12Bmidterm10_soln

# 12Bmidterm10_soln - Ph 12b Midterm Exam Solutions J....

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Ph 12b Midterm Exam Solutions J. Preskill – 10 February 2010 1. Two-state quantum dynamics — 35 total points Let | e 1 a and | e 2 a denote two normalized and mutually orthogonal states in a Hilbert space H : A e 1 | e 1 a = A e 2 | e 2 a = 1 , A e 1 | e 2 a = 0 . A certain quantum system has Hamiltonian ˆ H , and the two normalized states | ω 1 a = 1 2 | e 1 a + 3 2 | e 2 a , | ω 2 a = - 3 2 | e 1 a + 1 2 | e 2 a , are eigenstates of ˆ H with eigenvalues ¯ 1 , ¯ 2 respectively: ˆ H | ω 1 a = ¯ 1 | ω 1 a , ˆ H | ω 2 a = ¯ 2 | ω 2 a . At time t = 0 , the system is prepared in the state | ψ (0) a = | e 1 a . ( a ) (10 points) Express | ψ (0) a as a linear combination of energy eigenstates. | ψ (0) a = | e 1 a = 1 2 | ω 1 a - 3 2 | ω 2 a . ( b ) (5 points) Solve the time-dependent Schr¨ odinger equation i ¯ h d dt | ψ ( t ) a = ˆ H | ψ ( t ) a to Fnd the state | ψ ( t ) a at time t . Express your answer in the form | ψ ( t ) a = f 1 ( t ) | ω 1 a + f 2 ( t ) | ω 2 a . | ψ ( t ) a = 1 2 e - 1 t | ω 1 a - 3 2 e - 2 t | ω 2 a . ( c ) (10 points) Now re-express | ψ ( t ) a in the form | ψ ( t ) a = g 1 ( t ) | e 1 a + g 2 ( t ) | e 2 a . 1

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| ψ ( t ) a = p 1 4 e - 1 t + 3 4 e - 2 t P | e 1 a + ± 3 4 e - 1 t - 3 4 e - 2 t ² | e 2 a . ( d ) (10 points) The states | e 1 a and | e 2 a are eigenstates of an observable ˆ A : ˆ A | e 1 a = a 1 | e 1 a , ˆ A | e 2 a = a 2 | e 2 a , where a 1 n = a 2 , that is measured at time t . Find the probability P ( a 1 ) that the outcome of the measurement is a 1 and the probability P ( a 2 ) that the outcome is a 2 .
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## This note was uploaded on 01/28/2011 for the course PH 2 taught by Professor Dudko during the Spring '09 term at UCSD.

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12Bmidterm10_soln - Ph 12b Midterm Exam Solutions J....

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