Time-Convolutionless Master Equations

A time-convolutionless master equation is a time-local equation for reduced dynamics:

$$\frac{d}{dt}\rho_S(t) = \mathcal K_{\mathrm{TCL}}(t)\rho_S(t).$$

It is called “convolutionless” because the equation does not contain an explicit integral over earlier states. Such an equation can still describe revivals or other memory effects. Time dependence alone does not prove their presence: even a driven memoryless model can have a time-dependent generator.

Helpful background. Reduced maps describe evolution from a fixed initial time; memory kernels and Nakajima–Zwanzig projection give an alternative with an explicit history integral.

Exact Map Formula

Suppose the reduced dynamics from $t_0$ to $t$ is represented by a linear map

$$\rho_S(t) = \Phi(t,t_0)\rho_S(t_0).$$

In finite dimension, suppose $\Phi(t,t_0)$ is differentiable and invertible as a linear map on the full operator space over the interval considered. Then an exact time-local generator exists:

$$\mathcal K_{\mathrm{TCL}}(t) = \dot\Phi(t,t_0)\Phi^{-1}(t,t_0).$$

Indeed,

$$\frac{d}{dt}\rho_S(t) = \dot\Phi(t,t_0)\rho_S(t_0) = \dot\Phi(t,t_0)\Phi^{-1}(t,t_0)\rho_S(t).$$

This formula is exact. The algebraic inverse $\Phi^{-1}$ need not be positive or physically implementable. The formula does not establish Markovianity. It says the same reduced map can be written as a time-local differential equation while the map remains invertible.

If $\Phi(t,t_0)$ becomes noninvertible, the time-local generator can become singular even when the finite-time map remains well defined.

Relation to Memory Kernels

A memory-kernel equation has the form

$$\frac{d}{dt}\rho_S(t) = \int_{t_0}^t ds\, \mathcal K_{\mathrm{NZ}}(t,s)\rho_S(s).$$

A TCL equation has the form

$$\frac{d}{dt}\rho_S(t) = \mathcal K_{\mathrm{TCL}}(t)\rho_S(t).$$

Both can describe the same reduced dynamics when the needed objects exist. The difference is where the history is stored:

• in Nakajima–Zwanzig form, the history is explicit in $\rho_S(s)$;
• in TCL form, the history is compressed into $\mathcal K_{\mathrm{TCL}}(t)$.

This is why “time local” should not be read as “memoryless.” A time-local generator can describe recoherence, revivals, and information backflow through time-dependent coefficients.

Projection-Operator TCL Form

Projection methods can also derive TCL equations directly. With a projection $\mathcal P$, the desired structure is

$$\frac{d}{dt}\mathcal P\rho(t) = \mathcal K_{\mathrm{TCL}}(t)\mathcal P\rho(t) + \mathcal I_{\mathrm{TCL}}(t),$$

where $\mathcal I_{\mathrm{TCL}}(t)$ contains initially irrelevant information if $\mathcal Q\rho(t_0)\ne0$. This is the TCL version of the initial-correlation caveat.

For projection-compatible initial states, $\mathcal Q\rho(t_0)=0$, the homogeneous TCL equation is

$$\frac{d}{dt}\mathcal P\rho(t) = \mathcal K_{\mathrm{TCL}}(t)\mathcal P\rho(t).$$

The formal construction expands $\mathcal K_{\mathrm{TCL}}(t)$ in ordered cumulants or powers of the system-bath coupling. This is often more practical than computing a full memory kernel.

Weak-Coupling Expansion

Let the interaction-picture Liouvillian be

$$\mathcal L_I(t) = -\frac{i}{\hbar} [H_I(t),\,\cdot\,].$$

For a coupling-strength parameter $\lambda$, the TCL generator is expanded as

$$\mathcal K_{\mathrm{TCL}}(t) = \sum_{n=1}^{\infty} \lambda^n \mathcal K_n(t).$$

For an initially factorized state, a fixed stationary bath reference, and centered interaction operators in the interaction picture, the first-order term vanishes:

$$\mathcal K_1(t)=0.$$

The second-order TCL equation for the reduced state has the form

$$\frac{d}{dt}\rho_S^{I}(t) = -\frac{\lambda^2}{\hbar^2} \int_{t_0}^t ds\, \operatorname{Tr}_E \left[ H_I(t), \left[ H_I(s), \rho_S^{I}(t)\otimes\rho_E \right] \right].$$

The second-order memory-kernel equation instead contains $\rho_S^I(s)$. In this centered setting, $\rho_S^I(t)-\rho_S^I(s)=O(\lambda^2)$ at fixed times, so replacing one by the other in an already $\lambda^2$ kernel changes terms at order $\lambda^4$. The two second-order solutions can nevertheless differ substantially at long times because each resums different higher orders. No short-memory assumption is required to define TCL2, and positivity of its solution is not automatic.

Markov and Redfield Limits

If bath correlations decay quickly and the upper limit can be extended,

$$\int_{t_0}^t ds \longrightarrow \int_0^\infty d\tau,$$

the second-order TCL generator approaches the usual Born-Markov generator. Before full secularization, this is closely related to the Redfield Equation. After secularization and positivity-preserving assumptions, one obtains the standard weak-coupling Lindblad–GKSL form.

Canonical rates and complete positivity

A finite-dimensional trace-preserving, Hermiticity-preserving time-local generator can be separated into a Hamiltonian term and a Hermitian coefficient matrix on an orthonormal traceless operator basis. Diagonalizing that matrix gives a canonical form

$$\dot\rho=-\frac{i}{\hbar}[H(t),\rho] +\sum_\mu\gamma_\mu(t) \left[L_\mu(t)\rho L_\mu^\dagger(t) -\frac12\{L_\mu^\dagger(t)L_\mu(t),\rho\}\right].$$

For a differentiable invertible dynamical map with a regular generator, CP divisibility is equivalent to nonnegative canonical rates (apart from irrelevant isolated times). Nonnegative rates in any well-defined GKSL representation are sufficient. Negative coefficients in a redundant representation, however, are not a diagnostic: $2\gamma\mathcal D[Z]-\gamma\mathcal D[Z]=\gamma\mathcal D[Z]$ is ordinary positive-rate dephasing.

Canonical rates can be negative over an interval while the maps from the specified initial time remain CPTP. In that case the intermediate propagator need not be CP on arbitrary states, even though it carries the states actually reached from the initial time to physical later states. See Complete Positivity for the reference-system test and non-Markovian dynamics for the distinction from multi-time memory.

Example: Pure Dephasing

A qubit pure-dephasing map may have

$$\rho_{01}(t) = \eta(t)\rho_{01}(0), \qquad \rho_{00}(t)=\rho_{00}(0).$$

When $\eta(t)\ne0$, the coherence obeys a time-local equation

$$\frac{d}{dt}\rho_{01}(t) = \frac{\dot\eta(t)}{\eta(t)}\rho_{01}(t).$$

Writing

$$\frac{\dot\eta(t)}{\eta(t)} = -\gamma_\phi(t)-i\Omega(t),$$

With $Z=\operatorname{diag}(1,-1)$, this means

$$\dot\rho=-i[\Omega(t)Z/2,\rho] +\frac{\gamma_\phi(t)}2(Z\rho Z-\rho).$$

The factor of one-half is needed: $\gamma_\phi$ is the decay rate of the off-diagonal element, not the coefficient multiplying $\mathcal D[Z]$.

If $|\eta(t)|$ temporarily increases, then $\gamma_\phi(t)$ becomes negative. With $\eta(t_0)=1$, this map is CPTP exactly when $|\eta(t)|\le1$ at each time. On an invertible interval, growth of $|\eta|$ violates CP divisibility.

For the exact two-qubit example in Nakajima–Zwanzig projection, $\eta(t)=\cos(2gt)$ and

$$\gamma_\phi(t)=2g\tan(2gt),\qquad \Omega(t)=0$$

away from zeros of the cosine. At $gt=\pi/6$ the coherence factor is $1/2$ and $\gamma_\phi=2\sqrt3g$; at $gt=\pi/3$ it is $-1/2$ and $\gamma_\phi=-2\sqrt3g$. Both finite-time maps are valid dephasing channels. At $gt=\pi/4$, every input coherence is erased and the TCL rate diverges, while the reduced map remains perfectly finite. Subsequent revival cannot be predicted uniquely from that erased system state alone.

Example: Amplitude Damping

For a qubit with $|0\rangle$ ground and $|1\rangle$ excited, define a vacuum damping family by

$$K_0=|0\rangle\langle0|+G(t)|1\rangle\langle1|, \qquad K_1=\sqrt{1-|G(t)|^2}\,|0\rangle\langle1|,$$

with $G(t_0)=1$ and $|G(t)|\le1$. Then $\rho_{11}(t)=|G|^2\rho_{11}(t_0)$ and $\rho_{01}(t)=G^*\rho_{01}(t_0)$. The conjugation fixes the Hamiltonian sign. When $G(t)\ne0$, the exact time-local generator has coefficients of the form

$$\gamma(t) = -2\,\operatorname{Re} \frac{\dot G(t)}{G(t)}, \qquad S(t) = -2\,\operatorname{Im} \frac{\dot G(t)}{G(t)}.$$

With $\sigma_-=|0\rangle\langle1|$, the corresponding equation is

$$\dot\rho=-i\left[\frac{S(t)}2|1\rangle\langle1|,\rho\right] +\gamma(t)\mathcal D[\sigma_-]\rho.$$

The Hamiltonian is $\hbar S(t)|1\rangle\langle1|/2$; $S$ is twice the frequency shift in this convention. For $G=e^{-(\kappa/2+i\delta)t}$, these formulas give $\gamma=\kappa$ and $S=2\delta$, as a direct substitution checks. Zeros of $G$ can make the TCL representation singular while the finite-time damping channel stays well defined.

Exercises

Derive the exact TCL generator

Assume $\rho(t)=\Phi(t,t_0)\rho(t_0)$ and $\Phi(t,t_0)$ is invertible. Derive $\mathcal K_{\mathrm{TCL}}(t)=\dot\Phi(t,t_0)\Phi^{-1}(t,t_0)$.

Solution

Differentiate the map:

$$\dot\rho(t) = \dot\Phi(t,t_0)\rho(t_0).$$

Since $\rho(t)=\Phi(t,t_0)\rho(t_0)$, invertibility gives

$$\rho(t_0) = \Phi^{-1}(t,t_0)\rho(t).$$

Substitute:

$$\dot\rho(t) = \dot\Phi(t,t_0)\Phi^{-1}(t,t_0)\rho(t).$$

Thus the time-local generator is $\mathcal K_{\mathrm{TCL}}(t)=\dot\Phi(t,t_0)\Phi^{-1}(t,t_0)$.

Pure-dephasing rate

If $\eta(t)=e^{-\Gamma(t)}e^{-i\Theta(t)}$, find $\gamma_\phi(t)$ and $\Omega(t)$ in

$$\frac{\dot\eta(t)}{\eta(t)} = -\gamma_\phi(t)-i\Omega(t).$$

Solution

Take the logarithmic derivative:

$$\frac{\dot\eta}{\eta} = -\dot\Gamma(t)-i\dot\Theta(t).$$

Therefore

$$\gamma_\phi(t)=\dot\Gamma(t), \qquad \Omega(t)=\dot\Theta(t).$$

If $\dot\Gamma(t)\lt0$, the coherence magnitude is increasing and the instantaneous dephasing rate is negative.

Compare second-order NZ and TCL

In the second-order weak-coupling equations, what is the main difference between the memory-kernel and TCL forms?

Solution

The memory-kernel form contains the delayed state:

$$\rho_S^{I}(s)$$

inside the integral. The TCL form contains the current state:

$$\rho_S^{I}(t).$$

Both integrate over bath correlation history, but only the memory-kernel equation is explicitly time nonlocal in the reduced state.

Negative rate

A differentiable invertible TCL evolution has a canonical rate negative over an interval, but its finite-time maps from the chosen initial time remain CPTP. Is it automatically invalid? What if only a coefficient in a redundant representation is negative?

Solution

No. Under the stated regularity assumptions, the negative canonical rate shows failure of CP divisibility, not failure of the maps from the initial time to be physical. A negative coefficient in a redundant representation need not even show that: identical dissipators may combine to a positive net coefficient. Diagonalize the canonical coefficient matrix or test intermediate maps.

References

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