Selective and Nonselective Measurements

Selective and nonselective measurements answer different prediction questions.

A selective measurement is conditioned on a known outcome. A nonselective measurement is the state transformation after the measurement interaction when the outcome is not retained, not read, or deliberately averaged over.

Helpful background. Density operators describe mixed states; measurement as an operation introduces the instrument maps used below.

The General Instrument Formula

Let a measurement have outcomes $m$ and outcome-resolved operations $\mathcal I_m$. Each $\mathcal I_m$ is completely positive and trace nonincreasing, and their sum is trace preserving. We use finite-dimensional systems and discrete outcomes here. The probability of outcome $m$ is

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

If $m$ is observed and $p(m)\ne0$, the selective state is

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

If the measurement is performed but the outcome is ignored, the nonselective output is

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

The same apparatus can therefore lead to two different state assignments:

$$\text{known record }m \quad\Rightarrow\quad \rho_m, \qquad \text{ignored record} \quad\Rightarrow\quad \sum_m p(m)\rho_m.$$

The nonselective state is an ensemble average of the selective states, weighted by their probabilities.

Projective Special Case

For the ideal Lüders implementation of a projective measurement with projectors $\{P_a\}$, the operation for outcome $a$ is

$$\mathcal I_a(\rho)=P_a\rho P_a.$$

The selective update is

$$\rho_a = \frac{P_a\rho P_a} {\operatorname{Tr}(P_a\rho)}.$$

The nonselective update is

$$\rho' = \sum_a P_a\rho P_a.$$

These formulas look similar, but they represent different information states. The selective state is used after learning a definite outcome. The nonselective state is used when the outcome information is unavailable or irrelevant.

Record Kept Versus Record Ignored

It is often useful to represent the classical record explicitly. If the outcome is stored in a classical register $R$ with orthonormal labels $\{|m\rangle_R\}$, then the system and a classical, dephased outcome register have the joint state

$$\rho_{SR}' = \sum_m p(m)\rho_m\otimes |m\rangle_R\langle m|.$$

If the record is read and equals $m$, the conditional system state is $\rho_m$. If the record exists but is ignored, the system state is obtained by tracing out the record:

$$\operatorname{Tr}_R\rho_{SR}' = \sum_m p(m)\rho_m.$$

This is a classical–quantum record state: it contains no off-diagonal terms between different record labels. A coherent system–pointer state before readout need not have this form. Keeping the record permits sorting later data by $m$; tracing it out describes all runs together.

Qubit Example

Consider an ideal $\sigma_z$ measurement of a qubit with projectors

$$P_+=|0\rangle\langle0|, \qquad P_-=|1\rangle\langle1|.$$

Let the initial state be

$$\rho= \begin{pmatrix} a & c\\ c^* & 1-a \end{pmatrix}, \qquad 0\le a\le1,\qquad |c|^2\le a(1-a).$$

The outcome probabilities are

$$p(+)=a, \qquad p(-)=1-a.$$

If $+$ is observed, the selective state is

$$\rho_+ = |0\rangle\langle0| = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}.$$

If $-$ is observed, the selective state is

$$\rho_- = |1\rangle\langle1| = \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}.$$

If the outcome is ignored, the nonselective output is

$$\rho' = P_+\rho P_+ + P_-\rho P_- = \begin{pmatrix} a & 0\\ 0 & 1-a \end{pmatrix}.$$

The bound on $c$ ensures positivity of the input density operator. Conditional states are defined only for outcomes with positive probability. The off-diagonal coherence $c$ is gone from the unread output. The nonselective measurement has acted as a complete dephasing channel in the $\sigma_z$ basis.

Why the Average Is Not the Original State

A common mistake is to think that if the outcome is ignored, nothing has happened. That is generally false. Ignoring the outcome does not undo the physical system-apparatus correlation.

Before the measurement, the coherence between $|0\rangle$ and $|1\rangle$ appears in $c$. After an ideal unread $\sigma_z$ measurement, that coherence is not available in the system alone. It has been transferred to, or destroyed by coupling with, the record-bearing degrees of freedom.

This is the same mechanism described in decoherence. When alternatives become correlated with distinguishable records and those records are ignored, local interference terms are suppressed.

No-Click Is an Outcome

In realistic measurements, an apparatus may have a “no click,” “lost,” or “inconclusive” result. If that result is recorded, it is still an outcome. Conditioning on no-click is selective:

$$\rho_{\varnothing} = \frac{\mathcal I_{\varnothing}(\rho)} {\operatorname{Tr}\mathcal I_{\varnothing}(\rho)}.$$

The nonselective state includes all outcomes, including no-click:

$$\rho' = \sum_{\text{clicks }m}\mathcal I_m(\rho) + \mathcal I_{\varnothing}(\rho).$$

Forgetting to include inconclusive outcomes is a frequent way to produce a trace-decreasing “state” where a normalized unconditional state was intended.

Sequential Predictions

The selective/nonselective distinction becomes unavoidable in sequential experiments. Suppose a first measurement has outcomes $m$ and a later measurement has effects $F_n$. If the first outcome is known to be $m$, then

$$p(n|m) = \operatorname{Tr}(F_n\rho_m).$$

If the first outcome is ignored, the later probability is

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

The two are related by $p(n)=\sum_m p(m)p(n|m)$. For the first exercise below, measuring $z$ again gives $p(0|0)=1$, while ignoring the first result gives $p(0)=3/4$. See sequential measurements for the joint-probability calculation.

Exercises

1. A qubit in state

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

is measured projectively in the $\{|0\rangle,|1\rangle\}$ basis. Find the selective states and the nonselective state.

Solution

The outcome probabilities are $p(0)=3/4$ and $p(1)=1/4$. The selective states are $|0\rangle\langle0|$ and $|1\rangle\langle1|$. The nonselective state is

$$\rho' = \frac{3}{4}|0\rangle\langle0| + \frac{1}{4}|1\rangle\langle1|.$$

2. Suppose an instrument has two outcomes with operations $\mathcal I_0$ and $\mathcal I_1$, but a calculation keeps only $\mathcal I_0(\rho)$ and normalizes it. Is that selective or nonselective?

Solution

It is selective, conditioned on outcome $0$. The nonselective state is $\mathcal I_0(\rho)+\mathcal I_1(\rho)$, assuming those are all outcomes. Normalizing only $\mathcal I_0(\rho)$ describes the subensemble in which outcome $0$ occurred.

3. Show that the nonselective output can be written as $\sum_m p(m)\rho_m$ whenever $p(m)\ne0$ for all included outcomes.

Solution

By definition,

$$\rho_m = \frac{\mathcal I_m(\rho)} {p(m)}.$$

Therefore $\mathcal I_m(\rho)=p(m)\rho_m$. Summing over $m$ gives

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

Outcomes with $p(m)=0$ contribute the zero unnormalized state and may be omitted from the weighted sum.

4. A qubit detector has $K_{\rm click}=\sqrt\eta\,|0\rangle\langle1|$ and $K_{\varnothing}=|0\rangle\langle0|+\sqrt{1-\eta}|1\rangle\langle1|$, with $0\leq\eta\leq1$. Starting from $|+\rangle$, find the no-click probability and state. Does “no click” mean nothing happened?

Solution

The two effects sum to $I$. The unnormalized no-click vector is $(|0\rangle+\sqrt{1-\eta}|1\rangle)/\sqrt2$, so

$$p(\varnothing)=1-\eta/2,\qquad |\psi_{\varnothing}\rangle= \frac{|0\rangle+\sqrt{1-\eta}|1\rangle}{\sqrt{2-\eta}}.$$

For $\eta>0$ this differs from $|+\rangle$: absence of a detected decay changes the relative weights. The click probability is $\eta/2$, with output $|0\rangle$ when it occurs. Averaging both unnormalized branches restores unit trace, not the original input state.

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.