Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 343 / 1

The generator of every finite-dimensional, time-homogeneous Markovian quantum channel has Lindblad equation formHere $H=H^{†}$. The Lindblad operators $L_{\alpha }$ may be arbitrary bounded operators; positivity of their rates is already incorporated by rescaling them. They need not be Hermitian, traceless, normalized, or mutually orthogonal. Their description is nonunique: unitary mixing of the $L_{\alpha }$, and shifts by scalar multiples of the identity accompanied by a compensating change of $H$, leave the same Lindbladian.

The exact spectral condition is that $0$ be a simple eigenvalue of $\mathcal{L}$ and that every other eigenvalue satisfy $\mathrm{Re}\lambda <0$. Then $e^{t\mathcal{L}}\rho$ converges for every initial density operator to the unique stationary state ${\rho }_{\ast }$ satisfying $\mathcal{L}({\rho }_{\ast })=0$. This is the unique stationary state of a Lindbladian condition; the smallest nonzero value of $-\mathrm{Re}\lambda$ is the Lindbladian gap.

A standard operator criterion is irreducibility: the only subspaces invariant under $H$, all $L_{\alpha }$, and all $L_{\alpha }^{†}$ are the zero and full spaces, equivalently their common commutant consists only of scalar multiples of the identity. With no nonzero imaginary-axis eigenvalues, this makes the semigroup relaxing. The spectral statement is the safest general answer because uniqueness of a fixed point alone does not exclude persistent oscillatory modes.

Let $|s⟩=|{\psi }^{-}⟩$ be the spin-one-half singlet state, complete it to an orthonormal basis $\{|s⟩,|e_1⟩,|e_2⟩,|e_3⟩\}$, and chooseEach excited population decays into the singlet. On the matrix-unit basis, $|e_j⟩⟨e_k|$ has decay eigenvalue $-\gamma$ after separating off its contribution to the stationary population, while $|s⟩⟨e_j|$ and its adjoint have eigenvalue $-\gamma /2$. The sole zero mode is $|s⟩⟨s|$, so every initial state converges toThe nonzero eigenvalue closest to the origin is therefore

In a quantum-trajectory unravelling, a pure state evolves between jumps underand a jump of type $\alpha$ sends $|\varphi ⟩$ to $L_{\alpha }|\varphi ⟩/\Vert L_{\alpha }|\varphi ⟩\Vert$. Averaging $|{\varphi }_t⟩⟨{\varphi }_t|$ over the stochastic jump records recovers $\rho (t)$.

Equivalently, over a short interval $dt$ use the operators in a Kraus representationand the Stinespring dilation $V={\sum }_{\mu }{K}_{\mu }\otimes |\mu {⟩}_E$. Iterating with fresh environment systems produces a pure system-environment history whose partial trace is the Lindblad evolution.

If the fixed point is pure, stationarity requires $L_{\alpha }|{\psi }_{\ast }⟩={\ell }_{\alpha }|{\psi }_{\ast }⟩$ for every $\alpha$, together with preservation of its ray by the adjusted effective Hamiltonian. After shifting the jumps one may take it to be a dark state of a Lindblad equation. Once a trajectory reaches that ray, neither no-jump evolution nor a jump takes it away; in the example of part (c), a jump transfers any excited component directly into the singlet.