advanced30 min studyLesson 939 in Physics

Quantum-trajectory Monte Carlo estimators

Connect stochastic pure-state paths to density matrices while separating sampling noise from propagation bias.

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What you’ll understand

  • Explain no-jump norm loss and stochastic jump selection in a stated unraveling.
  • Average projectors and observables rather than wavefunction amplitudes.
  • Calculate finite-ensemble uncertainty under independent sampling assumptions.

Quantum trajectories replace one density-matrix evolution by a statistical ensemble of pure-state paths. Between jumps, an effective non-Hermitian equation changes the unnormalized state's norm; jump probabilities and state updates compensate so that the ensemble reproduces the chosen master equation. This can reduce storage demands, but introduces sampling uncertainty. A single trajectory is not the unconditional density matrix, and its physical measurement interpretation depends on the specified unraveling and monitoring model.

Define the no-jump generator and jump rule together

For the GKSL convention used previously, the no-jump Hamiltonian subtracts i hbar over two times the sum of rate-weighted L-adjoint L operators. Its norm decreases according to the instantaneous jump hazard. In a small-time-step implementation, the probability of channel r is gamma-r times timestep times the corresponding expectation in the normalized state, to leading order. A large timestep can invalidate that linear probability approximation or miss multiple jumps.

Exact waiting-time or carefully controlled small-step schemes update both the channel and post-jump state with the right conditional distribution. The normalized no-jump state and the jump weights must be treated consistently; deterministic renormalization without stochastic compensation gives a different ensemble. Different valid unravelings can yield different-looking paths and variances while reproducing the same density evolution. Numerical convergence should target the ensemble observable, not the visual appearance of an individual random path.

Hnj=H−iℏ2∑rγrLr†Lr,pr=γrΔt⟨Lr†Lr⟩+O(Δt2)H_{\rm nj}=H-\frac{i\hbar}{2}\sum_r\gamma_rL_r^\dagger L_r,\qquad p_r=\gamma_r\Delta t\langle L_r^\dagger L_r\rangle+O(\Delta t^2)
The probability expansion assumes a sufficiently small step and the stated finite generator. The non-Hermitian no-jump propagation is a conditional part of a complete stochastic algorithm.↔ Scroll wide equations horizontally.

Average physical objects rather than arbitrary phases

The density estimator is the average of normalized pure-state projectors. It is positive and has unit trace for every finite sample, while its expectation equals the master-equation density when the paths are correctly sampled. An observable estimator averages each path's expectation value. Averaging state amplitudes instead is not equivalent: independent global phases can change that average without changing any projector or physical member of the ensemble.

Finite sampling noise and algorithmic bias are distinct. Increasing the number of paths reduces random fluctuations around the distribution produced by the implemented algorithm, but does not fix an incorrect jump rule, finite timestep or too-small basis. Compare timestep and Hilbert dimension at sufficiently large sample size, and report random seeds or reproducibility information as appropriate. Common-random-number comparisons can reduce differences but introduce correlations that must be included in uncertainty interpretation.

ρ^N=1N∑a=1N∣ψa⟩⟨ψa∣,O^N=1N∑a=1N⟨ψa∣O∣ψa⟩\widehat\rho_N=\frac1N\sum_{a=1}^N|\psi_a\rangle\langle\psi_a|,\qquad \widehat O_N=\frac1N\sum_{a=1}^N\langle\psi_a|O|\psi_a\rangle
Each sampled state is normalized. These estimators average projectors or real expectation values of Hermitian observables, not coherent sums of unrelated trajectories.↔ Scroll wide equations horizontally.

State the sampling model behind an error bar

For independent identically distributed paths at one fixed evaluation time, an observable sample mean has variance equal to the single-path variance divided by N. For a basis-state indicator this reduces to the familiar binomial probability variance. If its true probability is unknown, estimating that variance from data adds its own finite-sample uncertainty. A standard error is a statistical scale, not a deterministic bound that every run must satisfy.

Trajectory samples at different times on the same path are correlated, as can be different observables or deliberately paired runs. Time averaging those samples does not automatically create the same number of independent paths. Rare jumps can also require many samples before a normal approximation is persuasive. Exact binomial examples offer a transparent benchmark for the simplest indicator, while general coherent paths require the variance of their actual per-path expectation values.

Var⁡(O^N)=Var⁡(Oa)N,Var⁡(p^N)=p(1−p)Nfor independent Bernoulli indicators\operatorname{Var}(\widehat O_N)=\frac{\operatorname{Var}(O_a)}{N},\qquad \operatorname{Var}(\widehat p_N)=\frac{p(1-p)}N\quad\text{for independent Bernoulli indicators}
The variance formula is about the chosen sampling distribution at a fixed time. It excludes propagation bias and requires independence for the stated N-inverse scaling.↔ Scroll wide equations horizontally.

Four exact damping trajectories

For zero-Hamiltonian amplitude damping start excited and choose time γt=ln⁡4\gamma t=\ln4. Exact trajectories end excited if no jump occurred and ground otherwise. Suppose four independent samples give one excited and three ground. Find the density estimate and its excited-population variance using the known exact probability.

  1. The exact no-jump probability is e−γt=1/4e^{-\gamma t}=1/4. A jump transfers the state permanently to ground in this zero-temperature model, so the final excited indicator has probability one quarter.

  2. The sample density is diagonal with ground entry three quarters and excited entry one quarter. Its excited estimate happens to equal the exact value in this particular sample, but that does not mean sampling variance is zero.

  3. With four independent indicators the true variance of the estimated population is (1/4)(3/4)/4=3/64(1/4)(3/4)/4=3/64. Its standard deviation is 3/8≈0.2165\sqrt3/8\approx0.2165. Other four-path samples can differ appreciably even though the sampling algorithm is exact.

The result

The sample density has excited entry one quarter, while its true population-estimator variance is three sixty-fourths under the stipulated independent sampling model.

Explore the model

Interactive lab

Interactive lab · sampling

An exact distribution for a sample mean

Reset and set success probability 0.25, sample size four; predict sample-mean variance three sixty-fourths. Compare a fresh Reset run at probability 0.25, size sixteen; predict variance three over 256. Playback highlights an exact binomial distribution and does not generate random paths, resample experiments or solve a quantum jump equation.

An exact distribution for a sample meanExact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). Current readings: Mean sample proportion: 0.5; Sample-mean variance: 0.0125; Uniform CDF error bound: ≤ 0.1062. All parameters are adjustable with the labeled controls below.Exact distribution of the mean of 20 independent Bernoulli variables with success probability 0.5; playback highlights outcome 0.00.250.50.75100.0580.1150.1730.23Sample mean · k / nProbability massNormal bin masses are an approximationk = 0 · exact probability = 9.54e-7
Exact binomial massNormal interval massHighlighted outcome / mean
Mean sample proportion
0.5
Sample-mean variance
0.0125
Uniform CDF error bound
≤ 0.1062
t = 0 s

Exact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). 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 forQuadrupling independent sample count halves the standard deviation, but the displayed distribution is computed exactly rather than sampled.

Check your understanding

Which operation reproduces the unconditional density from correctly sampled normalized trajectories?

Try it yourself

Take an equal two-member ensemble of plus and minus states, (∣0⟩+∣1⟩)/2(|0\rangle+|1\rangle)/\sqrt2 and (∣0⟩−∣1⟩)/2(|0\rangle-|1\rangle)/\sqrt2. Compute its projector average and purity. Compare with averaging the two vectors and then normalizing.

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

Give me a hint

Opposite off-diagonal projector terms cancel. Vector averaging cancels a different object and is not mixture formation.

Compare with the solution

The projector average is I/2I/2, with both populations one half, zero coherence and purity one half. The average vector is ∣0⟩/2|0\rangle/\sqrt2, whose squared norm is one half; normalizing it yields pure zero state with ground population one. Thus coherent vector averaging produces a different state from the intended mixture. Independent global-phase choices could change that vector average again while leaving the projector ensemble unchanged.

Take it with you

  • No-jump propagation and jump selection form one consistent stochastic model.
  • Density matrices arise from projector averages, not averaged amplitudes.
  • Sampling variance decreases with independent path count while numerical and physical biases remain.

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