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

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Physics 731 Assignment #5, Solutions 1. By inspection, the Hamiltonian H = H 11 | 1 1 | + H 22 | 2 2 | + H 12 | 1 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 = n 1 n ! - iHt ¯ h n = 1 - iH 12 t ¯ h | 1 2 | . (2) A general state | α = a 1 | 1 + a 2 | 2 (with | a 1 | 2 + | a 2 | 2 = 1 ) then evolves to | α ( t ) = a 1 ( t ) | 1 + a 2 ( t ) | 2 , 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 = 1 , and hence the total probability is not conserved. 2. (a) The initial state is | α = | ˆ n, + = cos β/ 2 | + + sin β/ 2 |- . 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 ) = cos β 2 e - iE + t/ ¯ h | + + sin β 2 e - iE - t/ ¯ h |- . (3) The probability that the electron is in the S x = +¯ h/ 2 state is given by | + , x | α ( t ) | 2 = 1 2 | ( + | + -| )(cos β 2 e - iE + t/ ¯ h | + + sin β 2 e - iE - t/ ¯ h |- ) | 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 α ( t ) | S x | α ( t ) = α ( t ) | ¯ h 2 ( | + -| + |- +) | α ( t ) = ¯ 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 is an eigenstate of the Hamiltonian. Hence, the probability of being in state S x = +¯ h/ 2 is equal to 1 / 2 and S x = 0 . For β = π/ 2 , the initial state is aligned in the + x direction. Since [ S x , H ] = 0 , the state at time t
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