advanced34 min studyLesson 414 in Physics

The structure of a GKSL master equation

Connect infinitesimal channel normalization to the Lindblad generator and nonnegative dissipative rates.

Before you begin

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The question for today

What you’ll understand

Identify the Hamiltonian, recycling, and anticommutator terms.
Check trace preservation and rate positivity.
Solve a simple qubit semigroup and distinguish stationary states from thermal states.

The Gorini–Kossakowski–Sudarshan–Lindblad structure characterizes generators of continuous completely positive, trace-preserving semigroups in finite dimension. It is a theorem about a class of maps, rather than a claim that every open system belongs to that class. Understanding how its terms fit together is more useful than memorizing an operator list: the apparent loss term and the recycling term cooperate to conserve probability while allowing purity and energy to change.

An infinitesimal channel has balanced terms

Take a short positive interval and represent its channel by one Kraus operator close to the identity plus jump operators proportional to the square root of the interval. Expanding the normalization identity through first order fixes the Hermitian part of the identity branch. Its anti-Hermitian part supplies a Hamiltonian commutator. The remaining terms form a dissipator. This explains why dropping the anticommutator while keeping the recycling term would continually create trace instead of describing a deterministic experiment.

Use dimensionless operators for the jump directions and put their physical rates explicitly outside the dissipator. Alternatively, absorb each square root of a rate into its operator, which then has units of inverse square root seconds. Both conventions are common, so check units before comparing equations. The effective Hamiltonian may include a bath-induced Lamb shift. Its commutator changes phase relations but, acting alone, preserves the spectrum and purity of a density matrix.

\dot\rho=\mathcal L\rho=-\frac{i}{\hbar}[H,\rho]+\sum_j\gamma_j\mathcal D[A_j]\rho,\quad \mathcal D[A]\rho=A\rho A^\dagger-\frac12\{A^\dagger A,\rho\},\quad\gamma_j\ge0

H is energy in joules, γj is in s⁻¹, dimensionless Aj gives a dimensionless dissipator, and braces denote an anticommutator. For a time-independent finite-dimensional generator with nonnegative rates, exp(tL) is a CPTP semigroup for t ≥ 0.↔ Scroll wide equations horizontally.

Positivity lives in the coefficient matrix

For the canonical criterion, choose a Hilbert–Schmidt orthonormal basis of traceless operators and separate the Hamiltonian and its gauge freedom. Dissipative terms then mix these basis directions through a Hermitian coefficient matrix, which must be positive semidefinite for a completely positive semigroup. Its eigenvalues are the canonical rates. Off-diagonal entries need not be positive. The necessity claim does not apply to arbitrary redundant lists or identity directions: an identity jump has a zero dissipator even with a negative coefficient.

Trace conservation follows directly by cycling factors inside each term. Complete positivity is deeper and is encoded by the full generator structure. For time-independent generators this structure also gives the composition law: evolving for two intervals is equivalent to evolving for their sum. An arbitrary time-local equation need not have a constant generator or this law. Negative canonical rates in a time-dependent representation can accompany valid maps from the initial time while preventing completely positive intermediate propagation.

\operatorname{Tr}\mathcal D[A]\rho=\operatorname{Tr}(A^\dagger A\rho)-\frac12\operatorname{Tr}(A^\dagger A\rho)-\frac12\operatorname{Tr}(\rho A^\dagger A)=0,\qquad e^{(t+s)\mathcal L}=e^{t\mathcal L}e^{s\mathcal L}

The trace identity uses finite-dimensional cyclicity and holds term by term. The composition law assumes the same time-independent L throughout; t and s are nonnegative times. Neither statement alone establishes positivity of an arbitrary proposed generator.↔ Scroll wide equations horizontally.

Stationary does not automatically mean thermal

A stationary density matrix is annihilated by the generator. Existence does not imply uniqueness: pure dephasing leaves every density matrix diagonal in its preferred basis unchanged. Nor does a stationary state necessarily equal a Gibbs state. Thermal detailed balance imposes an additional relation between upward and downward rates tied to a specified Hamiltonian and temperature. Reservoirs built for state preparation can deliberately violate that relation while still giving a legitimate GKSL equation.

For a bit-flip dissipator, the population difference decays exponentially while the total population remains fixed. That calculation exposes the factor of two arising from transitions in both directions. It also shows why a nonzero dissipative rate does not imply every input changes: states commuting with the bit-flip operator are stationary under this particular dissipator. To identify an attracting state, analyze all modes of the generator, not just the population equation for one convenient initial condition.

Work it through

Population mixing from a bit-flip dissipator

Set $H=0$ and $\dot\rho=\gamma(X\rho X-\rho)$ with $\gamma=0.5\,\mathrm{s}^{-1}$ . Start in $|0\rangle$ . Find the two populations at $t=\ln2\,\mathrm{s}$ and check their sum.

For a diagonal state, $\dot p_0=\gamma(p_1-p_0)$ and $\dot p_1=\gamma(p_0-p_1)$ . Adding gives zero, so the normalization stays one from its initial value.
Subtracting gives $\dot z=-2\gamma z$ for $z=p_0-p_1$ , with $z(0)=1$ . Hence $z(t)=e^{-2\gamma t}$ ; the two-way mixing rate is twice the individual flip rate.
At the specified time, $2\gamma t=\ln2$ , so $z=1/2$ . Using $p_0=(1+z)/2$ and $p_1=(1-z)/2$ gives $3/4$ and $1/4$ , respectively.

The result

The populations are 0.75 and 0.25 and sum to one. This initial diagonal state becomes mixed; the same generator also has stationary states with nonzero X coherence.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

A discrete population counterpart

Reset and set both controls to p = 0.25 and q = 0.25. Predict state-1 probability 0.25 after one Step and 0.375 after two Steps; the population difference halves each transition. This is a classical discrete counterpart of symmetric population relaxation, not a solver for arbitrary GKSL dynamics. It has no coherence, Hamiltonian, or Lindblad-operator inputs.

P(state 0)P(state 1)Stationary distribution

P(state 1) at step 0: 0%
Stationary P(state 1): 60%
Chain behavior: Monotone convergence

Transition p · 0 → 10.3Transition q · 1 → 00.2

t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forBoth models conserve total probability, but a population plot alone cannot determine a quantum generator.

Pause & reason

Check your understanding

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Read the checkpoint explanation

In the specified canonical basis, positive semidefiniteness gives nonnegative canonical rates. A Hermitian matrix may have negative eigenvalues, and a positive semidefinite matrix may have negative off-diagonal entries. Redundant or identity jump directions would require separate treatment.

Your turn

Try it yourself

A qubit has jump directions $|0\rangle\langle1|$ at rate $3\,\mathrm{s}^{-1}$ and $|1\rangle\langle0|$ at rate $1\,\mathrm{s}^{-1}$ , with no coherent drive. Find its stationary excited population and its population at $t=(\ln2)/4\,\mathrm{s}$ if it starts with excited population $3/4$ .

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The population derivative is an influx minus an outflux. Solve for its fixed point before evolving the deviation.

Compare with the solution

Writing $p=\rho_{11}$ gives $\dot p=1(1-p)-3p=1-4p$ . The stationary population is $1/4$ , and $p(t)=1/4+(3/4-1/4)e^{-4t}$ . At the stated time the exponential is one half, so $p=1/2$ . These positive rates define a valid dissipator; interpreting the stationary state as thermal additionally requires a specified gap and the corresponding detailed-balance ratio.

Keep the idea

Take it with you

The recycling and anticommutator terms conserve trace together.
Nonnegative canonical rates ensure a finite-dimensional CPTP semigroup.
Stationary, unique, attracting, and thermal are distinct properties.

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