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p137bp3sol

# p137bp3sol - Physics 137B Fall 2007 Moore Problem Set 3...

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Physics 137B, Fall 2007, Moore Problem Set 3 Solutions (Revised 10/7/2007) 1. (a) In the usual basis, the Hamiltonian is H = B ¯ h 2 1 0 0 - 1 . The initial state is | ψ (0) = 1 0 , so the time-evolved state is | ψ ( t ) = e - iHt/ ¯ h | ψ (0) = e - iBt/ 2 1 0 . (b) Now the Hamiltonian is H = B ¯ h 2 0 1 1 0 , whose eigenstates are | + = 1 2 1 1 and |- = 1 2 1 - 1 , with eigenvalues E ± = ± B ¯ h 2 . Since we can expand the initial state as | ψ (0) = 1 2 ( | + + |- ), its time evolution is | ψ ( t ) = e - iHt/ ¯ h | ψ (0) = 1 2 ( e - iBt/ 2 | + + e + iBt/ 2 |- ) = cos Bt 2 - i sin Bt 2 . 1

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The expectation value of S z for this state is therefore S z = ψ ( t ) | S z | ψ ( t ) = ( cos Bt 2 + i sin Bt 2 ) ¯ h 2 1 0 0 - 1 cos Bt 2 - i sin Bt 2 = ¯ h 2 cos 2 Bt 2 + - ¯ h 2 sin 2 Bt 2 = ¯ h 2 cos Bt . Remark: Once again, don’t forget to complex-conjugate your coefficients when you go from a ket to a bra. Many of you forgot to do this, and ended up with S z = ¯ h/ 2 instead of the answer above. 2. The unperturbed energy eigenstates of the infinite square well potential are ψ (0) n ( x ) = 2 L sin nπx L , with energies E (0) n = n 2 π 2 ¯ h 2 2 mL 2 . The first-order energy shifts for the perturbation H = ( x - L/ 2) are E (1) n = ψ (0) n | H | ψ (0) n = L 0 dx 2 L sin 2 nπx L ( x - L/ 2) = 2 A L sin 2 2 = 2 A/L n odd 0 n even .
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