Quantum Operations

Take a qubit in \(\ket\psi=a\ket0+b\ket1\) and an auxiliary qubit in \(\ket0\). Apply a CNOT from the qubit to the auxiliary. The pair becomes \(a\ket{00}+b\ket{11}\). Now suppose we keep only the first qubit. What is its state?

The partial trace over the auxiliary keeps terms whose auxiliary labels match. The diagonal entries survive as \(|a|^2\) and \(|b|^2\). The off-diagonal entry would pair auxiliary label 0 with label 1, so it vanishes:

\[\rho_{\text{out}}=\begin{pmatrix}|a|^2&0\\0&|b|^2\end{pmatrix}.\]

Removing every off-diagonal entry in the computational basis is called complete dephasing. For a plus input, the kept qubit is \(I/2\), so a later X measurement is fair.

Has anything irreversible happened? Not to the pair. If we keep both qubits and apply CNOT again, the pair returns to \(\ket+\ket0\), and the X measurement gives plus with certainty. The joint unitary has an inverse. What we lost is access to the auxiliary, which holds a record of the computational label.

An auxiliary system used this way is called an ancilla. It may be an apparatus we control, or a model of an environment we cannot reach. Discarding it means our predictions no longer use it: we neither act on it nor condition on it. The object need not be destroyed.

The CNOT example has three steps: prepare an ancilla, apply a joint unitary, and discard part of the output. We take this as the general recipe for a physical process on an open system. How drastic can its effect on the kept qubit be?

Consider the swap, which exchanges two qubits' labels: \(\ket{x,y}\mapsto\ket{y,x}\). It permutes the four configurations, so it is unitary. Swap a qubit with an ancilla in \(\ket0\), then discard the ancilla. Whatever the input, the kept qubit is now \(\ket0\). This is a reset. Many inputs give the same output, so no operation on the kept qubit can undo it. The input state still exists; it has moved into the discarded system.

What if, instead of discarding the ancilla, we measure it and keep the result? After the CNOT, the ancilla's Z value equals the qubit's. Conditioning on the record leaves the qubit in \(\ket0\) or \(\ket1\), with probabilities \(|a|^2\) and \(|b|^2\). That is the projective measurement of the measurement postulate. Ignoring the record averages the two outcomes and gives complete dephasing again.

Go back to the CNOT record and sort the output by the ancilla's label:

\[a\ket{00}+b\ket{11}=\bigl(a\ket0\bigr)\ket0_E+\bigl(b\ket1\bigr)\ket1_E=\bigl(K_0\ket\psi\bigr)\ket0_E+\bigl(K_1\ket\psi\bigr)\ket1_E,\]

where \(K_0=\ket0\bra0\) keeps the zero component and \(K_1=\ket1\bra1\) keeps the one component. Tracing out the ancilla removes the cross terms between different labels, just as in a purification. What remains is \(K_0\ket\psi\bra\psi K_0^\dagger+K_1\ket\psi\bra\psi K_1^\dagger\). Averaging over any input ensemble gives

\[\mathcal E(\rho)=\sum_jK_j\rho K_j^\dagger.\]

For our two projectors this sum removes exactly the off-diagonal entries. Does every ancilla construction reduce to a few matrices sandwiching \(\rho\)? Take any joint unitary \(U\) with the ancilla starting in \(\ket0_E\). The output can always be sorted by ancilla label as \(\sum_j(K_j\ket\psi)\ket j_E\). Here \(K_j\) collects the system amplitudes attached to ancilla output \(j\), written \(K_j=\bra j_EU\ket0_E\). The same partial trace gives the same formula.

This is the operator-sum representation of the process, and the \(K_j\) are its Kraus operators. A process described this way is called a channel. Each \(K_j\) belongs to one ancilla branch. Nobody need see which branch occurred; if the ancilla is discarded, the branches are only averaged.

Suppose someone hands us matrices \(K_j\). When does \(\sum_jK_j\rho K_j^\dagger\) describe a physical process? Its output must be a density operator, so it must have trace one and be positive.

For the trace we need one fact. For matrices \(A\) and \(B\), \(\mathrm{tr}(AB)=\sum_{i,k}A_{ik}B_{ki}\). Exchanging the names of the two summed indices shows this equals \(\mathrm{tr}(BA)\). Moving each \(K_j\) around the trace,

\[\mathrm{tr}\,\mathcal E(\rho)=\mathrm{tr}\Bigl[\rho\sum_jK_j^\dagger K_j\Bigr].\]

This equals \(\mathrm{tr}\,\rho=1\) for every input exactly when \(\sum_jK_j^\dagger K_j=I\). A channel with this property is trace preserving. The ancilla construction always has it: \(\sum_jK_j^\dagger K_j=\bra0_EU^\dagger U\ket0_E=I\), because \(U\) is unitary. A single branch can have trace below one. Its trace is the probability of that branch.

Positivity also comes free: \(\bra v K_j\rho K_j^\dagger\ket v=\bra w\rho\ket w\geq0\) with \(\ket w=K_j^\dagger\ket v\). But is positivity of every output enough to make a map physical? Try the transpose, \(T(\rho)=\rho^T\). It keeps the trace, and it keeps positivity: \(\bra v\rho^T\ket v=\bra{v^*}\rho\ket{v^*}\), where \(v^*\) conjugates every entry.

Now apply the transpose to Alice's half of a Bell pair while Bob's half is untouched. Transposing Alice's labels sends \(\ket{ab}\bra{a'b'}\) to \(\ket{a'b}\bra{ab'}\). The Bell matrix \(\tfrac12(\ket{00}\bra{00}+\ket{00}\bra{11}+\ket{11}\bra{00}+\ket{11}\bra{11})\) becomes

\[\rho'=\tfrac12\bigl(\ket{00}\bra{00}+\ket{10}\bra{01}+\ket{01}\bra{10}+\ket{11}\bra{11}\bigr).\]

Test it with \(\ket v=(\ket{01}-\ket{10})/\sqrt2\). Only the two middle terms contribute, and each gives \(-1/4\), so \(\bra v\rho'\ket v=-1/2\). A test would pass with probability \(-1/2\), which is impossible. So the transpose is not a physical operation on half of an entangled pair, even though it is harmless on a qubit by itself.

A map that keeps outputs positive even when it acts on one part of a larger system, with the rest untouched, is completely positive. Operator-sum maps have this property, because \(K_j\otimes I\) is again a sandwich of the same form. A quantum channel is a completely positive, trace-preserving map. We quote the converse without proof: every such map can be built by the ancilla recipe (a theorem due to Stinespring, with the operator-sum form due to Kraus).

Suppose a stray field applies \(Z\) to a qubit on a fraction \(p\) of uses, and does nothing otherwise. What happens to the coherences? The phase-flip channel is

\[\mathcal E(\rho)=(1-p)\rho+pZ\rho Z,\]

with Kraus operators \(\sqrt{1-p}\,I\) and \(\sqrt p\,Z\). Conjugating by \(Z\) flips the sign of the off-diagonal entries and leaves the diagonal alone. So the off-diagonal entries are multiplied by \(1-2p\). On plus, an X measurement then gives plus with probability \(1-p\).

At \(p=1/2\) the coherences vanish, and the channel is complete dephasing again. At \(p=1\) every qubit receives the same known \(Z\), and plus becomes pure minus. A flip probability of one is not maximal ignorance; it is a known operation.

Random flips and known flips

Predict first. Predict which phase-flip probability makes a plus input most mixed. Compare one half with one.

A different physical process is decay. An atom in its excited level \(\ket1\) drops to \(\ket0\) with probability \(\gamma\), giving its energy to the surroundings. The amplitude-damping channel has Kraus operators

\[K_0=\ket0\bra0+\sqrt{1-\gamma}\,\ket1\bra1,\qquad K_1=\sqrt\gamma\,\ket0\bra1.\]

The second one moves \(\ket1\) to \(\ket0\). They satisfy \(K_0^\dagger K_0+K_1^\dagger K_1=\ket0\bra0+(1-\gamma)\ket1\bra1+\gamma\ket1\bra1=I\). On \(\rho=\begin{pmatrix}a&c\\c^*&b\end{pmatrix}\) the output is

\[\begin{pmatrix}a+\gamma b&\sqrt{1-\gamma}\,c\\\sqrt{1-\gamma}\,c^*&(1-\gamma)b\end{pmatrix}.\]

For plus, \(a=b=c=1/2\), and the output is

\[\frac12\begin{pmatrix}1+\gamma&\sqrt{1-\gamma}\\\sqrt{1-\gamma}&1-\gamma\end{pmatrix}.\]

At \(\gamma=0\) it is plus, and at \(\gamma=1\) it is \(\ket0\). In between, the populations shift towards 0 and the coherence shrinks. Phase-flip noise, by contrast, never changes the populations.

Could the two channels be confused? Choose \(p\) so that \(1-p=(1+\sqrt{1-\gamma})/2\). Then both give the same X-measurement statistics on plus. A Z measurement still tells them apart: phase flips leave it fair, while damping favours 0. Two noise mechanisms can agree in one experiment and differ in another.

A qubit passes through a channel with Kraus operators \(K_j\), then through one with operators \(L_l\). What is the combined channel? Substitute one formula into the other:

\[\sum_lL_l\Bigl(\sum_jK_j\rho K_j^\dagger\Bigr)L_l^\dagger=\sum_{l,j}(L_lK_j)\,\rho\,(L_lK_j)^\dagger.\]

So the combined Kraus operators are the products \(L_lK_j\). The order matters, as it does for unitaries.

Can a channel ever help Bob tell two states apart? Suppose Alice sends \(\ket0\) or \(\ket+\) with equal probability, and Bob guesses which. With a Z measurement he guesses "zero" on outcome 0 and "plus" on outcome 1. He is right on every zero and on half of the pluses, so his success is \(\tfrac12\cdot1+\tfrac12\cdot\tfrac12=\tfrac34\). An X measurement, guessing "plus" on plus, also succeeds with probability \(3/4\).

Now let the qubit be completely dephased first. Zero stays \(\ket0\bra0\), and plus becomes \(I/2\). The Z statistics are unchanged, so the first strategy still succeeds with \(3/4\). The X strategy drops to \(1/2\), because both inputs now give plus half the time. The channel removed the coherence that the X strategy used. It added nothing that either strategy could use.

That is a general fact, and its proof is one sentence. Whatever Bob does after the channel, he could have done to the original input, by applying the channel himself and then doing it. So his best success after the channel is never more than his best success before. The information about Alice's label that the best measurement can extract is called accessible information. Processing cannot increase it. This is the quantum form of the data-processing principle. It concerns the best achievable decisions. A fixed measurement can do better after some channels than before: an X measurement cannot tell \(\ket0\) from \(\ket1\), but after a Hadamard it can.

A measurement is a channel whose branch record we keep instead of discarding. If Bob keeps that record, which states can he be left in, and which measurements become possible?