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 n denote a unit vector, i.e., n∈Rd2−1 and ∑d2−1i=1n2i=1, and define Fn=∑d2−1μ=1nμFμ. Let the minimum eigenvalue of each Fn be denoted m(Fn). The "Bloch space" B(Rd2−1) is the set of all Bloch vectors and is a closed convex set, since the set of states S(H) is closed and convex, and the map b↦ρ is linear homeomorphic. The Bloch space is characterized in the "spherical coordinates" determined by {Fn} as B(Rd2−1)={b=rn∈Rd2−1:r≤1|m(Fn)|}. 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 {Fμ} being the Pauli matrices; it is a simple sphere, since it so happens that for a qubit the minimum eigenvalues m(Fn) are 1 for all Fn.
Here Fμ, is a basis of Hermitean operators normalised as Tr(FμFν)=dδμν.
What's the reasoning behind the r≤1|m(Fn)| 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.
Answer
\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 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
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.
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