## Worse than quantum physics, part 2

This is Part 2 of a two part explanation — Part 1 is here. It won’t make much sense on its own!

In this post I’m going to get into the details of the analogy I set up last time. So far I’ve described how the PR box is ‘worse than quantum physics’ in a specific sense: it violates the CHSH inequality more strongly than any quantum system, pushing past the Tsirelson bound of $2\sqrt{2}$ to reach the maximum possible value of 4. I also introduced Piponi’s box example, another even simpler ‘worse than quantum physics’ toy system.

This time I’ll explain the connection between Piponi’s box and qubit phase space, and then show that a similar CHSH-inequality-like ‘logical Bell inequality’ holds there too. In this case the quantum system has a Tsirelson-like bound of $\sqrt{3}$, interestingly intermediate between the classical limit of 1 and the maximum possible value of 3 obtained by Piponi’s box. Finally I’ll dump a load of remaining questions into a Discussion section in the hope that someone can help me out here.

## A logical Bell inequality for the Piponi box

Here’s the table from the last post again:

 Measurement T F $a$ 1 0 $b$ 1 0 $a \oplus b$ 1 0

As with the PR box, we can use the yellow highlighted cells in the table to get a version of Abramsky and Hardy’s logical Bell inequality $\sum p_i \leq N-1$, this time with $N = 3$ cells. These cells correspond to the three incompatible propositions $a, b, a\oplus b$, with combined probability $\sum p_i = 3$, violating the inequality by the maximum amount.

Converting to expected values $E_i = 2p_i -1$ gives

$\sum E_i = 3 > N-2 = 1$.

So that’s the Piponi box ↔ PR box part of the analogy sorted. Next I want to talk about the qubit phase space ↔ Bell state part. But first it will be useful to rewrite the table of Piponi box results in a way that makes the connection to qubit phase space more obvious:

The four boxes represent the four ‘probabilities’ $P(a,b)$ introduced in the previous post, which can be negative. To recover the values in the table, add up rows, columns or diagonals of the diagram. For example, to find $p(\lnot a)$, sum up the left hand column:

$p(\lnot a) = P(\lnot a, b) + P(\lnot a, \lnot b) = \frac{1}{2} - \frac{1}{2} = 0$.

Or to find $p(a \oplus b)$, sum up the top-left-to-bottom-right diagonal:

$p(a \oplus b) = P(a, \lnot b) + P(\lnot a, b) = \frac{1}{2} + \frac{1}{2} = 1$.

I made the diagram below to show how this works in general, and now I’m not sure whether that was a good idea. It’s kind of busy and looking at the example above is probably a lot more helpful. On the other hand, I’ve gone through the effort of making it now and someone might find it useful, so here it is:

## Qubit phase space

That’s the first part of the analogy done, between the PR box and Piponi’s box model. Now for the second part, between the CHSH system and qubit phase space. I want to show that the same set of measurements that I used for Piponi’s box also crops up in quantum mechanics as measurements on the phase space of a single qubit. This quantum case also violates the classical bound of $\sum E_i = 1$, but, as with the Tsirelson bound for an entangled qubit system, it doesn’t reach the maximum possible value. Instead, it tops out at $\sum E_i = \sqrt{3}$.

The measurements $a, b, a\oplus b$ can be instantiated for a qubit in the following way. For a qubit $|\psi\rangle$, take

$p(a) = \langle \psi | Q_z | \psi \rangle$,

$p(b) = \langle \psi | Q_x | \psi \rangle$,

with $Q_i = \frac{1}{2}(I-\sigma_i)$ for the Pauli matrices $\sigma_i$. The $a\oplus b$ diagonal measurements then turn out to correspond to

$p(a\oplus b) = \langle \psi | Q_y | \psi \rangle$,

completing the set of measurements.

This is the qubit phase space I described in my second post on negative probability – for more details on how this works and how the corresponding $P(a,b)$s are calculated, see for example the papers by Wootters on finite-state Wigner functions and Picturing Qubits in Phase Space.

As a simple example, in the case of the qubit state $|0\rangle$ these measurements give

$p(a) = 0$

$p(b) = \frac{1}{2}$

$p(a\oplus b) = \frac{1}{2}$,

leading to the following phase space:

## A Tsirelson-like bound for qubit phase space

Now, we want to find the qubit state $|\psi\rangle$ which gives the largest value of $\sum p_i$. To do this, I wrote out $|\psi\rangle$ in the general Bloch sphere form $|\psi\rangle = \cos(\theta / 2) |0\rangle + e^{i\phi} \sin(\theta / 2) |1\rangle$ and then maximised the value of the highlighted cells in the table:

$\sum p_i = p(a) + p(b) + p(a\oplus b) = \frac{3}{2} - \frac{1}{2}(\cos\theta + \sin\theta\cos\phi + \sin\theta\sin\phi )$

This is a straightforward calculation but the details are kind of fiddly, so I’ve relegated them to a separate page (like the boring technical appendix at the back of a paper, but blog post style). Anyway the upshot is that this quantity is maximised when $\phi = \frac{5\pi}{4}$, $\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}$ and $\cos\theta = -\frac{1}{\sqrt{3}}$, leading to the following table:

 Measurement T F $a$ $\frac{1}{2}\left(1 + \frac{1}{\sqrt{3}} \right)$ 0 $b$ $\frac{1}{2}\left(1 + \frac{1}{\sqrt{3}} \right)$ 0 $a \oplus b$ $\frac{1}{2}\left(1 + \frac{1}{\sqrt{3}} \right)$ 0

The corresponding qubit phase space, if you’re interested, is the following:

Notice the negative ‘probability’ in the bottom left, with a value of around -0.183. This is in fact the most negative value possible for qubit phase space.

This time, adding up the numbers in the yellow-highlighted cells of the table gives

$\sum p_i = \frac{3}{2}\left(1 + \frac{1}{\sqrt{3}} \right)$,

or, in terms of expectation values,

$\sum E_i = \sum (2p_i - 1) = \sqrt{3}$.

So $\sqrt{3}$ is our Tsirelson-like bound for this system, in between the classical limit of 1 and the Piponi box value of 3.

## Further questions

As with all of my physics blog posts, I end up with more questions than I started with. Here are a few of them:

Is this analogy already described in some paper somewhere? If so, please point me at it!

Numerology. Why $\sqrt{3}$ and not some other number? As a first step, I can do a bit of numerology and notice that $\sqrt{3} = \sqrt{N/2}$, where $N=6$ is the number of cells in the table, and that this rule also fits the CHSH bound of $2\sqrt{2}$, where there are $N=16$ cells.

I can also try this formula on the Mermin example from my Bell post. In that case $N=36$, so the upper bound implied by the rule would be $3\sqrt{2}$ … which turns out to be correct. (I didn’t find the upper bound in the post, but you can get it by putting $\tfrac{1}{8}(2+\sqrt 2)$ in all the highlighted cells of the table, similarly to CHSH.)

The Mermin example is close enough to CHSH that it’s not really an independent data point for my rule, but it’s reassuring that it still fits, at least.

What does this mean? Does it generalise? I don’t know. There’s a big literature on different families of Bell results and their upper bounds, and I don’t know my way around it.

Information causality. OK, playing around with numbers is fine, but what does it mean conceptually? Again, I don’t really know my way around the literature. I know there’s a bunch of papers, starting from this one by Pawlowski et al, that introduces a physical principle called ‘information causality’. According to that paper, this states that, for a sender Alice and a receiver Bob,

the information gain that Bob can reach about the previously unknown to him data set of Alice, by using all his local resources and $m$ classical bits communicated by Alice, is at most $m$ bits.

This principle somehow leads to the Tsirelson bound… as you can see I have not looked into the details yet. This is probably what I should do next. It’s very much phrased in terms of having two separated systems, so I don’t know whether it can be applied usefully in my case of a single qubit.

If you have any insight into any of these questions, or you notice any errors in the post, please let me know in the comments below, or by email.