# Relevance of SIC-POVMs to quantum information

What is the real relevance of SIC-POVMs (symmetric informationally complete POVMs) to concrete tasks in quantum information theory? A lot of work has been put into giving explicit constructions, and yet it seems that in many concrete applications that at first glance require SIC-POVM-like objects, one can do without them using e.g. suitably chosen set of random projectors. So, my question is - if the problem of constructing SIC-POVMs in arbitrary dimensions was solved one day, would that enable solving problems which can't be solved or approximated using different techniques now?

SIC-POVMs seem to play a role in some approaches to quantum foundations (see e.g. Chris Fuchs, Bayesian approaches to QM that rely on the existence of SIC-POVMs), but it's unclear to me how "mainstream" such applications are.

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When you describe SIC-like objects do you mean Mutually Unbiased Bases (MUB)? –  Matty Hoban Oct 26 '11 at 18:19
@Matty - yes, that counts, too. –  Marcin Kotowski Oct 26 '11 at 19:55
MUBs are important for tomography. But the connection between SICs and MUBs is not that concrete to my knowledge. They have analogous structure if not a direct relationship. –  Matty Hoban Oct 26 '11 at 20:09
May be some examples are needed for clarification of the question, how suitably chosen set of random projectors are using instead of SIC-POVM-like objects for dimensions there SIC or MUB may be introduced? If SIC is analogue of orthonormal basis (see cite in my answer), it may be compared with using sets of random vectors instead of the orthonormal base. –  Alex 'qubeat' Oct 30 '11 at 11:43
While I can't say how mainstream "Quantum Bayesianism" is, I question your reservation. It's not as if they're proposing a multiverse, action at a distance or spontaneous collapse. An "interpretation-free" application of SICs would be whether one can write a state $\rho$ in terms of POVM elements $E_i$ as $$\rho = -I + d(d+1)\sum_{i=1}^{d^2}\mathrm{tr}(\rho E_i) E_i,$$ which seems to me an interesting question in itself and relevant to quantum state tomography. –  Mateus Araújo Nov 2 '11 at 16:36

Andrew Scott has shown here that SIC-POVMs are optimal (in a certain sense of the word, of course) for quantum state tomography. The intuition follows through these steps (take dim($\mathcal H$) $=d$):

1. If a POVM is optimal it ought to be informationally complete --- so it has $n\geq d^2$ elements.
2. It's elements also should be "as close as possible" to orthogonal so they are maximally unbiased --- this defines a special class of tight measurements.
3. Among all tight measurements, the one with the least number of elements $n =d^2$ is best --- this essentially defines the SIC-POVMs for each dimension $d$.
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Thanks for the answer, though my question should be interpreted as - are there scenarios where a "good enough" result can be obtained without having a SIC-POVM in hand? In other words, is the relative payoff of having a SIC-POVM, as opposed to using some other, perhaps approximate, methods, worth the difficulty of constructing them? –  Marcin Kotowski Oct 30 '11 at 1:37
Ah, well Zhu and Englert (arxiv.org/abs/1105.4561) have shown that just doing a product measurement with a SIC-POVM on each subsystem is "good enough". Since we have analytic constructions of SIC-POVMs for $d\leq 16$, I'm pretty sure we are covered for all conceivably sized "subsystems". –  Chris Ferrie Oct 30 '11 at 1:58
Somewhat related, Englert et al (arxiv.org/abs/1103.1025) have shown that $4$ MUBs probably do not exist in dimension $6$ but you can find $4$ (you can even find $7$) which are close enough to being unbiased that you probably couldn't tell given current experimental errors. –  Chris Ferrie Oct 30 '11 at 2:01