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12b-prob4-10

12b-prob4-10 - 1 Ph 12b Homework Assignment No 4 Due 5pm...

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1 Ph 12b Homework Assignment No. 4 Due: 5pm, Thursday, 4 February 2010 1. Weaker decoherence . In class we discussed the phase damping of a qubit that results when the qubit scatters a photon with probability p . The scattered photon is knocked into one of two mutually orthogonal states {| 0 a E , | 1 a E } , correlated with the qubit’s state, both of which are orthogonal to the state | un a E of the unscattered photon. If the initial state of the qubit is | ψ a S = a | 0 a S + b | 1 a S , then the joint state of the qubit and photon evolves as | ψ a S ⊗| un a E r 1 - p | ψ a S ⊗| un a E + p ( a | 0 a S ⊗ | 0 a E + b | 1 a S ⊗ | 1 a E ) . (1) Thus the qubit density operator ˆ ρ evolves as ˆ ρ = p ρ 00 ρ 01 ρ 10 ρ 11 P ˆ ρ p = p ρ 00 (1 - p ) ρ 01 (1 - p ) ρ 10 ρ 11 P . Now consider a diFerent model of decoherence, in which photon scat- tering does not perfectly resolve the state of the qubit. The scattered photon is knocked to the normalized state | γ a E if the qubit’s state is | 0 a S and it is knocked to the normalized state | η a E if the photon’s state is | 1 a S ; thus eq.(1) is replaced by | ψ a S ⊗| un a E r 1 - p | ψ a S ⊗| un a E + p ( a | 0 a S ⊗ | γ a E + b | 1 a S ⊗ | η a E ) . (2) Both | γ a E and | η a E are orthogonal to the state | un a E of the unscattered photon, but they are not necessarily mutually orthogonal; rather E A η | γ a E = 1 - e, where e is a real number. Thus for e = 1, the states | γ a E and | η a E are orthogonal, and we recover the model considered previously, while for e = 0, the scattered photon remains uncorrelated with the qubit, and there is no decoherence at all.

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12b-prob4-10 - 1 Ph 12b Homework Assignment No 4 Due 5pm...

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