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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 $s(t)-s_{\ast }=e^{At}(s(0)-s_{\ast })$.

A traceless Hermitian matrix in a Hilbert-Schmidt inner product orthonormal basis of the real vector space of traceless Hermitian $N$-by-$N$ matrices. One choice uses symmetric and imaginary antisymmetric off-diagonal matrix units, normalized by $1/\sqrt{2}$, and traceless diagonal generators. Together with $I/\sqrt{N}$ they form an orthonormal basis of all Hermitian matrices. The coordinates $s_k=\mathrm{Tr}(\rho {\sigma }_k)$ give the Generalized Bloch representation.

Use the real vector space of Hermitian matrices and its Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{Tr}(AB)$. A Hermitian $N$-by-$N$ matrix has $N$ real diagonal entries and two real parameters for each off-diagonal pair, hence real dimension $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$. Imposing zero trace removes one real dimension. Each supplied generalized Pauli matrix is Hermitian and traceless, and there are $N(N-1)$ off-diagonal matrices and $N-1$ diagonal matrices, totaling $N^2-1$.

Their assumed orthonormality makes them linearly independent. Since this equals the dimension of the traceless space, they are an orthonormal basis of the traceless Hermitian matrices. To extend the family, the required additional generator is ${\sigma }_0=I/\sqrt{N}$, which is implicit in the indexing of the question. It has unit Hilbert-Schmidt norm and is orthogonal to every traceless generator, so the extended family is an orthonormal basis of all Hermitian matrices.

For a density operator $\rho$, cyclicity of the trace givesThus $s\in {\mathbb{R}}^{N^2-1}$. Since $\mathrm{Tr}\rho =1$, the resulting Generalized Bloch representation is

By orthonormality in the Generalized Bloch representation,A pure state has density operator $\rho =|\psi ⟩⟨\psi |$ with $⟨\psi |\psi ⟩=1$, so ${\rho }^2=\rho$ and its purity of a density operator is one. ConsequentlyEvery pure state therefore maps to the specified radius-$\sqrt{1-1/N}$ sphere in ${\mathbb{R}}^{N^2-1}$, of dimension $N^2-2$. The expansion is unique, so it distinguishes pure-state density operators, while vectors differing only by global phase describe the same point. This proves that pure states lie on the sphere, without asserting that they fill it.

The eigenvalues $p_j$ of a density operator are nonnegative and sum to one. ThereforeFor a non-pure state at least two eigenvalues are positive, so the last expression is strictly positive. Combining this with the Generalized Bloch representation givesThus every non-pure quantum state lies strictly inside the radius sphere for $N\ge 2$. It need not be an interior point of the entire physical state body: for $N>2$, a rank-deficient mixed density operator can lie on that body's boundary while still lying inside the Euclidean sphere.

A density operator is Hermitian, positive semidefinite and has trace one. Part (a) therefore makes all its coordinates real, and the identity basis vector fixes the final coordinate:The remaining basis vectors are traceless by orthogonality to $I/\sqrt{N}$. The Generalized Bloch representation is consequentlyUsing orthonormality to expand the squared Hilbert-Schmidt norm gives the purity bound for generalized Bloch vectors:If $p_1,\dots ,p_N$ are its nonnegative eigenvalues, then ${\sum }_j{p}_j=1$ andIt follows thatEquality requires every pairwise product $p_j{p}_k$ to vanish, so exactly one eigenvalue equals one and all others are zero. Conversely, a rank-one projector has precisely that spectrum and attains equality. Thus equality holds exactly for a pure state. This normalization is different from the commonly used unit-radius qubit Bloch vector.

Write $\rho ={\sum }_n{r}_n{\sigma }_n$ in the Generalized Bloch representation. Linearity of the Lindbladian gives ${\dot{r}}_m={\sum }_n\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))r_n$. For the Hamiltonian term, cyclicity yieldsFor one Lindblad operator $V_d$, the jump term becomes $\mathrm{Tr}({\sigma }_m{V}_d{\sigma }_n{V}_d^{†})=\mathrm{Tr}(V_d^{†}{\sigma }_m{V}_d{\sigma }_n)$, and the two remaining terms giveThusEvery coefficient is real because $\mathcal{L}({\sigma }_n)$ is Hermitian. In addition, $L_{mn}=-L_{nm}$, although a dissipative $D^{(d)}$ need not be symmetric.

The Hamiltonian commutator has zero trace, and each dissipator obeys $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, so $\mathrm{Tr}\mathcal{L}(\rho )=0$. Consequently $\boxed{{\dot{r}}_{N^2}=0,r_{N^2}=1/\sqrt{N}}$ for normalized density operators. Orthogonality to $I/\sqrt{N}$ makes every other ${\sigma }_k$ traceless. Define $s=(r_1,\dots ,r_{N^2-1}{)}^T$. With $M=L+{\sum }_d{D}^{(d)}$, trace preservation gives the block formHere the indices of $A$ range from $1$ to $N^2-1$. Since $\mathcal{L}(I)={\sum }_d(V_d{V}_d^{†}-V_d^{†}{V}_d)$, the offset can also be written $c_m=N^{-1}\mathrm{Tr}({\sigma }_m{\sum }_d[V_d,V_d^{†}])$. This is the Affine Bloch equation; its offset is zero for unital dynamics, which preserve the identity.

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.