m/s,
(b) a proton with kinetic energy 1; 100 eV,
(c) an electron travelling at 10; 000 m/s.
(d) a 800 kg car moving at 60 km/h.
Solution: (a)
h h h 13
= = = 27 ) (437000)
= 9:065 10 m (10)
p mp v (1:673 10
(b)
r
1 (2) (1100 jqe j)
E= mp v 2 = 1100 jqe j ! v = = 4:59 105 m/s (11)
2 1:673 10 27
h h h 13
= = = = 8:631 10 m
p mp v (1:673 10 ) (4:59 105 )
27
(c)
h h h 8
= = = 31 ) (10000)
= 7:274 10 m (12)
p me v (9:1095 10
(d)
km 1m 1 hour 1 min 60
60 = = 16:67 m/s (13)
hour 10 3 km 60 min 60 s 10 3 (602 )
h h h
= = = = 4:969 10 38 m
p mv (800) (16:67)
2.7. Determine the wavelength of a 150 gram baseball traveling 90 miles/hour. Use this result to explain
why baseballs do not seem to di¤ract around baseball bats.
Solution:
90 miles 1 km 1m 1 hour 1 min
(14)
hour 0:6214 miles 10 km 60 min
3 60 s
90
= = 40:23 m/s (15)
0:6214 (10 3 ) (602 )
h h h
= = = = 1:098 10 34 m
p mv (150 10 3 ) (40:23)
The de Broglie wavelength is too small to observe di¤raction, since one would observe di¤raction on
size scales of the order of . The size scale of the bat is far to large.
2
, 2.8. How much would the mass of a ball need to be in order for it to have a de Broglie wavelength of 1 m
(at which point its wave properties would be clearly observable)? Assume that the ball is travelling 90
miles/hour.
Solution:
h h h h 35
= = = =1m!m= = 1:647 10 kg (16)
p mv m (40:23) 40:23
2.9. Determine the momentum carried by a 640 nm photon. Since a photon is massless, does this momentum
have the same meaning as the momentum carried by a particle with mass?
Solution:
h 9 h 27
= = 640 10 !p= 9
= 1:035 10 Js/m=kg m/s (17)
p 640 10
The momentum has essentially the same meaning as for a particle having mass: the photon momentum
exerts a force on objects (in general, force multiplied by time equals momentum) that can be used to,
for example, move objects.
2.10. Consider a 4 eV electron, a 4 eV proton, and a 4 eV photon. For each, compute the de Broglie
wavelength, the frequency, and the momentum.
Solution: For the photon,
h hc hc
= = = = 310:17 nm, (18)
p E 4 jqe j
E 4 jqe j
E = hf ! f = = = 9:672 Hz,
h h
E 4 jqe j
p= = = 2:136 10 27 kg m/s.
c c
For the electron,
h h
=p =p = 0:613 nm, (19)
2me E 2me 4 jqe j
E 4 jqe j
E = hf ! f = = = 9:672 Hz,
h h
p = me v = (me ) 1:186 106 = 1:080 10 24 kg m/s, since
s
1 (2) (4 jqe j)
E = me v 2 = 4 jqe j ! v = = 1:186 106 m/s
2 me
For the proton,
h h
=p =p = 0:0143 nm, (20)
2mp E 2mp 4 jqe j
E 4 jqe j
E = hf ! f = = = 9:672 Hz,
h h
p = mp v = mp (27683) = 4:630 10 23 kg m/s, since
s
1 2 (2) (4 jqe j)
E = mv = 4 jqe j ! v = = 27; 683 m/s
2 mp
Obviously, f is the same for all particles since E = hf . The momentum values are very small, but
smallest for the photon. The wavelength is far larger for the photon than for the electron, which itself
has a far larger wavelength than for the proton (the proton has far greater mass than the electron).
2.11. Determine the de Broglie wavelength of an electron that has been accelerated from rest through a
potential di¤erence of 1:5 volts.
3
, Solution:
h h
E = 1:5 eV, and thus =p =p = 1:001 nm. (21)
2me E 2me (1:5) jqe j
2.12. Calculate the uncertainty in velocity of a 1 kg ball con…ned to
(a) a length of 20 m,
(b) a length of 20 cm,
(c) a length of 20 m.
(d) What can you conclude about observing “quantum e¤ects” using 1 kg balls? What kind of objects
would you need to use to see quantum e¤ects on these length scales?
Solution: From
}
p x ; (22)
2
(a)
} }
v = = 2:637 10 30 m/s (23)
2m x 2 (1) (20 10 6 )
(b)
} } 34
v = 2)
= 2:637 10 m/s (24)
2m x 2 (1) (20 10
(c)
} } 36
v = = 2:637 10 m/s (25)
2m x 2 (1) (20)
(d) 1 kg balls are far too heavy to observe quantum e¤ects - one would need to have masses on
the order of an atomic particle to observe quantum e¤ects, since the value of h is so small.
2.13. If we know that the velocity of an electron is 40:23 0:01 m/s, what is the minimum uncertainty in
its position? Repeat for a 150 gram baseball travelling at the same velocity.
Solution: From
}
m v x ; (26)
2
for the electron,
} 3
x = 5:789 10 m. (27)
2me (0:01)
For the baseball,
} 32
x = 3:52 10 m. (28)
2 (:15) (0:01)
2.14. If a molecule having mass 2:3 10 26 kg is con…ned to a region 200 nm in length, what is the minimum
uncertainty in the molecule’s velocity?
Solution: From
}
m v x ; (29)
2
} }
v = 26 ) (200 9)
= 0:0115 m/s. (30)
2m x 2 (2:3 10 10
2.15. Determine the minimum uncertainty in the velocity of an electron that has its position speci…ed to
within 10 nm.
Solution: From
}
m v x ; (31)
2
} }
v = 9)
= 5788:5 m/s. (32)
2me x 2me (10 10
4
,2.16. Explain the di¤erence between a fermion and a boson, and give two examples of each.
Solution: Particles with integral (in units of }) spin are bosons. Examples are photons and phonons.
Particles with half-integral spin are fermions. Examples are electrons, protons, and neutrons.
3 Problems Chapter 3: Quantum Mechanics of Electrons
5 0
3.1. For the matrix operator L = , show that eigenvalues and eigenvectors are
1 2
0
= 2, x= ; (33)
7
= 5, x= ;
where ; 6= 0. That is, show that the preceding quantities satisfy the eigenvalue problem Lx = x.
Solution:
Lx = x (34)
5 0 0 0
=2
1 2
0 0 p
=
2 2
Lx = x (35)
5 0 7 7
= 5
1 2
35 35 p
=
5 5
n o
3.2. Consider the set of functions p1 einx ; n = 0; 1; 2; ::: .
2
(a) Show that this is an orthonormal set on the interval ( ; ).
Solution:
Z
1 1 1 ei(n m) e i(n m)
p einx p e imx dx = =0 if n 6= m (36)
2 2 2 i (n m)
Z
1 inx 1
p e p e inx dx = 1 if n = m
2 2
(b) On the interval ( =2; =2), is the set an orthogonal set, an orthonormal set, or neither?
Solution: Evaluating the integral one sees that the set is not orthogonal (and, hence, can’t be
orthonormal).
nq o
2
3.3. Consider the set of functions sin(nx); n = 1; 2; ::: on the interval (0; ).
(a) Show that this is an orthonormal set.
5
, Solution:
Z r r Z
2 2 1
sin(nx) sin(mx)dx = (cos (mx nx) cos (mx + nx)) dx
0 0
1 sin ((n m) ) sin ((n + m) )
= = 0 if n 6= m
n m n+m
Z Z
2 2 1 1
= sin2 (nx)dx = cos 2nx dx
0 0 2 2
2 1
= = 1 if n = m
2
(b) Determine an operator (including boundary conditions) for which the preceding set are eigenfunc-
tions. What are the eigenvalues?
Solutions: The operator is
d2
ob = 2 ; (37)
dx
the second derivative operator ( d2 =dx2 also works), acting on functions de…ned over 0 x .
Every function (x) = A sin (kx) + B cos (kx) is an eigenfunction with eigenvalue = k 2 , since
d2
(A sin (kx) + B cos (kx)) = k 2 (A sin (kx) + B cos (kx)) : (38)
dx2
If k is to be an integer, k = n, and B = 0, then the boundary condition is (0) = ( ) = 0. The
eigenvalues are simply n = 0; 1; 2; :::.
3.4. For the di¤erential operator L = d2 =dx2 , u (0) = u (a) = 0, determine eigenvalues and eigenfunc-
tions u. That is, solve
Lu = u; (39)
where u (x) is a nonzero function subject to the given boundary conditions. Normalize the eigenfunc-
tions, and show that the eigenfunctions are orthonormal.
Solution:
d2
u u=0 (40)
dx2 p p
u = A sin x + B cos x
d 2 p p p p
check: 2
A sin x + B cos x A sin x + B cos x =0
dx
u (0) = B = 0
p p
u (a) = A sin a=0! a = n , n = 0; 1; 2; :::
n 2
= :
a
Therefore
n
u = A sin x (41)
a
To normalize the eigenfunctions,
Z a
n
A2 sin2 ( x)dx = 1 (42)
0 a
Z a
1 1
A2 cos 2 nx dx =1
0 2 2 a
r
2 1 2
A a =1!A=
2 a
6
, Finally,
Z a
2 n m
sin( x) sin( x)dx = 0 if n 6= m (43)
a 0 a a
= 1 if n = m:
3.5. Repeat problem 3.4, but for boundary conditions u0 (0) = u0 (a) = 0, where u0 = du=dx.
Solution:
d2
u u=0 (44)
dx2 p p
u = A sin x + B cos x
0
u (0) = A = 0
p p p
u0 (a) = B sin a=0! a = n , n = 0; 1; 2; :::
n 2
= :
a
Therefore
n
u = B cos x: (45)
a
To normalize the eigenfunctions, set
Z a
n
B 2 cos2 x dx = 1 (46)
0 a
leading to r
"n
B= ; (47)
a
where
1; n=0
"n : (48)
2; n 6= 0
3.6. Assume that some observable of a certain system is measured and found to be n for some integer n.
By postulate 2, we know that immediately after the measurement the system is in state n , which is
an eigenstate of the measurement operator ob (i.e., where ob n = n n ).
(a) What can we conclude about the system’s state immediately before the measurement?
Solution: Nothing, other than that it was in a superposition with some content in n .
(b) Assume that the identical measurement is then performed on 100; 000 identical systems, and
each time the measurement result is the same, n . What can we infer about the system’s state
immediately before the measurement?
Solution: We can reasonably assume that before the measurement the system was in state n .
8
3.7. Assume that an electronic state has a lifetime of 10 s. What is the minimum uncertainty in the
energy of an electron in this state?
Solution: From
}
E t ; (49)
2
then
27
} } 5:273 10 8
E = 8)
= J=3:291 10 eV. (50)
2 t 2 (10 jqe j
7
,3.8. In the example of solving the one-dimensional Schrödinger equation on p. 65, we obtained the state
functions
(x; t) = (x) e iEn t=}
where
1=2
2 n
(x) = sin x ; n even, (51)
L L
1=2
2 n
= cos x ; n odd,
L L
are eigenfunctions of the second derivative operator d2 =dx2 , and where energy eigenvalues were found
to be
}2 n 2
En = : (52)
2m L
(a) Show that the odd eigenfunction (sine) can be written as
1 1 pn 1 i p}n x
(x) = p p ei } x p e (53)
2 i L i L
= + ;
and determine a similar expression for the even eigenfunction. The term + ( ) represents a
wave propagating with positive (negative) momentum. Thus, any state described by sine and
cosine can be thought of as representing a superposition of positive and negative momentum
states.
Solution:
1=2 1=2 n n
2 n 2 ei L x e i L x
(x) = sin x = ; (54)
L L L 2i
and using
n
k= ; (55)
L
it was shown in the example that
h hk n }
p= = = = pn ; (56)
2 L
so that
1=2
1 1 pn
i p}n x
(x) = ei } x
e : (57)
i 2L
(b) Although the decomposition of a standing wave into two counterpropagating waves, as in part (a),
is useful, it can be misinterpreted. Since the probability density (x; t) (x; t) is independent of
time, the expectation value of position, hxi, is independent of time, and so, really, we should not
think of the particle as “bouncing” back and forth in the con…ned space (otherwise, hxi would be
a function of t). Determine the expectation value of momentum, using either (3.217) or (3.219),
and discuss your answer in light of the above comment.
Solution: The expectation value for momentum is,
Z L=2
@
hpx i = i} dx (58)
L=2 @x
Z L=2
1 1 pn 1 pn
= p p e i}x p ei } x
L=2 2 i L i L
@ 1 1 pn 1 pn
i} p p ei } x p e i } x dx
@x 2 i L i L
Z L=2
i pn
= pn sin 2 x dx = 0:
L L=2 }
8
, Therefore, the average momentum is zero, meaning that there are an equal number of positive
and negative momentum states, and no net movement.
(c) Assume that the particle is in a state composed of the …rst two eigenfunctions,
!
1=2 1=2
1 2 i
E1 t 2 2 i
E2 t
(x; t) = p cos x e } + sin x e } : (59)
2 L L L L
Show that the expectation value of position as a function of time is
16 L 3 } 2
hxi = cos t : (60)
9 2 2 m L2
Interpret this solution, compared with the expectation value of position for a single stationary
state n , which is time-independent.
Solution:
Z L=2
hxi = x dx (61)
L=2
Z L=2
12 E1 t 2 E2 t
= x cos x ei } + sin x ei }
2L L=2 L L
iE1 t 2 i
E2 t
cos x e } + sin x e } dx
L L
Z L=2
1 2 (E1 E2 )t
= x cos2 x + cos x sin x ei }
L L=2 L L L
2 (E2 E1 )t 2
+ cos x sin x ei } + sin2 x dx
L L L
Z L=2
2 (E1 E2 ) t 2
= cos x cos x sin x dx
L } L=2 L L
16L (E1 E2 ) t
= cos ;
9 2 }
where we used the fact that the integral of an odd function over symmetric limits is zero. Since
!
2
(E1 E2 ) 1 }2 2 }2 2
= (62)
} } 2m L 2m L
!
2
} 2 2 3} 2
= = ;
2m L L 2m L2
then
16L 3} 2
hxi = cos t : (63)
9 2 2m L2
3.9. Since Schrödinger’s equation is a homogeneous equation, the most general solution for the state function
is a sum of homogeneous solutions (3.142),
X
(r; t) = an n (r) e iEn t=} : (64)
n
Show that if (r; 0) is known then an expression for the weighting amplitudes an can be determined.
Assume that the eigenfunction n form an orthonormal set. Hint: multiply
X
(r; 0) = an n (r) (65)
n
9
