Homework 3 Solutions
TA: Kevin Chiou (Original Solutions by Matt Block)
Problem 1
Let’s start by stating what we know. We have from Griffiths Eq. (1.33) that:
h
ˆ
p
i
=

i
¯
h
Z
Ψ
*
∂
Ψ
∂x
!
d
x
(1)
The limits of integration are implicitly from negative infinity to positive infinity. We have
Schr¨
odinger’s Equation (and its complex conjugate):
i
¯
h
∂
Ψ
∂t
=
H
Ψ
(2)

i
¯
h
∂
Ψ
*
∂t
=
H
Ψ
*
(3)
Here,
H
=
H
(
x
) because we are working in position space.
Recall that
ˆ
H
is Hermitian
so the complex conjugate of
H
is itself.
Lastly, we also have the generic form of the
Hamiltonian (in a position space representation):
ˆ
H
.
=
H
(
x
) =

¯
h
2
2
m
∂
2
∂x
2
+
V
(
x
)
(4)
We’ll be using integration by parts a bit in this problem and I won’t be writing out all
the steps.
You can look at the footnote on page 16 of Griffiths to recall the details.
In
particular, every time you integrate by parts, you get two terms; one still has an integral in
it and the other has no integral and is being evaluated at the original limits of integration.
This second term is called the
surface term
and since Ψ (and Ψ
*
) is (are) zero at
±∞
(for
normalizable wave functions), these surface terms will always evaluate to zero throughout
this problem. Bear in mind that this is not universally true; I don’t want you to forget
how to do integration by parts, but if you notice that I only write one term after doing it,
it’s because the surface term is zero. Additionally, the (generally) nonzero term with the
integral obeys a simple property if the original integral had a derivative in the integrand;
that is, often we integrate by parts to “move” a derivative from one term to another. The
price we pay is a minus sign. Consider the following example: assume
f
(
x
) and
g
(
x
) go to
zero at
±∞
. Then:
Z
∞
∞
f
(
x
)
d
g
(
x
)
d
x
d
x
=

Z
∞
∞
d
f
(
x
)
d
x
g
(
x
)d
x
(5)
1
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Notice the minus sign. This technique is very useful so become comfortable with it. Moving
on, we start by taking the time derivative of Eq. (22) above:
d
h
ˆ
p
i
d
t
=

i
¯
h
Z
∂
Ψ
*
∂t
∂
Ψ
∂x
+ Ψ
*
∂
∂x
∂
Ψ
∂t
!
d
x
(6)
Important notational clarification:
here we are assuming that
x
and
t
are independent
variables; that is,
x
does
not
depend on
t
.
Therefore, it is legitimate to bring the time
derivative inside the integral (as a partial derivative) and to interchange differentiation
with respect to
x
and with respect to
t
. Now we do our first integration by parts on the
second term:
d
h
ˆ
p
i
d
t
=

i
¯
h
Z
∂
Ψ
*
∂t
∂
Ψ
∂x

∂
Ψ
*
∂x
dΨ
d
t
!
d
x
(7)
d
h
ˆ
p
i
d
t
=
Z
"

i
¯
h
∂
Ψ
*
∂t
!
∂
Ψ
∂x
+
∂
Ψ
*
∂x
i
¯
h
∂
Ψ
∂t
!#
d
x
(8)
Now we use Eqs. (23) and (24) above:
d
h
ˆ
p
i
d
t
=
Z
"
H
Ψ
*
∂
Ψ
∂x
+
∂
Ψ
*
∂x
H
Ψ
#
d
x
(9)
Now we can insert Eq. (25) for
H
. I will collect the two terms coming from the kinetic
energy part of the Hamiltonian and separately collect the two terms with the potential
energy,
V
(
x
), in them:
d
h
ˆ
p
i
d
t
=

¯
h
2
2
m
Z
"
∂
2
Ψ
*
∂x
2
∂
Ψ
∂x
+
∂
Ψ
*
∂x
∂
2
Ψ
∂x
2
#
d
x
+
Z
"
V
(
x
)Ψ
*
∂
Ψ
∂x
+
∂
Ψ
*
∂x
V
(
x
)Ψ
#
d
x
(10)
d
h
ˆ
p
i
d
t
=

¯
h
2
2
m
Z
∂
∂x
"
∂
Ψ
*
∂x
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 Spring '03
 Nelson
 mechanics, Work, wave function, dx

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