hws07 - PHYS 115B Homework 7 Solutions Spring 2008 Problem...

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PHYS 115B Homework 7 Solutions Spring 2008 Problem 1 (a) Since ( σ z ) 2 = I , we have: If n is even, then ( σ z ) n = I . (Write even n =0 , 2 , 4 , 6 , . . . as n =2 j where j , 1 , 2 , 3 , 4 , . . . ) If n is odd, then ( σ z ) n = σ z . (Write odd n =1 , 3 , 5 , 7 , . . . as n k +1 where k , 1 , 2 , 3 , 4 , . . . ) U ( t )= e 0 z / 2 = ± n =0 i n ² ω 0 t 2 ³ n 1 n ! ( σ z ) n = ± n even i n ² ω 0 t 2 ³ n 1 n ! ( σ z ) n + ± n odd i n ² ω 0 t 2 ³ n 1 n ! ( σ z ) n = ´ ± n even i n ² ω 0 t 2 ³ n 1 n ! µ I + ± n odd i n ² ω 0 t 2 ³ n 1 n ! σ z = ± j =0 i 2 j ² ω 0 t 2 ³ 2 j 1 (2 j )! I + ´ ± k =0 i 2 k +1 ² ω 0 t 2 ³ 2 k +1 1 (2 k + 1)! µ σ z = ± j =0 ( - 1) j ² ω 0 t 2 ³ 2 j 1 (2 j )! I + i ´ ± k =0 ( - 1) k ² ω 0 t 2 ³ 2 k +1 1 (2 k + 1)! µ σ z = cos ² ω 0 t 2 ³· I + i sin ² ω 0 t 2 ³· σ z = cos ( ω 0 t/ 2) 10 01 · + i sin ( ω 0 t/ 2) 0 - 1 · = cos ( ω 0 t/ 2) + i sin ( ω 0 t/ 2) 0 0 cos ( ω 0 t/ 2) - i sin ( ω 0 t/ 2) · = e 0 t/ 2 0 0 e - 0 t/ 2 · 1
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(b) For | ψ (0) ± = | + ± y = 1 2 ± 1 i ² we have | ψ ( t ) ± = U ( t ) | ψ (0) ± = ± e 0 t/ 2 0 0 e - 0 t/ 2 ² 1 2 ± 1 i ² = 1 2 ± e 0 t/
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This note was uploaded on 07/15/2008 for the course PHYS 115b taught by Professor Cleland during the Spring '08 term at UCSB.

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hws07 - PHYS 115B Homework 7 Solutions Spring 2008 Problem...

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