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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.

Choose three pure qubit states whose unit Bloch vectors are coplanar, sum to zero and have pairwise angle $120$ degrees. The three effects $E_j=\frac{2}{3}|{\varphi }_j⟩⟨{\varphi }_j|$ sum to the identity, giving a POVM. If an input is the state orthogonal to ${\varphi }_1$ or to ${\varphi }_2$, outcome one excludes the first input and outcome two excludes the second. Each conclusive outcome has probability one half on its corresponding input; the third outcome is inconclusive with probability one half on either input. This gives unambiguous quantum state discrimination, rather than identification on every trial.

Three equally weighted pure qubit states with coplanar unit Bloch vectors separated by $120$ degrees. One example is $|t_j⟩=(|0⟩+{\omega }^j|1⟩)/\sqrt{2}$, with $\omega =e^{2\pi i/3}$ and $j=0,1,2$. The roots-of-unity identity $1+\omega +{\omega }^2=0$ cancels the average off-diagonal matrix entries, so $\frac{1}{3}{\sum }_j|t_j⟩⟨t_j|=I/2$.

This is an ensemble of preparation states, whereas a trine qubit POVM consists of measurement effects $E_j=\frac{2}{3}|t_j⟩⟨t_j|$. The POVM–ensemble duality for the maximally mixed state relates them but does not identify their operational roles.

The rotation-invariant average of pure qubit states is uniform area measure on the Bloch sphere, induced by Haar measure on unitary transformations. For a uniformly distributed unit Bloch vector $r$, rotational symmetry gives $\mathbb{E}{r}_i=0$ and $\mathbb{E}{r}_i{r}_j={\delta }_{ij}/3$. Uniform polar angle without the factor $\sin \theta$ is a different distribution.

A time-independent Hamiltonian evolves a state by the unitary operator $U(t)=e^{-iHt/\hslash }$.

For a closed quantum system, unitary time evolution changes a quantum state through a unitary operator, preserving norms and inner products. For a time-independent self-adjoint Hamiltonian operator $H$, choose the initial time as zero:Here $\hslash =h/(2\pi )$ is the reduced Planck constant. The same $U(t)$ acts on every initial quantum state; it depends on the Hamiltonian operator and elapsed time, rather than the chosen preparation.^[ref]

The self-adjointness of $H$ makes the exponential a unitary operator. In finite dimension, conjugating its matrix exponential givesThe two exponents commute because both are multiples of $H$. Hence, for any two initial vectors,This short proof explains why normalization survives, why orthogonality of initial vectors survives, and why the transition probability between two states evolved by the same $U$ is unchanged. It does not require that a fixed measurement in quantum mechanics have time-independent probabilities: its measurement basis need not move with the quantum states.

For an unbounded self-adjoint operator, define the exponential through the spectral theorem, rather than assuming that its power series acts on every vector. It is a unitary operator on the entire Hilbert space; the strong Schrödinger equation holds for initial vectors in the operator domain $D(H)$. For a density operator, unitary time evolution reads $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. It preserves eigenvalues and purity of a density operator, but can change its entries in a fixed basis.

The useful distinction is between conserving total probability and conserving each measurement's probabilities. A fixed energy spectrum supplies different ticking rates for different probability amplitudes. Their relative quantum phases can move the Bloch vector and create quantum interference even while every energy eigenstate population stays fixed. Qubit precession under a constant Hamiltonian makes this visible; Ramsey interferometry converts the accumulated relative quantum phase into a population difference that a simple readout can detect.

The ideal universal NOT map sends every qubit Bloch vector $\mathbf{s}$ to $-\mathbf{s}$, taking a pure state to its orthogonal pure state. It is a trace-preserving positive map but not a quantum channel: applied to one half of a Bell state, it produces an operator with a negative eigenvalue. A physical approximation can reverse all directions with factor $-1/3$ using the equal mixture of conjugations by the three Pauli matrices. The isotropic qubit map complete-positivity bound proves that a larger uniform reversal is impossible among such isotropic maps.

The isotropic Weyl Hamiltonian acts on a two-component spinor as $H=v\mathbf{\sigma }\cdot \mathbf{p}$. For a plane wave of momentum $\hslash \mathbf{k}$, its two energy eigenvalues are $E_{\pm }=\pm \hslash |v||\mathbf{k}|$. For $v>0$ the corresponding Bloch vectors point along and opposite $\mathbf{k}$, respectively. If the Pauli matrices describe physical spin one-half, the mean spin is $\pm (\hslash /2)\hat{\mathbf{k}}$; in an effective two-band model the same algebra can describe an internal pseudospin instead. The bands meet linearly at the origin. ^[ref]