advanced34 min studyLesson 419 in Physics

Non-Markovian memory kernels

Use an exact finite-environment model to separate valid initial-time channels, divisibility, and information revival.

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What you’ll understand

  • Solve a simple convolution equation and check its units.
  • Recognize a valid channel with non-positive intermediate propagation.
  • Use trace-distance revival as a specified memory witness without equating all definitions.

Memory means that discarding the environment can hide information needed for later predictions. It does not mean that quantum states stop obeying probability rules. With a fixed uncorrelated initial preparation, the exact map from that initial time remains completely positive and trace preserving even when a semigroup description fails. A small exactly solvable environment makes this distinction visible without an uncontrolled approximation to a large reservoir.

An exact reversible bath can return coherence

Take a system qubit and one environment qubit with interaction Hamiltonian HSE=ℏgZS⊗ZE/2H_{SE}=\hbar g Z_S\otimes Z_E/2, and prepare the environment in its plus state. Remove any commuting free Hamiltonians by changing frame. Conditional environment rotations give the overlap η(t)=cos⁡(gt)\eta(t)=\cos(gt), multiplying the system off-diagonal entries while leaving its populations unchanged. This is exact for the stated finite composite, with no Born or Markov approximation.

An equivalent channel is the equal mixture of system phase rotations generated by the two environment Z eigenvalues. Consequently every initial-time map is a valid channel even when the cosine becomes negative. A negative coherence factor is a phase reversal, not a negative population. The magnitude returns to one after reaching zero, so local purity can revive. A one-qubit environment is not a thermal reservoir causing irreversible decay; its finite information capacity makes recurrence central rather than negligible.

ρ01(t)=cos⁡(gt)ρ01(0),ρ˙01(t)=−g2∫0tρ01(s)ds,ρ01′(0)=0\rho_{01}(t)=\cos(gt)\rho_{01}(0),\qquad \dot\rho_{01}(t)=-g^2\int_0^t\rho_{01}(s)ds,\qquad \rho_{01}'(0)=0
g is an angular frequency in s⁻¹; density entries are dimensionless. The constant convolution coefficient g² has units s⁻². Differentiating once more gives a harmonic equation with initial value ρ01(0) and zero initial slope, yielding the exact cosine.↔ Scroll wide equations horizontally.

A time-local equation can still carry memory

Where the cosine is nonzero, divide its derivative by itself to obtain a time-local dephasing equation. The resulting rate becomes negative during intervals of coherence revival. This does not invalidate the original channel from the initial preparation. It instead signals that intermediate propagation cannot always be represented as a completely positive dephasing channel acting on arbitrary inputs. At the cosine zeros, the reduced map is noninvertible and the time-local rate becomes singular even though the exact composite evolution is smooth.

The convolution description avoids that singular inverse by retaining a history integral. More general projection-operator equations can include an inhomogeneous term from initial correlations, as well as complicated superoperator kernels. Merely choosing a decaying kernel does not guarantee complete positivity; an approximate memory equation must be checked just as carefully as an approximate local equation. Conversely, writing an equation locally in time does not establish that it represents a memoryless physical process.

ρ˙=γ(t)2(ZρZ−ρ),γ(t)=gtan⁡(gt)[cos⁡(gt)≠0],D(ρ+(t),ρ−(t))=∣cos⁡(gt)∣\dot\rho=\frac{\gamma(t)}2(Z\rho Z-\rho),\qquad \gamma(t)=g\tan(gt)\quad[\cos(gt)\ne0],\qquad D(\rho_+(t),\rho_-(t))=|\cos(gt)|
γ(t) has units s⁻¹ and the local form is singular at zero cosine. D = ½||ρ+ − ρ−||₁ is dimensionless trace distance for initial plus and minus states. Its revival witnesses failure of positive divisible propagation for this model; no negative probabilities are implied.↔ Scroll wide equations horizontally.

Define which meaning of memory is being tested

Completely positive divisibility asks whether each later map factors through an earlier map with a valid intermediate channel. Trace-distance backflow instead asks whether some initially distinct states become more distinguishable over an interval. Contractivity of positive trace-preserving propagation makes an increase a useful witness, but the two criteria are not equivalent for every system dimension and process. Multi-time conditional statistics provide still more information, so a label of non-Markovianity should name the operational or mathematical criterion.

For this finite-bath example, opposite equatorial inputs become the same local state at a zero of the cosine. Later they become distinguishable again. No map on that single identical intermediate state can recover two different outputs without additional information, demonstrating why the environment matters. Resetting the environment at the intermediate time changes the experiment and can eliminate this return. Parameter changes in a teaching instrument likewise should not be interpreted as implementing a continuous physical bath protocol.

Coherence loss followed by a complete revival

Set g=1 s−1g=1\,\mathrm{s}^{-1} in the finite-environment model and compare initial system states ∣+⟩|+\rangle and ∣−⟩|-\rangle. Calculate their trace distance at t=π/3 st=\pi/3\,\mathrm{s}, π/2 s\pi/2\,\mathrm{s}, and π s\pi\,\mathrm{s}. Explain the intermediate-map obstruction.

  1. The two reduced states have equal populations but opposite real coherences, each of magnitude ∣cos⁡t∣/2|\cos t|/2. Their difference has eigenvalues ±cos⁡t\pm\cos t, so trace distance is ∣cos⁡t∣|\cos t|.

  2. At the three times, the cosine is 1/21/2, zero, and −1-1. The distances are therefore 1/21/2, zero, and one. At the final time the coherence signs have reversed but the states are again perfectly distinguishable.

  3. At π/2\pi/2 both reduced states equal I/2I/2. A channel acting only on that common input must give one common output, whereas the actual later outputs differ. Correlations with the retained environment supply the missing distinction.

The result

The distances are 0.5, 0, and 1. The full initial-time maps stay CPTP, while an autonomous system-only intermediate channel cannot describe the revival through the zero-distance time.

Explore the model

Interactive lab

Interactive lab · markov

Contrast recurrence with classical periodicity

Reset and set both controls to p = 1 and q = 1. Predict state-1 probability 1 after one Step and 0 after two Steps, despite stationary probability 1/2. This discrete chain is perfectly Markovian and periodic. It has no quantum coherence, bath, or trace-distance comparison of two preparations. Its oscillation shows why a wavy single-population trace alone cannot diagnose quantum memory.

Contrast recurrence with classical periodicityExact 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. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
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 forA revival witness must specify what revives and under which maps; periodic populations alone are insufficient.

Check your understanding

A finite-bath dephasing channel has a temporarily negative time-local canonical rate but valid maps from its fixed initial product preparation. What follows?

Try it yourself

Set g=2 s−1g=2\,\mathrm{s}^{-1} and start the system in ∣+⟩|+\rangle. Find the coherence, purity, and local dephasing rate at t=π/6 st=\pi/6\,\mathrm{s} and t=π/3 st=\pi/3\,\mathrm{s}.

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 arguments gt are π/3 and 2π/3. Keep the sign of cosine for coherence, and square it for purity.

Compare with the solution

At the first time η=1/2\eta=1/2, so ρ01=1/4\rho_{01}=1/4; at the second η=−1/2\eta=-1/2, so ρ01=−1/4\rho_{01}=-1/4. Both states have purity (1+η2)/2=5/8(1+\eta^2)/2=5/8 and positive eigenvalues 3/4,1/43/4,1/4. The rates are 2tan⁡(π/3)=23 s−12\tan(\pi/3)=2\sqrt3\,\mathrm{s}^{-1} and 2tan⁡(2π/3)=−23 s−12\tan(2\pi/3)=-2\sqrt3\,\mathrm{s}^{-1}. The second lies in a revival interval, while the density matrices remain valid.

Take it with you

  • A valid initial-time channel need not admit CPTP intermediate maps.
  • A time-local equation can have memory and singular rates.
  • Trace-distance revival is a specified witness, not a universal synonym for every memory definition.

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