# 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?

• Literature examples: doi.org/10.1103/PhysRevA.94.022334 and doi.org/10.22331/q-2021-06-04-467 Commented Sep 20, 2021 at 19:08
• this might be of interest: arxiv.org/abs/quant-ph/0604189. Tbh, I clearly remember seeing a series of iff conditions listed to determine whether a given operator is an effect (i.e. an element of a povm) in terms of its Bloch-sphere representatoin parameters. It was an old "pre-latex" paper, related to contextuality I'm pretty sure, but I can't seem to find it now
– glS
Commented Sep 21, 2021 at 15:33

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.

• This answer can be improved by showing how to express the coefficients in terms of the operator (ie as the trace of a product with a Pauli matrix). Commented Sep 22, 2021 at 10:24
• @EmilioPisanty good point, I added the relations
– glS
Commented Sep 22, 2021 at 10:55

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 ($$\sum_i p_i=1$$).

• Eq. (25) is actually a PVM. In any case you defined a $M_i$ in which there is no arbitrary prefactor in front of the identity, is this the most general case? and the vector $m_i$ is not restricted, i. e. is a positive bounded vector? What constraints it must satisfy? Commented Sep 20, 2021 at 19:39
• Yes, in that case the neat thing is that the optimal POVM is in fact a PVM. I'll elaborate the definition of the most general $M_i$ in my edit Commented Sep 21, 2021 at 13:04