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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.

glS
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mavzolej
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1 Answers1

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$\newcommand{\bs}[1]{\boldsymbol{#1}}\newcommand{\on}[1]{\operatorname{#1}}$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 is readily verified 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 $f(\bs b)\ge 0$, which means that $f(\bs b)$ not always represents a state.

Define the "generalised Bloch sphere" (which isn't actually a sphere btw) as the set of $\bs b\in\mathbb R^{d^2-1}$ such that $f(\bs b)$ is a state, that is, $$B(\mathbb R^{d^2-1})\equiv\left\{ \bs b\in\mathbb R^{d^2-1}\text{ such that }f(\bs b)\ge0\right\}.$$ Now the problem is figuring out for what $\bs b$ we have $$d \,\,f(\bs b)=I+\sum_{\mu=1}^{d^2-1}b_\mu F_\mu\ge0.$$ Suppose $\bs b$ points in some direction $\hat{\bs n}$, $\|\hat{\bs n}\|=1$. Write $\|\bs b\|=r$, so that $\bs b=r\hat{\bs n}$. The condition then reads $$d\,\,f(\bs b)=I + \bs b \cdot \bs F = I + r \,\hat{\bs n}\cdot\bs F \ge 0,\tag{A}$$ where $\bs F=(F_1,...,F_{d^2-1})$. Note that $\bs F_{\hat{\bs n}}\equiv \hat{\bs n}\cdot\bs F$ is again traceless and Hermitian, which means that $\bs F_{\hat{\bs n}}$ can be unitarily diagonalised, and thus the same must hold for $I + r \,\bs F_{\hat{\bs n}}$.

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 $\bs F_{\hat{\bs n}}$. 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 $B^{(ij)}$, $i<j$, 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 $1$.

glS
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