Quantum Trajectories

Quantum-jump simulation replaces one density matrix by an ensemble of stochastically evolving statevectors. Each trajectory uses $O(d)$ state storage rather than $O(d^2)$, but observables acquire sampling uncertainty. Different unravelings can represent the same unconditioned master equation while producing different records.

Helpful background. No-jump propagation supplies survival probabilities; Lindblad solvers supplies an independent ensemble benchmark.

Jump and no-jump branches

During a small interval $\Delta t$, a normalized state has jump probabilities

$$p_\mu=\Delta t\,\langle L_\mu^\dagger L_\mu\rangle, \qquad p=\sum_\mu p_\mu.$$

With probability $p_\mu$ replace it by $L_\mu\psi/\sqrt{\langle L_\mu^\dagger L_\mu\rangle}$. Otherwise propagate with $I-iH_{\rm eff}\Delta t/\hbar$ and normalize. Expanding the branch-averaged projector to first order gives

$$\mathbb E[\rho(t+\Delta t)-\rho(t)] =\Delta t\left[-\frac{i}{\hbar}[H,\rho] +\sum_\mu\mathcal D[L_\mu]\rho\right]+O(\Delta t^2).$$

The jump branch contributes $L_\mu\rho L_\mu^\dagger$; loss of no-jump norm supplies the anticommutator. This verifies the connection to the master equation without assigning a unique physical reality to a chosen unraveling.

More explicitly, the normalized no-jump projector has the first-order correction $-i[H,\rho]\Delta t/\hbar-\{C,\rho\}\Delta t/2+\lambda\rho\Delta t$, where $C=\sum_\mu L_\mu^\dagger L_\mu$ and $\lambda=\operatorname{Tr}(C\rho)$. Multiplying by its probability $1-\lambda\Delta t$ cancels the last term. The jump probabilities cancel their own normalization denominators, leaving $\Delta t\sum_\mu L_\mu\rho L_\mu^\dagger$. A branch with zero rate must be excluded from sampling; dividing its zero vector by its norm is not a harmless limiting operation.

A fixed-step algorithm requires small $p$ and sufficiently accurate conditional propagation. Event-driven waiting-time sampling avoids the one-jump-per-step approximation but still needs converged event locations.

Ensemble estimators

For independent trajectories $\psi_r(t)$,

$$\widehat{\langle A\rangle}(t)=\frac1R\sum_{r=1}^R \langle\psi_r(t)|A|\psi_r(t)\rangle, \qquad {\rm SE}=\frac{s_A}{\sqrt R}.$$

Calculate the sample variance from trajectory observables. Averaging wavefunctions and then forming an expectation is wrong: arbitrary trajectory phases would change the answer. For mixed initial states, sample a valid initial ensemble with its physical probabilities, or propagate each component with correctly weighted estimators.

Time samples within one trajectory are correlated. They do not provide $R$ independent replicates. For a time-integrated quantity, first integrate each trajectory and then estimate variance across independent trajectories.

Spontaneous-emission benchmark

An initially excited undriven qubit with $L=\sqrt\gamma|0\rangle\langle1|$ has excited population $P_e(t)=e^{-\gamma t}$. Each trajectory is excited until its first jump and in the ground state thereafter. Therefore

$${\rm Var}[\widehat P_e(t)] =\frac{e^{-\gamma t}(1-e^{-\gamma t})}{R}.$$

At $t=\log2/\gamma$, the standard error is $1/(2\sqrt R)$: for $R=10^4$ it is $0.005$. A discrepancy much smaller than this can occur by chance and does not establish a more accurate integrator. Compare repeated ensembles or confidence intervals, not an exact equality to the curve.

Test the waiting-time distribution separately and verify that each trajectory has at most one jump in this undriven example. Introducing a drive then tests repeated collapse and re-excitation.

Separate bias from variance

Refine the propagation tolerance at fixed large $R$, then increase $R$ at fixed converged propagation settings. Independent ensembles should show sampling fluctuations scaling as $R^{-1/2}$; integration bias will not disappear as $R$ grows.

For strong comparisons between time-step levels, couple the random thresholds where possible, but retain independent trajectories within each ensemble. Parallel workers require independent random streams, not copies of the same seed. Record the seed policy and define whether uncertainty describes trajectories, parameter disorder, or both.

Rare jumps can make a bounded observable appear noise-free in a small ensemble. Report event counts and compare with expected rates. A trajectory method is not automatically cheaper when high accuracy is required for tiny probabilities.

Zero observed events do not imply zero uncertainty. For $R$ independent trajectories and a fixed-window event probability $q$, the probability of observing none is $(1-q)^R$. A one-sided 95% upper bound after zero events is $q_{\rm upper}=1-0.05^{1/R}\approx3/R$. This bound concerns whether an event occurred, not the number of events in a driven trajectory; repeated counts require their own distribution and estimator.

Exercise

For the undriven qubit, what trajectory count gives standard error at most $10^{-3}$ for every time?

Solution

The Bernoulli variance is at most $1/4$, so $R\geq1/(4\times10^{-6})=250{,}000$. This is a standard-error criterion, not a simultaneous confidence band over an entire time trace. It also excludes integration bias.

References

• Dalibard, J., Castin, Y., and Mølmer, K., “Wave-function approach to dissipative processes in quantum optics,” Physical Review Letters 68, 580–583 (1992), doi:10.1103/PhysRevLett.68.580.
• QuTiP developers, “Monte Carlo Solver”, QuTiP documentation — trajectory statistics and parallel execution.