Solutions to Problem Set 4
Physics 341
by: Callum Quigley
1
DeltaFunction Potentials
(i) and (ii)
See midterm solutions
(iii)
Away from
x
=
±
a
the potential vanishes, so for bound states (
E <
0), the wavefunctions
will go like (superpositions of)
ψ
(
x
)
’
e
±
κx
with energy
E
=

~
2
κ
2
2
m
. More precisely,
ψ
n
(
x
) =
A
n
e
κx
,
x
≤ 
a
B
n
e
κx
+
C
n
e

κx
,

a
≤
x
≤
a
D
n
e

κx
,
x
≥
a
(1.1)
where
n
labels the different possible energy eigenstates.
Imposing continuity of
ψ
(
x
) at
x
=
±
a
we have
A
n
=
B
n
+
C
n
e
2
κa
(1.2)
D
n
=
B
n
e
2
κa
+
C
n
.
However, as in the case of a single deltafunction,
ψ
0
(
x
) is discontinuous at
x
=
±
a
. Inte
grating the Schr¨odinger equation around those points yields
lim
→
0
(
ψ
0
(
±
a
+ )

ψ
0
(
±
a

)) =
2
mg
~
2
ψ
(
±
a
)
≡ 
κ
0
ψ
(
±
a
)
.
(1.3)
Note that
κ
0
=
2
m

g

~
2
>
0. Plugging in our ansatz for
ψ
(
x
) leads to the constraints
A
n
=
κ
κ
0
(
A
n

B
n
+
C
n
e
2
κa
)
(1.4)
D
n
=
κ
κ
0
(
D
n

C
n
+
B
n
e
2
κa
)
.
Combining (1.2) and (1.4), and performing a little algebra, we obtain the following transcen
dental relation on the allowed energies
e

2
κa
=
±
2
κ
κ
0

1
(1.5)
1
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which also constrains
B
n
=
±
C
n
. Thus, an even and an odd solution are possible
ψ
±
(
x
) =
A
±
e
κx
,
x
≤ 
a
B
±
(
e
κx
±
e

κx
)
,

a
≤
x
≤
a
±
A
±
e

κx
,
x
≥
a
.
(1.6)
Notice, however, that for
κ
0
a <
1 the odd solution disappears from the spectrum. This can
be seen as follows: we know
κ
= 0 is a trivial solution of
e

2
κa
= 1

2
κ/κ
0
, now consider the
derivatives of both sides at 0. If the RHS has a the greater slope, then we are guaranteed
that the two curves intersect again with
κ >
0. Otherwise, no solution exists (with
κ >
0).
In particular, as
a
→
0 the energies must satisfy
1 =
±
2
κ
κ
0

1
.
(1.7)
For the even solution, this basically reduces to the single deltafunction potential result:
κ
=
κ
0
(though now
g/ra
2
g
). In terms of the energy
E
+
=

2
m

g

~
2
.
(1.8)
For the odd solution, there is only the trivial solution
κ
=
E

= 0. This limit agrees with
our intuition, since
a
→
0 reduces to parts (
i
) and (
ii
) (with the modified
g
).
2
Transmission Through Multiple Barriers
Note:
I should begin by pointing out that when I discussed this problem with the professor,
the intention was to find
k
’s (in the region of each barrier) which are not integer multiples
of each other, as opposed to the energies (away from the barriers). Unfortunately, this isn’t
how the question was posed. I’ll attempt to answer both “interpretations” here.
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 Spring '10
 Sghy
 Physics, mechanics, Empty set, coherent, Boundary conditions, coherent states, BCH formula

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