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Unformatted text preview: PHYS 507 Answers to Homework IV (Fall ’05) 1. U ( t, 0) = • cosΩ t sinΩ t sinΩ t cosΩ t ‚ . (a) U ( t, 0) U ( t, 0) † = • cosΩ t sinΩ t sinΩ t cosΩ t ‚• cosΩ t sinΩ t sinΩ t cosΩ t ‚ = • 1 0 0 1 ‚ . Normally, we should also check if U ( t, 0) † U ( t, 0) = I is also satisfied, but this is only necessary for infinite dimensional Hilbert spaces. For finite dimensions (like n × n matrices) U ( t, 0) U ( t, 0) † = I necessarily implies U ( t, 0) † U ( t, 0) = I . (b) U ( t 2 ,t 1 ) = U ( t 2 , 0) U ( t 1 , 0) † = • cosΩ t 2 sinΩ t 2 sinΩ t 2 cosΩ t 2 ‚• cosΩ t 1 sinΩ t 1 sinΩ t 1 cosΩ t 1 ‚ = • cosΩ( t 2 t 1 ) sinΩ( t 2 t 1 ) sinΩ( t 2 t 1 ) cosΩ( t 2 t 1 ) ‚ . Therefore the system has “time translation symmetry” as U ( t 2 ,t 1 ) = U ( t 2 t 1 , 0) depends only on t 2 t 1 . This will imply that the Hamiltonian is time independent and the energy is conserved. (c) H ( t ) = i ¯ h ∂U ( t, 0) ∂t U ( t, 0) † = i ¯ h Ω • sinΩ t cosΩ t cosΩ t sinΩ t ‚• cosΩ t sinΩ t sinΩ t cosΩ t ‚ = i ¯ h Ω • 1 1 ‚ = ¯ h Ω σ y .....
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This note was uploaded on 02/11/2012 for the course MATH 435 taught by Professor Starg during the Spring '11 term at Al Ahliyya Amman University.
 Spring '11
 starg

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