# Characterisation of the generalised Bloch space in spherical coordinates

I'm so confused by the following definition in the "Quantum Error Correction" by Lidar and Brun that not even sure how to formulate the question properly.

Let $\mathbf n$ denote a unit vector, i.e., $\mathbf n\in \mathbb R^{d^2-1}$ and $\sum_{i=1}^{d^2-1} n_i^2 = 1$, and define $F_\mathbf n = \sum_{\mu=1}^{d^2-1} n_\mu F_\mu.$ Let the minimum eigenvalue of each $F_\mathbf n$ be denoted $m(F_\mathbf n)$. The "Bloch space" $B(\mathbb R^{d^2-1})$ is the set of all Bloch vectors and is a closed convex set, since the set of states $\mathscr S(\mathscr H)$ is closed and convex, and the map $\mathbf b\mapsto \rho$ is linear homeomorphic. The Bloch space is characterized in the "spherical coordinates" determined by $\{F_\mathbf n\}$ as $$B(\mathbb R^{d^2-1}) = \left\{ \mathbf b = r\mathbf n \in \mathbb R^{d^2-1} : r\leq \frac{1}{|m(F_\mathbf n)|} \right\}. \tag{1.11}$$ This result is useful for visualization of quantum states. For example, for two qubits the Bloch space is given by Eq. (1.11) with $d=4$, which corresponds to a certain 15-dimensional convex set. The Bloch space of a qubit is defined with the $\{F_\mu\}$ being the Pauli matrices; it is a simple sphere, since it so happens that for a qubit the minimum eigenvalues $m(F_\mathbf n)$ are $1$ for all $F_\mathbf n$.

Here $F_\mu$, is a basis of Hermitean operators normalised as $\mathrm{Tr}(F_\mu F_\nu)=d\delta_{\mu\nu}$.

What's the reasoning behind the $r\leq\dfrac{1}{|m(F_{\bf{n}})|}$ requirement? Why does it give the boundary corresponding to the pure states?

Why for a qubit "the minimum $m$ is $1$"? Feels like it's $-1$ (the eigenvalues of Pauli matrices!).

Any explanations would be greatly appreciated.

$$\newcommand{\bs}{\boldsymbol{#1}}\newcommand{\on}{\operatorname{#1}}$$Let $$f:\mathbb R^{d^2-1}\to\mathscr B(\mathscr H)$$ be the mapping from points in $$\mathbb R^{d^2-1}$$ to bounded operators on the Hilbert space $$\mathscr H$$, defined by $$f(\bs b)\equiv \frac{1}{d}\left(I+\sum_{\mu=1}^{d^2-1}b_\mu F_\mu\right).$$ It can be easily checked that, for all $$\bs b\in\mathbb R^{d^2-1}$$, $$\,f(\bs b)$$ is normalised and Hermitian. It is however not always the case that $$f(\bs b)>0$$, which means that $$f(\bs b)$$ not always represents a state.

The Bloch sphere is defined as the set of all those vectors $$\bs b\in\mathbb R^{d^2-1}$$ such that $$f(\bs b)$$ is a state, that is, $$B(\mathbb R^{d^2-1})\equiv\left\{ \bs b\in\mathbb R^{d^2-1}\text{ such that }f(\bs b)\ge0\right\}.$$ The nontrivial problem is that of figuring out for what $$\bs b$$ we have $$d \,\,f(\bs b)=I+\sum_{\mu=1}^{d^2-1}b_\mu F_\mu\ge0.$$ Consider a $$\bs b$$ pointing in an arbitrary direction $$\bs n$$, with $$\|\bs n\|=1, \|\bs b\|=r$$, so that $$\bs b=r\bs n$$. The condition thus becomes $$d\,\,f(\bs b)=I + \bs b \cdot \bs F = I + r \,\bs n\cdot\bs F \ge 0,\tag{A}$$ where $$\bs F=(F_1,...,F_{d^2-1})$$. Note that $$\bs F_{\bs n}\equiv \bs n\cdot\bs F$$ is again traceless and Hermitian, which means that $$\bs F_{\bs n}$$ can be unitarily diagonalised, and thus the same must hold for $$I + r \,\bs F_{\bs n}$$.

Consider (A) in its eigenbasis. Positivity of a Hermitian operator is equivalent to all its eigenvalues being positive. Let us denote with $$\lambda_i$$ the eigenvalues of $$\bs F_{\bs n}$$. We then see that (A) is equivalent to the following set of $$d^2-1$$ inequalities: $$1+r \lambda_i \ge 0,\text{ for all }i=1,...,d^2-1.$$ Note that if a matrix is traceless and Hermitian then it must have negative eigenvalues, that is, $$\lambda_i<0$$ for some $$i$$. If $$\lambda_i\ge0$$ the inequality is trivially satisfied, so let us assume $$\lambda_i<0$$. In this case we want $$r\le1/(-\lambda_i)$$ for all $$i$$, that is $$r\le\frac{1}{\lvert\min_i\lambda_i\rvert}.$$

Another interesting point is that the Bloch representation of qudits, for $$d>2$$, is not, in general, a simple sphere.

To see this, let us fix as basis for the traceless Hermitian operators the matrices $$A^{(ij)}$$ and $$B^{(ij)}$$, $$i, defined to be zero everywhere except in the two-dimensional blocks spanned by the indices $$i$$ and $$j$$, and equal on these blocks to the Pauli matrices $$\sigma_x$$ and $$\sigma_y$$, respectively. In other words, $$A^{(ij)}$$ and $$B^{(ij)}$$ are defined component-wise as $$A^{(ij)}=\sqrt{d/2}(\lvert i\rangle\!\langle j\rvert+\lvert j\rangle\!\langle i\rvert), \qquad B^{(ij)}=\sqrt{d/2}i(\lvert i\rangle\!\langle j\rvert-\lvert j\rangle\!\langle i\rvert)$$ Let us also define $$C^{(\ell)}$$, $$\ell=1,...,d-1$$, as the diagonal matrices $$C^{(\ell)}\equiv\frac{\sqrt d}{\sqrt{\ell(\ell+1)}}\left(\sum_{k=1}^\ell\lvert k\rangle\!\langle k\rvert-\ell\lvert \ell+1\rangle\!\langle \ell+1\rvert\right)$$

As can be easily verified, all of these matrices are mutually orthogonal and normalised as $$\operatorname{Tr}(A^2)=d$$, so to satisfy the assumptions of the first part of the answer.

The smallest eigenvalue of $$A^{(jk)}, B^{(jk)}$$ is $$-\sqrt{d/2}$$, while the smallest eigenvalue of $$C^{(\ell)}$$ is $$-\ell\sqrt{\frac{d}{\ell(\ell+1)}}$$, so clearly the distance between the boundary of the state space and the center is not constant.

Regarding your second question, I would guess that the authors simply meant to say instead the absolute value of the minimum eigenvalue is $$1$$.