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 ² . The initial state is | ψ (0) i = ± 1 0 ² , so the time-evolved state is | ψ ( t ) i = e - iHt/ ¯ h | ψ (0) i = e - iBt/ 2 ± 1 0 ² . (b) Now the Hamiltonian is H = B ¯ h 2 ± 0 1 1 0 ² , whose eigenstates are | + i = 1 2 ± 1 1 ² and |-i = 1 2 ± 1 - 1 ² , with eigenvalues E ± = ± B ¯ h 2 . Since we can expand the initial state as | ψ (0) i = 1 2 ( | + i + |-i ), its time evolution is | ψ ( t ) i = e - iHt/ ¯ h | ψ (0) i = 1 2 ( e - iBt/ 2 | + i + e + iBt/ 2 |-i ) = ± cos Bt 2 - i sin Bt 2 ² . 1

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The expectation value of S z for this state is therefore h S z i = h ψ ( t ) | S z | ψ ( t ) i = ( cos Bt 2 + i sin Bt 2 ) ¯ h 2 ± 1 0 0 - 1 ²± cos Bt 2 - i sin Bt 2 ² = ¯ h 2 cos 2 Bt 2 + ± - ¯ h 2 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 h S z i = ¯ h/ 2 instead of the answer above. 2. The unperturbed energy eigenstates of the inﬁnite square well potential are ψ (0) n ( x ) = r 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 0 = ( x - L/ 2) are E (1) n = h ψ (0) n | H 0 | ψ (0) n i = Z L 0 dx 2 L sin 2 nπx L ( x - L/ 2) = 2 A L sin 2 2 = ³ 2 A/L n odd 0 n even .
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This note was uploaded on 08/01/2008 for the course PHYSICS 137B taught by Professor Moore during the Fall '07 term at Berkeley.

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

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