Bloch vector

Every qubit state has the unique formIt is pure exactly when $|r|=1$, and
$r_i=2⟨S_i⟩/\hslash$.

Unit-length Bloch vectors form the Bloch sphere; mixed-state vectors occupy the ball’s interior.

The ensemble Bloch vector of a two-level quantum state. With $\rho =(I+\mathbf{P}\cdot \mathbf{\sigma })/2$, full polarization has $|\mathbf{P}|=1$ and is pure, while zero polarization has $\rho =I/2$ and is maximally mixed. This internal-state ensemble polarization differs from the polarization of a light wave used to drive it.

The qubit density operators $\rho =(I+\mathbf{s}\cdot \mathbf{\sigma })/2$ correspond exactly to the closed unit ball: their eigenvalues are $(1\pm |\mathbf{s}|)/2$, so positivity is equivalent to $|\mathbf{s}|\le 1$. The boundary is the Bloch sphere of pure states, and the centre is $I/2$. A depolarizing channel scales this ball isotropically, possibly with a sign reversal in the larger completely positive parameter interval.

The convex hull of the six Pauli eigenstates in the Bloch ball. A mixture has a Bloch vector obeying the displayed inequality. Conversely, assign weights $|s_x|,|s_y|,|s_z|$ to the corresponding signed vertices and distribute the leftover weight over any opposite pair; this realizes every vector satisfying the inequality. Single-qubit Clifford gates permute its vertices, while a T gate can take a vertex to a pure state outside it.

For a qubit density matrix $\rho =(I+n\cdot \sigma )/2$ with nonzero Bloch vector, $\hat{n}=n/\Vert n\Vert$ and the multiplication rule for the Pauli matrices give these orthogonal rank-one spectral projectors, with eigenvalues $(1\pm \Vert n\Vert )/2$. Their Bloch vectors are opposite unit vectors $\pm \hat{n}$. At $n=0$ the state is maximally mixed and the direction is undefined; any orthonormal eigenbasis gives an antipodal pair, but no pair is distinguished by the state.

Choose a Hilbert-Schmidt inner product orthonormal Hermitian operator basis with final element $I/\sqrt{N}$. A density operator has real coefficients $r_k=\mathrm{Tr}({\sigma }_k\rho )$ and fixed final coefficient $1/\sqrt{N}$. The remaining coefficients form its generalized Bloch vector. Positivity restricts these real vectors to a compact convex body; for $N>2$ this body is not the entire Euclidean ball allowed by a purity bound.

In a Hilbert-Schmidt orthonormal operator basis, the generator $-i[H,\cdot ]$ is represented on the traceless Hermitian subspace by a real skew-symmetric matrix. Its propagator is therefore in the special orthogonal group and preserves the purity of a density operator. For $N>2$, these rotations form the adjoint image of the accessible unitary group, generally a proper subgroup of the full Euclidean rotation group; preservation of the full spectrum of the density operator imposes further constraints.

For a Hilbert-Schmidt orthonormal traceless Hermitian basis, the Generalized Bloch representation gives the displayed identity by orthogonality. Nonnegative eigenvalues of a density operator sum to one, so their squared sum is at most one, with equality precisely for a rank-one projector. For $N>2$, the radius bound alone does not guarantee positivity: $(2/N)I-P$ for a rank-one projector $P$ has trace one and squared norm one, but one negative eigenvalue.

A time-independent Lindblad equation becomes an affine linear equation for the Generalized Bloch representation. If $\mathcal{L}(I)={\sum }_d[V_d,V_d^{†}]$, then $A_{mn}=\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))$ and $c_m=\mathrm{Tr}({\sigma }_m\mathcal{L}(I))/N$ for traceless basis indices. The offset vanishes exactly when the dynamics are unital. Equilibria solve $A{s}_{\ast }=-c$, and deviations evolve as $s(t)-s_{\ast }=e^{At}(s(0)-s_{\ast })$.

Decay of the off-diagonal coherence of a two-level density operator. Population relaxation contributes half its total rate; additional dephasing contributes ${\gamma }_{\varphi }\ge 0$ in the usual completely positive two-level model. Thus transverse relaxation should not automatically be identified with pure dephasing alone.

For a two-level density operator, rates ${\gamma }_{12}$ from level one to two and ${\gamma }_{21}$ in the reverse direction give ${\dot{p}}_1=-{\gamma }_{12}{p}_1+{\gamma }_{21}{p}_2$. Their sum sets the longitudinal population relaxation rate, and their ratio fixes the stationary population imbalance. If both rates vanish, populations are conserved.

An arbitrary real affine equation has steady states exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. Its solution set is $s_{\ast }+\ker A$. For a finite-dimensional time-independent Lindblad equation, physical stationary states always exist by time averaging trajectories. If $A$ is invertible the steady state is unique. Conversely, a nonzero traceless Hermitian stationary direction can be decomposed into the difference of two equally weighted density operators; time averaging their trajectories yields two distinct stationary density operators. Thus singularity of the physical Bloch matrix also obstructs physical uniqueness.

The Bloch sphere is the unit sphere of Bloch vectors, corresponding to pure qubit states through $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$. A local unitary change of qubit basis induces a rotation of this sphere, with its common phase physically irrelevant. Mixed states occupy the interior of the unit ball.

Write each pure density matrix as $(I+r\cdot \sigma )/2$. The trace identities of the Pauli matrices, $\mathrm{Tr}{\sigma }_i=0$ and $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$, make the trace of their product equal to the displayed overlap. Antipodal unit vectors represent orthogonal states; vectors at right angles correspond to squared overlap one half. For mixed states the same trace formula gives Hilbert–Schmidt overlap, not in general quantum fidelity.

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.

Three Hermitian expectation values determine an arbitrary qubit state exactly when the traceless parts of the three observables span the three-dimensional space generated by the Pauli matrices. Mere linear independence as Hermitian matrices is insufficient: $I,{\sigma }_x,{\sigma }_z$ are independent but cannot detect the sign or magnitude of the Bloch $y$ component.