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Physics 512
Winter 2003
Homework Set #8 – Solutions
1. This is based on Sakurai, Chapter 7, Problem 9. Consider scattering by a repulsive
δ
shell potential
2
mV
(
r
)
¯
h
2
=
γδ
(
r
−
a
)
,γ
>
0
a
) Set up an equation that determines the
s
wave phase shift,
δ
0
, as a function of
k
(where
E
=¯
h
2
k
2
/
2
m
).
Because of the
δ
function, we expect the Frst derivative of the wavefunction to
jump at
r
=
a
. ±or this reason, we cannot use the standard logarithmic derivative
matching conditions. However this problem is easy to solve for the
`
=0
phase
shift. ±or
r
6
=
a
, all we have are free spherical waves. So we may write the exact
solution for the radial wavefunction as
R
(
r
)=
±
j
0
(
kr
)
,r
<
a
B
[
h
(2)
0
(
)+
e
2
iδ
0
h
(1)
0
(
)]
≥
a
(1)
where
B
is a normalization constant, and
δ
0
is the
s
wave phase shift. Actually,
for the
δ
function jump, it is better to work with the function
u
(
r
rR
(
r
)
, which
satisFes the simple onedimensional equation
²
d
2
dr
2
−
`
(
`
+1)
r
2
−
2
mV
¯
h
2
+
k
2
³
u
(
r
)=0
±or
`
and the
δ
shell potential, this becomes simply
u
0
+
k
2
u
=
(
r
−
a
)
u
(2)
In terms of
u
(
r
)
, the radial wavefunction
(1)
becomes
u
(
r
±
k
−
1
sin(
)
<
a
ik
−
1
B
[
e
−
ikr
−
e
2
iδ
0
e
ikr
]
≥
a
±or convenience, we may choose a di²erent normalization in order to eliminate
the
k
−
1
factor. Thus we write
u
(
r
±
sin(
)
<
a
B
sin(
+
δ
0
)
≥
a
(where we have additionally absorbed a phase
e
iδ
0
into
B
). This clearly solves
(2)
for
r
6
=
a
, and moreover retains the correct deFnition of the phase shift
δ
0
.
±rom
(2)
, we now match the wavefunction and Frst derivative at
r
=
a
. The two
conditions give
B
sin(
ka
+
δ
0
) = sin(
)
kB
cos(
+
δ
0
k
cos(
γ
sin(
)
1
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View Full Document Dividing the two eliminates
B
, and yields the result
cot(
ka
+
δ
0
) = cot(
)+
γ
k
Using the relation
cot(
α
+
β
) = (cot
α
cot
β
−
1)
/
(cot
α
+ cot
β
)
, and solving for
cot
δ
0
, we Fnd the equivalent condition
cot
δ
0
=
−
cot(
)
−
k
γ
csc
2
(
)
(3)
b
) Assume that
γ
is very large
γ
±
1
a
,γ
±
k
Show that if tan
is not close to zero, the
s
wave phase shift resembles the
hardsphere result,
δ
0
=
−
.
Since
tan
is not close to zero,
sin
is not close to zero, and
csc
2
(
)
is
bounded. Thus for
γ
±
k
, we may drop the last term in
(3)
and Fnd
cot
δ
0
≈
−
cot(
)
. This demonstrates that
δ
0
≈−
, which is the hard sphere result.
Note that this result does not depend on
being small, (although it does depend
on the condition
²
γa
being satisFed). A comparison of the exact result with
the hard sphere approximation is shown by
5
10
15
20
3
2.5
2
1.5
1
0.5
δ
0
δ
0
ka
= 
ka
(mod
π
)
exact
Note, of course, that the phase shift is only deFned up to factors of
nπ
. This
shows that, for the most part, the spherical shell behaves just like a hard sphere.
However part
c
)
below shows us that it is not just a hard sphere
...
c
) Also show that for tan
close to (but not exactly equal to) zero, resonance
behavior is possible; that is, cot
δ
0
goes through zero from the positive side as
k
increases (so that
δ
0
is increasing as
k
is increasing). Determine approximately
the positions of the resonances keeping terms of order 1
/γ
.
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This note was uploaded on 02/15/2011 for the course PHYS 512 taught by Professor Unknow during the Winter '03 term at Cornell College.
 Winter '03
 Unknow
 Physics, Work

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