Quantum Measurement

A system in \(\ket\psi\) meets an ancilla in \(\ket0\), and a joint unitary \(U\) acts. Then we read the ancilla in its computational basis. Which outcome do we see, and what is the system's state afterwards?

Sort the joint output by the ancilla's label, exactly as for a channel:

\[U\ket\psi\ket0=\sum_j\bigl(M_j\ket\psi\bigr)\ket j,\qquad M_j=\bra jU\ket0.\]

Reading the ancilla is a projective measurement on it. The projector for outcome \(j\) keeps the term \((M_j\ket\psi)\ket j\), so outcome \(j\) has probability \(p_j=\|M_j\ket\psi\|^2\). Given that outcome, the system is left in \(M_j\ket\psi/\sqrt{p_j}\).

For a mixed input we average over an ensemble, as before. The unnormalized branch is \(M_j\rho M_j^\dagger\). Its trace is the probability, and dividing by it gives the conditional state:

\[p_j=\mathrm{tr}\bigl(M_j\rho M_j^\dagger\bigr),\qquad\rho_j=\frac{M_j\rho M_j^\dagger}{p_j}\quad(p_j>0).\]

The \(M_j\) are the measurement operators of the procedure. They are the Kraus operators of the channel we would get by discarding the ancilla. The only difference is that we now keep the record.

Does the general rule agree with the measurement postulate when it should? Take the CNOT record. Its measurement operators are \(M_0=\ket0\bra0\) and \(M_1=\ket1\bra1\), the projectors of the Z measurement. So reading the ancilla carries out a Z measurement of the system, with the probabilities and updates of the postulate.

Follow a plus input. CNOT produces a Bell pair. Reading the ancilla gives 0 or 1, each with probability \(1/2\). On the 0 branch the system is \(\ket0\), and on the 1 branch it is \(\ket1\). What we should predict next depends on what we keep. If we keep the label, we use the branch state. If we ignore the label, we average the branches and get \(I/2\).

The mixed-input rule behaves the same way. Dividing \(M_j\rho M_j^\dagger\) by its trace is the density-operator version of dividing a projected vector by its length. A branch with zero probability has no conditional state, because nothing is left to normalize.

A Z detector misreports its result one time in four. It might, for instance, measure Z perfectly and then flip the displayed result on a quarter of the trials. On input \(\rho\) it reports 0 with probability \(\tfrac34\rho_{00}+\tfrac14\rho_{11}\). Is this a projective measurement in some basis?

The probability is linear in \(\rho\), so we can write it with a matrix. Let \(E_0=\mathrm{diag}(3/4,1/4)\), where diag lists the diagonal entries of a matrix that is zero elsewhere:

\[p_0=\mathrm{tr}(E_0\rho),\qquad p_1=\mathrm{tr}(E_1\rho),\qquad E_1=I-E_0.\]

The probabilities for all inputs fix \(E_0\): the inputs \(\ket0\) and \(\ket1\) give its diagonal entries, and inputs such as \(\ket+\) give the rest. A projector \(P\) satisfies \(P^2=P\), because keeping a component twice is the same as keeping it once. But \(E_0^2=\mathrm{diag}(9/16,1/16)\neq E_0\). So no projective measurement, in any basis, has these statistics.

How does such a matrix arise from measurement operators? Moving \(M_j^\dagger\) around the trace, as we did for channels, gives \(p_j=\mathrm{tr}(M_j\rho M_j^\dagger)=\mathrm{tr}(M_j^\dagger M_j\rho)\). So the probabilities use only

\[E_j=M_j^\dagger M_j.\]

This matrix is the effect of outcome \(j\). Different measurement operators can share an effect. Replacing \(M_j\) by \(VM_j\) for any unitary \(V\) leaves \(E_j\) unchanged, because \(V^\dagger V=I\).

Which matrices can be effects? Each probability is nonnegative, and \(\bra v E_j\ket v=\|M_j\ket v\|^2\geq0\), so every effect is positive. The probabilities add to one for every input. The ancilla construction guarantees this, since \(\sum_jM_j^\dagger M_j=\bra0U^\dagger U\ket0=I\). A list of positive matrices adding to \(I\) is called a positive-operator-valued measure, or POVM.

A projective measurement is the special case in which every effect is a projector. A projector has eigenvalues 0 and 1 only. The eigenvalues of an effect can lie anywhere from 0 to 1, like the \(3/4\) and \(1/4\) of our detector.

Every ancilla construction gives a POVM. Can every POVM be built from an ancilla? Try the noisy detector. The flip-the-display version measures Z fully and leaves \(\ket0\) or \(\ket1\). We look for a device with the same effects that disturbs the qubit less.

First we need measurement operators with \(M_j^\dagger M_j=E_j\). For this diagonal effect we can take square roots of the diagonal entries: \(M_0=\mathrm{diag}(\sqrt3/2,1/2)\) and \(M_1=\mathrm{diag}(1/2,\sqrt3/2)\). Next we need a unitary that produces \(\sum_j(M_j\ket\psi)\ket j\). Look at what it must do to each system label:

\[\ket0\ket0\mapsto\ket0\bigl(\tfrac{\sqrt3}2\ket0+\tfrac12\ket1\bigr),\qquad\ket1\ket0\mapsto\ket1\bigl(\tfrac12\ket0+\tfrac{\sqrt3}2\ket1\bigr).\]

This is a controlled rotation. If the system is 0, the ancilla is turned by the angle \(\pi/6\), and if it is 1, by \(\pi/3\), where a turn by \(\theta\) is the real unitary \(\bigl(\begin{smallmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{smallmatrix}\bigr)\). Each rotation is unitary, so the whole operation is unitary. Reading the ancilla then gives exactly \(M_0\) and \(M_1\).

The general case follows the same plan. For any POVM choose \(M_j\) with \(M_j^\dagger M_j=E_j\). The map \(\ket\psi\mapsto\sum_j(M_j\ket\psi)\ket j\) keeps inner products, because \(\sum_j\bra\phi M_j^\dagger M_j\ket\psi=\braket\phi\psi\). It is defined only on inputs with the ancilla in \(\ket0\). Extending it to a unitary on all inputs means completing a set of orthonormal columns, which is always possible; we take that standard linear-algebra fact as given. So every POVM can be realized physically.

What does the noisy detector do to plus? Outcome 0 has probability \(\tfrac34\cdot\tfrac12+\tfrac14\cdot\tfrac12=\tfrac12\). The unnormalized branch is \(M_0\ket+=(\sqrt3\ket0+\ket1)/(2\sqrt2)\), with squared length \(1/2\). Normalizing gives \((\sqrt3\ket0+\ket1)/2\). The detector has nudged the state towards \(\ket0\) without projecting onto it.

Suppose that after outcome \(j\) the device applies a unitary \(V_j\) of its own. The measurement operator becomes \(V_jM_j\), but the effect \((V_jM_j)^\dagger V_jM_j=M_j^\dagger M_j\) is unchanged. Every outcome probability is the same. Is the measurement the same?

Compare two Z detectors. The first is the ordinary one: after outcome \(j\) it leaves \(\ket j\). The second applies \(X\) after outcome 1, so it always leaves \(\ket0\). Its measurement operators are \(\ket0\bra0\) and \(\ket0\bra1\). On any input the two detectors report 0 and 1 with identical probabilities. Yet one leaves the observed state and the other resets the qubit.

So outcome probabilities do not determine the state change. The full list of branch maps \(\rho\mapsto M_j\rho M_j^\dagger\), one for each outcome, is called an instrument. A POVM is what the instrument tells us about probabilities alone.

How could anyone tell the two Z detectors apart? Measure twice. On input \(\ket1\), the ordinary detector followed by another Z measurement reports 1 and then 1. The resetting detector reports 1 and then 0. The second measurement sees what the first left behind.

Does a weakly informative measurement escape disturbance? Return to the noisy detector on plus, and ignore its outcome. Each branch contributes an off-diagonal entry \(\sqrt3/8\), so the averaged output is

\[\frac12\begin{pmatrix}1&\sqrt3/2\\\sqrt3/2&1\end{pmatrix}.\]

A later plus test succeeds with probability \((1+\sqrt3/2)/2\approx0.933\), instead of 1. The detector revealed only a little about the Z value, and it disturbed the state only a little, but the disturbance is observable.

The instrument matters here too. Append \(Z\) to every branch. The first outcome statistics are identical, but the off-diagonal entries reverse sign, and the plus test now succeeds with probability \((1-\sqrt3/2)/2\approx0.067\). Recording an outcome, ignoring it, and acting on it are different operations, and each must be stated.

We can now describe any measurement, its probabilities and its aftermath. Suppose Bob knows that Alice sent one of two known states. Which measurement should he choose, and how often can he be right?