Normally we will not bother to include that the spins are one half since that’s always true for
electrons. We will (and must) keep track of the intermediateℓandsquantum numbers. As can be
seen above, we need them to identify the states.
In the atomic physics section, we will even deal with more than two electrons outside a closed shell.
21.7.5 The parity of the pion fromπd→nn.
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We can determine the internal parity of the pion by studyingpion capture by a deuteron,
π+d→n+n.The pion is known to have spin 0, the deuteron spin 1, and the neutron spin^12.
The internal parity of the deuteron is +1. The pion is captured by the deuteron from a 1S states,
implyingℓ= 0 in the initial state. So the total angular momentum quantum number of the initial
state isj= 1.
So the parity of the initial state is
(−1)ℓPπPd= (−1)^0 PπPd=PπThe parity of the final state is
PnPn(−1)ℓ= (−1)ℓTherefore,
Pπ= (−1)ℓ.Because the neutrons are identical fermions, the allowed states of two neutrons are^1 S 0 ,^3 P 0 , 1 , 2 ,^1 D 2 ,
(^3) F 2 , 3 , 4 ...The only state withj= 1 is the (^3) P 1 state, soℓ= 1
⇒Pπ=− 1.
21.8 Derivations and Computations
21.8.1 Commutators of Total Spin Operators
S~ = S~(1)+S~(2)
[Si,Sj] = [S(1)i +Si(2),Sj(1)+S(2)j ]= [S(1)i ,S(1)j ] + [S(1)i ,S(2)j ] + [S(2)i ,S(1)j ] + [S(2)i ,Sj(2)]
= i ̄hǫijkS(1)k + 0 + 0 +i ̄hǫijkSk(2)=i ̄hǫijkSk