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852quiz2_Solution

# 852quiz2_Solution - V 1 = I 1 V I 1 = V | 00 aA 1 | | 1 aA...

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NAME: QUIZ 2 PHYS852 Quantum Mechanics II, Spring 2009 A two-spin quantum system has four states. Given in terms of total-spin quantum number s and its z- component m s , these states are the singlet state | 0 , 0 ) , and the three triplet states | 1 , 1 ) , | 1 , 0 ) , and | 1 , 1 ) . Let us assume that the bare Hamiltonian is H 0 = E 0 planckover2pi1 2 S 2 z , where S z is the z-component of the total spin. 1. Give the energy levels of the system and their degeneracies S 2 z | s, m s ) = planckover2pi1 2 m 2 s H 0 | 0 , 0 ) = 0 H 0 | 1 , 0 ) = 0 H 0 | 1 , 1 ) = E 0 H 0 | 1 , 1 ) = E 0 E (0) 1 = 0 , d 1 = 2 E (0) 2 = E 0 , d 2 = 2 2. Write the projector I n onto each degenerate subspace I 1 = | 0 , 0 )( 0 , 0 | + | 1 , 0 )( 1 , 0 | I 2 = | 1 , 1 )( 1 , 1 | + | 1 , 1 )( 1 , 1 | 3. Assume the interaction takes the form V = V 0 ( | 1 , 0 )( 0 , 0 | + | 0 , 0 )( 1 , 0 | + | 1 , 1 )( 1 , 1 | − | 1 , 1 )( 1 , 1 | ) . Identify the ’good’ basis for perturbation theory. V 1 = I 1 V I
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Unformatted text preview: V 1 = I 1 V I 1 = V ( | 00 aA 1 , | + | 1 , aA , | ) = V p 0 1 1 0 P V 2 = I 2 V I 2 = V ( | 1 , − 1 aA 1 , − 1 | + | 1 , 1 aA 1 , 1 | ) = V p 1 0 0 1 P Since V 1 and V s are of familiar forms, we know that the eigenstates of V 1 are: | 1 , 1 (0) a = 1 √ 2 ( | , a − | 1 , a ) | 1 , 2 (0) a = 1 √ 2 ( | , a + | 1 , a ) | 2 , 1 (0) a = | 1 , − 1 a | 2 , 2 (0) a = | 1 , 1 a These four states form a ‘good’ basis for perturbation theory....
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