Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 52 / 3 / c

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- Solution c

Solution

Take the driving and feedback strengths to be real, so the feedback Quantum Hamiltonian is Hermitian. The lowering operator satisfies

σ^{2} = (σ^{†})^{2} = 0

. Using the supplied convention for

σ_{y}

,

σ^{†} σ_{y} = i σ^{†} σ, σ_{y} σ = - i σ^{†} σ .

The additional Quantum Hamiltonian term is consequently

\frac{1}{2} (M^{†} F + FM) = \frac{λ}{2} (σ^{†} σ_{y} + σ_{y} σ) = 0.

With no drift Quantum Hamiltonian, the total Quantum Hamiltonian is just

α σ_{y}

. The feedback-modified Lindblad operator is

K = σ - iλ σ_{y} = (1 + λ) σ - λ σ^{†} = (0 - λ 1 + λ 0) .

Substitution into the measurement-feedback equation gives

\overset{ρ}{˙} = - i [α σ_{y}, ρ] + D [K] ρ .

Thus the feedback affects the Lindblad dissipator even though its extra coherent Quantum Hamiltonian contribution cancels. This is a concrete mechanism of Markovian reservoir engineering, rather than merely a classical control force added to an unchanged dissipative equation.

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