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# ph135cS5 - Ph135c Solution set#5 1 F ~ q = Z d 3 r± ~ r e...

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Ph135c. Solution set #5, 5/6/10 1. F ( ~ q ) = Z d 3 ( ~ 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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sin( 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 spin-0. 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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ph135cS5 - Ph135c Solution set#5 1 F ~ q = Z d 3 r± ~ r e...

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