I started writing the previous post as an introduction to another subject, but it got too long and I decided to publish it separately. What I actually wanted to talk about is the following question: what happens if you are doing device-independent quantum key distribution (DIQKD) experimentally, and you violate the Tsirelson bound?

I don’t meant in the sense of violating quantum mechanics, but doing it in the way quantum mechanics allows. If you do the experiment perfectly, then the probability of winning each round of the CHSH game is exactly the Tsirelson bound $\omega = \frac{2+\sqrt{2}}{4}$. Then the probability of getting a number of victories $v$ out of $n$ rounds of the CHSH game such that

\[\frac{v}{n} > \frac{2+\sqrt{2}}{4}\] is given by

\[ \sum_{v = \lceil n \omega \rceil}^n \binom{n}{v} \omega^v(1-\omega)^{n-v}. \] This is not small at all, it is equal to $\omega$ for $n=1$, and goes to 1/2 for large $n$.

So yeah, it’s perfectly possible to violate the Tsirelson bound, and it is not a matter of experimental error or doing too few rounds1. On the contrary, experimental error is precisely what makes it very unlikely to win the CHSH game too often. This is very unsatisfactory, though, we are relying on experimental error to sweep the problem under the rug. Clearly DIQKD must also work in the ideal case.

Even if you only care about the realistic case, there’s a different scenario where this matters: as proposed by Brown and Fawzi$^{\otimes 2}$, one can use an estimate of the whole probability distribution to do DIQKD instead of only the probability of winning the CHSH game. This makes it harder for the eavesdropper to cheat, and thus gives us better key rates. The problem is that we’re now dealing with a high-dimensional object instead of a one-dimensional parameter, and we need the estimates of all the parameters to land in the quantum-allowed region. The probability that at least one falls outside is appreciable. It’s hard to give a precise statement about this, because it will depend on the quantum state and the measurements you are doing, but the fact of the matter is that experimentalists routinely get estimates outside of the quantum-allowed region2. For simplicity, though, we’ll focus on the one-dimensional case.

Why wouldn’t it work, though? What is the problem with violating the Tsirelson bound? The idea of DIQKD is that Alice and Bob play the CHSH game, calculate the frequency of winning $\frac{v}n$, and do an optimization over all quantum states with winning probability equal to $\frac{v}n$, picking up the worst, that is, the one that gives Eve the most information about the key they’re trying to generate. Well, if $\frac{v}n > \frac{2+\sqrt{2}}{4}$ there’s just no quantum state with this winning probability, so you can’t figure out how much information Eve can have in this way.

What can we do then? One obvious solution is to say that the winning probability $p$ is equal to $\frac{2+\sqrt{2}}{4}$. After all, this is the closest we can get to the frequency $\frac{v}n$ while staying in the range allowed by quantum mechanics. That’s not a good idea though. We would be assuming that Eve has no information whatsoever about Alice and Bob’s key, while it is perfectly possible that $p$ is slightly smaller than $\frac{2+\sqrt{2}}{4}$, which would give her a bit of information. In fact, the probability that $p$ is exactly $\frac{2+\sqrt{2}}{4}$ is zero, just because $p$ is a continuous parameter. It is very likely, on the other hand, that $p$ is *close* to $\frac{2+\sqrt{2}}{4}$. This is what you should assume. And this is not even related to violating the Tsirelson bound. Even if you find that $\frac{v}n = 0.8$, it would be stupid to assume that $p=0.8$, as it’s almost certainly not. Assuming a flat prior over the quantum-allowed region, the probability density of $p$ is given by

\[f(p|v,n) = \frac{p^v(1-p)^{n-v}}{\int_{1-\omega}^{\omega}\mathrm{d}q\, q^v(1-q)^{n-v}},\] for $p \in [1-\omega,\omega]$ and zero otherwise.

Which finally brings us to the DIQKD papers I mentioned in the previous post. How did they deal with this problem? It turns out, they did something completely different. They set some expected winning probability $p_0$ and some tolerance $\delta$, and if the measured frequency $v/n$ is at least $p_0-\delta$, they assume that the actual probability $p$ is also at least $p_0-\delta$.

I find that very strange, they are not using the measured frequency for anything other than this test, the key rate is calculate solely based on $p_0-\delta$. This is not *wrong*, I must emphasize, the probability that the winning probability is at least $p_0-\delta$ given that the frequency is at least $p_0-\delta$ is indeed very high, and they have a proper security proof. I just find it bizarre that they are discarding valuable information, using the measured frequency can give you a much better idea on what the actual winning probability is. For example, if the measured frequency is very close to $p_0-\delta$, then the probability that the winning probability is at least $p_0-\delta$ is close to 1/2. Not as high as we’d like. On the other hand, if the measured frequency is much higher than $p_0-\delta$, the winning probability is likely much higher, and you’re needlessly lowering your key rate by being so pessimistic.