Is there a nice formula to represent POVMs $\Pi$ in a Bloch-vector-like form?

Question

A way to write a quantum state is to use the Bloch vector representation, i.e., $$ \varrho = \frac{1}{2}\left(\mathbb{I}_2 + \boldsymbol{r} \cdot \boldsymbol{\sigma}\right) $$ In general a POVM for a qubit is the set $$ \Pi = \left\{\Pi_k \geq 0, \,\sum_{i=1}^n \Pi_i = \mathbb{I}_2 \right\} $$ where $n$ is arbitrary.

My question is: is there a nice formula for representing the POVM $\Pi$ in a Bloch-vector-like form? Is there in the literature someone that use this representation? Is it useful for some reason?

Answer 1

Any Hermitian operator $A$ can be decomposed as $\newcommand{\bs}[1]{{\boldsymbol{#1}}}A(\bs\alpha)=\alpha_0 I + \sum_k \alpha_k \sigma_k$ for some basis of traceless orthogonal operators $\sigma_k$ (well, it can be done even if the basis is not orthogonal, but let's assume it here), and real coefficients $\alpha_j\in\mathbb R$.

In the case of a single qubit, the canonical choice is to use the Pauli matrices as basis, so that $\sigma_1,\sigma_2,\sigma_3$ are what you'd expect. A generic Hermitian operator can thus be written as $$A(\bs\alpha) = \alpha_0 I + \vec\alpha\cdot\vec\sigma, \qquad \vec\alpha\cdot\vec\sigma\equiv\sum_{k=1}^3\alpha_k\sigma_k.$$ An Hermitian operator $A(\bs\alpha)$ can be a component of a POVM iff $0\le A(\bs\alpha)\le I$, that is, its eigenvalues are in the interval $[0,1]$. This can be readily seen to be equivalent, in terms of $\alpha$, to the conditions $$\|\vec\alpha\|\le \alpha_0 \le 1-\|\vec\alpha\|,$$ which can also only be verified for $\|\vec\alpha\|\le 1/2$.

If we also want $\{A(\bs\alpha_k)\}_k$ to be a POVM, we need $\sum_k A(\bs\alpha_k)=I$. This gives you the additional conditions $$\sum_k (\bs\alpha_k)_0 = 1,\qquad\sum_k\vec\alpha_k=0.$$ A similar discussion, with some extensions in the context of joint measurability, can be found in quant-ph:0811.0783.

It's worth noting that the conditions depend on the parametrisation of $A(\bs\alpha)$. The parameters are obtainable from the operator via the relations $${\rm Tr}(A(\bs\alpha)\sigma_k) = 2\alpha_k,$$ for $k=0,1,2,3$ and $\sigma_0\equiv I$.

If we were instead to use the parametrisation $A(\bs\beta)\equiv\beta_0(I+\vec\beta\cdot\vec\sigma)$, we'd get the conditions for each component of the POVM $$\|\vec\beta\|\le 1, \qquad 0\le\beta_0 \le \frac{1}{1+\|\vec\beta\|},$$ while the additional conditions for $\{A(\bs\beta_k)\}$ to be a POVM would be $$\sum_k(\bs\beta_k)_0 = 1,\qquad\sum_k \vec\beta_k=0.$$ With this parametrisation the coefficients are obtained from the operator via $2\beta_0\beta_k={\rm Tr}(A(\bs\beta)\sigma_k)$. These conditions should be consistent with the ones given in the other answer, with the constraints between $(\bs\beta_k)_0$ and $\vec\beta_k$ probably automatically satisfied when the other conditions are.

Answer 2

Equation (25) in this recent reference says that the locally-optimal POVM for measuring the angle $\theta$ of a unitary $U=\exp(-i\theta\pmb{n}\cdot\pmb{\sigma}/2)$ that rotates a qubit $$\rho=\frac{1}{2}\left(\mathbb{I}+ \pmb{a}\cdot\pmb{\sigma}\right)$$ about axis $\pmb{n}$ is defined by two projection operators $$M_\pm=\frac{1}{2}\left(\mathbb{I}\pm \frac{\pmb{n}\times\pmb{a}}{\left|\pmb{n}\times\pmb{a}\right|}\cdot\pmb{\sigma}\right),$$ where $\pmb{\sigma}$ is a vector of Pauli operators. So indeed the answer is yes!

A general POVM could be defined by any set of operators like $$M_i=\frac{p_i}{2}\left(\mathbb{I}+ \pmb{m}_i\cdot\pmb{\sigma}\right),\qquad i\in(1,n),\qquad|\pmb{m}_i|\leq 1,\qquad \sum_i p_i=1,\qquad \sum_i\pmb{m}_i=\pmb{0}.$$ This is the most general set that is positive ($|\pmb{m}_i|\leq 1$) and complete ($\sum_i p_i=1$).