Ph135c. Solution set #5, 5/6/10
1.
F
(
~
q
) =
Z
d
3
r±
(
~
r
)
e
i~
q
±
~
r
For spherically symmetric densities, we can write:
F
(
~
q
) =
Z
1
0
r
2
dr
Z
1
²
1
d
cos
²
Z
2
±
0
d³±
(
r
)
e
irq
cos
²
= 2
´
Z
1
0
r
2
dr±
(
r
)
±
1
irq
e
irq
cos
²
²³
³
³
³
1
²
1
=
4
´
q
Z
1
0
rdr±
(
r
) sin(
qr
)
(a)
±
(
~
r
) =
´
±
0
= const
r < R
0
r > R
F
(
~
q
) =
4
´
q
Z
R
0
rdr±
0
sin(
qr
)
=
4
´±
0
q
3
(sin(
qR
)
±
(
qR
) cos(
qR
))
(b)
±
(
~
r
) =
±
0
e
²
r=R
F
(
~
q
) =
4
´
q
Z
1
0
rdr±
0
e
²
r=R
sin(
qr
)
=
8
´R
3
±
0
(
q
2
R
2
+ 1)
2
2.
Although not explicitly stated in the problem, the notation
F
(
j
~
q
j
) implies we are looking at a
spherically symmetric density
±
(
r
). Then, from the last problem, we can write:
F
(
q
) =
4
´
q
Z
1
0
rdr±
(
r
) sin(
qr
)
We can expand this for small
q
using:
1
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View Full Documentsin(
qr
)
qr
= 1
±
(
qr
)
2
6
+
:::
So that:
F
(
q
) = 4
±
Z
1
0
r
2
dr²
(
r
)
±
1
6
q
2
4
±
Z
1
0
r
4
dr²
(
r
) +
:::
Assuming the density is properly normalized, this gives:
= 1
±
1
6
q
2
< r
2
>
+
:::
Thus we’ve shown:
6
±
±
±
±
dF
dq
2
±
±
±
±
q
=0
=
< r
2
>
3.
Note that we’re dealing with identical fermions, so all states must be antisymmetric under exchange
of any two particles (where, to emphasize, we treat protons and neutrons as di±erent isospin states of
the same underlying particle). Also, the intrinsic parity of the neutron and proton is +1, so the parity
of each state is given by (
±
1)
‘
.
(a)
Antisymmetry under exchange of the neutrons says that, together, they have spin0. They
also have no orbital angular momentum. Thus the total spin of the system is the same as the spin of
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 Spring '11
 Smith
 Physics

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