advanced33 min studyLesson 369 in Physics

Quantum instruments and postmeasurement states

Specify both an outcome probability and its conditional output state, and compare distinct state updates with identical measurement effects.

Before you begin

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The question for today

What you’ll understand

Compute probabilities and conditional states using Kraus operators.
Distinguish an instrument's selective branches from its unconditioned channel.
Demonstrate that identical POVM effects can accompany different disturbances.

A POVM tells us how often each outcome occurs, but it does not say what remains of the measured system. A quantum instrument provides that extra information by assigning an output operation to each recorded outcome. Its unnormalized branches keep probability and state together in a linear description. Conditioning divides a branch by its probability, while ignoring the record sums the branches. These are distinct operations with different operational meanings.

A branch is a completely positive operation

For finite-dimensional systems, a measurement outcome can be represented by one or more Kraus operators acting on the input. Sum their output sandwiches to get an unnormalized positive state. Complete positivity guarantees that this remains physically valid when the measured system is correlated with an untouched auxiliary system. Positivity on isolated input matrices alone is weaker and would permit maps that fail on some entangled extensions.

The trace of a branch is its outcome probability. Summing the adjoint-product effects over every outcome and hidden Kraus label gives identity, making the total operation trace preserving. The output Hilbert space can differ from the input, so an instrument can include loss, conversion, or re-preparation rather than preserving a particle in an eigenstate. A one-Kraus-per-outcome formula is convenient but is not the most general instrument with the same classical outcome alphabet.

\mathcal I_a(\rho)=\sum_\mu K_{a\mu}\rho K_{a\mu}^\dagger,\qquad E_a=\sum_\mu K_{a\mu}^\dagger K_{a\mu},\qquad \sum_aE_a=I

Kraus operators are dimensionless maps from the input to the specified output space, and μ labels unobserved branches within one reported outcome a. The effects Ea act on the input space and determine its POVM. This finite-dimensional representation defines completely positive outcome operations.↔ Scroll wide equations horizontally.

Conditioning and forgetting use different normalization rules

If outcome a is reported with nonzero probability, normalize its branch to obtain the conditional state. This normalization makes the selected state depend nonlinearly on the initial density operator, even though the unnormalized branch map is linear. For a zero-probability outcome the conditional state is undefined by this rule, since the supposed event does not occur in the specified ideal model. Dividing by zero and choosing an arbitrary output would add unsupported information.

When no outcome is retained, add the unnormalized branches. Weighting already normalized conditional states by their probabilities gives the same channel output. Adding the conditional states without weights generally breaks normalization and predicts the wrong disturbance. An unread measurement can still alter the system: forgetting a record is not the same as undoing its interaction. The averaged effect on an untouched distant subsystem remains constrained by trace preservation and no signalling.

p(a)=\operatorname{tr}\mathcal I_a(\rho)=\operatorname{tr}(\rho E_a),\qquad \rho_a=\frac{\mathcal I_a(\rho)}{p(a)}\ \ (p(a)>0),\qquad \rho'=\sum_a\mathcal I_a(\rho)

Each p(a) is dimensionless, each conditional ρa has unit trace, and the unconditioned output ρ′ also has unit trace for a complete instrument. The branch trace combines selection probability with output normalization; the formulas do not assume a particular interpretation of measurement.↔ Scroll wide equations horizontally.

Same effects, different physical outputs

For a projective computational measurement, one implementation leaves each detected eigenstate unchanged. Another measures the same outcome but resets the output to zero regardless of that outcome. Both return the same zero/one frequencies on every input because their effects agree, yet their conditional states and their unread outputs differ. Repeatability is therefore an additional implementation property, not a conclusion obtainable from POVM probabilities alone.

Likewise, choosing the positive square root of each effect as its Kraus operator specifies one particular implementation, often a useful idealization. Extra outcome-dependent unitaries preserve the effects while rotating output states. More general hidden branches can mix those outputs. To predict a sequence of measurements, use the actual instrument before applying the next probability rule. A list of effects is enough for one round's counts but insufficient for the joint statistics of an unspecified measurement sequence.

Work it through

A readout followed by a reset

Compare instruments N with K0=|0⟩⟨0|, K1=|1⟩⟨1| and R with J0=|0⟩⟨0|, J1=|0⟩⟨1|. For input |+⟩, find their outcome probabilities, state after reported outcome 1, and output if the record is discarded.

Both instruments have effects $E_0=|0\rangle\langle0|$ and $E_1=|1\rangle\langle1|$ , since $J_1^\dagger J_1=|1\rangle\langle1|$ . The input has equal computational populations, so both instruments give p(0)=p(1)=1/2.
For N, the outcome-1 branch is one half $|1\rangle\langle1|$ , giving conditional output |1⟩. For R, that branch is one half $|0\rangle\langle0|$ , giving conditional output |0⟩ despite the reported label 1.
Sum the branches without conditioning. N gives $I/2$ , while R gives $|0\rangle\langle0|$ . A subsequent computational measurement therefore sees zero with probability 1/2 after unread N and with probability one after unread R.

The result

The first-round effects and probabilities agree, but the conditional outcome-1 states and unread outputs differ. A detector label does not uniquely determine the surviving quantum state without its instrument model.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

Compare two possible conditional outputs

Reset and set both controls to θ=180° and φ=0°; predict computational-one probability 1, representing the calculated N outcome-1 output. For a fresh comparison, Reset and set both controls to θ=0° and φ=0°; predict computational-zero probability 1, representing the calculated R output. These are separate pure-state preparations for comparison, not measurement updates performed by the lab.

Bloch vector / P(0)P(1)Phase latitude

Probability |0⟩: 71.1%
Probability |1⟩: 28.9%
Bloch (x, y, z): 0.78, 0.45, 0.42

Mixing angle θ65°Relative phase φ30°

t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. 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 forTwo instruments with the same first-round effects can leave different pure outputs. The instrument plots those states but contains no Kraus controls, random outcomes, conditional update, or unread mixed-state display.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The shared effects fix p(a)=tr(ρEa). They do not fix the maps taking ρ to output branches. Outcome-dependent unitaries or re-preparation can preserve the effects while changing conditional and unconditioned output states.

Your turn

Try it yourself

Apply K0=diag(√3/2,1/2) and K1=diag(1/2,√3/2) to input |+⟩. Find p(0), its conditional pure state, and the unconditioned density matrix.

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

Give me a hint

Compute each unnormalized ket, square its norm for the probability, then normalize or form both outer products before summing.

Compare with the solution

The outcome-0 ket is $(\sqrt3|0\rangle+|1\rangle)/(2\sqrt2)$ , with norm squared 1/2. Its normalized version is $(\sqrt3|0\rangle+|1\rangle)/2$ . The other normalized branch exchanges the two positive coefficients. Adding the unnormalized projectors gives $\rho'=\begin{pmatrix}1/2&\sqrt3/4\\\sqrt3/4&1/2\end{pmatrix}$ . The unread operation reduces the original off-diagonal entry from 1/2 to √3/4 while preserving the diagonal populations. Its purity is 7/8, consistent with partial rather than complete dephasing.

Keep the idea

Take it with you

A quantum instrument specifies completely positive outcome branches as well as probabilities.
Condition on a nonzero-probability outcome by normalization; discard the record by summing branches.
Equal POVM effects do not guarantee equal disturbance or equal sequential statistics.

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