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Unformatted text preview: Physics 731 Assignment #5, Solutions 1. By inspection, the Hamiltonian H = H 11  1 ih 1  + H 22  2 ih 2  + H 12  1 ih 2  (1) is not Hermitian. Hence, the time evolution operator is not unitary, leading to a nonconservation of probability. To see this explicitly, set H 11 = H 22 = 0 for simplicity as suggested, and calculate the time evolution operator U ( t, 0) = e iHt/ ¯ h , as follows: U ( t, 0) = e iHt/ ¯ h = X n 1 n ! iHt ¯ h n = 1 iH 12 t ¯ h  1 ih 2  . (2) A general state  α i = a 1  1 i + a 2  2 i (with  a 1  2 +  a 2  2 = 1 ) then evolves to  α ( t ) i = a 1 ( t )  1 i + a 2 ( t )  2 i , in which a 1 ( t ) = a 1 and a 2 ( t ) = a 2 iH 12 ta 1 / ¯ h . Clearly,  a 1 ( t )  2 +  a 2 ( t )  2 6 = 1 , and hence the total probability is not conserved. 2. (a) The initial state is  α i =  ˆ n, + i = cos β/ 2  + i + sin β/ 2 i . The Hamiltonian is H = ~μ · ~ B = ( g s μ B σ z B ) / 2 , with energy eigenvalues E ± = ± g s μ B B/ 2 . The timeevolved state is  α ( t ) i = cos β 2 e iE + t/ ¯ h  + i + sin β 2 e iE t/ ¯ h i . (3) The probability that the electron is in the S x = +¯ h/ 2 state is given by h + ,x  α ( t ) i 2 = 1 2  ( h +  + h )(cos β 2 e iE + t/ ¯ h  + i + sin β 2 e iE t/ ¯ h i )  2 = 1 2  cos β 2 + sin β 2 e iωt  2 = 1 2 (1 + sin β cos ωt ) , (4) in which ω = ( E + E ) / ¯ h . (b) The expectation value of S x as a function of time is h α ( t )  S x  α ( t ) i = h α ( t )  ¯ h 2 (  + ih + ih +)  α ( t ) i = ¯ h 2 sin β cos ωt. (5) (c) For β = 0 , the initial state is aligned along the + z direction; it remains so at later times since it...
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This note was uploaded on 12/14/2009 for the course QUANTUM I 731 taught by Professor Everett during the Fall '09 term at University of Wisconsin.
 Fall '09
 Everett

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