# ph135cS3 - Ph135c Solution set#3 1 First recall that n p =...

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Ph135c. Solution set #3, 4/22/10 1. First recall that: ˆ s n · ˆ s p = 1 2 ( J 2 - ˆ s 2 n - ˆ s 2 p ) Since ˆ s 2 n = ˆ s 2 p = s ( s + 1) = 3 4 (since the proton and neutron are spin- 1 2 ), and since J 2 = j ( j + 1) is zero for the singlet and 2 for the triplet, we ﬁnd: ˆ s n · ˆ s p = ± - 3 4 singlet 1 4 triplet Thus the operator we want is given by: ˆ a = a s + 3 a t 4 + ( a t - a s ) ˆ s n · ˆ s p The expectation value of this operator in an unpolarized beam is given by: < ˆ a 2 > = 1 4 X spins < f | ˆ a 2 | i > where the factor of 1 4 comes from averaging over the initial spins. The only non-zero terms in the sum are: = 1 4 ² < 0 , 0 | ˆ a 2 | 0 , 0 > + < 1 , 1 | ˆ a 2 | 1 , 1 > + < 1 , 0 | ˆ a 2 | 1 , 0 > + < 1 , - 1 | ˆ a 2 | 1 , - 1 > = 1 4 ( a s 2 + 3 a t 2 ) giving the correct total cross section. Now we replace ˆ s p with ˆ s H 2 , where H 2 represents para-hydrogen with total spin zero. This operator is actually zero, so ˆ a becomes: ˆ a H 2 = a s + 3 a t 4 And, proceeding as before, we ﬁnd: 4 π < ˆ a 2 > = 4 π × 1 2 × ² < ↑ | ˆ a 2 | ↑ > + < ↓ | ˆ a 2 | ↓ > ³ = π ( a s + 3 a t ) 2 4 1

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2. Since the proton and neutron both have parity +1, the parity of the deuteron is ( - 1) . So if it were parity odd, the orbital angular momentum would have to be odd. The only way to make this consistent with a total spin of zero is to have = 1 and to have the spins combined into the triplet of total spin 1. Then we can take the spin zero combination of these two spin-1 multiplets. Also, since is odd and the spin state is symmetric, to get overall antisymmetry, we must take isospin to be 1, so that it is symmetric. Usually a ground state with 6 = 0 is disfavored, since the centrifugal repulsion term tends to raise the energy. However, this could conceivably be countererd by a large L · S term in the nuclear potential, since such a term is larger in the J = 0 state, and could lower the energy suﬃciently to make this state the ground state. 3.
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## This note was uploaded on 12/04/2011 for the course PHY 7070 taught by Professor Smith during the Spring '11 term at University of Wisconsin.

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ph135cS3 - Ph135c Solution set#3 1 First recall that n p =...

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