Stephen Gull wrote this in his webpage

"Many years ago (about 1984), I used to give a Mathematical Physics course to the Part II students. I illustrated the quantum paradox covered by Bell's theorem by showing that you can't program two independently running computers to mimic the results of spin measurements on two spin-1/2 particles in a singlet state. I believe this demonstration is actually better than Bell's original argument: bell.pdf"

My Questions

I have two questions (other than please don't close it due to me obviously not being a scientist, let alone a physicist)

  1. It's a bit embarrassing to say, but I'm simply trying to understand it. English is not may native language, and I simply have a problem to understand some of the handwritten words, and although this is not an english language community, I'm sure a background in physics will help understand it from context.

  2. Are there any resources out there that can help me, (who has a youtube and wikipedia level of understanding of Bell's theorem, in other words, non whatsoever), to fully understand Gull's explanation?

My attempt to understand

In the risk of making a fool of myself, but to show I at least tried a little, from what I can understand, he's sketching a proof to Bell's theorem, and using Fourier transform, to prove that two (obviously deterministic, and using hidden variables, by definition) distant computers can't ever use any formula that will let them agree on some result, that mimics the observed (and theoretical) quantum behavior around entanglement, where the timing of the "measurement" of both computer can't be known (if it was known, you could use something like a time based temporary password generator, but these have a time window where the two remote systems are correlated, but if I understand correctly, the quantum behavior is that once "one side" was observed, the other's observation is "locked" forever and won't change over time. e.g. even if they agree on some internal state using a time based PRNG, they can't know when the other will "lock" (that's my logical explanation, not sure if it sounds like I don't know what I'm taking about, as I don't feel that I know what I'm talking about) but it seems he's using some other proof using Fourier transform, and due to me not understanding the handwriting (no offense, it's me, not him), I don't feel I even have a chance to try to figure out the math (that's an excuse I'm telling myself probably, so help explaining the Fourier part will be appreciated as well)

tl;dr I need help with both English, Math and Physics, I will understand if you will choose to close this question, as I do ask for a lot. But I really want to be able to understand this (I emailed him as well. And just to clarify, I'm not a student of his, or a student in general, I painfully did my masters in CS in my 40s and realized that both the academic world and myself will be much better off if I don't quit my day job).

  • $\begingroup$ This might be relevant: arxiv.org/pdf/1312.6403.pdf $\endgroup$
    – Eran Medan
    Commented Apr 27, 2020 at 2:44
  • 1
    $\begingroup$ In my opinion, this is a good question: the title is descriptive, I did not find any duplicate questions about Gull's approach on Physics SE, and an explanation of this approach would probably be interesting to a wide audience. (It's a clever approach, but the handwriting is difficult to read even for a native english speaker.) If somebody posts a good explanation of Gull's approach in their own words, would that satisfy question 1? $\endgroup$ Commented Apr 27, 2020 at 3:56
  • $\begingroup$ @ChiralAnomaly definitely! and thank you! it feels like a modern case of "I have discovered a truly marvelous proof of this, which this margin is too narrow to contain" although Bell's theorem was already proven and probably a book won't be written about it, a detailed solution expanding it would help probably a lot of people :) $\endgroup$
    – Eran Medan
    Commented Apr 27, 2020 at 19:55
  • $\begingroup$ p.s. I emailed the author, I'll update here if I get a reply (in case he'll be ok with making it public) $\endgroup$
    – Eran Medan
    Commented Apr 27, 2020 at 20:00

1 Answer 1


This answer is based on the handwritten notes [1] that are linked in the question, but I wrote it in my own words and used slightly different conventions: the outcomes are labelled $0,1$ instead of $\pm 1$, and the outcomes in the same-angles case are always equal to each other instead of always opposite. (This can be arranged simply by changing the reference-direction with respect to which one of the angles is defined.)

The phenomenon

Suppose we have a device with one button. Each time we press the button, the device shoots out a pair of particles, $A$ and $B$, travelling in opposite directions. After the particles are far away from each other, we can measure an observable of particle $A$ along an axis with orientation $\alpha$, and we measure an observable of particle $B$ along an axis with orientation $\beta$. The particles and the observables are chosen so that each measurement has only two possible outcomes, which we can label $1$ and $0$.

Quantum theory predicts (and experimentes confirm) that we can design the device to produce a special state with the following properties: First, when the angles are equal, the outcomes are either both $1$ or both $0$. Second, no matter what angles $\alpha,\beta$ we choose, the fraction of trials in which both outcomes are $1$ approaches $$ p(\alpha-\beta) := \frac{1+\cos(\alpha-\beta)}{4} \tag{1} $$ if the number of trials is large enough. Loosely speaking, (1) is the "probability" that both outcomes are $1$ when the angles are $\alpha,\beta$.

An impossible project

The project is to write a pair of computer programs, $A$ and $B$, that satisfy the following requirements.

  • Program $A$ runs on computer $A$, and program $B$ runs on computer $B$. The programs may have as many common features as we want, but once the experiment begins, they are not allowed to communicate with each other.

  • Each program takes a sequence of $N$ angles as input and returns a sequence of $N$ binary digits. Let $A_n(\alpha)$ and $B_n(\beta)$ denote the binary digit that will be returned by programs $A$ and $B$, respectively, if the angles used in the $n$th trial are $\alpha$ and $\beta$ (for $A$ and $B$, respectively).

  • The programs should be designed so that $$ A_n(\theta)=B_n(\theta) \tag{2a} $$ and $$ \lim_{N\to\infty}\frac{1}{N}\sum_n A_n(\alpha)B_n(\beta) = p(\alpha-\beta) \tag{2b} $$ where $p(\alpha,\beta)$ is the function defined in equation (1).

This is impossible. Bell gave one proof, which is reviewed in [2]. I'll review Gull's proof instead because that's what the question is about.

Gull's proof that the project is impossible

Use (2a) in (2b) to get $$ \lim_{N\to\infty}\frac{1}{N}\sum_n A_n(\alpha)A_n(\beta) = p(\alpha-\beta) \tag{3} $$ Let $\tilde A_n$ denote the Fourier transform of $A_n$, so that $$ A_n(\alpha) = \sum_a e^{ia\alpha}\tilde A_n(a) \tag{4} $$ where the sum is over all integers $a\in\mathbb{Z}$. Insert this expression for $A_n$ into equation (3) to get the equivalent requirement $$ \lim_{N\to\infty}\frac{1}{N}\sum_{n,a,b} e^{ia\alpha}e^{ib\beta}\tilde A_n(a)\tilde A_n(b) = p(\alpha-\beta). \tag{5} $$ Rewrite the angles as $$ \alpha = \phi+\theta \hskip2cm \beta = \phi-\theta \tag{6} $$ to get $$ \lim_{N\to\infty}\frac{1}{N}\sum_{n,a,b} e^{i(a+b)\phi}e^{i(a-b)\theta} \tilde A_n(a) \tilde A_n(b) = p(2\theta). \tag{7} $$ Now take the Fourier transform $(2\pi)^{-1}\int_0^{2\pi} d\phi\ \cdots$ of both sides of (7) with respect to $\phi$ to get $$ \lim_{N\to\infty}\frac{1}{N}\sum_{n,a} e^{i2a\theta} \big|\tilde A_n(a)\big|^2 = p(2\theta) \tag{8} $$ using $\tilde A_n(-a) = \tilde A_n(a)^*$. This can also be written $$ \sum_a e^{i2a\theta} f(a) = p(2\theta) \tag{9} $$ with $$ f(a) := \lim_{N\to\infty}\frac{1}{N}\sum_{n} \big|\tilde A_n(a)\big|^2. \tag{10} $$ The function $p(2\theta)$ defined in (1) can be written $$ p(2\theta)=\frac{1}{4}+\frac{1}{8}\big(e^{i2\theta}+e^{-i2\theta}\big) \tag{11} $$ so the condition (9) implies that $f(a)$ is non-zero only for $a\in\{-1,0,1\}$. Since $f(a)$ is the average over $n$ of a non-negative quantity $\big|\tilde A_n(a)\big|^2$, this implies that $\tilde A_n(a)$ can only be non-zero for $a\in\{-1,0,1\}$ for every $n$. However, the original function $A_n(\alpha)$ is equal to either $0$ or $1$ for every $\alpha$, so either it is independent of $\alpha$ (which obviously contradicts the requirement (2b)) or it has a discontinuous jump at one or more values of $\alpha$, which contradicts the preceding statement that its Fourier transform has only a finite number of non-zero values. This completes the proof that the requirements (2a) and (2b) cannot be satisfied, so the project is impossible.


This demonstrates that quantum theory's prediction cannot be reproduced by any "local hidden variables" model. The name "hidden variables" refers to the functions $A_n(\theta)$ and $B_n(\theta)$, and the word "local" here refers to the fact that computer $A$ does not know what angles we're giving to computer $B$, and conversely.$^\dagger$

$^\dagger$ This is what "local" typically means in the context of hidden-variables models. The same word has other meanings in other contexts.


[1] http://www.mrao.cam.ac.uk/~steve/maxent2009/images/bell.pdf

[2] Section 3.1 in "Bell's theorem: experimental tests and implications" (http://physics.oregonstate.edu/~ostroveo/COURSES/ph651/Supplements_Phys651/RPP1978_Bell.pdf)

  • $\begingroup$ I can't thank you enough, I'm going to spend all the time it takes to fully understand it. $\endgroup$
    – Eran Medan
    Commented Apr 29, 2020 at 1:52
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    $\begingroup$ @EranMedan You're welcome! Beware that I found two minor mistakes: in equation (4), the symbols $a$ and $\alpha$ should be switched, and below equation (11), the words "strictly positive" should be changed to "non-negative." I'll make these changes in the answer soon. $\endgroup$ Commented Apr 29, 2020 at 2:34
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    $\begingroup$ @EranMedan I corrected equation (4) and the text below equation (11). $\endgroup$ Commented Apr 29, 2020 at 16:51

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