Measurement as an Operation

A measurement produces a classical outcome and changes the quantum state used to predict later experiments. A useful model therefore answers two questions: what is the probability of each outcome, and what state remains after it? The first answer alone does not determine the second. We begin with finite-dimensional systems and finitely many reported outcomes; the same distinction extends to continuous records.

Helpful background. Density operators describe both pure preparations and statistical mixtures. You will use matrix multiplication and the trace to calculate probabilities.

An outcome operation carries probability and state

Write the unnormalized output for outcome $m$ as $\mathcal I_m(\rho)$. Its trace gives the probability; dividing by that trace gives the conditional state:

$$p(m)=\operatorname{Tr}\mathcal I_m(\rho),\qquad \rho_m=\frac{\mathcal I_m(\rho)}{p(m)},\quad p(m)>0.$$

The family $\{\mathcal I_m\}$ is called an instrument. Each operation is linear, completely positive, and trace nonincreasing. Complete positivity ensures that the operation remains physically valid when the input is entangled with an untouched auxiliary system. For a complete list of outcomes, the sum $\Phi=\sum_m\mathcal I_m$ preserves the trace.

If the detector acts but its result is ignored, the state is

$$\rho'=\sum_m\mathcal I_m(\rho)=\sum_m p(m)\rho_m.$$

This is a probability-weighted average, not an equal average unless the outcomes are equally likely. Zero-probability branches contribute zero and have no conditional state. The normalized selective update generally depends nonlinearly on $\rho$ because its denominator does; the unnormalized operation is linear.

Effects predict outcomes; instruments predict subsequent experiments

In a Kraus representation,

$$\mathcal I_m(\rho)=\sum_\alpha K_{m\alpha}\rho K_{m\alpha}^\dagger, \qquad F_m=\sum_\alpha K_{m\alpha}^\dagger K_{m\alpha}.$$

Cyclicity of the trace then gives $p(m)=\operatorname{Tr}(F_m\rho)$. The effects obey $F_m\geq0$ and $\sum_mF_m=I$. Together they form a POVM. The index $\alpha$ describes alternatives not distinguished by the reported outcome; it is not an additional measured label.

The operators $K_{m\alpha}$ may map the input Hilbert space into a different output space, as for a destructive detector. A no-click or failure result must be included when describing a complete measurement. Discarding it gives a trace-decreasing operation, not a trace-preserving channel.

See quantum instruments for the general formulation. Here is a concrete reason to keep effects and operations distinct.

Two detectors with the same error probability

Let $P_j=|j\rangle\langle j|$ for a qubit, and suppose the reported bit is wrong with probability $\epsilon$, where $0\leq\epsilon\leq1/2$:

$$F_0=(1-\epsilon)P_0+\epsilon P_1,\qquad F_1=\epsilon P_0+(1-\epsilon)P_1.$$

One device performs a sharp $z$ measurement and then sometimes flips its classical label. Its outcome-0 operation is

$$\mathcal I_0^{\rm sharp}(\rho) =(1-\epsilon)P_0\rho P_0+\epsilon P_1\rho P_1.$$

The outcome-1 operation interchanges $\epsilon$ and $1-\epsilon$. Even if the label is unreliable, the physical sharp measurement has already removed all $z$-basis coherence when outcomes are averaged.

A second device uses $K_m=\sqrt{F_m}$, with no additional hidden alternative. Its unread output is

$$\Phi_{\rm gentle}\!\begin{pmatrix}a&c\\c^*&1-a\end{pmatrix} =\begin{pmatrix}a&\lambda c\\\lambda c^*&1-a\end{pmatrix}, \qquad \lambda=2\sqrt{\epsilon(1-\epsilon)}.$$

To obtain $\lambda$, multiply each off-diagonal entry by the two diagonal elements of $K_0$, then add the identical contribution from $K_1$. Both devices give $p(0)=(1-\epsilon)a+\epsilon(1-a)$, but the second can retain coherence.

For $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$ and $\epsilon=1/4$, both report each bit with probability $1/2$. A later $x$ measurement gives $+$ with probability $1/2$ after the sharp device and

$$p_x(+)=\frac{1+\lambda}{2}=\frac{2+\sqrt3}{4}\simeq0.933$$

after the second device. At $\epsilon=1/2$ the label conveys no information: the second device leaves every state unchanged, while the sharp device still dephases it. Readout quality alone does not determine disturbance.

How an apparatus produces an instrument

Assume a fixed apparatus state $\eta_A$ initially uncorrelated with the system, a unitary coupling $U$, and a projective pointer readout $\{R_m\}$. Then

$$\mathcal I_m(\rho)=\operatorname{Tr}_A\big[(I_S\otimes R_m) U(\rho\otimes\eta_A)U^\dagger \times(I_S\otimes R_m)\big].$$

The coupling creates system–apparatus correlations; selecting a pointer result and tracing out the apparatus produces the conditional operation. The fixed initial apparatus state matters: arbitrary initial correlations cannot automatically be encoded by one channel acting on every possible system input.

The von Neumann measurement model develops this picture. For an ideal Lüders apparatus, $\mathcal I_m(\rho)=P_m\rho P_m$. Such an apparatus is immediately repeatable, but repeatability is an additional property, not part of the definition of every measurement.

These formulas predict records and later statistics. The unitary apparatus model by itself does not select one observed result or settle how to interpret state reduction; see what decoherence does not solve.

Exercises

1. Suppose $\mathcal I_m(\rho)=K_m\rho K_m^\dagger$ and $\sum_mK_m^\dagger K_m=I$. Show that the probabilities sum to one and that the unread output has unit trace.

Solution

Cyclicity gives $\sum_mp(m)=\operatorname{Tr}[\rho\sum_mK_m^\dagger K_m]=\operatorname{Tr}\rho=1$. The unread output is $\sum_mK_m\rho K_m^\dagger$, whose trace is the same sum. For a retained outcome with $p(m)>0$, divide its term by $p(m)$.

2. In the noisy-detector example, find the two conditional output states for reported outcome 0 when the input is $|+\rangle$ and $\epsilon=1/4$.

Solution

Both probabilities are $1/2$. The sharp device leaves

$$\rho_0^{\rm sharp}=\tfrac34|0\rangle\langle0|+\tfrac14|1\rangle\langle1|.$$

The second leaves the pure state $|\psi_0\rangle=(\sqrt3|0\rangle+|1\rangle)/2$. Both have the same $z$ populations, but only the second retains the off-diagonal element $\sqrt3/4$. Their purities are $5/8$ and $1$.

3. An unread Lüders $z$ measurement acts on $|+\rangle$. Compare a later $x$ measurement with the experiment in which the detector never acted.

Solution

The unread output is $(P_0+P_1)/2=I/2$, so $p_x(+)=1/2$. Without the detector, the state remains $|+\rangle$ and $p_x(+)=1$. Ignoring the record does not undo the interaction.

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), chapter 1. Publisher.