Fix some finite-dimensional space $\mathcal X$. Define a POVM as a collection of positive operators summing to the identity: $\mu\equiv \{\mu(a):a\in\Sigma\}\subset{\rm Pos}(\mathcal X)$ such that $\sum_{a\in\Sigma}\mu(a)=I_{\mathcal X}$, where $\Sigma$ denotes some finite register.

A POVM $\mu$ is said to be extremal if it cannot be written as a (nontrivial) convex combination of other POVMs. Moreover, we say that a POVM is projective if each $\mu(a)$ is a (not necessarily unit trace) projection.

As discussed in Watrous' book, one can prove that any projective POVM is extremal.

Does the implication work in the other direction as well? If not, what are examples of extremal non-projective POVMs?

(Impossible for binary-outcome POVMs) At least in the case of two-element POVMs, this doesn't seem to be possible: if $\mu(1)+\mu(2)=I$, then they are mutually (unitarily) diagonalisable. In an appropriate choice of basis, we can write $$\mu(1)= \operatorname{diag}(s_1,...,s_n), \qquad \mu(2)=\operatorname{diag}(1-s_1,...,1-s_n),$$ for some $s_i\in[0,1]$. Such a POVM is projective iff $s_i\in\{0,1\}$ for all $i=1,...,n$. Suppose that the POVM is not projective, and thus there is some $s_i\in(0,1)$. Let us assume without loss of generality that $s_1\in(0,1)$. Define the Hermitian operators $$\theta(1)\equiv \operatorname{diag}(\epsilon,0,...,0), \qquad \theta(2)\equiv \operatorname{diag}(-\epsilon,0,...,0),$$ for some $\epsilon< \min(s_1,1-s_1)$. Then $\mu\pm\theta$ are again two POVMs, and $$\mu=\frac{1}{2}((\mu+\theta)+(\mu-\theta)),$$ and thus $\mu$ is not extremal. However, this argument relies on the POVM elements being mutually diagonalisable, which is only the case for two-outcome POVMs.

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    $\begingroup$ Any SIC-POVM will be extremal and non-projective. I guess you want a proof of extremality, though, so I'll let somebody else post an answer. $\endgroup$ Commented May 5, 2021 at 7:46
  • $\begingroup$ @MateusAraújo great tip, thanks! One can indeed show easily that any POVM whose components have rank one is extremal iff the components are linearly independent, which is always the case for SIC-POVMs. $\endgroup$
    – glS
    Commented May 5, 2021 at 11:04

1 Answer 1


It is indeed possible to have extremal, non-projective POVMs. Examples can be drawn from SIC-POVMs, as suggested in a comment.

For example, as mentioned in the Wiki page, the only possible SIC-POVM in $d=2$ dimensions is $\{\tilde\Pi_i\}_{i=1}^4$ where $\tilde\Pi_i\equiv \frac12 \Pi_i$, $\Pi_i\equiv \lvert\psi_i\rangle\!\langle\psi_i\rvert$, and $$|\psi_1\rangle\equiv |0\rangle, \qquad |\psi_2\rangle\equiv \frac1{\sqrt3}(|0\rangle+\sqrt2 |1\rangle), \\ |\psi_3\rangle\equiv \frac1{\sqrt3}(|0\rangle+\sqrt2 \omega_3 |1\rangle), \qquad |\psi_4\rangle\equiv \frac1{\sqrt3}(|0\rangle+\sqrt2 \omega_3^2 |1\rangle),$$ where $\omega_3 \equiv \exp(2\pi i/3)$.

To observe extremality, we can use the following characterisation, given e.g. in Watrous' book:

a POVM $\mu\equiv \{\mu(a):a\in\Sigma\}\subset\mathrm{Pos}(\mathcal X)$ is extremal iff there is a collection of Hermitian operators $\{\theta(a):a\in\Sigma\}\subset\mathrm{Herm}(\mathcal X)$ such that (1) $\sum_a \theta(a)=0$ and (2) $\mathrm{im}(\theta(a))\subseteq\mathrm{im}(\mu(a))$ for all $a\in\Sigma$.

In the case at hand, $\mathrm{im}(\tilde\Pi_i)=\mathbb C|\psi_i\rangle$, and thus any collection of Hermitians satisfying the above conditions must have the form $\theta_i= a_i \tilde\Pi_i$ for some $a_i\in\mathbb R$. But then, the condition $\sum_i \theta_i=0$ amounts to $\sum_i a_i \tilde\Pi_i=0$. But $\{\tilde \Pi_i\}_{i=1}^4$ is linearly independent, and thus $a_i=0$ and $\theta_i=0$ for all $i$. This means that the POVM $\{\tilde\Pi_i\}_{i=1}^4$ is extremal.

More generally, as again discussed in Watrous' book, a POVM whose components are all unit-rank is extremal iff the components are linearly independent, and this is always the case for SIC-POVMs (they consist of $d^2$ operators spanning $\mathrm{Herm}(\mathcal X)$ with $\dim(\mathcal X)=d$, thus they are linearly independent, and they are rank-1 by definition).

For an example of a rank-1 extremal POVM that is not a SIC-POVM consider: $$\mu(1) \equiv \frac12\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad \mu(2) \equiv \frac12\begin{pmatrix}0&0\\0&1\end{pmatrix}, \\ \mu(3) \equiv \frac14\begin{pmatrix}1&1\\1&1\end{pmatrix}, \qquad \mu(4) \equiv \frac14\begin{pmatrix}1&-1\\-1&1\end{pmatrix}.$$

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    $\begingroup$ I later found this 2005 paper by D'Ariano et al., which discusses, among other things, the inequivalence between a POVM being extremal and it being rank-1 $\endgroup$
    – glS
    Commented Jul 8, 2022 at 6:59
  • $\begingroup$ Just to be sure, your last sentence should be "an example of a rank-1 POVM that is not extremal" right? Since $\mu(1)+\mu(2)-\mu(3)-\mu(4)=0$, it means that the set $\left\{\mu(i)\right\}_i$ isn't linearly independent, and thus isn't an extremal POVM per the characterization right? $\endgroup$
    – Tristan Nemoz
    Commented Jun 10 at 16:01
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    $\begingroup$ @TristanNemoz indeed, you're right! I'll edit the answer with a correct example $\endgroup$
    – glS
    Commented 21 hours ago

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