hw_solutions_4

# hw_solutions_4 - Homework Solutions 4(Liboff Chapter 6 6.7...

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Unformatted text preview: Homework Solutions # 4 (Liboff Chapter 6) 6.7 Note that ψ ( x, 0) is an eigenstate of ˆ H . (a) ψ ( x, t ) = ψ ( x, 0) e- iωt where ω = E/ ¯ h = ¯ hk 2 / 2 m . (b) ψ ( x, t ) = 1 2 i e ik x- e- ik x e- iωt in terms of ˆ p eigenfunctions. Therefore, ± ¯ hk can be observed, with equal probability ( 1 2 ). (c) at t = 3 s, ψ = e ik x . So ψ ( x, t ) = e ik x e- iω ( t- 3) (The 3 s part is irrelevant; just an overall phase.) 6.13 At t = 0, the state is | ψ i , with ˆ A | ψ i = a | ψ i . For t > 0, using the time evolution operator ˆ U = exp(- it ¯ h ˆ H ), | ψ i = e- it ¯ h ˆ H | ψ i Then, since ˆ U † = exp( it ¯ h ˆ H ) = ˆ U- 1 , h A i = h ψ | e it ¯ h ˆ H ˆ Ae- it ¯ h ˆ H | ψ i We have to move ˆ A to the other side of ˆ U , so we need the commutator [ ˆ U, ˆ A ]. Start with using ˆ H ˆ A = ˆ A ˆ H + c , and finding [ ˆ H n , ˆ A ] = ˆ H n ˆ A- ˆ A ˆ H n = ˆ H n- 1 ˆ A ˆ H + c ˆ H n- 1- ˆ A ˆ H n = · · · = nc ˆ H n- 1 after interchanging...
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hw_solutions_4 - Homework Solutions 4(Liboff Chapter 6 6.7...

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