More Quantum

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 rounds1Often referred to as “finite statistics”, which is a really stupid expression, as all statistics are finite.. 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 region2One example where this essentially always happens is when evaluating whether some probability distribution obeys the no-signalling condition, just because non-signalling correlations are a measure zero subset of the set of all correlations.. 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.

This has been a momentous week for quantum information science. The long-awaited for experimental demonstration of device-independent quantum key distribution (DIQKD) is finally here! And not one only demonstration, but three in a row. First, the Oxford experiment came out, which motivated the München and the Hefei experiments to get their data out quickly to make it clear they did it independently.

To give a bit of context, for decades the community had been trying to do a loophole-free violation of a Bell inequality. To the perennial criticism that such an experiment was pointless, because there was no plausible physical model that exploited the loopholes in order to fake a violation, people often answered that a loophole-free Bell test was technologically relevant, as it was a pre-requisite for DIQKD.3This is essentially true, as I find it inconceivable that someone would be able to do DIQKD without being able to do a loophole-free Bell test, but there are some important differences between the experiments. That was finally achieved in 2015, but DIQKD had to wait until now. It’s way harder, you need less noise, higher detection efficiency, and much more data in order to generate a secure key.

Without further ado, let’s look at the experimental results, summarized in the following table. $\omega$ is the probability with which they win the CHSH game, distance is the distance between Alice and Bob’s stations, and key rate is the key rate they achieved.

Experiment	$\omega$	Distance	Key rate
Oxford	0.835	2 m	3.4 bits/s
München	0.822	700 m	0.0008 bits/s
Hefei	0.756	220 m	2.6 bits/s

I’ve highlight the München and the Hefei key rates in red because they didn’t actually generate secret keys, but rather estimated that this is the rate they would achieve in the asymptotic limit of infinitely many rounds. This is not really a problem for the Hefei experiment, as they were performing millions of rounds per second, and could thus easily generate a key. I suspect they simply hadn’t done the key extraction yet, and rushed to get the paper out. For the München experiment, though, it is a real problem. They were doing roughly 44 rounds per hour. At this rate it would take years to gather enough data to generate a key.

Why is there such a drastic difference between Hefei and München? It boils down to the experimental technique they used to get high enough detection efficiency. Hefei used the latest technology in photon detectors, superconducting nanowire single-photon detectors,2Don’t ask me what this is, I only know it’s good stuff. which allowed them to reach 87% efficiency. München, on the other hand, used a completely different technique: they did the measurement on trapped atoms, which has efficiency of essentially 100%. The difficulty is to entangle the atoms. To do that you make the atoms emit photons, and do an entangled measurement on the photons, which in turns entangles the atoms via entanglement swapping. This succeeds with very small probability, and is what makes the rate so low.

What about Oxford? Their experimental setup is essentially the same as München, so how did they get the rate orders of magnitude higher? Just look at the distance: in Oxford Alice and Bob were 2 metres apart, and in München 700 metres. The photon loss grows exponentially with distance, so this explains the difference very well. That’s cheating, though. If we are two metres apart we don’t need crypto, we just talk.

One can see this decay with distance very well in the Hefei paper: they did three experiments, with a separation of 20, 80, and 220 metres, and key rates of 466, 107, and 2.6 bits/s. In the table I only put the data for 220 metres separation because that’s the only relevant one.

It seems that the Hefei experiment is the clear winner then, as the only experiment achieving workable keyrates over workable distances. I won’t crown them just yet, though, because they haven’t done a standard DIQKD protocol, but added something called “random post-selection”, which should be explained in a forthcoming paper and in the forthcoming Supplemental Material. Yeah, when it appears I’ll be satisfied, but not before.

Another day working with MATLAB, another nasty surprise. I was solving a SDP that was already working well, did a minor change, and suddenly it started taking minutes to reach a solution, when usually it took seconds.

After a long investigation, I realized that the problem was that my input was not exactly Hermitian anymore. I had switched from writing manually the components of a matrix to defining the matrix as $A = \ket{\psi}\bra{\psi}$ for a complex vector $\ket{\psi}$. Now how could that possibly be a problem? It is a very simple theorem to prove that the outer product of a vector with itself is always Hermitian. Well, not with MATLAB. I can’t figure out why, but probably MATLAB uses some smart algorithm for the outer product that ruins this property due to floating-point arithmetic. What I could determine is that if you compute the outer product the naïve way, then even with MATLAB the result will be exactly Hermitian. Also, with Python the output is exactly Hermitian.

To solve this problem, one can simply redefine $A$ as $(A+A^\dagger)/2$, which is very fast and gives an exactly Hermitian output. I don’t like doing that, though, as $(A+A^\dagger)/2$ is always exactly Hermitian, even when $A$ is not even close to Hermitian. If I fucked up the definition of $A$ this will hide my mistake, and make debugging harder. Instead, what I did was write my own outer product function, computing it as $\ket{\psi}\otimes\bra{\phi}$. It is slower than whatever black magic MATLAB is doing, sure, but it is fast enough, and leaves my mistakes easily visible.

It dawned on me that this subtlety was probably the source of many bugs and slowdowns in my codes over the years. I decided to go hunting, and find out the fascinating properties of MATLAB algebra that doesn’t preserve Hermiticity where it should. It turns out that if $A$ and $B$ are exactly Hermitian, then both $A+B$ and $A \otimes B$ are exactly Hermitian, as they should. The problem is really when you do matrix multiplication.

Which shouldn’t produce further problems, right? After all, $AB$ is in general not Hermitian, so we don’t have anything to worry about. Except, of course, that the Hilbert-Schmidt inner product $\operatorname{tr}(A^\dagger B)$ is real for Hermitian $A,B$, and this is not true in MATLAB. Argh! $\operatorname{tr}(A^\dagger B)$ appears very often in the objective of a SDP, and it really needs to be a real number, as you can’t minimize a complex function. It turns out that this is not MATLAB’s fault, it is a fundamental problem of floating-point arithmetic.

An well-known but often-forgotten fact is the floating-point addition is not associative. Let $a=10^{20}$ and $b=10^{-20}$. Then with floating point numbers we see that $b+(a-a) = b$ but $(b+a)-a = 0$. Which is the issue with the Hilbert-Schmidt inner product: we get stuff like $z+\bar{z}+w+\bar{w}+\ldots$, but not in order. $z+\bar{z}$ is of course real, and $w+\bar{w}$ as well, but $z+(\bar{z}+w)+\bar{w}$? Nope, not in floating-point arithmetic.

Here I don’t think a proper solution is possible. One can of course write $\Re[\operatorname{tr}(A^\dagger B)]$, but that will hide your mistakes when $A$ and $B$ are not in fact Hermitian. A better idea is to vectorize $A$ and $B$ to avoid matrix multiplication, writing $\Re[\langle A|B\rangle]$. This only helps with speed, though, it doesn’t touch the real problem.

I’ve just finished reading Andy Weir’s new book “Project Hail Mary”. Great story, I heartily recommend it to anyone who is into hard sci-fi. There’s minor plot point that I found hilarious: an alien civilization managed to do interstellar travel before discovering relativity (it makes perfect sense in the story). The poor bastards were very confused: the trip was shorter and took much less fuel than they had planned for!

I was aware that relativity make trips shorter from the point of view of the traveller, but fuel? Never heard about relativity saving fuel as well. I just had to check it out!

First let’s loot at the time, though, as it is also rather interesting. Let’s say you want to travel to a star $L$ light-years away, and you have an extremely powerful and efficient rocket, so the only limitation you have is the acceleration your body can take. What you do is then accelerate at a constant rate $a$ up to half-way point, where you start decelerating at the same rate $a$, so that you actually stop at your destination. How long does that take? Luckily the Usenet Physics FAQ already has all the calculations, so I won’t do them again, I’ll just copy their results.

From the point of view of the Earth the number of years3These equations are valid for a system of units where $c=1$, which I’m doing here by measuring time in years and distance in light-years. This choice is specially nice because it makes $g \approx 1$, making calculations even easier. it takes is
\[t = \sqrt{L^2+4L/a}.\] You see it is always larger than $L$, which is the familiar result that you can’t travel faster than light. More interesting is that time it takes from the point of view of the traveller, the proper time:
\[\tau = \frac2a \cosh^{-1}(aL/2 +1 ).\] You’re probably not so familiar with the hyperbolic cosine, so it’s worth noting that $\cosh(x) \approx e^x/2$ for large $x$, which results in
\[\tau \approx \frac2a \log(aL+2)\] for large $L$. Yep, that’s a logarithm! It’s much, much faster from the point of view of the traveller. And those poor bastards that didn’t known relativity? They’ll calculate the trip time to be
\[t_\text{Newton} = 2 \sqrt{L/a}, \]which is indeed faster than the time from the point of view of the Earth $t$, but much slower than the time from the point of view of the traveller, which is what matters here.

For concreteness, let’s look at the time it takes to reach Tau Ceti, which is roughly 12 light-years away, at a nice comfy acceleration of one $g$. We have that $t=13.8$ years, $\tau=5.2$ years, and $t_\text{Newton} = 6.8$. Ok, not so brutal, but you’d definitely notice if you’d arrive at your destination one and a half years before schedule. This is a bit misleading, though, because this distance too small for the power of the logarithm to really kick in. If instead you go to the centre of the Milky Way, roughly 30,000 light-years away, you get $\tau=20$ years, $t_\text{Newton}=341$ years, and $t = 30,002$ years. Now that’s brutal.

How about fuel, though? I don’t think we’ll ever do interstellar travel with rockets, much more plausible is a laser-on-light-sail propulsion as investigated by Breakthrough Starshot, but it’s fun to do the calculations anyway. The Newtonian calculation is an easy application of Tsiolkovsky’s equation: for each kilogram of payload, we need
\[M_\text{Newton} = \exp\left(\frac2{v_e}\sqrt{aL}\right)-1 \]kilograms of fuel. Here $v_e$ is the effective exhaust velocity, which we take to be 1, as we are using an amazing photon rocket. The relativistic rocket equation is surprisingly simple: we just replace the time in Tsiolkovsky’s equation with the proper time. The result is that we need
\[M = \exp\left(\frac2{v_e}\cosh^{-1}(aL/2 +1 )\right)-1\] kilograms of fuel for each kilogram of payload. Using the same approximation for the hyperbolic cosine as before, we get
\[ M \approx (aL+2)^{2/v_e}.\] Now this is fucking amazing. We have an exponential in the Newtonian case, but in the relativistic case it’s just a polynomial! The exponential from Tsiolkovsky gets negated by the logarithm in the proper time. Even for Tau Ceti the difference is already extremely large: $M=204$ kg, whereas $M_\text{Newton} = 1,138$ kg! And for the centre of the Milky Way? $M \approx 10^9$ kg, and $M_\text{Newton} \approx 10^{152}$ kg. Ok, both are impossible, but the Newtonian one is much more impossible!

So don’t complain about the lightspeed barrier. Relativity actually makes life much easier for would-be interstellar travellers. A Newtonian universe would be a shit place to live.

I was arguing on the internet about the possibility of building a space elevator; someone was claiming that carbon nanotubes were going to make it possible and we’d build one in the near future. I’m very skeptical, first because we’re nowhere near the capability of building high-strength carbon nanotube bundles tens of thousands of kilometres long, and secondly because it needs to be cheap, and this absurdly long and absurdly strong cable seems anything but cheap. Furthermore, if we postulate that some magical technology will allow that to happen, I think it’s dishonest to not allow some magical technology to also help the competition. For instance, we could have fully reusable rockets which cost little other than fuel to launch; and this technology seems much more plausible.

But this is not what I’m going to talk about, space elevator cables have been discussed to death. What I’m interested in is a point that another poster brought up: how much does the energy needed to climb the space elevator costs? They were arguing that the proposed method of powering the climber, laser beaming, had an absurdly low efficiency of 0.005, and paying for that energy alone already made the space elevator uneconomical, we didn’t even need to consider the astronomical costs of constructing it in the first place.

Now wait, come on, so even if we did have the magical cable for free it still wouldn’t be worth using the space elevator? That’s going too far. If laser beaming has such a low efficiency, this just means we should use some other method for powering the climber. Delivering the power via copper wire is out of question: the wire would be extremely massive, and over such a long distance the electrical resistance would also make it uneconomical. Of course we could use a superconductor to get rid of resistance, but then we have to cool the damn thing as well, making the weight and the cost even worse. No no no, I want to find a solution for powering the climber without needing any more magical technology, the only magic I’ll allow is the space elevator cable itself.

The obvious solution is solar panels. In space they work very reliably, there’s no atmosphere to filter the sun, and even the shadow of the Earth quickly stops being a problem as you get higher2Calculating that exactly is a bit annoying, but there are two main effects at play here: first the orbit gets wider, but the width of Earth’s shadow is constant, and second because of orbital inclination, you’ll not cross Earth’s shadow at all once you’re high enough (except during equinoxes).. They don’t generate much power, though, so it would take a long time to climb to geostationary orbit. How long exactly?

It turns out this is an interesting physics problem. We’re not moving with a constant speed, or a constant force, but with a constant power. I didn’t study this problem either in school or university, so I wanted to find out how to do it. The principle is easy: kinetic energy is given by
\[ E = \frac12 mv^2, \] so we differentiate that and get
\[ \frac{d}{dt}E = P = m v a,\]which results in the differential equation
\[a = \frac{P}{mv},\] that even I can solve.

Life gets hard when we put that together with the forces acting on the climber: gravitational $GMm/r^2$ and centrifugal $m\omega^2 r$ (the famous “fictitious” force that hurts as much as any “real” force). The resulting acceleration is
\[ a = \frac{P}{mv} – \frac{GM}{r^2} + \omega^2 r,\]a really nasty differential equation that I have no idea how to solve. No matter, my laptop can solve it numerically without even turning on the cooling fan, I just need the values for the parameters. Most of them are fixed: $G$ is the gravitational constant, $M$ is the mass of the Earth, and $\omega$ is its angular speed. I just need to determine $P/m$, the power per mass available from the solar panels.

Annoyingly enough, I couldn’t find solid numbers anywhere, but a good estimate can be obtained by considering the solar panels of the ISS. They produce roughly 30 kW of power (before degradation) and weight roughly 1.1 tons, giving as a power density of 27 W/kg. To compensate for degradation, they’re about to install new ones, that promise a 20% weight reduction for the same performance. Again, I couldn’t find solid numbers, but given that they are going to be installed next month, and do have a plausible method for reducing weight, I’ll believe that 20% and put a power density of 34 W/kg solidly in the “not magic” category.

The climber is not made only of solar panels, though. Let’s reserve 80% of its mass for payload, structure, and so on, and use 20% for solar panels, which lowers the power density to 6.8 W/kg. Solving the equation numerically, letting the initial position be the equatorial radius of the Earth and initial velocity 1 m/s2We can’t set it to zero, as the constant power assumption gives a singularity, but the initial velocity makes very little difference., we see that 82.5 days are needed to reach geostationary orbit. But wait! At geostationary orbit the gravitational and centrifugal forces balance out, so we’ll just have the motors working with no resistance, and the speed of the climber will quickly diverge! We have to check if we’re not accidentally breaking relativity. Solving numerically for speed, we see that at geostationary height it is roughly 300 m/s. Yeah, no danger of breaking relativity here. Still, that’s a little too fast to be plausible, so let’s apply the brakes at 100 m/s. This only lengthens the journey to 82.7 days, so it’s really not worth considering; there’s a bigger error anyway from not taking into account the time the climber spends in Earth’s shadow, which will make the journey a couple of days longer.3I’m not in the mood to calculate exactly how much longer, it’s an easy calculation but very annoying to do.

That takes a lot of patience, but it’s really not an issue for geostationary satellites or interplanetary missions that take years anyway. It is an issue for manned missions, we really don’t want to spend 3 months loitering in space getting bathed in radiation. Can we speed it up somehow?

Well, we can send power up via a copper wire for a short distance, say 100 km. In fact we need to do this, as we can’t deploy this massive solar array inside the atmosphere. Transmission over such distances is completely routine, so I’m not going to bother calculating the specifics. Just assume that we send enough power for a constant speed of 100 m/s, and solve the differential equation numerically again with these new initial conditions. The result is 81 days. Well, yeah, we’ll leave the space elevator for cargo and send people up in rockets.

It has happened to all of us. You spend months refining your latest result, carefully writing it up as a well-polished manuscript. You argue with your coauthors about everything: from which version of the proof is more enlightening, to the order the ideas should be presented, and even whether $\epsilon$ or $\varepsilon$ has more typographical merit. Finally content with the manuscript, you submit it to the journal of your desires, and the long wait begins for the referee reports. When they finally arrive, there’s a short and shallow report from Referee #1 saying it’s fine, publish it, just add a citation to Refereey McRefereeFace. But then comes the long, thorny report of Referee #2. This shrivelled soul writes that your bombastic claims in the abstract are not supported by the results in the text. This arrogant asshole thinks the problem you solved is of no interest whatsoever. This heartless pedant says that your proof is nothing but a handwavy argument. This presumptuous prick argues that the whole research program was misguided to start with. And anyway it was already done a long time ago in Russia.

Well, I have a confession to make. I am Referee #2. Contrary to common misconception, I am not a sadist. I have received myself countless negative referee reports, and I know very well how much it hurts to see your work trashed like this. It breaks my heart to reject a paper, specially when I see it’s the first paper a student has ever written. Nevertheless, it is my duty. I do believe in peer-review, and I think readers should be able to trust what they read in a reputable journal. This is specially important in our age of widespread disinformation.

The most common problem a paper has is claiming too much in the abstract, what is known as overselling the result4Exceedingly rare, but happens, is underselling the result. The most hilarious example I known is this paper, which claimed in the abstract a trivial observation that had been known for decades, but proved something much more interesting in the body. The authors quickly realised their mistake and wrote a more reasonable abstract in the subsequent version.. This is what I find most frustrating as a referee because almost always the authors know what they wrote is bullshit, they just think it’s harmless, and I should just let it slide. The argument is that anybody that reads the paper will understand what the result is actually about and disregard the abstract, which is just colourful marketing. This is true, but misses the point that almost nobody reads the paper; the vast majority of the readers will be lay people that know of the result through a press release, that credulously repeats the claims of the abstract. Great, now you’re fooling the general public that paid for the research. Even among the physicists that read the paper, most of them will not be experts, and have to take your claims at face value. So no, I will not let it slide, and I will tell the authors to cut the crap without the least bit of remorse.

When there is a technical mistake in the paper there is usually very little drama; of course it’s bad news, but the mistake was already there, and I helped find it. The authors usually just fix it and thank me in the acknowledgments. Once the authors refused to fix the mistake, the paper was rejected, and they simply resubmitted it to a different journal. Which asked me again to referee the paper. Then I got furious about wasting my time trying to help the authors, and simply copy-pasted my previous referee report.

But in the other cases? Yes, sometimes the problem you have worked on indeed interests nobody but you. I do feel sorry for the authors, but I will write that in the referee report if that’s my opinion. And when the result was in fact done a long time ago in Russia? It’s very frustrating, I know, I have also rediscovered stuff that was published in the Siberian Journal of Mathematics. Life sucks, but what can I do? We have to respect the priority of Soviet physicists. And when the research program was a terrible idea in the first place? Well, I wrote that in my referee report only once, and it was really heartbreaking, as it was also the first paper of a PhD student. I decided to say it anyway, because otherwise the student would probably spent the rest of their PhD working on a nonsensical idea, and their career would go down the drain. I tried to be gentle about it, but sometimes being gentle doesn’t help when the news is that your mother has died.

Perhaps you’re not convinced, and want anyway to punish Referee #2 for the negative reports. Well, I have a feeling that editors know that I’m strict with my referee reports, so they send me papers they want to be rejected. Every day I see really nice papers getting published, but somehow they never come to me for refereeing. So rest assured, being Referee #2 is its own punishment.

EDIT: And just after I posted this, I got rejected from TQC with this paper. The killing blow came in fact from Referee #2: “Overall, I enjoyed reading this work. It starts with a clear and well-motivated question and mixes up wisely several ingredients from the theory of non-locality in order to tackle it. However, I did not find the results that exciting.” Oh, that hurts!

There are a lot of derivations of the Born rule in the literature. To cite just a few that I’m most familiar with, and illustrate different strategies: by Everett, Deutsch, Wallace, Żurek, Vaidman, Carroll and Sebens, me, and Mandolesi. I’m intentionally leaving out frequentist derivations, who should just be ashamed of themselves, and operationalist derivations, because I don’t think you can explain anything without going into ontology.

I’m not happy with any of them. Before going into detail why, I need to say what are my criteria for a satisfactory derivation of the Born role:

1. It needs to use postulates that either come from the non-probabilistic part of quantum mechanics, or are themselves better motivated than the Born rule itself.
2. The resulting Born rule must predict that repeated measurements on a state $\alpha\ket{0}+\beta\ket{1}$ will often yield relative frequencies of 0 close to $|\alpha|^2/(|\alpha|^2+|\beta|^2)$.

The first criterion is hopefully uncontroversial: if the derivation of the Born rule depends on the probabilistic part of quantum mechanics it is circular, and if it uses more arcane postulates than the Born rule itself, why bother?

The second criterion tries to sidestep the eternal war about what is the definition of probability, focussing on the one point that everyone should agree about: probabilities must predict the frequencies. Should, but of course they don’t. Subjectivists insist that probabilities are the degrees of belief of a rational agent, leaving the connection with frequencies at best indirect. To the extent that this is a matter of semantics, fine, I concede the point, define probabilities to be subjective, but then I don’t care about them. I care about whatever objective property of the world that causes the frequencies to come out as they do. Call it propensity, call it chance, call it objective probability, doesn’t matter. This thing is what the Born rule is about, not about what is going on in the heads of rational agents.

And there is such a thing, there is a regularity in Nature that we have to explain. If you pass 1000 photons through a polariser, then a half-wave plate oriented at $\pi/4$ radians, and then measure their polarisation, you will often find the number of photons with horizontal polarisation to be close to 500.2Well you should lose half of the photons in the polariser and the photodetector should miss 70% of the remaining ones, so more realistically you would get about 75 photons with horizontal polarisation. But that’s beside the point. To simulate this experiment,2I would be delighted if an experimentalist with too much time in their hands would generate real data for me. I generated 8000 hopefully random bits from /dev/urandom, divided them into sets of 1000, and counted the number of ones in each set, obtaining $504, 495, 517, 472, 504, 538, 498$, and $488$. We often got something close to 500. This is the regularity in Nature, this is the brute fact that a derivation of the Born rule must explain. Everything else is stamp collection.3Even this objectivist argument makes it look like the Born rule is about our predictions for our carefully controlled experiments. It’s not. For a less narcissistic argument, consider plants that died 5,730 years ago. Often about half of their carbon-14 atoms will have decayed to nitrogen-14. Independently of whether we put them in a mass spectrometer to find out.

I’ve emphasized the second criterion because it’s where most derivations fail. Consider the most fashionable one, by Deutsch and Wallace, which I reviewed here. They have some axioms about what it means to be a rational agent, some axioms about what is a measurement, and some axioms about how rational agents should be indifferent to some transformations done on the quantum state. From these axioms they conclude that rational agents should bet on the outcomes of a quantum experiment with probabilities given by the Born rule. Who cares? Should we really believe that the statistics of an experiment will be constrained by rationality axioms? And conversely, if you can show that the statistics of a quantum experiment follow the Born rule, doesn’t it become obvious that rational agents should bet that they do, making the whole decision-theoretic argument superfluous? It’s worth noting that this same criticism applies to my derivation, as it is just a cleaned up version of the Deutsch-Wallace argument.

Let’s move on to Vaidman, Carroll, and Sebens. Their derivations differ on several important points, but I’m interested here in their common point: they passionately argue that probability is about uncertainty, that a genuine source of uncertainty in Many-Worlds is self-locating uncertainty, and that locality implies that your self-locating uncertainty must be given by the Born rule. Arguing about whether probability is uncertainty is a waste of time4I’m speaking from experience, I wasted a lot of time doing that., but their second point is well-taken: after a measurement has been done and before you know the outcome, you are genuinely uncertain about in which branch of the wavefunction you are. I just don’t see how could this be of fundamental relevance. I can very well do the experiment with my eyes glued to the screen of the computer, so that I’m at first aware that all possible outcomes will happen, and then aware of what the outcome in my branch is, without ever passing through a moment of uncertainty in between. Decoherence does work fast enough for that to happen.5Incomprehensibly, Carroll and Sebens argue that because decoherence works so fast there will always be uncertainty. What now? No probability anymore? And then it appears when I close my eyes for a few seconds? That makes sense if probability is only in my head, but then you’re not talking about how Nature works, and I don’t care about your notion of probability.

Żurek’s turn, now. His proof follows the pattern of all the ones I talked about until now: it argues that probabilities should be invariant under some transformations on entangled states, and from a bunch of equalities gets the Born rule. An important different is that he refuses to define what probabilities are, perhaps in order to avoid the intractable debate. I think it’s problematic for his argument, though. One can coherently argue about what agents can know about, what they should care about, as the other authors did. But an undefined notion? It can do anything! Without saying what probability is how can you say it is invariant under some transformation? Also, how can this undefined notion explain the regularity of the frequencies? Maybe it has nothing to do with frequencies, who knows?

Let’s go back to Everett’s original attempt. It is of a completely different nature: he wanted to derive a measure $\mu$ on the set of worlds, according to which most worlds would observe statistics close to what the Born rule predicted. I think this makes perfect sense, and satisfies the second criterion. The problem is the first. He assumed that $\mu$ must be additive on a superposition of orthogonal worlds, i.e., that if $\ket{\psi} = \sum_i \ket{\psi_i}$ then
\[ \mu(\ket{\psi}) = \sum_i \mu(\ket{\psi_i}). \] I have no problem with this. But then he also assumed that $\mu$ must depend only on the 2-norm of the state, i.e., that $\mu(\ket{\psi}) = f(\langle \psi|\psi\rangle)$. This implies that
\[ f\left( \sum_i \langle \psi_i|\psi_i\rangle \right) = \sum_i f(\langle \psi_i|\psi_i\rangle), \] which is just Cauchy’s equation. Assuming that $\mu$ is continuous or bounded is enough to imply that $\mu(\ket{\psi}) = \alpha \langle \psi|\psi\rangle$ for some positive constant $\alpha$, that can be taken equal to 1, and we have a measure that gives rise to the Born rule.

But why assume that $\mu$ must depend only on the 2-norm of the state? This unjustified assumption is what is doing all the work. If we assume instead that $\mu$ depends only on the $p$-norm of the state, the same argument implies that $\mu(\ket{\psi}) = \|\ket{\psi}\|_p^p$, an absolute value to the $p$ rule, instead of an absolute value squared rule. Everett would be better off simply postulating that $\mu(\ket{\psi}) = \langle \psi|\psi\rangle$. Had he done that, it would be clear that the important thing was replacing the mysterious the Born rule, with its undefined probabilities, with the crystal clear Born measure. Instead, the whole debate was about how proper was his derivation of the Born measure itself.

The last proof I want to talk about, by Mandolesi, agrees with Everett’s idea of probability as measure theory on worlds, and as such satisfies the second criterion. The difficulty is deriving the Born measure. To do that, he uses not only the quantum states $\ket{\psi}$, but the whole rays $R_\psi = \{c \ket{\psi}; c \in \mathbb C \setminus 0\}$. He then considers a subset $R_{\psi,U} = \{c \ket{\psi}; c \in U\}$ of the ray, for some measurable subset $U$ of the complex numbers, to which he assigns the Lebesgue measure $\lambda(R_{\psi,U})$. Given some orthogonal decomposition of a ray $c \ket{\psi} = \sum_i c \ket{\psi_i}$ – which is most interesting when the states $\ket{\psi_i}$ correspond to worlds – the fraction of worlds of kind $i$ is given by
\[ \frac{\lambda(R_{\psi_i,U})}{\lambda(R_{\psi,U})} = \frac{\langle \psi_i|\psi_i \rangle}{\langle \psi|\psi \rangle}, \] which does not depend on $U$, and coincides with the Born measure.

I think he’s definitely using postulates more arcane than the Born measure itself. He needs not only the rays and their subsets, by also the Lebesgue measure on the complex line, for which his only justification is that it is natural. No argument here, it is very natural, but so is the Born measure! It is just an inner product, and is there anything more natural than taking inner products in a Hilbert space?

To conclude, I’m skeptical that there can even be a satisfactory derivation of the Born rule. It obviously doesn’t follow from the non-probabilistic axioms alone; we need anyway an additional postulate, and it’s hard to think of a more natural one than $\mu(\ket{\psi}) = \langle \psi|\psi\rangle$. Perhaps we should focus instead in showing that such an axiomatization already gives us everything we need? In this vein, there’s this curious paper by Saunders, which doesn’t try to derive the Born rule, but instead carefully argues that the mod-squared amplitudes play all the roles we require from objective probabilities. I find it curious because he never mentions the interpretation of the mod-squared amplitudes as the amounts of worlds with a given outcome, which is what makes the metaphysics intelligible.

I’m feeling bad about having used $p$-values in my paper about Bell inequalities, despite considering them bullshit. The reason we decided to do it is because we wanted the paper to be useful for experimentalists, and all of them use $p$-values instead of Bayes factors6That is, when they are not making the obvious mistake of using the variance of the observed data as a measure of significance.. We are already asking them to abandon their beloved Bell inequalities and use nonlocal games instead, and that is the main point of the paper. We thought that also asking them to give up on $p$-values would be too much cognitive dissonance, and distract from the main issue. Good reasons, no?

Still, still. The result is that there’s yet another paper perpetuating the use of $p$-values, and moreover my colleagues might see it and think that I defend them! The horror! My reputation would never recover. To set things right, I’m writing this post here to explain why $p$-values are nonsense, and restate our results in terms of Bayes factors.

First of all, what is a $p$-value? It is the probability of observing the data you got, or something more extreme, under the assumption that the null hypothesis is true2I’m only considering simple hypotheses here, for simplicity.. If this probability is less than some arbitrary threshold, often set at $0.05$ or $10^{-5}$, the null hypothesis is said to be rejected.

The very definition already raises several red flags. Why are we considering the probability of “something more extreme”, instead of just the probability of the data we actually got? Also, what counts as “more extreme”? It doesn’t only sound ambiguous, it is ambiguous. And what about this arbitrary threshold? How should we set it, and why should we care about this number? More worrisome, how can we possibly regard the null hypothesis in isolation? In practice we always have another hypothesis in mind, but come on. What if we do reject the null hypothesis, but the alternative hypothesis attributes only slightly higher probability to the observed outcome? Obviously the experiment was just inconclusive, but apparently not if you take $p$-values seriously.

Perhaps these are just counterintuitive properties of a definition that has a solid basis in probability theory, and we need to just grow up and accept that life is not as simple as we would want. Well, as it turns out, there’s absolutely no basis for using $p$-values. Pearson just pulled the definition out of his ass, for use in his $\chi^2$-test. People then ran with it and started applying it everywhere.

In stark contrast, a Bayes factor is defined simply as the ratio of the probabilities of observing the data you got, given the null or the alternative hypotheses:
\[ K := \frac{p(D|H_1)}{p(D|H_0)}.\] It suffers none of the problems that plague the definition of $p$-value: We only consider the actually observed data, no “more extreme” events are relevant. It explicitly depends on both hypotheses, there’s no attempt to reject an hypothesis in isolation. There’s no arbitrary threshold to set, it is just a number with a clear operational interpretation.

More importantly, it doesn’t come out of anybody’s ass, but from Bayes’ theorem: it is the data we need to calculate the posterior probability $p(H_0|D)$. In the case of two hypotheses, we have that
\[ p(H_0|D) = \frac{p(D|H_0)p(H_0)}{p(D|H_0)p(H_0) + p(D|H_1)p(H_1)} = \frac1{1+K\frac{p(H_1)}{p(H_0)}}.\] How do our results look when written in terms of the Bayes factor then? Our goal was to minimize the expected $p$-value from a Bell test. The expression for the expected $p$-value is a hideous expression that we have no hope of optimizing directly, so with a lot of work we showed that the expected $p$-value after $n$ rounds $\langle p_n \rangle$ is upperbounded by
\[ \langle p_n \rangle \le (1-\chi^2)^n, \] where $\chi := \omega_q-\omega_\ell$ is the gap of the nonlocal game, and $\omega_q$ and $\omega_\ell$ are the Tsirelson and local bounds. We can easily maximize the gap, and a large gap is a sufficient condition for having a small $\langle p_n \rangle$, so that’s we did. We suspect that a large gap is also a necessary condition, and that the lower bound $\frac12(1-\chi)^n \le \langle p_n \rangle$ holds, but we couldn’t prove it.

With the Bayes factor we have pure bliss. The expression for the expected Bayes factor after $n$ rounds $\langle K_n \rangle$ is simple, no bound is needed:
\[ \langle K_n \rangle = \left( 1 + \frac{\chi^2}{\omega_\ell(1-\omega_\ell)} \right)^n. \] Maybe we can even optimize that directly. In any case, a large gap is also a sufficient condition for a large $\langle K_n \rangle$, as we have the bound
\[ \frac1{(1-\chi)^n} \le \langle K_n \rangle, \] valid for $\chi \ge 1/2$3 For $\chi < 1/2$ the lower bound is instead $(1+4\chi^2)^n$, but we don’t really care about the small-gap regime. In contrast to the $p$-value case, it’s obvious from the expression that a large gap is not necessary for a large $\langle K_n \rangle$. We even have an explicit example of a nonlocal game with arbitrarily small gap and arbitrarily large $\langle K_n \rangle$, the Khot-Vishnoi game. Ironically enough, we use this same example in the paper to argue for the opposite conclusion, as it has $\langle p_n \rangle$ close to 1.

The Khot-Vishnoi game is of widespread interest in the literature because its ratio $\omega_q/\omega_\ell$ is arbitrarily large, which implies a large resistance to noise. We wanted to argue that this is useless if $\omega_q$ is too small, as the favourable outcome would happen too seldom in a finite experiment. As pointed out above, this is not true for this game, but the reasoning is still sound: if we have a nonlocal game with $\omega_q = 1/d^{3/4}$ and $\omega_\ell = 1/d$, the ratio is $d^{1/4}$, which does get arbitrarily large with $d$, but $\langle K_n \rangle$ goes roughly like $1+n/\sqrt{d}$ for large $d$, so the large ratio doesn’t help at all.

It’s easy to show that
\[ \langle K_n \rangle \le \left( \frac{\omega_q}{\omega_\ell} \right)^n, \] so having a large ratio is a necessary condition for a large $\langle K_n \rangle$. We have then a nice duality between the ratio and the gap: the gap is sufficient, but not necessary, and the ratio is necessary, but not sufficient.

I don’t think that beauty implies true; otherwise neutrinos would be massless. But it’s a joy when you have an independent argument for something to be true, and it turns out to be quite beautiful as well.

A friend of mine, Flavien Hirsch, invented a fascinating paradox in probability, called Hilbert’s Hotel and Gödel’s Freezer. Technically speaking it is the same thing as the Sleeping Beauty problem, which I’ve written about here, but I find it much nicer because it’s more symmetrical, you actually care about the answer, and intuition leads you astray. Without further ado, here it is:

Gödel has a Freezer which contains 4 people in a state of suspended animation, with you among them. He picks 2 of them randomly, with uniform probability, to send to Hilbert’s Hotel, a nice place which is never full. The other 2 are doomed to stay in the Freezer forever. Being a bit mad and a bit evil, Gödel decides to do the transfer in a capricious way. He sends them via two tram trips, each with 3 people: one of the chosen ones, who stays at Hilbert’s Hotel at the end of the trip, and the doomed ones, who are taken back to the Freezer. Not being too evil, Gödel erases the memory of the doomed ones after each trip so that they won’t be tortured by their dream of freedom. You wake up in the tram, are told about the whole setup, and ask yourself: what’s the probability that I’m going to Hilbert’s Hotel?

There are two answers that seem obviously correct, but conflict with each other:4These answers mirror the $1/2$ and $1/3$ possibilities of the original Sleeping Beauty problem.

• $1/2$ – As 2 out of 4 people are going to Hilbert’s Hotel, and by assumption they were picked randomly and uniformly, the probability of each person must be $1/2$. Since everybody will wake up with certainty and ask themselves the question, you don’t learn anything by waking up, so the probability stays $1/2$.
• $1/3$ – Out of the 3 people in the tram only one is going to Hilbert’s Hotel. Since there’s nothing distinguishing anyone there, they must all have the same probability of being the chosen one, which must therefore be $1/3$.

Now suppose there are $n$ people instead of 4, with still only 2 doomed ones. The probability $1/2$ becomes $(n-2)/n$, while $1/3$ stays $1/3$. The probability that you will go to Hilbert’s Hotel is therefore rather high. But you wake up in the tram – which will happen in any case – and now suddenly you should be sad that your probability is only $1/3$? How can that possibly be true?

The difficulty here is that we’re trying to work with an informal, intuitive notion of probability. That’s not good enough. If we want to have actual answers we need to figure out what probability means. We should first notice that the paradox is about subjective probability. The probability that Gödel used to pick the chosen ones can be taken to be objective, but that’s in the past, now the objective probability that you’ll go to Hilbert’s Hotel is either zero or one, you just don’t know what it is. So even if we had an amazing and universally agreed upon definition of objective probability that wouldn’t help us at all.

What definition of subjective probability do we have then? I’m happy with the decision-theoretical one: you say that an event $E$ happens with subjective probability $p$ if $p$ is the fair price you assign to a bet that pays 1€ if $E$ happens, and nothing otherwise. Fair here means that you don’t expect to win or lose money from this bet, so you’d be fine buying or selling the bet for this this price. For example, if $E$ happens with objective probability $q$, and you pay $p$ for the bet, then with probability $q$ you win $(1-p)$€, and with probability $1-q$ you win $-p\,$€. The expected value is $q(1-p)-(1-q)p=q-p$, so you will expect to win money if $p < q$, and you will expect to lose money if $p>q$, so the fair price is $p=q$.

You wake up in the tram then, and consider, what is the fair price of a bet that pays 1€ if I end up at Hilbert’s Hotel? There are two cases to consider: if you are a chosen one, you’ll pay $p$€ and win 1€, so your net gain is $(1-p)$€. But if you’re a doomed one, you’ll pay $p$€ and win nothing, with a net gain of $-p$€. In this trip. But if you are a doomed one you will certainly do another trip (or will have already done it). So you’ll do the same bet twice, paying in total $-2p$€. The bet just became less attractive, and it stands to reason that you’ll be willing to pay less for it! The expectation value is $(1-p)/2 -2p/2 = (1-3p)/2$, which is zero for $p=1/3$.

But wait. Is it fair to make the doomed one pay twice for the bet? Well, the bet is imaginary in any case, but the doomed one is who wakes up twice and asks themselves the question twice. Paying for the bet twice is just a formalisation of being wrong twice. Intuitively, you might not want to guess you’re a doomed one because even though the probability is $1/2$, when you’re wrong you’ll be wrong twice.

Maybe we shouldn’t count the mistake of the doomed one twice, because their memory will be erased between the trips? Wiseman suggested this would be natural in my original post about the Sleeping Beauty. I find it rather artificial. If we erase their memory after the second trip as well (as the statement of the paradox says), should we then never count their mistake? And thus assign subjective probability 1 to going to Hilbert’s Hotel? Or in a completely different situation, should I assign probability 1 to winning a game of Russian roulette, as my memory will be erased if I lose? I think this is really changing the question from being wrong about your prediction than knowing that you were wrong.

It does raise an interesting question, though, of what does it take to make the subjective probability of going to Hilbert’s Hotel match the objective probability $1/2$. One could get rid of the multiple awakenings, by saying that Gödel will only let the passengers of the tram wake up in the first trip. But this is not enough, because now the probability that you will wake up if given that you are a chosen one is $1/2$, while the probability that you will wake up given that you are a doomed one is still 1, so the expectation value is $(1-p)/4 – p/2 = (1-3p)/4$, so the fair price is still $1/3$.

It does work if Gödel decides to wake up each of the four people exactly once, say all passengers of the tram wake up once, in the first trip, and the remaining chosen one wakes up alone in the second trip (and somehow doesn’t notice that the other two passengers are still asleep). Now everybody will wake up once, and only once, with probability 1, and the expectation value is $(1-p)/2 – p/2 = 1/2 – p$, and finally we have $p=1/2$. But the problem clearly changed.

I’ve recently proved a theorem that is probably well-known to anyone that studied number theory. I didn’t know it before, though, and it was a lot of fun, so I decided to write it up here in case there are other people around that like number theory a bit but not enough to actually learn it.

The question I wanted to answer was which beautiful angles have beautiful cosines? My completely objective definitions are that a beautiful angle is of the form $\frac{p}{q}\pi$ for small integers $p,q$, and a beautiful cosine is of the form $\frac{a}{b}+\sqrt{\frac{c}{d}}$, again for small integers. Surprisingly, the question can be answered completely, and it is not even difficult if you know number theory (I don’t).

The first thing to notice is that any algebraic number uniquely defines its minimal polynomial, which is the polynomial with rational coefficients (and leading coefficient one) of least degree that has this algebraic number as a root. So we only need to compute the minimal polynomial of $\alpha$ and see if it matches the minimal polynomial of $\cos(\frac{p}{q}\pi)$ for any $p,q$. Which doesn’t really help, because what the hell is the minimal polynomial of $\cos(\frac{p}{q}\pi)$?

Luckily though the minimal polynomial of $e^{i\pi\frac{p}{q}}$ instead is very well characterized: it is known as a cyclotomic polynomial, and it is given simply by $\Phi_n(x) := \prod_k (x-e^{i2\pi\frac{k}{n}})$, over all $k$ that are coprime to $n$.

Since $\cos(\frac{p}{q}\pi) = \alpha$ iff $e^{i\frac{p}{q}\pi} = \alpha \pm i\sqrt{1-\alpha^2}$ what we do instead is to compute the minimal polynomial of $\alpha + i\sqrt{1-\alpha^2}$ (which is equal to the minimal polynomial of $\alpha – i\sqrt{1-\alpha^2}$), and see if it matches some cyclotomic polynomial.

Let’s start by disregarding the trivial cases where $\frac{a}{b}+\sqrt{\frac{c}{d}}$ is $1,0,$ or $-1$. We know that $\cos(0),\cos(\pi/2),\cos(-\pi/2),$ and $\cos(\pi)$ give them, there’s no need for the big guns. We’ll assume from here on that $\frac{a}{b}+\sqrt{\frac{c}{d}}$ is never an integer.

The first interesting case is when $c=0$, so we’re just asking which beautiful angles have rational cosines. To answer that we need the minimal polynomial of $z=\frac{a}{b}+i\sqrt{1-\frac{a}{b}^2}$. It’s clear that $(x-z)(x-z^*)$ works, and we can’t get rid of the root $z^*$, so that’s the minimal one. Explicitly it is
\[ x^2 -2 \frac{a}{b} x +1, \]so it can only match cyclotomic polynomials of degree 2. Well, the degree of a cyclotomic polynomial $\Phi_n(x)$ is simply the number of integers $k \le n$ that are coprime to $n$, so we need to find which integers $n$ have exactly 2 smaller integers that are coprime to it.

This is kind of a strange question, but the reverse is rather straightforward: the number of integers coprime to $n$ is the definition of Euler’s totient function $\varphi(n)$. It has the simple lower bound of $\varphi(n) \ge \sqrt{n/2}$, which allows us to brute-force our problem: we just need to look at $n$ up to 8 and we’ll have all integers with exactly 2 coprimes. They are 3,4, and 6, so we just need to check whether our minimal polynomial matches $(x-e^{i2\pi\frac{1}{3}})(x-e^{i2\pi\frac{2}{3}})$, $(x-e^{i2\pi\frac{1}{4}})(x-e^{i2\pi\frac{3}{4}})$, or $(x-e^{i2\pi\frac{1}{6}})(x-e^{i2\pi\frac{5}{6}})$. Which is kind of overkill, because we already have the angles, so we can simply check instead which of them have rational cosines, which turn out to be $\cos(\frac{2}{3}\pi) = \cos(\frac{4}{3}\pi) = -\frac{1}{2}$, and $\cos(\frac{1}{3}\pi) = \cos(\frac{5}{3}\pi) = \frac{1}{2}$. And these are the only ones.

Moving on to the next interesting case, we set $a=0$, so that we are asking which beautiful angles have cosines equal to $\sqrt{\frac{c}{d}}$. We need then to figure out the minimal polynomial of $z = \sqrt{\frac{c}{d}} +i\sqrt{1-\frac{c}{d}}$. Now the polynomial $(x-z)(x-z^*)$ doesn’t work anymore, as it has irrational coefficients, but this is easy to fix by considering instead $(x-z)(x-z^*)(x+z)(x+z^*)$, which does work and is also clearly the minimal polynomial. Working it out, we get
\[x^4 + \left(2-4\frac{c}{d}\right)x^2 + 1.\] Calculating which integers have exactly four coprimes, like before, we see that they are 5, 8, 10, and 12, which is way too much work to check by hand, so we’ll do something else instead: another nice property of cyclotomic polynomials is that they only have integer coefficients, so we only need to look for $c,d$ such that $4\frac{c}{d}$ is integer. The only nontrivial cases are $c=1,d=2$ and $c=3,d=4$, which tells us that the only beautiful cosines that are square roots of rationals are $\pm\sqrt{1/2}$ and $\pm\sqrt{3/4}$. The angles that give these cosines are the well-known $\pi/4,7\pi/4,3\pi/4,5\pi/4$ and $\pi/6,11\pi/6,5\pi/6,5\pi/6$.

To finalize let’s consider cosines of the form $\frac{a}{b}+\sqrt{\frac{c}{d}}$ for $a \neq 0$ and $c \neq 0$. We need the minimal polynomial for the hideous thing
\[z = \frac{a}{b}+\sqrt{\frac{c}{d}}+i\sqrt{1-\left(\frac{a}{b}+\sqrt{\frac{c}{d}}\right)^2}.\] The previous case does not work, but if you think about it, it is easy to see that with
\[w = \frac{a}{b}-\sqrt{\frac{c}{d}}+i\sqrt{1-\left(\frac{a}{b}-\sqrt{\frac{c}{d}}\right)^2}\] the polynomial $(x-z)(x-z^*)(x-w)(x-w^*)$ will do the job. Working it out with Mathematica, because that’s already above what I can do by hand, we get
\[x^4 – 4\frac{a}{b}x^3 + \left(2+4\frac{a^2}{b^2}-4\frac{c}{d}\right)x^2 – 4\frac{a}{b}x + 1.\] At this point it’s better to just look up what are the cyclotomic polynomials for $n=5,8,10,12$, which as we said before are the only ones with degree 4. They are
\begin{align*}
\Phi_5(x) &= x^4 + x^3 + x^2 + x +1 \\
\Phi_8(x) &= x^4 + 1 \\
\Phi_{10}(x) &= x^4 – x^3 + x^2 – x + 1 \\
\Phi_{12}(x) &= x^4 – x^2 + 1,
\end{align*}which is not that bad. $\Phi_8(x)$ and $\Phi_{12}(x)$ are right out because they require $a=0$, which we analysed before. To make our minimal polynomial equal to $\Phi_5(x)$ we need $a=-1$ and $b=4$, which tells us that $c/d = 5/16$. We have a match! It implies that the cosines of the angles $2\pi/5, 4\pi/5, 6\pi/5, 8\pi/5$ must be $-\frac{1}{4}+\sqrt{\frac{5}{16}}$ or $-\frac{1}{4}-\sqrt{\frac{5}{16}}$. A similar matching occurs with $\Phi_{10}(x)$, the only difference is that $a=1$ instead of $-1$. It implies that the cosines of the angles $\pi/5, 3\pi/5, 7\pi/5, 9\pi/5$ must be $\frac{1}{4}+\sqrt{\frac{5}{16}}$ or $\frac{1}{4}-\sqrt{\frac{5}{16}}$. And that’s it! There can be no more.

To summarize, we found out that the only possible beautiful cosines of beautiful angles are $0, \pm1, \pm\frac{1}{2},\pm\frac{1}{\sqrt{2}},\pm\frac{\sqrt{3}}{2},\frac{1\pm\sqrt{5}}{4},\frac{-1\pm\sqrt{5}}{4}$. And the beautiful angles with these cosines are all the integer multiples of $\pi,\pi/2,\pi/3,\pi/4,\pi/5,\pi/6$. But be careful, this holds really only for cosines, the sines of integer multiples of $\pi/5$ are never beautiful.2Because that would imply that the cosines of odd multiples of $\pi/10$ are beautiful, as e.g. $\cos(\pi/10) = \sin(2\pi/5)$, and we know that isn’t true.

Funnily enough, these are precisely the angles and cosines we learn in school, except for $\pi/5$. Usually we only learn a couple of examples that are good enough, and leave behind an incredible wealth of interesting cases. But here no, out teachers taught us all the angles and cosines that were worth knowing! $\pi/5$ is an understandable exception, it is not as beautiful as the others, and we should anyway demand both the sines and cosines to be beautiful.