Steady States

What a steady state is

A steady state (stationary state, fixed point) of a Lindblad equation is a density matrix $\rho_{ss}$ that does not change in time:

\frac{d\rho_{ss}}{dt} = -\frac{i}{\hbar}[H,\rho_{ss}] + \sum_k\!\left(L_k\rho_{ss}L_k^\dagger - \tfrac12\{L_k^\dagger L_k,\rho_{ss}\}\right) = 0.

It is the kernel of the Lindblad generator $\mathcal{L}$ : $\mathcal{L}[\rho_{ss}]=0$ . Unlike a closed system — whose stationary states are simply the energy eigenstates — an open system relaxes toward a state fixed by the bath, regardless (often) of where it started. This is the quantum analogue of thermal equilibration.

Existence and uniqueness

A Lindblad semigroup always has at least one steady state. It is unique (and globally attracting) when the jump operators, together with the Hamiltonian, leave no proper subspace invariant — formally, when no nontrivial subspace is preserved by all $L_k$ and $H$ . In that common case every initial state flows to the same $\rho_{ss}$ as $t\to\infty$ . Degenerate or multiple steady states arise when there are decoherence-free subspaces or conserved quantities.

Worked example: a thermal qubit

A qubit coupled to a bath at temperature $T$ has both de-excitation and excitation channels,

L_- = \sqrt{\gamma_\downarrow}\,\sigma_-, \qquad L_+ = \sqrt{\gamma_\uparrow}\,\sigma_+ .

Writing the excited population $p_1=\rho_{11}$ and ground population $p_0=\rho_{00}=1-p_1$ , the dissipator gives the rate equation

\dot p_1 = \gamma_\uparrow\,p_0 - \gamma_\downarrow\,p_1 .

Setting $\dot p_1=0$ yields the steady-state population

p_1^{\,ss} = \frac{\gamma_\uparrow}{\gamma_\uparrow+\gamma_\downarrow}.

Off-diagonal coherences decay to zero (no Hamiltonian to sustain them), so the steady state is diagonal — a classical mixture of $|0\rangle$ and $|1\rangle$ .

Detailed balance gives the Gibbs state

Microscopic bath physics fixes the ratio of rates by detailed balance,

\frac{\gamma_\uparrow}{\gamma_\downarrow} = e^{-\hbar\omega/k_BT},

where $\hbar\omega$ is the qubit splitting. Substituting,

p_1^{\,ss} = \frac{e^{-\hbar\omega/k_BT}}{1+e^{-\hbar\omega/k_BT}} = \frac{1}{1+e^{+\hbar\omega/k_BT}},

which is exactly the Boltzmann population. The steady state is the Gibbs state $\rho_{ss} = e^{-H/k_BT}/Z$ . At $T\to 0$ the rates give $p_1^{ss}\to0$ — pure ground state; at $T\to\infty$ , $p_1^{ss}\to\tfrac12$ — maximally mixed.

Try it

This is a numerical exercise — your code should return a number. With detailed balance fixing $\gamma_\uparrow/\gamma_\downarrow = 0.25$ , compute the steady-state excited-state population $p_1^{ss} = \gamma_\uparrow/(\gamma_\uparrow+\gamma_\downarrow)$ and return it.

Run your code to see the quantum state.