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A quantum state ensemble of normalized states $|{\psi }_i⟩$, with probabilities $p_i\ge 0$ and ${\sum }_i{p}_i=1$, has density operatorIt is a positive semidefinite Hermitian operator with trace one, and the Born rule assigns a measurement outcome with projector $\Pi$ the probability $\mathrm{Tr}(\rho \Pi )$.

A pure state has $\rho =|\psi ⟩⟨\psi |$, hence ${\rho }^2=\rho$ and purity of a density operator $\mathrm{Tr}({\rho }^2)=1$. Its only nonzero eigenvalue is one. A mixed state has at least two nonzero eigenvalues and $\mathrm{Tr}({\rho }^2)<1$. The spectral theorem proves the converse criteria, since the nonnegative eigenvalues sum to one. Different quantum state ensembles may give the same density operator; the mere presence of several terms in a preparation list does not make the state mixed if all positive-weight terms represent the same quantum state ray.

Taking the outer product for A and averaging the two orthogonal projections for B givesThe off-diagonal entries in ${\rho }_A$ retain quantum coherence between the $z$-basis states. They are absent from B's classical quantum state ensemble. In particular, $\mathrm{Tr}{\rho }_A^2=1$ and $\mathrm{Tr}{\rho }_B^2=1/2$: A is a pure state, B a mixed state.

The opposite $x$-axis pure states have projectors $(I\pm {\sigma }_x)/2$. Averaging them givesTherefore C and B cannot be distinguished by any measurement on the atoms alone. Every POVM outcome has the same probability $\mathrm{Tr}(E\rho )$ for the two preparations; this is the operational equivalence of different quantum state ensembles. Accessible preparation labels or additional correlations would be extra information. C can be distinguished from A by the $x$-axis spin measurement in part (c).

A POVM $\{E_i\}$ in dimension $d$ corresponds to a quantum state ensemble with $p_i=\mathrm{Tr}(E_i)/d$ and ${\rho }_i=E_i/\mathrm{Tr}(E_i)$, omitting zero effects. Its average is the maximally mixed state $I/d$. Conversely any ensemble with that average defines a POVM by $E_i=d{p}_i{\rho }_i$. These transformations are inverse. A pure POVM corresponds exactly to a pure-state ensemble decomposition of $I/d$.

A quantum state describes a preparation through its predicted measurement in quantum mechanics probabilities. In the Hilbert space formulation, a pure state is a quantum state ray: a normalized vector representative $|\psi ⟩$ satisfies $⟨\psi |\psi ⟩=1$, and representatives differing by a global phase describe the same quantum state. ^[ref]

Both pure states and mixed states are represented by a density operator on the chosen complex Hilbert space:Thus $\rho$ is Hermitian, positive semidefinite and of unit trace; in infinite dimension it must also be a trace-class operator. For a normalized pure state, $\rho =|\psi ⟩⟨\psi |$ is a rank-one orthogonal projection. A mixed state has no single such vector representative. Its density matrix is the matrix of $\rho$ in a chosen orthonormal basis. ^[ref] For a discrete POVM with effects $E_a\ge 0$ and ${\sum }_a{E}_a=I$, the Born rule readsThese constraints and Born rule probabilities describe arbitrary finite-dimensional preparations. ^[ref]

Randomly choosing normalized pure states with probabilities $w_j$ and forgetting the label givesThis quantum state ensemble averages probabilities; a quantum superposition adds probability amplitudes. Several ensembles can give the same density operator. Quantum entanglement can make a reduced density matrix mixed while the joint quantum state is pure. ^[ref]

The preparation, the choice of basis, and the measurement are separate ingredients. A wavefunction $\psi (x)=⟨x|\psi ⟩$ is the position representation of a pure state; a general mixed state requires a density operator. For a closed quantum system, a unitary operator evolves it as $\rho \mapsto U\rho U^{†}$, preserving its eigenvalues and purity of a density operator. ^[ref] The Bloch sphere and Bloch ball give the qubit geometry; quantum state tomography connects repeated preparations to an estimated density matrix; projective measurement treats the conditional state after an outcome.

For normalized pure states $|u⟩,|v⟩$, a two-member quantum state ensemble gives the maximally mixed state of a qubit exactly when its members are orthogonal and equally probable. To prove necessity, take expectation values of its density matrix in the two members: $1/2=p+(1-p)|⟨u|v⟩{|}^2\ge p$ and $1/2=1-p+p|⟨u|v⟩{|}^2\ge 1-p$. Hence $p=1/2$ and the overlap vanishes. Sufficiency is the resolution of the identity in an orthonormal basis.