State Update Rules

A state-update rule describes what a specified apparatus leaves behind, conditioned on the information retained from its readout. This guide compares common finite-dimensional models. In every case, effects determine probabilities; instruments determine the output states.

Helpful background. Selective and nonselective measurements explains the difference between a conditional subensemble and an average over all outcomes.

Measurement updates by apparatus

The entries in the middle column are unnormalized operations. Their trace is the reported outcome probability. Divide by that trace only when conditioning on an outcome of positive probability.

Apparatus and retained result	Operation	Physical assumption
Ideal Lüders measurement, outcome $a$	$P_a\rho P_a$	No extra disturbance inside the eigenspace.
Refined projective measurement, only $a$ retained	$\sum_\lambda P_{a\lambda}\rho P_{a\lambda}$	The device resolves finer labels and they are discarded; $P_a=\sum_\lambda P_{a\lambda}$.
One measurement operator for result $m$	$M_m\rho M_m^\dagger$	A single Kraus operator describes that outcome.
General result $m$	$\mathcal I_m(\rho)=\sum_\alpha M_{m\alpha}\rho M_{m\alpha}^\dagger$	Unresolved alternatives may mix the output state.
Complete measurement, result ignored	$\Phi(\rho)=\sum_m\mathcal I_m(\rho)$	Include no-click and failure branches; no additional normalization is needed.

For Lüders and refined instruments the coarse effect is $P_a$ in both cases, although their output states differ. For the general instrument, $F_m=\sum_\alpha M_{m\alpha}^\dagger M_{m\alpha}$.

Universal Instrument Form

A measurement instrument is a family of completely positive, trace-nonincreasing maps $\{\mathcal I_m\}$ indexed by outcomes. For input state $\rho$,

$$p(m) = \operatorname{Tr}\mathcal I_m(\rho).$$

If outcome $m$ is known and $p(m)\ne0$, the conditional state is

$$\rho_m = \frac{\mathcal I_m(\rho)} {\operatorname{Tr}\mathcal I_m(\rho)}.$$

If the outcome is ignored, the output state is

$$\rho' = \sum_m\mathcal I_m(\rho).$$

The sum over all outcomes must be trace preserving for an unconditional measurement with no discarded failure branch:

$$\operatorname{Tr} \left[ \sum_m\mathcal I_m(\rho) \right] = \operatorname{Tr}\rho.$$

This instrument form contains projective measurements, Kraus updates, detector inefficiency, coarse graining, postselection, and continuous-measurement time steps as special cases or limits.

Same POVM, Different Updates

The reason effects do not determine updates is visible in one line. If $F_m=M_m^\dagger M_m$, then

$$N_m=U_mM_m$$

with a unitary $U_m$ gives the same effect:

$$N_m^\dagger N_m = M_m^\dagger U_m^\dagger U_mM_m = F_m.$$

The probabilities are unchanged. The conditional state can change:

$$\frac{N_m\rho N_m^\dagger} {\operatorname{Tr}(N_m\rho N_m^\dagger)} = U_m \left( \frac{M_m\rho M_m^\dagger} {\operatorname{Tr}(M_m\rho M_m^\dagger)} \right) U_m^\dagger.$$

For example, start from $|+\rangle$ and compare $M_0=P_0$, $M_1=P_1$ with $N_0=XP_0$, $N_1=P_1$, where $X$ exchanges $|0\rangle$ and $|1\rangle$. Both report 0 and 1 with equal probability. The first unread output is $I/2$; the second is always $|1\rangle\langle1|$. An outcome-dependent kick has turned a dephasing operation into a reset without changing the outcome probabilities. Even a rank-one PVM does not determine the output state unless its implementation is specified.

For degeneracy, compare the explicit Lüders and refined operations. Replacing $P_a$ by a chosen basis vector is not justified merely because the observed eigenvalue is degenerate.

Bayesian Analogy and Its Limit

Classically, conditioning on a result updates probabilities by Bayes’ rule:

$$p(h|m) = \frac{p(m|h)p(h)} {p(m)}.$$

Quantum selective update has a similar conditional structure: divide the unnormalized conditional state by the probability of the outcome. But the analogy is limited. Quantum measurements can disturb the state, remove coherence, or apply outcome-dependent transformations. The state update is not determined only by prior probabilities and likelihoods; it depends on the physical instrument.

When the state and all measurement operators are diagonal in one common basis and the operators have nonnegative diagonal entries, the diagonal-state update reduces to Bayes’ rule. Conditional unitary kicks or off-diagonal coherence require the additional quantum description. For the full comparison, see Bayesian Quantum Measurement.

Continuous-Time Preview

In continuous measurement, the update is applied over small time steps and conditioned on a measurement record. A schematic discrete step has the same structure:

$$\rho(t+dt) = \frac{\mathcal I_{dr}(\rho(t))} {\operatorname{Tr}\mathcal I_{dr}(\rho(t))},$$

where $dr$ denotes a small record increment and $\mathcal I_{dr}$ is understood as an operation density with respect to a specified reference measure for a continuous record. Taking a continuous-time limit can lead to stochastic master equations or quantum trajectories. If the record is averaged over, one recovers an unconditional master equation.

The conceptual distinction is already present here: conditioned states follow records; unconditioned states average over records.

Exercises

1. Let $P_0=|0\rangle\langle0|+|1\rangle\langle1|$ and $P_2=|2\rangle\langle2|$. For

$$|\psi\rangle = \frac{|0\rangle+|1\rangle+|2\rangle}{\sqrt3},$$

find the Lüders-updated state after outcome $0$.

Solution

The probability of outcome $0$ is

$$p(0) = \langle\psi|P_0|\psi\rangle = \frac{2}{3}.$$

The projected state is $(|0\rangle+|1\rangle)/\sqrt3$. Normalizing gives

$$|\psi_0\rangle = \frac{|0\rangle+|1\rangle}{\sqrt2}.$$

The coherence between $|0\rangle$ and $|1\rangle$ is preserved.

2. For the same input state, suppose the device distinguishes $|0\rangle$ and $|1\rangle$ but reports only the coarse outcome $0$. What conditional state results after outcome $0$?

Solution

The refined operation for coarse outcome $0$ is

$$\mathcal I_0(\rho) = |0\rangle\langle0|\rho|0\rangle\langle0| + |1\rangle\langle1|\rho|1\rangle\langle1|.$$

For the given state, after normalization by $p(0)=2/3$ this gives

$$\rho_0 = \frac{1}{2}|0\rangle\langle0| + \frac{1}{2}|1\rangle\langle1|.$$

This differs from the Lüders pure state because the refined measurement erased the coherence between $|0\rangle$ and $|1\rangle$.

3. Let $M_0=|0\rangle\langle0|$ and $N_0=X M_0$, where $X$ swaps $|0\rangle$ and $|1\rangle$. Do $M_0$ and $N_0$ define the same effect for outcome $0$? Do they give the same conditional output state?

Solution

They define the same effect:

$$N_0^\dagger N_0 = M_0^\dagger X^\dagger X M_0 = M_0^\dagger M_0 = |0\rangle\langle0|.$$

But the conditional output differs. The $M_0$ update outputs $|0\rangle\langle0|$ when outcome $0$ occurs, while the $N_0$ update outputs $|1\rangle\langle1|$. The same effect gives the same probability, not the same backaction.

4. A calculation uses $\rho_m=\mathcal I_m(\rho)/\operatorname{Tr}\mathcal I_m(\rho)$ and then averages over $m$ with equal weights. What is wrong unless all outcomes are equally probable?

Solution

The nonselective state is the probability-weighted average:

$$\rho' = \sum_m p(m)\rho_m, \qquad p(m)=\operatorname{Tr}\mathcal I_m(\rho).$$

Equal weights are correct only if the outcome probabilities are equal. Otherwise the calculation describes a different ensemble from the one produced by the measurement.

References

• P. Busch, P. J. Lahti, and P. Mittelstaedt, The Quantum Theory of Measurement, 2nd ed., Springer (1996). Publisher.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), chapter 2. Author’s text.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010). Publisher.