3
$\begingroup$

Consider the following general formulation of the standard quantum state tomography problem: given an unknown state $\rho$, a set of (known) observables $\{\mathcal O_k\}_k$ (generally the elements of some POVM), and a corresponding vector of measured probabilities (or more realistically, frequencies) $\mathbf Y$, we want to retrieve a description of $\rho$, that is, the coefficients of the decomposition of $\rho$ with respect to some "canonical" operatorial basis $\{\sigma_k\}$ (the typical example being the basis built with Pauli matrices).

This amounts to solving the linear problem $\mathbf Y = \mathbf X\boldsymbol \theta$ for $\boldsymbol\theta$. Here, $\mathbf Y$ vector of measured frequencies, $\mathbf X$ the matrix whose elements are the coefficients of the decomposition of the observables $\mathcal O_k$ in terms of $\sigma_j$, and $\boldsymbol\theta$ the vector of coefficients obtained decomposing $\rho$ with respect to $\{\sigma_j\}$ (this notation is from (Granade et al. 2017)).

The naive solution to this linear problem is $\boldsymbol\theta=\mathbf X^+\mathbf Y$, with $\mathbf X^+$ the pseudoinverse of $\mathbf X$. However, this method is known to be problematic. For one thing, it is not guaranteed to produce a positive semidefinite estimate for the state. As mentioned by (Granade et al. 2017), possible workarounds include performing constrained least squares, or using a "two-step approach that outputs the closest physical state to a given matrix".

I'm aware of the plethora of alternative approaches to state tomography. However, I'm specifically looking for references discussing the issues with this "naive" linear reconstruction method, and in particular the related problems of numerical (in)stability and lack of positive semidefiniteness of the estimate. The only thing I found was the brief mention in the paper above, the discussion in (Qi et al. 2013), and some discussion of the numerical stability problems in Appendix A of (Opatrný et al. 1997).

$\endgroup$
1
  • $\begingroup$ Interesting question. I don't know the answer, but here are some thoughts. A straightforward solution to the psd problem is that you simply project onto the psd cone, see doi.org/10.1088/1751-8121/ab8111 I don't see any problems with numerical stability there. Linear inversion can be tricky, but in some cases, the pseudoinverse can be computed analytically (e.g. for tight IC-POVMs such as 2-designs), see the same paper. $\endgroup$ Oct 6 '21 at 14:09
3
$\begingroup$

The lack of positive semidefiniteness is very easy to see. Suppose your quantum state is $|0\rangle\langle 0|$, and you do tomography by measuring in the $X, Y,$ and $Z$ bases. Furthermore, assume there is no experimental error whatsoever. Then after $n$ measurement rounds in each basis, the estimate for $\operatorname{tr}(\rho X)$ will be $2k/n-1$ with probability $2^{-n}\binom{n}{k}$, the estimate for $\operatorname{tr}(\rho Y)$ will be $2l/n-1$ with probability $2^{-n}\binom{n}{l}$, and the estimate for $\operatorname{tr}(\rho Z)$ will be 1 with probability 1.

Now, the problem is that if you do linear inversion, the density matrix will be positive semidefinite only if $k=l=n/2$. If $n$ is odd this is impossible. If $n$ is even this happens with probability $$\left( 2^{-n} \binom{n}{n/2}\right)^2 \sim \frac2{n\pi},$$ which we might as well call zero.

Of course, this particular example is very contrived to make the calculations simple, but should make it clear that whenever your quantum state is near the border of the state space, ordinary statistical fluctuations will be enough to put the estimate outside the border with appreciable probability.

$\endgroup$
6
  • 1
    $\begingroup$ very neat example, thanks. I wonder if one even needs to talk about "linear inversion" in such a situation though. After all, you assume the experimenter got estimates for $\langle\sigma_i,\rho\rangle$, so the corresponding vector of frequencies arguably is the estimated state (at least in the standard choice of representation). So we're essentially saying: the estimated state in this case is $(2k^X/n-1,2k^Y/n-1,1)$ with $k^X,k^Y\sim{\rm Binom}(n,1/2)$, which is almost never a physical state. The problem is then in using estimators for probabilities, rather than for the state itself $\endgroup$
    – glS
    Oct 6 '21 at 14:46
  • $\begingroup$ Well, yes, but "linear inversion" is precisely this, pretending that an estimator is the probability being estimated. $\endgroup$ Oct 6 '21 at 15:20
  • 1
    $\begingroup$ Not really. First of all, "error bounds" are not strict bounds, in general any value will have a nonzero probability. To get the usual +-0.3 bounds you have to make an arbitrary choice of what your confidence interval is. Most importantly, though, the problem remains to come up with a quantum state that is compatible with your estimators. Linear inversion cannot do that. You need a different method. Common choices are maximum likelihood, least-squares, or Bayesian tomography. $\endgroup$ Oct 6 '21 at 19:29
  • 1
    $\begingroup$ +1 The difficulty lies in the attempt to combine an expression of uncertainty arising from experimental observations with the certainty of the constraints imposed by the physics (specifically, that $\rho$ be positive semi-definite with unit trace). Error bounds are ill-suited to summarizing uncertainty on a bounded set especially in the vicinity of its boundary which, as this answer makes clear, is a common and interesting case (e.g. high purity states live near the boundary of the set of positive semi-definite operators). $\endgroup$ Oct 14 '21 at 21:52
  • 1
    $\begingroup$ One can have a meaningful "error bound" in a bounded set, though. You set the confidence $1-\alpha$ you're interested, and integrate your posterior distribution from the peak until you reach $1-\alpha$. The limits of integration will give you the confidence interval $[a,b]$. Of course, this won't give symmetric bounds in general, and your estimate might not even be inside $[a,b]$. One can do the same thing for a quantum state, calculate the posterior distribution on the state space and integrate to find the confidence region. This can even be done analytically for one qubit. $\endgroup$ Oct 15 '21 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.