Chapter 03

Density matrices and the state of a subsystem

Source C of Chapter 1 flips a coin and emits $|0\rangle$ or $|1\rangle$. Another source flips a coin and emits $|+\rangle$ or $|-\rangle$. Can any measurement on the qubits tell the two sources apart? And what describes one qubit of a pair when its partner is out of reach?

Before you start: Bloch vectors and Pauli expectation values (Chapter 1); effects and the averaging rule for unread outcomes (Chapter 2). Outer products $|a\rangle\langle b|$ and the trace of a matrix. Notation guide.

Two coin-flip sources

Call the second source D. Each run, it flips a fair coin and emits $|+\rangle$ or $|-\rangle$, without saying which. Source C does the same with $|0\rangle$ and $|1\rangle$. Chapter 2's rule for unread outcomes applies to both: a prediction is the average of the predictions for the two possible states.

Try a few measurements. In the computational basis, source C gives 0 with probability $\tfrac12\cdot1+\tfrac12\cdot0=\tfrac12$; source D gives $\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac12=\tfrac12$. In the $+/-$ basis, source C gives $+$ with probability $\tfrac12$, and so does source D, now as $\tfrac12\cdot1+\tfrac12\cdot0$. The numbers agree, for different reasons. Agreement in two bases proves nothing about the others, and there are infinitely many. What we need is one object that holds the predictions of every measurement at once.

Derive the matrix from an ensemble

The coin is choosing which state to prepare, not which result our detector must report. There are therefore two probabilities in each prediction: the chance of preparing a state, and the chance of a reading given that state. Multiplying those probabilities and adding over the possible preparations is the averaging rule we have just used.

We could repeat that average for every new detector. A density matrix lets us do the preparation part once, then reuse it for any measurement.

Suppose a preparation chooses $|\psi_j\rangle$ with classical probability $p_j$, but the label $j$ is withheld. Such a list of states and probabilities is called an ensemble. For a measurement effect $E_m$, average the conditional probabilities:

$$p(m)=\sum_jp_j\langle\psi_j|E_m|\psi_j\rangle=\operatorname{Tr}(\rho E_m),\qquad \rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|.$$

Read the definition of $\rho$ from right to left. Form one outer-product matrix $|\psi_j\rangle\langle\psi_j|$ for each possible preparation, multiply it by the chance $p_j$ of making that preparation, then add the matrices. The effect $E_m$ describes the measurement we choose afterwards. Taking the trace, the sum of the diagonal entries, returns the outcome probability.

The trace identity follows by inserting any orthonormal basis: $\operatorname{Tr}(|\psi\rangle\langle\psi|E)=\langle\psi|E|\psi\rangle$. The density matrix is Hermitian, positive in the sense of Chapter 2 ($\langle v|\rho|v\rangle\ge0$ for every $v$), and has trace one. Conversely, a matrix with these properties has an eigenvector decomposition with nonnegative eigenvalues summing to one, so it is a valid ensemble state.

For a pure qubit, multiplying the column by its conjugate row gives

$$|\psi\rangle\langle\psi|=\begin{pmatrix}|\alpha|^2&\alpha\beta^*\\\alpha^*\beta&|\beta|^2\end{pmatrix}.$$

The diagonal entries are computational-basis probabilities. The off-diagonal entries affect measurements in other bases. They depend on the basis: “diagonal” does not mean “classical” independently of which observables are being considered.

Now compare the two sources. For source C, $\rho_C=\tfrac12|0\rangle\langle0|+\tfrac12|1\rangle\langle1|=I/2$. For source D, adding the two outer products gives

$$\rho_D=\frac12\cdot\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}+\frac12\cdot\frac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\frac I2.$$

The matrices are equal, so $\operatorname{Tr}(\rho E)$ is the same for every effect $E$. No measurement on the qubits, of any kind, can tell source C from source D. What distinguishes them is the coin's record, which is not in the qubit.

Coherence is measurable; ensemble labels may not be

Compare source A of Chapter 1 with source C:

$$\rho_+=|+\rangle\langle+|=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix},\qquad \rho_C=\frac12|0\rangle\langle0|+\frac12|1\rangle\langle1|=\frac I2.$$

Both have the same diagonal. But $\operatorname{Tr}(\rho_+X)=1$ whereas $\operatorname{Tr}(\rho_CX)=0$. Thus $X$ measurements distinguish these preparations statistically. The off-diagonal entries of $\rho_+$ are what the $X$ measurement detects; source C has none. Source D has none either: its ensemble averages to $I/2$, like source C. The density matrix keeps exactly what measurements can see: coherence, when it is present, but not the list of states a coin chose from.

Purity and the Bloch ball

Sources C and D have the same matrix, while source A has a different one. We now want to recognise from a matrix whether it describes a single pure state or a mixture. Squaring the matrix and taking its trace gives a useful test, called the purity.

For a pure state, $\rho=|\psi\rangle\langle\psi|$ and $\langle\psi|\psi\rangle=1$, so $\rho^2=\rho$ and its purity is one. For our coin source, $\rho=I/2$, squaring gives $I/4$; adding its two diagonal entries gives purity $1/2$. The following general result tells us that purity one characterises pure states.

If $\lambda_j$ are the eigenvalues of $\rho$, then $\operatorname{Tr}\rho^2=\sum_j\lambda_j^2\le1$. Equality requires one eigenvalue equal to one and all others zero: exactly a pure state. For a qubit, the Pauli matrices and identity form a basis for Hermitian matrices, so

$$\rho=\frac12(I+r_xX+r_yY+r_zZ)=\frac12\begin{pmatrix}1+r_z&r_x-ir_y\\r_x+ir_y&1-r_z\end{pmatrix}.$$

Its eigenvalues are $(1\pm|\mathbf r|)/2$, which are nonnegative precisely when $|\mathbf r|\le1$. Pure states lie on the sphere; mixed states lie inside. Writing $(\sigma_1,\sigma_2,\sigma_3)=(X,Y,Z)$, each Pauli matrix is traceless and squares to $I$, so $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$. Using this gives $\operatorname{Tr}\rho^2=(1+|\mathbf r|^2)/2$. The centre $I/2$ has purity $1/2$.

Since $\rho$ depends linearly on $\mathbf r$, the Bloch vector of a mixture is the probability-weighted average of its members' Bloch vectors. A mixture therefore sits inside the ball, inside the convex hull of its members (a segment, polygon or solid), and any two ensembles with the same average are the same density matrix. Sources C and D both average to the centre. Away from the centre, the ensemble $\tfrac34|0\rangle,\tfrac14|1\rangle$ and the equal mixture of $(\sqrt3|0\rangle\pm|1\rangle)/2$ both give $\mathbf r=(0,0,\tfrac12)$, that is $\rho=\operatorname{diag}(\tfrac34,\tfrac14)$. The second pair's off-diagonal entries, $\pm\sqrt3/4$, cancel in the average.

Derive the partial trace from local predictions

So far the missing information was the coin's record. Now imagine that a source sends you one qubit of a pair and keeps the other. You can change your detector as often as you like, but it acts only on your qubit. We need a density matrix that predicts those local readings, using the description of the pair as its starting point.

The operation that constructs it is the partial trace. Its purpose is to preserve every prediction you can test on your qubit. The inaccessible partner need not be measured or destroyed.

Now take a joint state $\rho_{AB}$. An apparatus acting only on $A$ has effect $E_m\otimes I_B$, so its probability is $\operatorname{Tr}[\rho_{AB}(E_m\otimes I_B)]$. We seek a matrix $\rho_A$ giving the same number as $\operatorname{Tr}(\rho_AE_m)$ for every local effect. In product-basis indices the required matrix is

$$(\rho_A)_{ik}=\sum_j(\rho_{AB})_{ij,kj},\qquad \rho_A=\operatorname{Tr}_B\rho_{AB}.$$

Equivalently, $\operatorname{Tr}_B(|i\rangle\langle k|\otimes|j\rangle\langle\ell|)=\delta_{j\ell}|i\rangle\langle k|$. The repeated index is summed because the partner's outcome is not distinguished. Partial trace discards access to a subsystem; it does not require anyone to physically measure that subsystem.

Why this is the right matrix: the index calculation

For a concrete index check, expand a joint matrix as $\rho_{AB}=\sum_{i,j,k,\ell}r_{ij,k\ell}|i\rangle\langle k|\otimes|j\rangle\langle\ell|$. A local effect is blind to the second subsystem, so

$$\begin{aligned}p(m)&=\operatorname{Tr}[\rho_{AB}(E_m\otimes I)]\\ &=\sum_{i,j,k,\ell}r_{ij,k\ell}\langle k|E_m|i\rangle\langle\ell|j\rangle\\ &=\sum_{i,k}\left(\sum_jr_{ij,kj}\right)\langle k|E_m|i\rangle. \end{aligned}$$

The quantity in parentheses is precisely the matrix element $(\rho_A)_{ik}$. This establishes the partial trace by the predictions it must preserve, rather than by an arbitrary rule for deleting indices. Terms with unequal partner labels vanish because their partner states are orthogonal. Coherences between system labels $i,k$ can remain when the partner label is the same.

For comparison, $|00\rangle\langle10|$ becomes $|0\rangle\langle1|$: the partner is zero on both sides. But $|00\rangle\langle11|$ becomes zero, because it would require the overlap $\langle1|0\rangle$.

A pure joint state that is not a product of one-qubit states is called entangled; Chapter 7 gives a test for it and the separate definition needed for mixed states. For the Bell state $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$, which is entangled, the projector has four terms. The terms $|00\rangle\langle11|$ and $|11\rangle\langle00|$ vanish on tracing $B$, because their $B$ indices differ. Hence $\rho_A=I/2$, even though $\rho_{AB}$ is pure. One qubit of this pair is described by the same matrix as source C and source D, and no measurement on that qubit alone can tell the three apart.

Local mixedness does not by itself prove entanglement

The classically correlated state $\rho_{\rm cc}=(|00\rangle\langle00|+|11\rangle\langle11|)/2$ has exactly the same two reduced states as $|\Phi^+\rangle$. Both give matching $Z$ results. To tell the joint preparations apart, compute $\langle X\otimes X\rangle$: it is one for $|\Phi^+\rangle$ and zero for $\rho_{\rm cc}$. The latter has no off-diagonal terms to contribute.

If the joint state is known to be pure, a mixed reduced state certifies entanglement. Without that assumption, it does not. Nor does every interaction create entanglement: a controlled-NOT leaves $|00\rangle$ unchanged. The input and the interaction both matter.

Where this leaves us

No measurement on the qubits can tell source C from source D: both have $\rho=I/2$, and every outcome probability is $\operatorname{Tr}(\rho E)$. The density matrix holds everything a measurement can reveal about a preparation, and nothing more. It records coherence, which separates $|+\rangle$ from source C, but not which list of states a coin chose from. The same matrix describes one qubit of a Bell pair, so a mixed reduced state alone cannot say whether it came from a coin, from entanglement, or from classical correlation; only with the joint state known to be pure does it certify entanglement.

Every preparation here was a coin flip or a pair that was already correlated. The next chapter asks how states are changed and correlated in the first place: what a sequence of gates, including one qubit controlling another, does to the amplitudes.

Problems

1. Trace either subsystem out of $\sqrt{3/4}|00\rangle+\sqrt{1/4}|11\rangle$. Find the purity.

Worked answer

Write the coefficients as $\sqrt3/2$ and $1/2$, then form the outer product:

$$\rho_{AB}=\tfrac34|00\rangle\langle00|+\tfrac{\sqrt3}{4}(|00\rangle\langle11|+|11\rangle\langle00|)+\tfrac14|11\rangle\langle11|.$$

Tracing $B$ keeps the diagonal terms and multiplies the cross terms by $\langle1|0\rangle$ or $\langle0|1\rangle$, both zero. Hence $\rho_A=\operatorname{diag}(3/4,1/4)$, and tracing $A$ gives the same matrix. Squaring yields $\rho_A^2=\operatorname{diag}(9/16,1/16)$, so the purity is $10/16=5/8$. The pure joint state has a mixed reduced state and is therefore entangled.

The tempting wrong turn is to keep the cross terms $\sqrt3/4$ in $\rho_A$. They vanish because their partner labels differ; the same reduced state, $\operatorname{diag}(3/4,1/4)$, is the point at height $\tfrac12$ in the figure, where it could equally have come from a coin.

2. Someone assigns $\rho=\left(\begin{smallmatrix}1/2&3/4\\3/4&1/2\end{smallmatrix}\right)$ to a qubit. It is Hermitian with trace one. Why is it invalid?

Worked answer

Apply the proposed matrix to the two candidate eigenvectors. On the unnormalised column $(1,1)^T$ it gives $(5/4,5/4)^T$; on $(1,-1)^T$ it gives $(-1/4,1/4)^T$. Thus the normalised $|+\rangle,|-\rangle$ are eigenvectors with eigenvalues $5/4,-1/4$.

$$p(-)=\langle-|\rho|-\rangle=-\tfrac14\langle-|-\rangle=-\tfrac14.$$

This negative measurement probability excludes the matrix. Hermiticity and unit trace check two state conditions, but they do not guarantee the third, positivity.

References and further reading

Nielsen & Chuang, §2.4.1–2.4.3 (ensembles, density-operator properties, reduced states).