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Given a quantum state, the Born rule lets us compute probabilities. Conversely, given probabilities, can we reconstruct the quantum state? I think the answer is almost trivially positive but how simple can the reconstruction formula be?

Let me illustrate with the archetypical example of the spin 1/2 system. I give myself a density operator $D$ and $n$ projectors $P_i$, and then the associated probabilities of measurement

$$\newcommand{\tr}{\mathop{\rm tr}}p_i=\tr DP_i.$$

I only require that “all cases are covered”, i.e. that

$$\sum_{i=1}^n p_i = 1.$$

Can I express $D$ as an expansion on those projectors? I.e.

$$D=\sum_{i=1}^n a_iP_i,$$

where the coefficients $a_i$ can be computed from the probabilities $p_i$? It turns out that not only the answer is positive but that there is a very simple formula,

$$D = \sum_{i=1}^4 \left(\frac{3}{2}p_i-\frac{1}{4}\right)P_i,$$

by choosing the projectors as

$$P_i=\frac{1}{2}\left(I+\frac{u_i\cdot\sigma}{\sqrt{3}}\right),$$

where $\sigma$ is the usual Pauli vector and where,

$$\begin{align} u_1 &= (+1,+1,+1),\\ u_2 &= (+1,-1,-1),\\ u_3 &= (-1,+1,-1),\\ u_4 &= (-1,-1,+1). \end{align}$$

This result took me quite a bit of trials and errors but it is surely well known! However I have not encountered it in lectures and textbooks. A reference would be appreciated!

Let me emphasise that my question is not whether such a reconstruction of $D$ from the probabilities exists: that is not suprising, as the coefficients $a_i$ can be obtained as solution of the system of equations

$$\sum_{j=1}^n \tr(P_iP_j)\ a_j = p_i,\ i=1,\cdots,n.$$

But that one can choose the projectors so that the coefficients are as simple as in the reconstruction above is really suprising to me. So my question is whether this generalises to quantum systems more complex than the spin 1/2 system. Are there other systems where such a simple reconstruction holds? Would there even be a general framework to find such reconstructions?

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  • $\begingroup$ You want to do a keyword search on “quantum tomography”. For the specific case you have in mind, the keyword is “mutually unbiased bases”. $\endgroup$ – ZeroTheHero Nov 19 '17 at 19:53
  • $\begingroup$ @ZeroTheHero Having just found out I can now comment, thank you! Your suggestion gave me plenty to read. As I expected, I reinvented the wheel but that was formative… $\endgroup$ – frapadingue Nov 22 '17 at 21:36
  • $\begingroup$ well... only a $2\times 2$ wheel so not so bad... ;) $\endgroup$ – ZeroTheHero Nov 22 '17 at 21:46

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