# Experimental test of the non-statisticality theorem?

Context: The paper On the reality of the quantum state (Nature Physics 8, 475–478 (2012) or arXiv:1111.3328) shows under suitable assumptions that the quantum state cannot be interpreted as a probability distribution over hidden variables.

In the abstract, the authors claim: "This result holds even in the presence of small amounts of experimental noise, and is therefore amenable to experimental test using present or near-future technology."

The claim is supported on page 3 with: "In a real experiment, it will be possible to establish with high confidence that the probability for each measurement outcome is within $\epsilon$ of the predicted quantum probability for some small $\epsilon> 0$."

Something felt like it was missing so I tried to fill in the details. Here is my attempt:

First, without even considering experimental noise, any reasonable measure of error (e.g. standard squared error) on the estimation of the probabilities is going to have worst case bounded by:

$$\epsilon\geq\frac{2^n}{N},$$

(Tight for the maximum likelihood estimator, in this case) where $n$ is the number of copies of the system required for the proof and $N$ is the number of measurements (we are trying to estimate a multinomial distribution with $2^n$ outcomes).

Now, they show that some distance measure on epistemic states (I'm not sure if it matters what it is) satisfies:

$$D \ge 1 - 2\epsilon^{1/n}.$$

The point is, we want $D=1$. So, if we can tolerate an error in this metric of $\delta=1-D$ (What is the operational interpretation of this?), then the number of measurements we must make is:

$$N \ge \left(\frac4\delta\right)^n.$$

This looks bad, but how many copies do we really need? Note that the proof requires two non-orthogonal qubit states with overlap $|\langle \phi_0 |\phi_1\rangle|^2 = \cos^2\theta$. The number of copies required is implicitly given by:

$$2\arctan(2^{1/n}-1)\leq \theta.$$

Some back-of-the-Mathematica calculations seems to show that $n$ scales at least quadratically with the overlap of the state.

Is this right? Does it require (sub?)exponentially many measurements in the system size (not surprising, I suppose) and the error tolerance (bad, right?).

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Removed some comments (some offensive and some not) which combined to form a debate which was peripheral to this technical and concrete question. There are plenty of places to have debates on this paper and its deep meaning, an issue which attracted some attention. Here, I'd hope, is a place where people can engage with the specifics of the question asked. – user566 Nov 19 '11 at 6:42
I think the distance chosen does matter as $D(\mu_0,\mu_1) = 1$ iff $\mu_0$ and $\mu_1$ have disjoint support, that is, if your ontological model is actually $\psi$-ontic. So $1-D$ is just a measure of how $\psi$-epistemic is your ontological model. – Mateus Araújo Nov 29 '11 at 17:31
Of course, this is not the only distance with this property (consider fidelity-derived distances). Probably it was just the more convenient to prove their bound. But it is a mystery to me why they called it "total variation distance" instead of the more common "trace distance". – Mateus Araújo Nov 29 '11 at 18:37
My understanding is that numerical tests by Pusey support the fact that the bounds in the paper are as tight as they can be or at least that their scaling is. Thus my feeling is that your question is open: any result significantly improving them would be a worthwhile addition to the paper's math, as would a result proving that they are tight. – Emilio Pisanty Jul 2 '12 at 21:24
May be tangential to your question, but here's another question about the same article: physics.stackexchange.com/q/17170/3998 – Siva Mar 20 '13 at 5:37