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The purity of a density operator is $\gamma =\mathrm{Tr}({\rho }^2)$. It equals one exactly for a pure state; in finite dimension $d$ it lies between $1/d$ and $1$. The minimum occurs exactly at the maximally mixed state $I/d$.

A worked qubit family interpolates between the mixed state $I/2$ and the pure state ${\rho }_{+}=|+⟩⟨+|$, where $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$:The eigenvectors are $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, with eigenvalues ${\lambda }_{\pm }=(1\pm p)/2$. Squaring and summing these eigenvalues givesHere $P_X(+)$ is the Born rule probability of the positive outcome in an $X$ projective measurement. The Bloch vector is $(p,0,0)$, since ${\rho }_p=(I+p{\sigma }_x)/2$ with ${\sigma }_x$ one of the Pauli matrices. At $p=0$ the purity is $1/2$; at $p=1/2$ it is $5/8$, with $P_X(+)=3/4$; at $p=1$ it is one. Within this family, $p=1$ is a pure state and every $p<1$ is a mixed state. Extending the formula to $p>1$ produces a negative eigenvalue $(1-p)/2$, so it ceases to be a density operator despite remaining Hermitian with trace one.

The Bloch vector length measures polarization: for any qubit density operator, $\gamma =(1+|\mathbf{r}{|}^2)/2$. This connects the interior of the Bloch ball with mixed states and its boundary, the Bloch sphere, with pure states. ^[ref]

For a short proof of the general bound, use the spectral theorem: the nonnegative eigenvalues ${\lambda }_j$ sum to one, so $\gamma ={\sum }_j{\lambda }_j^2$. ThenThe lower bound is attained only when all eigenvalues equal $1/d$; the upper bound is attained only when exactly one eigenvalue is nonzero, and hence equals one. Purity is therefore a simple quantitative test of mixedness, invariant under a unitary operator. It does not specify the entire density operator: even distinct pure states all have purity one. ^[ref]

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.

Quantum state tomography estimates a density operator from measurement frequencies on many independently and identically prepared systems. One fixed projective measurement reveals only its basis populations, so further measurement settings are generally needed. For a qubit, writeThe three Pauli matrices thus turn measured expectation values into the Bloch vector. This reconstruction uses the Bloch ball representation of arbitrary qubit density operators. ^[ref].

Divide the prepared ensemble into groups measured in the $X$, $Y$ and $Z$ bases. For example, positive-outcome frequencies $0.8,0.5,0.9$ give the estimate $\mathbf{r}=(0.6,0,0.8)$ andHere $|\mathbf{r}|=1$, so the reconstructed state is the pure state $(3|0⟩+|1⟩)/\sqrt{10}$. Its $Z$ frequencies alone also fit the diagonal mixed state with populations $0.9,0.1$; the $X$ data distinguish them. For the angular pure state in measurement in quantum mechanics, $r_x=\sin \theta \cos \varphi$, $r_y=\sin \theta \sin \varphi$ and $r_z=\cos \theta$, so $Y$ resolves the sign of the phase that $X$ alone cannot.

Finite samples introduce statistical uncertainty: with $N$ independent binary trials in one basis, the frequency has variance $p(1-p)/N$. At $p=1/2$ and $N=100$, its standard deviation is $0.05$. Fluctuations can place a naive Bloch vector estimate outside the Bloch ball; a physical reconstruction must enforce positivity of the density operator. Repeated measurements on one unprepared specimen sample its changing conditional state, rather than independent copies of its original state. Multiple bases are needed in this projective protocol; a single fixed informationally complete POVM, such as the tetrahedral qubit POVM, is an alternative. It still requires an ensemble of trials.

For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.