HW5_Solutions

# HW5_Solutions - PHYS851 Quantum Mechanics I Fall 2008...

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PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 5: Two Level Systems: 1. [20 pts each] Cohen-Tannoudji pp 476-480, problems 4.1 and 4.5 Answer: 4.1 a. In {| + ) , |−)} basis, the operator S x is given by S x = planckover2pi1 2 parenleftbigg 0 1 1 0 parenrightbigg it has eigenvalues ± planckover2pi1 / 2 and eigenstates ,x ) = ( | + ) ± |−) ) / 2. Thus a measurement of S x will yield planckover2pi1 / 2 with probability |( + ,x | + )| 2 = 1 / 2, and planckover2pi1 / 2 with probability 1 / 2. b. Let the system evolves under H = γB 0 S y for time t . The eigenstates of S y are | + ,y ) = ( | + ) + i |−) ) / 2 and |− ,y ) = ( | + ) − i |−) ) / 2, with eigenvalues planckover2pi1 / 2 and planckover2pi1 / 2, respectively. The state of the system at time t is therefore | ψ ( t ) ) = | + ,y ) e iγB 0 t/ 2 ( + ,y | ψ (0) ) + |− ,y ) e - iγB 0 t/ 2 (− ,y | ψ (0) ) = 1 2 ( | + ,y ) e iγB 0 t/ 2 + |− ,y ) e - iγB 0 t/ 2 ) = | + ) cos( γB 0 t/ 2) − |−) sin( γB 0 t/ 2) c. If at time t we measure S z , we will obtain planckover2pi1 / 2 with probability cos 2 ( γB 0 t/ 2) and planckover2pi1 / 2 with probability sin 2 ( γB 0 t/ 2). This will yield certain results for γB 0 t/ 2 = nπ/ 2, with n being any integer. This can be understood because the spin vector is precessing around the y axis with a period of T = 4 π/γB 0 . The results for a measurement of S x will be similar, since the rotation about the y- axis starts at the z-axis, it will align with the x-axis every quarter and three-quarters of a period. Thus we will obtain planckover2pi1 / 2 with probability cos 2 ( γB 0 t/ 2 π/ 4) and planckover2pi1 / 2 with proba- bility sin 2 ( γB 0 t/ 2 π/ 4). A certain measurement occurs whenever γB 0 t/ 2 = π/ 4 + nπ/ 2, for reasons we have already explained. A measurement of S y will yield planckover2pi1 / 2 with probability 1 / 2 and planckover2pi1 / 2 with probability 1 / 2 At no time will the results be certain. This is because the spin vector lies in the x z plane as it rotates around the y -axis. So it never aligns with the y -axis itself. 1

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4.5 a. The Hamiltonian is H = γ vector B · vector S = B 0 ( ω x S x + ω y S y + ω z S z ) The evolution operator is U ( t ) = e - iHt/ planckover2pi1 = e - i ( ω x S x + ω y S y + ω z S z ) t/ planckover2pi1 = e - iMt M = 1 2 bracketleftbigg ω x parenleftbigg 0 1 1 0 parenrightbigg + ω y parenleftbigg 0 i i 0 parenrightbigg + ω z parenleftbigg 1 0 0 1 parenrightbiggbracketrightbigg = 1 2 parenleftbigg ω z ω x y ω x + y ω z parenrightbigg M 2 = 1 4 parenleftbigg ω z ω x y ω x + y ω z parenrightbigg parenleftbigg ω z ω x y ω x + y ω z parenrightbigg = ω 2 x + ω 2 y + ω 2 z 4 parenleftbigg 1 0 0 1 parenrightbigg with ω 2 0 = ω 2 x + ω 2 y + ω 2 z
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