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HW14_Solutions - PHYS851 Quantum Mechanics I Fall 2008...

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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 14: Solutions 1. [15 pts] Consider the two-state quantum system described by H = H + V , where H = planckover2pi1 δ 2 ( | 1 )( 1 | − | )( | ) , and V = planckover2pi1 Ω 2 ( | )( 1 | + | 1 )( | ) . In the Schr¨odinger picture we have | ψ S ( t ) ) = c ( t ) | ) + c 1 ( t ) | 1 ) . Solve Schr¨odinger’s equation for the coefficients c ( t ) and c 1 ( t ) with the initial conditions c (0) = 1 and c 1 (0) = 0. Let the operator W be defined in the Schr¨odinger picture as W = | 1 )( 1 | − | )( | . Compute ( W ) as a function of time. Answer: Starting from Schr¨odinger’s equation, i planckover2pi1 d dt | ψ ) = H | ψ ) , and applying ( | and ( 1 | from the left gives, respectively i planckover2pi1 d dt c = − planckover2pi1 δ 2 c + planckover2pi1 Ω 2 c 1 i planckover2pi1 d dt c 1 = planckover2pi1 δ 2 c 1 + planckover2pi1 Ω 2 c Putting these equations in matrix form gives: d dt parenleftbigg c c 1 parenrightbigg = − i 2 parenleftbigg − δ Ω Ω δ parenrightbiggparenleftbigg c c 1 parenrightbigg From det vextendsingle vextendsingle vextendsingle vextendsingle − δ − 2 ω Ω Ω δ − 2 ω vextendsingle vextendsingle vextendsingle vextendsingle = 0 we find the characteristic equation 4 ω 2 − δ 2 − Ω 2 = 0 so that the eigenvalues are ω ± = ± 1 2 radicalbig δ 2 + Ω 2 The solution is then of the form parenleftbigg c c 1 parenrightbigg = a parenleftbigg e +0 e +1 parenrightbigg e- iω + t + b parenleftbigg e- e- 1 parenrightbigg e- iω- t , 1 where a and b are free constants to be determined by initial conditions. Due to the presence of the free constants, it is not necessary to find normalized eigenvectors, so we can safely take e ± = 1. The eigenvalue equation is then: − δ − 2 ω ± + Ω e ± 1 = 0 so that e ± 1 = δ + 2 ω ± Ω Putting this together gives c = ae- iω + t + be- iω- t c 1 = a δ + 2 ω + Ω e- iω + t + b δ + 2 ω- Ω e- iω- t From the initial conditions c = 1 and c 1 = 0 we find a + b = 1 ( δ + 2 ω + ) a + ( δ + 2 ω- ) b = 0 the latter can be rewritten using ω- = − ω + as δ ( a + b ) + 2 ω + ( a − b ) = 0 which gives a − b = − δ 2 ω + with b = 1 − a this becomes 2 a = 1 − δ 2 ω + or a = ω + − δ/ 2 2 ω + and therefore b = ω + + δ/ 2 2 ω + so that finally we arrive at c = cos( ω +...
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HW14_Solutions - PHYS851 Quantum Mechanics I Fall 2008...

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