The Density Operator

The density operator $\rho$ is the single object that represents any quantum state — pure or mixed — on equal footing. It replaces the state vector as the most general description of a system.

From a pure state to its density operator

Given a pure state $|\psi\rangle$ , its density operator is the outer product

\rho = |\psi\rangle\langle\psi|.

For a qubit $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ , this is a $2\times 2$ matrix:

\rho = \begin{pmatrix}\alpha\\\beta\end{pmatrix} \begin{pmatrix}\alpha^* & \beta^*\end{pmatrix} = \begin{pmatrix} |\alpha|^2 & \alpha\beta^* \\ \alpha^*\beta & |\beta|^2 \end{pmatrix}.

The diagonal entries $|\alpha|^2$ and $|\beta|^2$ are the populations — the probabilities of measuring $0$ and $1$ . The off-diagonal entries are the coherences; they encode the phase relationship that distinguishes a superposition from a mixture.

The density operator of a mixture

For an ensemble $\{(p_i, |\psi_i\rangle)\}$ the density operator is the probability-weighted average of the individual pure-state operators:

\rho = \sum_i p_i\, |\psi_i\rangle\langle\psi_i|.

This is a convex combination of pure-state density operators. Every physically distinct preparation maps to exactly one $\rho$ , and any two ensembles with the same $\rho$ are physically indistinguishable — which is precisely the ambiguity we wanted the formalism to absorb.

A worked example

Take the equal superposition $|+\rangle = \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle)$ , so $\alpha = \beta = \tfrac{1}{\sqrt2}$ . Then

\rho_+ = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}.

Contrast this with the 50/50 mixture of $|0\rangle$ and $|1\rangle$ :

\rho_{\text{mix}} = \frac12|0\rangle\langle 0| + \frac12|1\rangle\langle 1| = \frac12\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.

Same diagonal (same measurement statistics in the $z$ -basis), but the superposition has nonzero off-diagonal coherences while the mixture has none. The matrix makes the distinction from the previous lesson explicit.

Try it

Build the density matrix $\rho = |\psi\rangle\langle\psi|$ for the parametrized real qubit state $|\psi\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle$ at $\theta = \pi/3$ , and return the top-left population $\rho_{00} = \cos^2(\theta/2)$ . Compute it from the outer product — do not just type the number.

Run your code to see the quantum state.