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.

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 by an ideal projective measurement with projectors $\{Q_b\}$:

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

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

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

This is the subsystem version of the projective state-update rule.

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

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

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 Lueders 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 is the bookkeeping heart of the distinction between conditioning and signaling.

General Measurement Preview

A general measurement on $B$ is not specified only by its outcome probabilities. To define conditional states, one needs a measurement operation, or instrument. In finite dimensions, a common representation uses operators $M_{b\mu}$ 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 $p(b)>0$.

This page uses this formula only as a preview. POVMs, instruments, channels, weak measurements, and realistic detector models belong in the measurement, decoherence, and open-systems volume.

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 is powerful because it can isolate correlations that are invisible in the unconditioned average. It is also probabilistic: discarded runs matter for resource accounting, and the success information must be communicated before the selected ensemble is known.

Steering Preview

Entanglement can allow measurements on $B$ to prepare different ensembles for the same reduced state of $A$. For $\lvert\Phi^+\rangle$, a $z$-basis measurement on $B$ gives the ensemble

$$\left\{ \frac12,\lvert0\rangle; \frac12,\lvert1\rangle \right\},$$

while an $x$-basis measurement gives

$$\left\{ \frac12,\lvert+\rangle; \frac12,\lvert-\rangle \right\}.$$

Both average to $I/2$. The fact that remote measurement choices can realize different ensemble decompositions of the same local density operator is the basic idea behind quantum steering.

This page only gives the algebraic preview. Steering inequalities, EPR arguments, and device-dependent assumptions belong in foundations and quantum information.

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 Mistakes

• Confusing $\rho_{A\vert b}$ with $\rho_A$. The first is conditioned on an outcome; the second is unconditioned.
• Forgetting to divide the unnormalized conditional state by $p(b)$.
• Treating postselection as if discarded trials did not matter.
• Thinking a remote measurement physically sends a controllable signal to $A$.
• Assuming POVM effects alone determine the post-measurement conditional state. Conditional states require an instrument, not only effects.
• Applying a selective update while saying the remote outcome is unknown.

Cross-Links

• Quantum Teleportation
• Local Measurement Statistics
• Purification
• Reduced Density Operators
• Partial Trace
• Bell States
• Classical Correlation versus Entanglement
• LOCC Preview
• Entanglement in Foundations
• Operators on Composite Systems
• State Update Rule
• Projective Measurement
• Born Rule

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• E. Schrodinger, “Discussion of Probability Relations between Separated Systems,” Mathematical Proceedings of the Cambridge Philosophical Society 31, 555-563, 1935.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• H. M. Wiseman, S. J. Jones, and A. C. Doherty, “Steering, Entanglement, Nonlocality, and the Einstein-Podolsky-Rosen Paradox,” Physical Review Letters 98, 140402, 2007.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.

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. Show that the two conditional states in the previous exercise average to $\rho_A=I/2$.

Solution

The average over outcomes is

$$\sum_b p(b)\rho_{A\vert b} = \frac12 \lvert0\rangle\langle0\rvert + \frac12 \lvert1\rangle\langle1\rvert = \frac12 I.$$

This equals the reduced density operator of either qubit in $\lvert\Phi^+\rangle$.

3. For the same Bell state, measure $B$ in the $x$ basis. What conditional states are assigned to $A$?

Solution

Use

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

The outcomes $+$ and $-$ each occur with probability $1/2$. The corresponding conditional states are

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

Their average is still $I/2$.

4. Explain why postselection on a rare outcome cannot be ignored in resource accounting.

Solution

If a desired outcome occurs with probability $p$, then only a fraction $p$ of the experimental runs remain after postselection. The selected state may have useful properties, but it is not produced deterministically. Any claim about rates, efficiencies, or signaling must include the discarded runs and the classical information telling which runs succeeded.

5. Why do POVM effects determine probabilities but not, by themselves, conditional states?

Solution

The effect $F_b$ determines the probability

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

The conditional state also depends on how the measurement physically updates the joint state. That update is specified by an instrument, for example by Kraus operators $M_{b\mu}$, not by $F_b$ alone. Different instruments can have the same effects but different post-measurement states.