Quantum regression and two-time correlators
Propagate operator-weighted states and identify the extra assumptions behind a regression formula.
On this page
The question for today
What you’ll understand
- Distinguish a two-time correlation from a product of means.
- Use a Markov regression formula with correct operator ordering.
- Relate exponential correlation decay to a spectral linewidth and state its limitations.
A density operator at one time gives all one-time observable averages, but spectroscopy and photon statistics depend on correlations between different times. These cannot generally be reconstructed by multiplying separate averages. In the appropriate Markov quantum-optical regime, the regression formula reuses the reduced generator to propagate an operator-weighted initial object. Its convenience should not hide the stronger assumptions needed to predict multi-time statistics from one-time dynamics.
The order of operators fixes the initial object
Let a time-homogeneous master equation generate the reduced propagator. For an ordered correlation with the later operator on the left, the regression prescription starts by multiplying the earlier operator onto the left of the earlier density matrix. Propagate that product, then trace with the later observable. The initial object can be non-Hermitian, have zero trace, and need not be positive. Linearity permits its evolution even though it is not itself a physical preparation.
Reversing operator order changes that initial product and can change the correlation. This matters for emission versus absorption: lowering an excitation and then raising it probes a different occupancy than raising first and lowering later. At zero delay, always check that the expression reduces to the expectation of the ordered operator product. This provides a quick diagnostic against an incorrect initialization, especially when both one-time dipole means vanish.
A decaying dipole gives a Lorentzian line
For a stationary undriven qubit, take the raising operator at the later time and the lowering operator at the earlier time. In a rotating frame, the off-diagonal matrix unit decays at the transverse rate. The coefficient multiplying it is the stationary excited population, not its square. Thus the two-time dipole correlation can be nonzero even though each dipole mean is zero. The amplitude is a population factor, while the decay constant carries the linewidth information.
Fourier transforming the exponential delay dependence gives a Lorentzian around the transition frequency. Defining the spectrum with twice the real part of the one-sided transform fixes the normalization used here; other spectral conventions distribute factors differently. The half width in angular frequency is the transverse rate and the full width is twice that rate. A measured non-Lorentzian shape can indicate several transitions, slow frequency noise, driving, or memory rather than a single changed decay constant.
Regression requires a multi-time justification
The underlying system and bath become correlated even if their initial preparation is a product. In the regression approximation, their relevant correlations after the earlier operator insertion are neglected or controlled in a way consistent with the Markov bath limit. A constant-rate fit to density matrices does not by itself establish this property. Even completely positive divisibility of all one-time maps is weaker than the full regression hierarchy for a general microscopic process.
At delays comparable to bath memory, or for structured reservoirs and correlated preparations, extra terms can modify two-time correlations. An exact one-time reduced equation may then fail if inserted into the standard regression formula. This is not an algebraic defect in propagating a nonpositive operator product; it is missing process information. Comparing a measured correlation with the prediction from independently determined relaxation times is therefore a substantive test of assumptions, rather than a redundant measurement of the same density matrix.
Work it through
A nonzero correlation with zero dipole mean
A stationary qubit has upward and downward rates and , no drive, and no pure dephasing. In its Markov regression regime, find in the rotating frame at and the line-centre spectrum.
The stationary excited population is . The transverse rate is half the sum of transition rates, .
Initialize with . Propagation multiplies this by , and tracing with returns its coefficient. At the stated delay the exponential is one half, giving .
At zero detuning the spectrum is . The dipole means individually vanish, so their product is zero and would miss this entire nonzero correlation.
The correlation is 1/8 at the specified delay and the line-centre spectrum is 0.25 s under the declared convention. The corresponding full width is 4 rad/s.
The living notebook
Explore the model
Interactive lab · markov
Conditional populations as a classical comparison
Reset and set both controls to p = 0.20 and q = 0.20. Predict state-1 probability 0.20 after one Step and 0.32 after two Steps. The chain starts in state 0, so these are conditional finite-step probabilities for that preparation. This lab can illustrate propagating a conditional distribution, but has no operator ordering, quantum dipole correlation, or spectral readout.
- P(state 1) at step 0
- 0%
- Stationary P(state 1)
- 60%
- Chain behavior
- Monotone convergence
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 forRegression needs the appropriate initial conditional object; an unconditioned stationary number alone is insufficient.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the same stationary qubit, let . Use the population regression law to calculate and its connected part at .
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Given excitation at the earlier time, the later excitation probability is p* + (1 − p*)exp(−Γ1τ). Multiply by the probability of that earlier excitation.
Compare with the solution
Here and . The correlation is . At the specified delay this gives . Subtracting the product of stationary means, , leaves connected correlation . Its decay rate is the population rate, distinct from the dipole transverse rate.
Keep the idea
Take it with you
- Two-time correlations require operator ordering and an operator-weighted initial object.
- Transverse exponential decay gives a Lorentzian under a declared transform convention.
- One-time Markovianity alone does not prove microscopic quantum regression.
Read further
Original sources and deeper explanations to follow at your own pace.
- Castin, Dalibard and Mølmer, A Wave Function Approach to Dissipative Processes, introduction and §§1–2: master-equation correlations and trajectories
- Guarnieri, Smirne and Vacchini, Quantum Regression Theorem and Non-Markovianity of Quantum Dynamics, §III: regression condition and its stronger requirements
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