Mixed States

A device flips a fair coin and prepares \(\ket0\) on heads and \(\ket1\) on tails. Is the result the same as preparing \(\ket+\), which also gives 0 and 1 equally often?

A Z measurement cannot tell: both give fair results. Try an X measurement. On \(\ket+\) it gives plus with certainty. On the coin's output, each trial is \(\ket0\) or \(\ket1\), and each of those passes the plus test with probability \(|\braket+0|^2=|\braket+1|^2=1/2\). Averaging over the coin gives \(1/2\). So the two preparations differ.

The general rule follows the same steps. A device that prepares \(\ket{\psi_j}\) with probability \(p_j\) defines an ensemble. The probability that its output passes a test for \(\ket\phi\) is the average of the separate probabilities,

\[p(\text{pass})=\sum_jp_j\,|\braket\phi{\psi_j}|^2.\]

This averages probabilities over trials. A superposition instead adds amplitudes within each trial and then squares. That is why plus can interfere in the X measurement and the coin's output cannot.

The rule raises a practical question. Must we carry the whole list of states and probabilities to make predictions? Two different lists might give the same prediction for every test. If so, the list holds more than any measurement on the system can reveal.

What part of the ensemble does a test probability actually use? Rewrite one term. A squared magnitude is a number times its conjugate, and the conjugate of \(\braket\phi\psi\) is \(\braket\psi\phi\). So

\[|\braket\phi\psi|^2=\braket\phi\psi\braket\psi\phi=\bra\phi\bigl(\ket\psi\bra\psi\bigr)\ket\phi.\]

The middle factor is a column times a row, which is a matrix. We met this outer product when building projectors. For \(\ket\psi=a\ket0+b\ket1\),

\[\ket\psi\bra\psi=\begin{pmatrix}|a|^2&ab^*\\a^*b&|b|^2\end{pmatrix}.\]

The test vector now appears only on the outside. Averaging over the ensemble therefore averages the matrices:

\[\rho=\sum_jp_j\ket{\psi_j}\bra{\psi_j},\qquad p(\text{pass})=\bra\phi\rho\ket\phi.\]

The matrix \(\rho\) is the density operator, or density matrix, of the ensemble. It is all we need for every test, so two ensembles with the same \(\rho\) can never be told apart by measuring the system.

Compare our two preparations. Plus has \(\rho=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\). The coin mixture has \(\tfrac12(\ket0\bra0+\ket1\bra1)=I/2\). The diagonal entries are the Z-measurement probabilities, and they agree. The off-diagonal entries differ. The plus test computes \(\bra+\rho\ket+\), which is half the sum of all four entries: 1 for plus, \(1/2\) for the mixture. The off-diagonal entries carry the relative phase that interference needs. They are often called coherences.

Which matrices can be density operators? Two conditions follow from what \(\rho\) predicts. Every test probability is nonnegative, so \(\bra v\rho\ket v\geq0\) for every vector \(v\). A matrix with this property is positive. The outcomes of a basis measurement exhaust the possibilities, so their probabilities \(\bra k\rho\ket k\) add to one. That sum is the sum of the diagonal entries, called the trace, \(\mathrm{tr}\,\rho\). Its value does not depend on the basis, because each outer product \(\ket\psi\bra\psi\) contributes \(\sum_k|\braket k\psi|^2=1\) in any basis. So a density operator is positive with trace one.

The matrix also mirrors itself: the entry \(a^*b\) is the conjugate of \(ab^*\). A matrix equal to its own adjoint, \(\rho^\dagger=\rho\), is called Hermitian. A state whose density operator is a single outer product is pure, as before. Any other density operator describes a mixed state.

Alice tests her half of \(\ket{\Phi^+}\) for a state \(\ket\phi=\phi_0\ket0+\phi_1\ket1\). Bob does nothing. What is the probability that she passes?

In the joint description, "Alice passes" includes every possible Bob label. It is the projector onto the two orthogonal directions \(\ket{\phi,0}\) and \(\ket{\phi,1}\). By the measurement postulate its probability is the sum of the two squared amplitudes, just as a marginal sums a joint table:

\[p(\text{pass})=|\braket{\phi,0}{\Phi^+}|^2+|\braket{\phi,1}{\Phi^+}|^2=\tfrac12|\phi_0|^2+\tfrac12|\phi_1|^2=\tfrac12.\]

The answer is \(1/2\) for every \(\phi\). Which single-qubit matrix gives \(1/2\) for every test? Only \(I/2\). So Alice's half behaves exactly like the coin mixture of 0 and 1.

Now do the same for any joint state \(\ket\Psi=\sum_{a,b}\alpha_{ab}\ket{ab}\). The amplitude to find \(\phi\) on Alice's side and \(b\) on Bob's is \(\sum_a\phi_a^*\alpha_{ab}\). Squaring and adding over \(b\),

\[p(\text{pass})=\sum_{a,a'}\phi_a^*\Bigl(\sum_b\alpha_{ab}\alpha_{a'b}^*\Bigr)\phi_{a'}=\bra\phi\rho_A\ket\phi,\qquad(\rho_A)_{aa'}=\sum_b\alpha_{ab}\alpha_{a'b}^*.\]

In terms of the joint density matrix, \((\rho_A)_{aa'}=\sum_b(\rho_{AB})_{ab,a'b}\): keep the entries whose Bob labels match on both sides, and add over that label. This is the partial trace over Bob, written \(\rho_A=\mathrm{tr}_B\,\rho_{AB}\). The result is Alice's reduced state. By construction it gives the same prediction as the joint state for every Alice-only test. For a mixed joint state the same formula holds, because both sides are averages of the pure-state case.

Let us watch the partial trace remove the coherences. Write the Bell density matrix in basis order \(00,01,10,11\):

\[\rho_{AB}=\frac12\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}.\]

Alice's 0 probability adds the 00 and 01 diagonal entries, giving \(1/2\). Her 1 probability adds the 10 and 11 entries. Her off-diagonal entry adds \(\rho_{00,10}+\rho_{01,11}\), both zero. The large corner entries belong to \(\ket{00}\bra{11}\), whose Bob labels differ, so the partial trace never collects them. The result is \(I/2\).

So the Bell pair is pure as a whole and mixed in each part. Does Alice's local uncertainty mean that some hidden pure state was secretly chosen for her qubit? Nothing in the construction chose one. The randomness of her results comes from the entanglement, not from a coin.

What does the reduced state miss? Compare the Bell pair with the equal mixture of \(\ket{00}\) and \(\ket{11}\), whose joint matrix is \(\tfrac12(\ket{00}\bra{00}+\ket{11}\bra{11})\). Both have reduced states \(I/2\) on each side. Yet their joint matrices differ in the corner coherences, and the X-measurement correlations showed that difference. A reduced state is complete for one party's predictions. Correlations between the parties need the joint state.

Suppose a coin now chooses between \(\ket+\) and \(\ket-\). Is that a different source from the coin choosing \(\ket0\) or \(\ket1\)? Compute its matrix:

\[\tfrac12\bigl(\ket+\bra++\ket-\bra-\bigr)=\tfrac14\begin{pmatrix}1&1\\1&1\end{pmatrix}+\tfrac14\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\frac I2.\]

The off-diagonal entries cancel. Both sources have the density operator \(I/2\). Since every test probability depends only on \(\rho\), no measurement on the qubit alone can reveal which coin procedure was used.

The procedures are still different physical histories. Someone holding the coin's record could tell them apart. That knowledge lives in the record, not in the qubit. So a density operator generally has many ensemble decompositions, and it does not say which one, if any, took place.

The Bell half was mixed but came from a pure pair. Can every mixed state be viewed that way? Try \(\rho=\tfrac34\ket0\bra0+\tfrac14\ket1\bra1\). Add an auxiliary qubit and prepare

\[\ket\Psi=\tfrac{\sqrt3}2\ket{00}+\tfrac12\ket{11}.\]

The partial trace over the auxiliary gives diagonal entries \(3/4\) and \(1/4\). The off-diagonal entry is \(\alpha_{00}\alpha_{10}^*+\alpha_{01}\alpha_{11}^*=0\), because each term pairs different auxiliary labels. So the reduced state is exactly \(\rho\). A pure joint state whose reduced state is \(\rho\) is a purification of \(\rho\).

The same construction works for any ensemble. Give each member its own orthogonal label in an auxiliary system \(E\):

\[\ket\Psi=\sum_j\sqrt{p_j}\,\ket{\psi_j}\ket j_E.\]

Tracing out \(E\) keeps only terms with matching labels, and \(\braket jk=0\) for \(j\neq k\). What remains is \(\sum_jp_j\ket{\psi_j}\bra{\psi_j}=\rho\). A purification is a mathematical representation. It does not claim that anyone holds or controls the purifying system, and the system alone stays mixed.

How unique is a purification? Apply any unitary to \(E\) alone. The labels \(\ket j\) become another orthonormal set, the cross terms still vanish, and the reduced state is unchanged. But measuring \(E\) in different bases leaves the system in different ensembles. The Bell pair shows this. If Bob measures Z, Alice is left with \(\ket0\) or \(\ket1\), each with probability \(1/2\). If he measures X, then since \(\ket{\Phi^+}=(\ket{++}+\ket{--})/\sqrt2\) she is left with \(\ket+\) or \(\ket-\). Both ensembles of \(I/2\) come from one purification, read in two ways.

So far a system has either evolved alone, by a unitary, or been described as part of something larger. What happens to a qubit that interacts with another system which is then discarded?