852HW3_Solutions

# 852HW3_Solutions - PHYS852 Quantum Mechanics II Spring 2009...

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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2009 HOMEWORK ASSIGNMENT 3: Solutions 1. Cohen-Tannoudji problem 10.1, page 1086 Answer: a.) For the 1 s ground state, we have ℓ = 0, s = 1 / 2, and i = 1. Thus the possible values of j run from j min = | ℓ − s | = 1 / 2 to j max = ℓ + s = 1 / 2. So only the value j = 1 / 2 is allowed. For f , the possible values range from f min = | j − i | = 1 / 2 to f max = j + i = 3 / 2. Thus the total angular momentum eigenvalues for 1 s deuterium are f = 1 / 2 , 3 / 2 . b.) For deuterium in the 2 p state, we have ℓ = 0, s = 1 / 2, and i = 1. Thus j min = | ℓ − s | = 1 / 2 and j max = ℓ + s = 3 / 2, so that the allowed j values are j = 1 / 2 , 3 / 2 . For j = 1 / 2, we have f min = | j − i | = 1 / 2 and f max = j + 1 = 3 / 2. For j = 3 / 2, we have f min = | j − i | = 1 / 2 and f max = j + i = 5 / 2, so the allowed f values are f = 1 / 2 , 3 / 2 , 5 / 2 . 1 2. Cohen-Tannoudji problem 10.5, page 1087 Answer: The original basis for the three-spin system is {| m 1 ,m 2 ,m 3 )} . We begin by defining vector S ′ = vector S 1 + vector S 2 . The eigenstates of S ′ 2 and S z are formed by combining the different m 1 ,m 2 states in the usual way to form a singlet and three triplet states. Listing the new states in the basis {| s ′ ,m ′ ,m 3 |} on the r.h.s, and their definition in terms of states in the old basis on the l.h.s. gives: | 1 , 1 ,m 3 ) = | 1 2 , 1 2 ,m 3 ) | 1 , ,m 3 ) = 1 √ 2 ( | 1 2 , − 1 2 ,m 3 ) + |− 1 2 , 1 2 ,m 3 ) ) | 1 , − 1 ,m 3 ) = |− 1 2 , − 1 2 ,m 3 ) | , ,m 3 ) = 1 √ 2 ( | 1 2 , − 1 2 ,m 3 ) − |− 1 2 , 1 2 ,m 3 ) ) Now we can let vector S = vector S ′ + vector S 3 , with the eigenstates of S ′ 2 , S 2 and S z forming the basis {| s ′ ,s,m )} . For the case s ′ = 0, we can only have s = 1 / 2. The selection rule is m = m ′ + m 3 . Since m ′ = 0, we have m = m 3 . There are only two s ′ = 0 states, distinguished by the value of m 3 , so we can immediately identify the s ′ = 0 states: (listed as | s ′ ,s,m ) = | s ′ ,m ′ ,m 3 ) = | m 1 ,m 2 ,m 3 ) ) | , 1 2 , 1 2 ) = | , , 1 2 ) = 1 √ 2 ( | 1 2 , − 1 2 , 1 2 ) − |− 1 2 , 1 2 , 1 2 ) ) | , 1 2 , − 1 2 ) = | , , − 1 2 ) = 1 √ 2 ( | 1 2 , − 1 2 , − 1 2 ) − |− 1 2 , 1 2 , − 1 2 ) ) For the case s ′ = 1, we have s = 3 / 2 , 1 / 2. In fact, the Clebsch Gordan coefficients for adding angular momenta 1 and 1 / 2 were computed in class in lecture 6, on 1/26/09. The table we constructed was: m ′ , m 3 1, 1 2 1, − 1 2 0, 1 2 0, − 1 2-1, 1 2-1, − 1 2 3 2 , 3 2 1 3 2 , 1 2 1 / √ 3 √ 2 / √ 3 s , m 3 2 , − 1 2 √ 2 / √ 3 1 / √ 3 3 2 , − 3 2 1 1 2 , 1 2 √ 2 / √ 3-1 / √ 3 1 2 , − 1 2 1 / √...
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852HW3_Solutions - PHYS852 Quantum Mechanics II Spring 2009...

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