HW5sol - Physics 731 Assignment #5, Solutions 1. By...

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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 non-conservation 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 time-evolved 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.

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HW5sol - Physics 731 Assignment #5, Solutions 1. By...

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