Chapter 02

Measurement: probabilities and state changes

Two measuring devices give exactly the same statistics on every qubit you send them. Both report 0 three times in four for our test source. But after one device reports 0, the qubit is left in state 0; after the other reports 0, it is left in state 1. How can the readings agree while the states left behind differ?

Before you start: Chapter 1: amplitudes, the Born rule, the computational basis, and two-qubit tensor products. For a slower account of measurement in general, see the measurement essay in the companion guide. Notation guide.

One preparation, two branches

Take the state from Chapter 1, $|\psi\rangle=(\sqrt3|0\rangle+i|1\rangle)/2$, and measure it in the computational basis. The Born rule gives two branches: the reading 0 with probability $3/4$, and the reading 1 with probability $1/4$. That answers one question. A second question is what the qubit is afterwards, if it still exists: a photon counter, for example, absorbs the photon it counts.

For an ideal measurement, the qubit is left in $|0\rangle$ after the reading 0 and in $|1\rangle$ after the reading 1. The evidence is operational. Repeat the same measurement immediately and it gives the same reading every time. The state after a reading is called the conditional state: conditional on that outcome having occurred and been recorded.

The probabilities come from the preparation and the question asked. The conditional state also depends on what the apparatus does to the system, so it is a separate piece of information. This chapter makes that separation precise, first for ideal measurements and then for general ones.

The same reading, a different state

Call the ideal device M. Make a second device, N, by adding one operation: after recording the result, it swaps the qubit's states $|0\rangle$ and $|1\rangle$. The recorded result stays as it was. A reading of 0 therefore leaves $|0\rangle$ under M and $|1\rangle$ under N.

The added swap happens after the reading, so it cannot change how often that reading occurred. Both devices give 0 with probability $3/4$ on our source. Yet a second measurement can tell their outputs apart. Send the qubit through the same device again: M repeats its reading, while N gives the opposite reading and swaps the state once more.

There are two things to describe: which result appears, and what remains available for the next operation. We will now put numbers to both. Keeping M and N in mind gives each new piece of notation a job.

Projectors and conditional states

For the reading 0, keep the $|0\rangle$ component of the input and remove the $|1\rangle$ component. In column-vector form this is just

$$\begin{pmatrix}1&0\\0&0\end{pmatrix} \begin{pmatrix}\sqrt3/2\\i/2\end{pmatrix} =\begin{pmatrix}\sqrt3/2\\0\end{pmatrix}.$$

The matrix is $P_0=|0\rangle\langle0|$. Applying it again changes nothing: the unwanted component has already been removed. This is the defining action of a projector. In general a projector satisfies $P^\dagger=P$ and $P^2=P$ and keeps the component in a chosen subspace.

For the other reading, $P_1=|1\rangle\langle1|$ keeps the other component. Together they account for the entire input: $P_0+P_1=I$. Their outputs are orthogonal, since the two retained components point along different basis directions.

An ideal projective measurement generalises this pair to projectors $P_m$, one per outcome. They satisfy $\sum_mP_m=I$ and $P_mP_n=\delta_{mn}P_m$, where $\delta_{mn}$ is one for equal labels and zero otherwise. The retained vector tells us both how likely the outcome was and what state remains:

$$p(m)=\langle\psi|P_m|\psi\rangle,\qquad |\psi_m\rangle=\frac{P_m|\psi\rangle}{\sqrt{p(m)}}.$$

In the example, $P_0|\psi\rangle=(\sqrt3/2)|0\rangle$ has squared norm $3/4$, which is $p(0)$. Dividing by $\sqrt{3/4}$ gives the conditional state $|0\rangle$. The second expression is defined only for an outcome with nonzero probability; its denominator restores norm one. This is the ideal (Lüders) update. The same outcome statistics can be accompanied by additional physical disturbance, as device N shows.

Repeating the ideal measurement after outcome 0 gives 0 with certainty because $P_0^2=P_0$. A degenerate measurement, one with a projector of rank greater than one, behaves differently from measuring a complete basis. On two qubits, even parity uses $P_{\rm even}=|00\rangle\langle00|+|11\rangle\langle11|$. It preserves $\alpha|00\rangle+\beta|11\rangle$ unchanged; separate measurements of the two bits would destroy the coherence between those terms.

Recorded and ignored outcomes

A conditional state is the description for someone who knows the outcome. Suppose instead that the measurement is made and its reading discarded unread. Later predictions must then average over the branches: each branch's probability times the prediction made from its conditional state. Probabilities are averaged, not amplitudes, because the apparatus now holds a record that distinguishes the branches, whether or not anyone reads it.

Take $|+\rangle$, source A of Chapter 1. Measured in the $+/-$ basis it gives $+$ every time. Now let it first pass through a computational-basis measurement whose reading is thrown away. The branch $|0\rangle$ gives $+$ with probability $1/2$, and so does the branch $|1\rangle$, so the average is $1/2$. For every later measurement on the qubit alone, the unread measurement has made source A equivalent to source C: the relative phase that produced the certain outcome is gone from the qubit. The qubit and apparatus together are a different matter. The apparatus still holds a record correlated with the qubit, and a measurement that included the apparatus could detect that correlation. No single state vector describes the qubit on its own. Chapter 3 constructs the object that does; this chapter needs only the averaging rule.

Introducing a measuring apparatus

To see where conditional states come from, model the apparatus as a second quantum system. Start with a concrete case. A pointer qubit $A$ starts in $|0\rangle_A$ and interacts with the system through a unitary, a reversible evolution that preserves inner products (Chapter 4 treats these). This one flips the pointer when the system is 1:

$$|0\rangle|0\rangle_A\mapsto|0\rangle|0\rangle_A,\qquad |1\rangle|0\rangle_A\mapsto|1\rangle|1\rangle_A.$$

By linearity, the running example becomes $(\sqrt3/2)|0\rangle|0\rangle_A+(i/2)|1\rangle|1\rangle_A$. Reading the pointer gives 0 with probability $3/4$ and leaves the system in $|0\rangle$: the ideal measurement, obtained from an interaction and a reading of the pointer. Chapter 4 calls this interaction controlled-NOT.

From a joint interaction to a measurement operator

In general, prepare an auxiliary system in $|0\rangle_A$, couple it to the system by a unitary $U$, and read its orthonormal pointer basis $\{|m\rangle_A\}$. Define an operator on the original system by taking the apparatus matrix element,

$$M_m=(I\otimes\langle m|_A)U(I\otimes|0\rangle_A).$$

There are three spaces to distinguish in this expression. $I\otimes|0\rangle_A$ attaches the prepared apparatus to any system vector. $U$ acts on the resulting joint vector. Finally, $I\otimes\langle m|_A$ extracts the component with pointer value $m$, leaving a vector in the system space. Their composition is therefore a linear map from the system space back to itself: the operator $M_m$.

For example, if the joint output has the form $|v_0\rangle_S|0\rangle_A+|v_1\rangle_S|1\rangle_A$, applying $I\otimes\langle0|_A$ leaves $|v_0\rangle_S$. The other term vanishes because $\langle0|1\rangle=0$. The vectors $v_0,v_1$ need not be normalised: their squared norms are the two pointer probabilities. This is why the selected branch must later be divided by its norm. In the running example, $|v_0\rangle=(\sqrt3/2)|0\rangle$ and $|v_1\rangle=(i/2)|1\rangle$, so $M_0=|0\rangle\langle0|$ and $M_1=|1\rangle\langle1|$.

We can describe the effect on the qubit without carrying the pointer through every line. For each reading $m$, a measurement operator $M_m$ returns the qubit's unnormalised branch vector. For M, these are $M_0=P_0$ and $M_1=P_1$: they keep the component associated with the reading. The optional calculation above obtains these operators directly from the joint interaction.

The branch vector carries two pieces of information. Its squared length is the probability of getting that reading. Its direction, after restoring length one, is the conditional state. These are the same two steps we used for the projector.

The joint output is $\sum_m M_m|\psi\rangle\otimes|m\rangle_A$. Its norm is one for every normalised input, so $\sum_mM_m^\dagger M_m=I$. Projecting the apparatus onto $m$ selects the vector $M_m|\psi\rangle$. Therefore

$$p(m)=\langle\psi|M_m^\dagger M_m|\psi\rangle,\qquad |\psi_m\rangle=\frac{M_m|\psi\rangle}{\sqrt{p(m)}}.$$

The operators $E_m=M_m^\dagger M_m$ are called effects. Each is positive, written $E_m\ge0$, meaning $\langle v|E_m|v\rangle\ge0$ for every vector $v$. Their collection is a positive operator-valued measure, or POVM: $E_m\ge0$ and $\sum_mE_m=I$. These conditions guarantee nonnegative probabilities summing to one. Effects need not be orthogonal projectors, and there may be more effects than the system dimension.

Completeness from the apparatus basis, without probabilities

Insert the full apparatus basis between $U^\dagger$ and $U$:

$$\begin{aligned}\sum_mM_m^\dagger M_m &=(I\otimes\langle0|)U^\dagger\left(I\otimes\sum_m|m\rangle\langle m|\right)U(I\otimes|0\rangle)\\ &=(I\otimes\langle0|)U^\dagger U(I\otimes|0\rangle)=I. \end{aligned}$$

The first simplification uses an exhaustive list of pointer outcomes. The second uses reversible joint evolution and a normalised initial apparatus. The general measurement rule thus follows from ordinary evolution and projective readout on a larger system.

Why effects do not determine disturbance

Return to device N, which swaps the qubit after recording the reading. The swap is the bit flip $X$ introduced in Chapter 1. Its operators are $N_m=XM_m$: $N_0=|1\rangle\langle0|$ and $N_1=|0\rangle\langle1|$. The effects are unchanged, since $N_m^\dagger N_m=M_m^\dagger X^\dagger XM_m=M_m^\dagger M_m$. Every input therefore gives the same outcome probabilities under N as under M. For the running example N reads 0 with probability $3/4$, as M does, but leaves $|1\rangle$. An immediate second reading by N gives 1, not 0.

This answers the opening question. The probability of reading 0 was fixed by the preparation and the effect $E_0=|0\rangle\langle0|$. The state left behind was fixed by the apparatus, and two apparatuses with the same effects can leave different states. A POVM specifies outcome probabilities, not a unique state update.

In general $M_m=V_m\sqrt{E_m}$, where $\sqrt{E_m}$ is the positive matrix whose square is $E_m$ and $V_m$ is an outcome-dependent unitary; every such choice gives the same effects and potentially different conditional states. If one recorded outcome groups several unobserved apparatus outcomes, it requires several operators rather than one; Chapter 13 gives that update.

A measurement allowed to answer “inconclusive”

General effects allow measurements that no projective measurement on the qubit can perform. Suppose the input is either $|0\rangle$ or $|+\rangle$, with equal prior probabilities. These are not orthogonal, so no measurement identifies them every time without error. We want some outcomes that identify the state without error, while allowing other runs to report failure. Choose

$$E_+=c|1\rangle\langle1|,\quad E_0=c|-\rangle\langle-|,\quad E_?=I-E_+-E_0.$$

The label on each effect is the answer the device reports. $E_+$ has zero probability on $|0\rangle$, so seeing it identifies $|+\rangle$ without a mistake. Similarly, $E_0$ rules out $|+\rangle$. The remaining effect collects the runs where the device cannot decide.

The factor $c$ sets how often we get a conclusive answer. It cannot be arbitrarily large: all three effects must give nonnegative probabilities on every possible input. The calculation below gives $c=2-\sqrt2$, for which the correct answer occurs with probability $1-1/\sqrt2\approx0.293$. The remaining probability, about $0.707$, belongs to the inconclusive answer. We optimise this symmetric construction, without proving optimality over all measurements.

How positivity sets the success probability (used in Problem 2)

The label on an effect is the conclusion. $E_+$ cannot occur on input $|0\rangle$, and $E_0$ cannot occur on input $|+\rangle$. Each conclusive result has conditional probability $c/2$ on the input it identifies. To find an allowed $c$, write

$$|1\rangle\langle1|+|-\rangle\langle-|=\begin{pmatrix}1/2&-1/2\\-1/2&3/2\end{pmatrix}.$$

Its characteristic polynomial is $\lambda^2-2\lambda+1/2$, so its eigenvalues are $1\pm1/\sqrt2$. Positivity of $E_?$ requires $1-c(1+1/\sqrt2)\ge0$. Taking the largest allowed value in this family gives $c=2-\sqrt2$, and hence $p(\text{success})=1-1/\sqrt2\approx0.293$. Failure is explicitly reported; it is not an incorrect identification. We have optimised this symmetric construction, without proving optimality over all measurements.

Every finite POVM has an apparatus construction

The apparatus model produced a POVM from an interaction and a pointer reading. The converse also holds: any finite collection of positive effects summing to $I$ can be realised by some interaction with an auxiliary system followed by a projective reading of that system. So the three-effect measurement above is physically allowed, not only mathematically consistent.

The construction

Given effects, define $V|\psi\rangle=\sum_m\sqrt{E_m}|\psi\rangle\otimes|m\rangle$. Then $V^\dagger V=\sum_mE_m=I$, so $V$ preserves inner products. In finite dimensions it can be extended to a unitary on a sufficiently large space by completing an orthonormal basis. Reading the apparatus realises the required POVM. This is a dilation construction. The original system is a tensor factor; identifying it with the initial subspace $\mathcal H_S\otimes|0\rangle_A$ is part of the construction, not a new physical postulate.

Where this leaves us

After reading 0, the running example is in $|0\rangle$ if the measurement is ideal, and in $|1\rangle$ under apparatus N, although both apparatuses read 0 with the same probability $3/4$. The outcome probabilities are fixed by the effects; the conditional state needs the measurement operators, which is more information. When a reading is made but not kept, predictions average over the branches. For measurements on the qubit alone, $|+\rangle$ then behaves like a preparation that no single state vector describes, while the apparatus keeps a correlated record.

That is source C again. It is also what a subsystem looks like when its partner is out of reach, as later chapters on entanglement need. The next chapter gives both situations one description.

Problems

1. Apply even/odd parity measurement to $|++\rangle$. Find both outcome probabilities and normalised conditional states.

Worked answer

Expand the input before applying either projector: $|++\rangle=(|00\rangle+|01\rangle+|10\rangle+|11\rangle)/2$. Even parity keeps the equal-bit terms:

$$P_{\rm even}|++\rangle=(|00\rangle+|11\rangle)/2,\qquad p_{\rm even}=1/4+1/4=1/2.$$

Divide this vector by $\sqrt{1/2}$ to obtain $(|00\rangle+|11\rangle)/\sqrt2$. Odd parity similarly gives unnormalised $(|01\rangle+|10\rangle)/2$, probability $1/2$, and normalised $(|01\rangle+|10\rangle)/\sqrt2$. Both branches retain a superposition because parity does not distinguish the two terms within the selected subspace.

The tempting wrong turn is to conclude that after the reading “even” the bits are individually $00$ or $11$. The measurement reports only that they agree. The conditional state keeps both terms with their amplitudes, and a later $X\otimes X$ measurement on it gives $+1$ with certainty, which neither $|00\rangle$ nor $|11\rangle$ alone would do.

2. Is $c=1$ allowed in the three-effect example?

Worked answer

Let $A=|1\rangle\langle1|+|-\rangle\langle-|$, whose eigenvalues are $1\pm1/\sqrt2$. The inconclusive effect is $E_?=I-cA$, so on an eigenvector of $A$ with eigenvalue $\lambda$ its eigenvalue is $1-c\lambda$.

$$\lambda_{\min}(E_?)=1-c(1+1/\sqrt2)\xrightarrow{c=1}-1/\sqrt2.$$

A normalised eigenvector with that eigenvalue would have outcome probability $\langle v|E_?|v\rangle=-1/\sqrt2$. Thus $c=1$ is invalid even though the three effects still sum to identity: positivity is the failing condition.

References and further reading

Nielsen & Chuang, §2.2.3–2.2.6 (measurement operators, state discrimination, projective measurements, POVMs).