HW4 Solutions
Notice numbers may change randomly in your
assignments and you may have to recalculate solutions for
your specific case.
1. T&L 6.P.03
In a region of space, a particle has a wave function given by
! (x) = Ae" x
2
/(2 L2 )
and energy E =
!2
.
2mL2
(a) Find the potential energy V as a function of x. Sketch V(x)
vs. x.
(b) What is the classical potential that has this dependence?
Solution:
a) To find the potential energy, we use the time-independent
Schrodinger equation: !
wave function, we have
! 2 d 2" (x)
+ V (x)" (x) = E" (x) . For the given
2m dx 2
2
2
d! (x)
Ax 2 2
x
# 1 &
= A % " 2 ( 2xe" x /(2 L ) = " 2 e" x /(2 L ) = " 2 ! (x) , and thus
$ 2L '
dx
L
L
2
2
d ! (x)
1
x
= " 2 ! (x) + 2 ! (x) . Therefore, the Schrodinger equation
2
dx
L
L
2
! " 1 x2 %
!2
becomes !
! +
( (x) + V (x)( (x) =
( (x) , resulting in
2m $# L2 L4 '&
2mL2
!
!2 x 2
!2 x 2
"
(x)
+
V
(x)
"
(x)
=
0
V
(x)
=
.
Hence,
. This is a parabola
2mL4
2mL4
centered at x=0.
b) The classical potential with this dependence is the simple
harmonic oscillator potential.
2. T&L 6.P.22
Show that the wave functions of a particle in a one-dimensional
infinite square well are orthogonal: i.e., for n ! m , " ! n (x)! m (x)dx = 0 .
Solution:
As the wave functions of the infinite square well are given by
2
n" x
sin
inside the well and zero outside, the integral of
L
L
L
2
n# x
m# x
interest takes the following form: " ! n (x)! m (x)dx = " sin
sin
dx .
L0
L
L
1
Using the identity sin a sin b = ( cos(a ! b) ! cos(a + b)) , the integral
2
! n (x) =
becomes
1 #
(n ! m)" x
(n + m)" x &
1
(n ! m)" x
1
(n + m)" x
! cos
sin
!
sin
%$ cos
(' dx =
)
L0
L
L
" (n ! m)
L
" (n + m)
L
0
L
L
for n ! m .
3. T&L 6.P.33
Compute < x > 0 = " ! 0* (x)x! 0 (x)dx and < x 2 > 0 = " ! 0* (x)x 2! 0 (x)dx for the
$ m" '
% # ! )(
ground state of a harmonic oscillator: ! 0 (x) = &
1/ 4
e* m" x
2
/(2!)
.
Solution:
For the ground state of the harmonic oscillator, the expectation
value of the position operator x is given by
< x > 0 = " ! 0* (x)x! 0 (x)dx =
&
"
%&
m# % m# x 2 /!
xe
dx = 0 . The fact that this
$!
expression vanishes can be seen either by brute force
calculation, or by symmetry (the integrand is odd and the limits
of integration are symmetric).
The expectation value of x2, however, doesn’t vanish:
$
< x 2 >0 =
%
#$
m! 2 # m! x 2 /!
x e
dx .
"!
Integrals of this type are given in Appendix B1 of the textbook.
However, here is a useful trick: knowing the fundamental integral
L
=0
0
"
#e
!"
"
#
! cx 2
dx =
$
, we can take a derivative with respect to c to obtain
c
x 2 e! cx dx = !
2
!"
"
2
d
d $
$
x 2 e! cx dx = !
= 3/2 .
#
dc !"
dc c 2c
$
m!
m! 2 # m! x 2 /!
Here c =
, and thus < x 2 > 0 = %
xe
dx =
!
"!
#$
m! 1 " ! 3
!
.
=
3 3
"! 2 m !
2m!
4. T&L 6.P.33
$ m" '
a) Using ! 0 (x) = &
% # ! )(
1/ 4
e* m" x
2
/(2!)
, write down the total wave function
! 0 (x,t) for the ground state of a harmonic oscillator.
b) Compute the expectation value of the momentum operator
< p > 0 = $ ! 0* (x)("i!)
#! 0 (x)
#x
dx for this state.
Solution:
a) The ground state energy is E0 =
! 0 (x,t) = " 0 (x)e
#iE0 t /!
& m$ )
=(
' % ! +*
!!
, and hence
2
1/ 4
e# m$ x
2
/(2!) #i$ t /2
e
.
b) The expectation value of the momentum is given by
< p > 0 = $ ! 0* (x)("i!)
d! 0 (x) $ m" '
=&
% # ! )(
dx
# m! &
< p >0 = %
$ " ! ('
1/ 4
1/2
#! 0 (x)
#x
dx . As
$ *m" x ' * m" x 2 /(2!) $ *m" x '
=&
! (x) ,
&%
)e
% ! )( 0
! (
m!
* m! x 2 /!
(i!)
e
xdx = 0 , again by symmetry.
! )
The expectation value of p2 doesn’t vanish. We have
d 2! 0 (x) "m#
m 2# 2 x 2
=
!
(x)
+
! 0 (x) . Therefore,
0
dx 2
!
!2
1/2
*
m"
m" x 2 '
2
2 $ m" '
! m" x 2 /! $
< p > 0 = !! &
+e
&% !1 + ! )( dx . The first term is
% # ! )(
! !*
# m! &
!m! %
$ " ! ('
1/2 *
+e
) m! x 2 /!
)*
# m! &
dx = !m! %
$ " ! ('
$ m" '
term is !(m" ) &
% # ! )(
1/2 *
1/2
# "! &
%$
(
m! '
1/2
= !m! , and the second
(m" )2 $ m" '
dx = !
&
)
+xe
2 % #! (
!*
!m!
Hence, the final answer is < p 2 > 0 =
.
2
2
2 ! m" x 2 /!
1/2
$ # !3 '
&% m 3" 3 )(
1/2
=!
!m"
.
2
5. T&L 6.P.54
A particle of mass m is in an infinite square potential given by
V = !, x < ! L 2,
V = 0, ! L 2 < x < L 2,
V = !, x > L 2.
Since the potential is symmetric about the origin, the probability
density | ! (x) |2 must also be symmetric.
a) Show that this implies either ! ("x) = ! (x) or ! ("x) = "! (x) .
b) Show that the proper solutions of the time-independent
Schrodinger equation can be written
! (x) =
2
n" x
cos
, n = 1, 3, 5, 7,… and ! (x) =
L
L
2
n" x
sin
, n = 2, 4, 6, 8,… .
L
L
n 2! 2 ! 2
.
c) Show that the allowed energies are En =
2mL2
Solution
a) Since ! 2 (x) = ! 2 ("x) = ! ("x)! ("x) , we must have ! ("x) = ±! (x) .
b) The analysis of the time-independent Schrodinger equation
is similar to that of the infinite well located between x=0
and x=L, as done in class.
The main differences are that the wave function is nonvanishing
only for ! L 2 < x < L 2 , and the boundary conditions on the wave
(
) ( )
functions are ! " L 2 = ! L 2 = 0 .
More precisely, the wave function is given by ! ( x ) = A cos kx + Bsin kx ,
where A and B are constants to be determined by the boundary
conditions and normalization.
The potential is symmetric, and hence the solutions can be
classified as even, ! ("x) = ! (x) , or odd, ! ("x) = "! (x) . For the even
solutions, B=0, and thus the wave function takes the form
! (x) = A cos kx . The boundary conditions ! (x = ± L 2) = A cos
kL
= 0 then
2
n!
, with n = 1, 3, 5,… . The wave function for the even
L
n" x
solutions is therefore given by ! (x) = A cos
, n = 1, 3, 5,… .
L
imply that k =
For the odd solutions, A=0, and the boundary conditions
kL
n!
, with n = 2, 4, 6,… . The
= 0 imply that k =
2
L
n" x
wave function for the odd solutions is ! (x) = Bsin
, n = 2, 4, 6,… .
L
2
! 2 k 2 ! 2 " n! %
n 2! 2 ! 2
=
=
c) The energy is given by E =
.
$ '
2m 2m # L &
2mL2
! (x = ± L 2) = ±Bsin
.