advanced34 min studyLesson 418 in Physics

Quantum trajectories and unravelings

Connect conditional environment records to jump updates while recovering the unconditional master equation.

Before you begin

On this page

The question for today

What you’ll understand

Compute jump probabilities and normalized conditional states.
Explain information gained from a no-click record.
Distinguish a computational unraveling from an actual detector model.

A master equation predicts a density matrix after an environment record is ignored. If that record is monitored, a conditional state can instead follow a stochastic trajectory. Averaging the trajectory projectors recovers the unconditional state when the monitoring model is consistent. Individual trajectories are therefore not uniquely determined by the density matrix equation: the environment measurement, detector efficiency, and available record decide which conditional story is appropriate.

Jump probabilities come from the state

Write each Lindblad jump operator with its rate absorbed into it. Over a sufficiently short interval, the probability of a particular jump is the interval multiplied by the expectation of its operator product. It is not generally a fixed Poisson probability independent of the current state. A downward jump from a qubit is impossible when that qubit is already in its ground state, while excitation and coherent driving can replenish the probability of later emissions.

Between jumps, the unnormalized state evolves under an effective non-Hermitian Hamiltonian. Its norm loss equals the total jump probability to first order. Normalize only after choosing the no-jump branch. This gives both the correct branch probability and a conditional state that changes when no photon is detected. Treating non-Hermitian evolution alone as a normalized deterministic dynamics would discard the recycling terms that are needed to recover the full master equation.

H_{\rm eff}=H-\frac{i\hbar}{2}\sum_jL_j^\dagger L_j,\qquad dp_j=dt\,\langle L_j^\dagger L_j\rangle,\qquad |\psi\rangle\longmapsto\frac{L_j|\psi\rangle}{\sqrt{\langle L_j^\dagger L_j\rangle}}

H and Heff are in joules, Lj in s⁻¹/², and dt in seconds. The jump formula applies only to a branch of nonzero probability. The small-step construction requires total jump probability much less than one and a valid underlying Markov generator.↔ Scroll wide equations horizontally.

A missing click is still information

For an ideal photon-counting monitor of vacuum damping, a click identifies the emitted-photon branch and leaves the qubit in its ground state. A no-click record instead suppresses the excited amplitude relative to the ground amplitude. Starting in a superposition, the normalized no-click state therefore drifts toward the ground state even when no energy quantum is registered. This is a conditional inference from a measurement record, not evidence that every unobserved qubit follows that same state vector.

For finite intervals with no drive, the exact amplitude-damping Kraus operators give a particularly clear calculation. The no-click probability is the squared norm of the no-click vector, and the click probability is the decay probability times the initial excited population. Weighting the two normalized projectors by these probabilities reproduces the unconditional channel, including its residual coherence. Averaging vectors themselves would be wrong because relative trajectory phases are arbitrary; the physical ensemble average acts on projectors.

\rho(t)=\mathbb E[|\psi_c(t)\rangle\langle\psi_c(t)|],\qquad P(0)=|a|^2+(1-\lambda)|b|^2,\qquad |\psi_0\rangle=\frac{a|0\rangle+\sqrt{1-\lambda}\,b|1\rangle}{\sqrt{P(0)}}

The finite-interval expressions assume initial pure state a|0⟩ + b|1⟩, vacuum damping without drive, and ideal monitoring resolving the no-emission versus emission record. λ is dimensionless decay probability. The stochastic average is over the declared conditional measurement process.↔ Scroll wide equations horizontally.

The measurement scheme selects the unraveling

Direct photon counting gives discrete jumps. Mixing the emitted field with a strong phase reference and measuring a quadrature can instead produce a diffusive conditional equation. Different allowed measurements of the same output field can yield different trajectories while sharing the same unconditional master equation. This is why a master equation alone does not specify an objectively unique sequence of jumps. A numerical jump simulation may be useful even when no detector is physically installed.

Unit-efficiency monitoring of all relevant channels can preserve conditional purity for pure initial states. Missed photons or unobserved channels generally leave residual conditional mixing, requiring a density matrix rather than one fully informed state vector. Detector dark counts, dead time, bandwidth, and finite time resolution require further modeling. The microscopic Markov approximation also limits interpretation below the bath correlation time. A simulated infinitesimal step is a mathematical device, not a promise of arbitrarily fine physical measurement resolution.

Work it through

Exact click and no-click branches

An initially pure qubit is $|+\rangle$ and undergoes vacuum damping with $\lambda=3/4$ . An ideal environment monitor distinguishes emission from no emission. Compute both probabilities and conditional states, then recover the unconditional matrix.

The emission probability is $\lambda|b|^2=(3/4)(1/2)=3/8$ . The emitted-photon branch leaves the normalized system state $|0\rangle$ .
The no-emission probability is $1/2+(1/4)(1/2)=5/8$ . Its unnormalized vector is $(|0\rangle+|1\rangle/2)/\sqrt2$ , giving normalized state $(2|0\rangle+|1\rangle)/\sqrt5$ . Its excited population is $1/5$ .
The no-click projector has entries $4/5,2/5,2/5,1/5$ . Weight it by $5/8$ and add the ground projector weighted by $3/8$ : the result has populations $7/8,1/8$ and coherences $1/4$ .

The result

Emission has probability 3/8 and leaves the ground state. No emission has probability 5/8 and leaves (2|0⟩ + |1⟩)/√5. Their projector average matches the amplitude-damping channel exactly.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

Independent repetitions of a fixed monitoring experiment

For a separate finite-interval monitoring experiment with click probability 1/4, Reset and set both controls to success probability p = 0.25 and trials n = 8. Predict mean click count 2 and variance 1.5 from the readouts. Each trial here means a fresh, independent preparation of that experiment, not successive time bins of one decaying atom. The exact binomial bars do not generate detector records or conditional wavefunctions.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forRepeated fresh experiments can be binomial even when a single continuously monitored atom has state-dependent jump probabilities.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The excited component is more likely to emit, so conditioning on no emission suppresses it relative to the ground component. Perfect detection and the declared channel are assumptions; other monitoring schemes give different conditional trajectories.

Your turn

Try it yourself

Start in $\sqrt3|0\rangle/2+|1\rangle/2$ and use vacuum damping with $\lambda=1/2$ . For an ideal no-emission record, find its probability and the normalized conditional excited population. Also give the unconditional excited population.

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

Give me a hint

First compute the unnormalized surviving excited weight. Divide it by the no-emission probability only for the conditional answer.

Compare with the solution

The no-emission probability is $3/4+(1/2)(1/4)=7/8$ . The surviving excited weight is $1/8$ , so its conditional population is $(1/8)/(7/8)=1/7$ . A normalized no-click vector is $\sqrt{6/7}|0\rangle+\sqrt{1/7}|1\rangle$ . Ignoring the record instead gives excited population $1/8$ , because the emission branch, of probability $1/8$ , is ground. Confusing these populations would mix conditioned and unconditional experiments.

Keep the idea

Take it with you

Average trajectory projectors to recover an unconditional density matrix.
A no-click record can update a conditional quantum state.
A detector model and efficiency are part of a physical unraveling.

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