Is there a nice formula to represent POVMs $\Pi$ in a Bloch-vector-like form? 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?
 A: 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.
A: 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 ($\sum_i p_i=1$).
