The Lindblad Master Equation

The equation

The Born–Markov–secular approximations produce a time-local equation of motion for $\rho_S$ in the Lindblad (Gorini–Kossakowski–Sudarshan–Lindblad, GKSL) form:

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

The first term is ordinary unitary (Hamiltonian) evolution. The second — the dissipator $\mathcal{D}[\rho]$ — encodes the irreversible effect of the bath through a set of Lindblad (or jump) operators $L_k$ . The braces denote the anticommutator $\{A,\rho\} = A\rho + \rho A$ .

Why this exact structure

The GKSL form is not arbitrary. A theorem of Lindblad (and Gorini–Kossakowski–Sudarshan) states that this is the most general generator of a continuous one-parameter family of maps $\rho(0)\mapsto\rho(t)$ that is

trace preserving — $\mathrm{Tr}\,\rho = 1$ at all times (probabilities sum to one), and
completely positive — $\rho(t)$ stays a valid density matrix even when the system is entangled with an untouched ancilla.

Any other functional form risks producing negative "probabilities". This is why the secular approximation, which forces the master equation into Lindblad form, is so important.

Reading the dissipator

Each $L_k$ describes one physical decay channel. The two pieces play complementary roles:

The recycling term $L_k\,\rho\,L_k^\dagger$ feeds population into the states $L_k$ maps to.
The anticommutator $-\tfrac12\{L_k^\dagger L_k,\rho\}$ drains population out of the states $L_k$ acts on and, crucially, also damps coherences.

Trace preservation is automatic: $\mathrm{Tr}\,\mathcal{D}[\rho] = \mathrm{Tr}(L_k\rho L_k^\dagger) - \tfrac12\mathrm{Tr}(L_k^\dagger L_k\rho) - \tfrac12\mathrm{Tr}(\rho L_k^\dagger L_k) = 0$ , using cyclicity of the trace.

Worked example: amplitude damping

Spontaneous emission of a two-level atom into the vacuum uses a single jump operator

L = \sqrt{\gamma}\,\sigma_- , \qquad \sigma_- = |0\rangle\langle1| ,

where $\gamma$ is the decay rate and $\sigma_-$ takes the excited state $|1\rangle$ to the ground state $|0\rangle$ . With $H=0$ , write $\rho = \begin{pmatrix}\rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11}\end{pmatrix}$ . Then $L^\dagger L = \gamma\,|1\rangle\langle1|$ , and the dissipator gives

\dot\rho_{11} = -\gamma\,\rho_{11}, \qquad \dot\rho_{00} = +\gamma\,\rho_{11}, \qquad \dot\rho_{01} = -\tfrac{\gamma}{2}\,\rho_{01}.

The excited population decays as $\rho_{11}(t) = \rho_{11}(0)\,e^{-\gamma t}$ , draining into the ground state, while the coherence decays at half the rate, $\rho_{01}(t) = \rho_{01}(0)\, e^{-\gamma t/2}$ .

Try it

This is a numerical exercise — your code should return a number. Using the amplitude-damping result $\rho_{11}(t)=\rho_{11}(0)e^{-\gamma t}$ , with $\rho_{11}(0)=1$ , $\gamma=1.5$ , and $t=2.0$ , compute and return the excited-state population at time $t$ .

Run your code to see the quantum state.