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852quiz1_Solution

# 852quiz1_Solution - = 0 we have s = 1 only ±or s ′ = 1...

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NAME: QUIZ 1 PHYS852 Quantum Mechanics II, Spring 2009 Consider a system of three spins, with s 1 = 1, s 2 = 1 and s 3 = 1. The total angular momentum is then v S = v S 1 + v S 2 + v S 3 . What are the allowed s values for the total angular momentum, and what are the degeneracies of each s level? Hint: Break it up into v S = p v S 1 + v S 2 P + v S 3 and Frst compute v S = v S 1 + v S 2 . Answer: We start from the basis | s 1 s 2 s 3 m 1 m 2 m 3 a and Frst use v S = v S 1 + v S 2 to replace the eigenvalues m 1 and m 2 with eigenvalues s and m . The allowed values of s run from s min = | s 1 s 2 | = 0 to s max = s 1 + s 2 = 2. The eigenstates of S 2 1 , S 2 2 , S 2 3 , S 3 z , S 2 and S z form a new basis, | s 1 s 2 s 3 s m m 3 a . We now want to use v S = v S + v S 3 to replace the eigenvalues m and m 3 with s and m . The new eigenstates of S 2 1 , S 2 2 , S 2 3 , S 2 , S 2 and S z are labeled | s 1 s 2 s 3 s sm a The allowed values of s run from s min = | s s 3 | = | s 1 | to s max = s + s 3 = s + 1. So we see that the allowed values of s depend on the value of s . ±or s
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Unformatted text preview: = 0, we have s = 1 only. ±or s ′ = 1 we have s = 0 , 1 , 2. ±or s ′ = 2 we have s = 1 , 2 , 3. Thus the allowed values of s are 0 , 1 , 2 , 3. The degeneracies are: ±or s = 0 we have d = 1, since there is only a single s ′ = 1, m = 0 state. ±or s = 1 we have d = 3 + 3 + 3 = 9, corresponding to s ′ = 0, m = − 1 , , 1; s ′ = 1 m = − 1 , , 2; and also s ′ = 2, m = − 1 , , 1. ±or s = 2 we have d = 5 + 5 = 10, corresponding to s ′ = 1, m = − 2 , − 1 , , 1 , 2; and also s ′ = 2, m = − 2 , 1 , , 1 , 2. Last, for s = 3 we have d = 7 corresponding to s ′ = 2, m = − 3 , − 2 , − 1 , , 1 , 2 , 3....
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