Conditional States

A conditional state is the state assigned to one subsystem after an outcome has been obtained on another subsystem and that outcome is used as information. If a measurement on $B$ produces outcome $b$, the state assigned to $A$ can change from the reduced state $\rho_A$ to a conditional state $\rho_{A\vert b}$.

This is not the same as signaling. Before the outcome on $B$ is known, the local state on $A$ remains $\rho_A$. Conditioning changes the description available to someone who has learned the remote outcome. No-Cloning and No-Signaling proves the corresponding channel-level no-signaling theorem.

Helpful background. Partial trace explains contraction over the measured factor; local measurement statistics explains the unconditioned marginal.

Projective conditioning on the other subsystem

Let the joint state be $\rho_{AB}$ on

$$\mathcal H_A\otimes\mathcal H_B.$$

Suppose subsystem $B$ is measured with projectors $\{Q_b\}$ and the joint state follows the Lüders update:

$$Q_bQ_{b'} = \delta_{bb'}Q_b, \qquad \sum_b Q_b=I_B.$$

The probability of outcome $b$ is

$$p(b) = \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(I_A\otimes Q_b) \bigr] = \operatorname{Tr}_B(\rho_B Q_b),$$

where $\rho_B=\operatorname{Tr}_A\rho_{AB}$.

The unnormalized conditional state of $A$ is

$$\widetilde\rho_{A\vert b} = \operatorname{Tr}_B \bigl[ (I_A\otimes Q_b)\rho_{AB}(I_A\otimes Q_b) \bigr].$$

Its trace is the outcome probability:

$$\operatorname{Tr}_A\widetilde\rho_{A\vert b} = p(b).$$

If $p(b)>0$, the normalized conditional state is

$$\rho_{A\vert b} = \frac{\widetilde\rho_{A\vert b}}{p(b)}.$$

The same conditional state of the untouched subsystem $A$ results from any local instrument on $B$ having these effects. The joint state can depend on the instrument, as the general calculation below shows.

Rank-One Shortcut

If the outcome on $B$ is represented by a rank-one projector

$$Q_b = \lvert b\rangle\langle b\rvert,$$

then the unnormalized conditional state can be written as

$$\widetilde\rho_{A\vert b} = (I_A\otimes\langle b\rvert) \rho_{AB} (I_A\otimes\lvert b\rangle).$$

For a pure joint state

$$\lvert\Psi\rangle_{AB} = \sum_b \lvert\psi_b\rangle_A \lvert b\rangle_B,$$

where the vectors $\lvert\psi_b\rangle_A$ need not be normalized, measurement of $B$ in the $\{\lvert b\rangle\}$ basis gives

$$p(b) = \langle\psi_b|\psi_b\rangle, \qquad \rho_{A\vert b} = \frac{ \lvert\psi_b\rangle\langle\psi_b\rvert }{ \langle\psi_b|\psi_b\rangle }.$$

For a nonzero-probability branch, the conditional state is pure in this rank-one projection case when the joint pre-measurement state is pure.

Bell-State Example

For

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr),$$

the reduced state of either qubit is $I/2$. If qubit $B$ is measured in the computational basis, then

$$p(0)=p(1)=\frac12.$$

Conditioned on the outcome,

$$\rho_{A\vert0} = \lvert0\rangle\langle0\rvert, \qquad \rho_{A\vert1} = \lvert1\rangle\langle1\rvert.$$

If the outcome is not known or is ignored, the average state is

$$\frac12 \lvert0\rangle\langle0\rvert + \frac12 \lvert1\rangle\langle1\rvert = \frac12 I.$$

Thus the conditional state can be sharp even though the unconditioned reduced state is maximally mixed.

Different Remote Measurements, Same Local Average

The same Bell state can also be written in the $x$ basis:

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert++\rangle+\lvert--\rangle \bigr),$$

where

$$\lvert\pm\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle\pm\lvert1\rangle \bigr).$$

If $B$ is measured in the $x$ basis, then

$$\rho_{A\vert +} = \lvert+\rangle\langle+\rvert, \qquad \rho_{A\vert -} = \lvert-\rangle\langle-\rvert,$$

with probabilities $1/2$ and $1/2$. The average is again

$$\frac12 \lvert+\rangle\langle+\rvert + \frac12 \lvert-\rangle\langle-\rvert = \frac12 I.$$

The remote measurement choice changes which ensemble decomposition of $\rho_A$ is available after the outcome is communicated. It does not change the unconditioned state $\rho_A$.

Nonselective Measurement

If a projective measurement on $B$ is performed but its outcome is not selected, the post-measurement joint state in the ideal Lüders model is

$$\rho'_{AB} = \sum_b (I_A\otimes Q_b)\rho_{AB}(I_A\otimes Q_b).$$

The reduced state of $A$ is unchanged:

$$\operatorname{Tr}_B\rho'_{AB} = \operatorname{Tr}_B\rho_{AB} = \rho_A.$$

Equivalently,

$$\sum_b p(b)\rho_{A\vert b} = \rho_A.$$

This identity explains why different selected states can coexist with an unchanged unconditioned marginal. Terms with zero probability are omitted rather than assigned a normalized conditional state.

Conditional states from general measurement effects

A measurement instrument specifies how a device changes the measured and joint states. In finite dimensions, write its branch operators as $M_{b\mu}$ on $\mathcal H_B$, satisfying

$$\sum_{b,\mu} M_{b\mu}^\dagger M_{b\mu} = I_B.$$

The effect associated with outcome $b$ is

$$F_b = \sum_\mu M_{b\mu}^\dagger M_{b\mu}.$$

The probability of outcome $b$ is

$$p(b) = \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(I_A\otimes F_b) \bigr].$$

The unnormalized conditional state of $A$ is

$$\widetilde\rho_{A\vert b} = \operatorname{Tr}_B \left[ \sum_\mu (I_A\otimes M_{b\mu}) \rho_{AB} (I_A\otimes M_{b\mu}^\dagger) \right].$$

Then

$$\rho_{A\vert b} = \frac{\widetilde\rho_{A\vert b}}{p(b)}$$

when $p(b)>0$.

Partial-trace cyclicity on $B$ gives a simplification:

$$\begin{aligned} \widetilde\rho_{A|b} &=\operatorname{Tr}_B\left[\rho_{AB}\left(I_A\otimes\sum_\mu M_{b\mu}^\dagger M_{b\mu}\right)\right]\\ &=\operatorname{Tr}_B[\rho_{AB}(I_A\otimes F_b)]. \end{aligned}$$

For a fixed initial joint state, the conditional state of the untouched partner therefore depends only on $F_b$. The measured subsystem’s state, or the full postmeasurement joint state, generally needs the instrument as well.

For $|\Phi^+\rangle$, the particularly useful identity is

$$\widetilde\rho_{A|b}=\frac{F_b^{\mathsf T}}2,\qquad p(b)=\frac{\operatorname{Tr}F_b}{2},\qquad \rho_{A|b}=\frac{F_b^{\mathsf T}}{\operatorname{Tr}F_b}.$$

The transpose is in the computational basis used to define $|\Phi^+\rangle$. For a rank-one effect proportional to $|b\rangle\langle b|$, the remote vector is $|b^*\rangle$, not generally $|b\rangle$. This complex conjugation explains why the Bell pair has equal $X$ outcomes but opposite $Y$ outcomes.

Postselection

Postselection means keeping only runs in which a specified outcome or set of outcomes occurred. If $S$ is a set of outcomes on $B$, then

$$p(S) = \sum_{b\in S}p(b).$$

The conditional state of $A$ after postselecting on $S$ is

$$\rho_{A\vert S} = \frac{ \sum_{b\in S}p(b)\rho_{A\vert b} }{ p(S) }, \qquad p(S)>0.$$

Postselection isolates a subensemble that can differ strongly from the average. It is probabilistic: include the success probability when predicting preparation rates, and communicate the successful-run labels before comparing the selected data.

Conditioning and quantum steering

Different remote settings can select different ensembles of the same marginal. For a Bell pair, $Z_B$ selects $|0\rangle,|1\rangle$ while $X_B$ selects $|+\rangle,|-\rangle$, each with equal weights. Both average to $I/2$.

This motivates steering, but different conditional ensembles alone do not demonstrate it. Even the separable state $\rho_{\mathrm{cc}}=(|00\rangle\langle00|+|11\rangle\langle11|)/2$ yields pure $Z$ states after a $Z_B$ measurement and $I/2$ after either $X_B$ outcome.

To demonstrate steering from $B$ to $A$, the collection of conditional states must rule out a single local-state explanation across all tested settings $y$:

$$\widetilde\rho_{A|b,y} =\sum_\lambda q_\lambda\,p(b|y,\lambda)\,\tau_A^{(\lambda)}.$$

Here $q_\lambda$ is a probability distribution and the states $\tau_A^{(\lambda)}$ are independent of $y$. The model says that $A$ already carries one of these local states, and $B$ merely supplies setting-dependent classical labels. Separable states admit such a model. The Bell example can rule it out with suitable incompatible measurements; a formal test must include the measurement and trust assumptions. This is the distinction formalized by Wiseman, Jones, and Doherty (2007).

Conditioning Is Not Signaling

Conditioning requires knowing the outcome. Before the outcome on $B$ is communicated, an observer with access only to $A$ assigns the unconditioned state

$$\rho_A = \sum_b p(b)\rho_{A\vert b}.$$

Every local measurement on $A$ is computed from this $\rho_A$. Therefore the local observer cannot tell which measurement was chosen on $B$, nor which outcome occurred, from local data alone.

After classical communication, the observer can sort the data into subensembles labeled by $b$. Those subensembles can have different states and different statistics. The communication is ordinary classical communication, not faster-than-light signaling.

Common pitfalls

Effects suffice for the untouched partner, not for every postmeasurement state. Trace out the measured output before deciding which details of the instrument matter.

A zero-probability result has no normalized conditional state. Keep the unnormalized branch, which is zero, and omit it from averages over normalized states.

Classical conditioning is not automatically steering. Correlated separable states also have outcome-dependent states. Steering requires excluding a common local-state model across measurement settings.

References

• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, chapters 2 and 8.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, section 2.3.1, equations (2.226)–(2.228), for conditional states of unmeasured registers; chapter 6 for entanglement. Author-hosted text.
• Howard M. Wiseman, S. J. Jones, and Andrew C. Doherty, “Steering, entanglement, nonlocality, and the Einstein–Podolsky–Rosen paradox,” Physical Review Letters 98, 140402 (2007). DOI; preprint.

Exercises

1. Compute the conditional states of $A$ when $B$ is measured in the computational basis for $\lvert\Phi^+\rangle$.

Solution

The state is

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr).$$

If $B$ gives outcome $0$, the unnormalized state of $A$ is $\lvert0\rangle/\sqrt2$, so $p(0)=1/2$ and

$$\rho_{A\vert0} = \lvert0\rangle\langle0\rvert.$$

If $B$ gives outcome $1$, then $p(1)=1/2$ and

$$\rho_{A\vert1} = \lvert1\rangle\langle1\rvert.$$

2. Let $|\Psi_p\rangle=\sqrt p\,|00\rangle+\sqrt{1-p}\,|11\rangle$, with $0\le p\le1$. Measure $B$ using the effects $F_\pm=(I\pm\eta X)/2$, where $0\le\eta\le1$. Find the outcome probabilities, the conditional states of $A$, and their average. When is each conditional state pure?

Solution

The reduced state of $B$ is diagonal, so $\operatorname{Tr}(\rho_B X)=0$ and both probabilities are $1/2$. Expanding the joint projector in $\operatorname{Tr}_B[\rho_{AB}(I\otimes F_\pm)]$ gives

$$\widetilde\rho_{A|\pm}=\frac12 \begin{pmatrix} p&\pm\eta\sqrt{p(1-p)}\\ \pm\eta\sqrt{p(1-p)}&1-p \end{pmatrix}.$$

Normalization removes the factor $1/2$. Averaging the two normalized states cancels their opposite coherences and returns $\rho_A=\operatorname{diag}(p,1-p)$. Their determinants are $p(1-p)(1-\eta^2)$, so they are pure exactly when $\eta=1$ or $p\in\{0,1\}$. At $\eta=0$ the outcome carries no information and both conditional states equal the marginal. An imperfect measurement of one part of a pure entangled state can therefore leave the other part mixed even after its outcome is known.

3. For $|\Phi^+\rangle$, measure $B$ in the $Y$ basis $|y_\pm\rangle=(|0\rangle\pm i|1\rangle)/\sqrt2$. Find the conditional states on $A$.

Solution

For outcome $y_+$, contract with $\langle y_+|=(\langle0|-i\langle1|)/\sqrt2$:

$$(I\otimes\langle y_+|)|\Phi^+\rangle =\frac{|0\rangle-i|1\rangle}{2} =\frac{|y_-\rangle}{\sqrt2}.$$

Thus the probability is $1/2$ and the conditional state is $|y_-\rangle$. The other outcome gives $|y_+\rangle$ with probability $1/2$. The average is $I/2$ and the outcomes are anticorrelated.

4. A desired postselection outcome has probability $p=0.01$ per independent run. What is the expected number of trials until the first success, and the probability of at least one success in 100 trials?

Solution

The waiting time is geometrically distributed, so its mean is $1/p=100$. The probability of at least one success in 100 trials is $1-(1-p)^{100}=1-0.99^{100}\approx0.634$. An expected waiting time of 100 trials is not a guarantee of success by trial 100.

5. On $B$, compare branch operators $M_0=|0\rangle\langle0|$ and $N_0=|1\rangle\langle0|$ applied to $|\Phi^+\rangle$. They have the same effect. Which conditional states agree and which differ?

Solution

Both satisfy $M_0^\dagger M_0=N_0^\dagger N_0=|0\rangle\langle0|$ and occur with probability $1/2$. The normalized joint outputs are respectively $|00\rangle$ and $|01\rangle$. Thus the measured $B$ outputs differ, but the untouched $A$ output is $|0\rangle\langle0|$ in both cases. This directly illustrates why the effect determines the remote conditional state but not the full state update.