I read this description of Bell's theorem. I understand he's restating it slightly, so there may be incorrect assumptions there, or I may have some. I think Bell's theorem should lead to FTL communication, and I'll try to lay out my assumptions as to how, and propose an experiment. So:

  1. A and B are entangled photons.
  2. With 100% probability, if I measure A at X°, and you measure B at X°, our measurements will be opposite (mine will go through iff yours is absorbed). 3.

So this is the experiment:

  1. Generate A and B, and send B lightyears away.
  2. Whoever's manning the experiment at B does the following:

    1. If they want to send binary 1, measure B at 20°.

    2. If they want to send binary 0, don't do anything.

  3. Now whoever is at A measures A first at 0°, then (if it goes through), at 40°.

  4. The results should be (it seems to me):

    1. If 1 above, then B was absorbed with 50% probability and was transmitted with 50% probability. In either case, the probability of seeing A get transmitted at 0° and get absorbed at 40° is less than or equal to 11.6% (because the probability of case 1 or 2 in the article is 50% for the measurement at 20° times either the probability of A transmitted at 0°, or B transmitted 40° (which under assumption 2, is the same as A not transmitted 40°) so the probability of both must be less than or equal to 11.6%).

    2. If 2 above, then the probability of seeing A get transmitted at 0° and get absorbed at 40° is 20.7%.

So A can determine a binary message from B, violating the no-communication theorem. If done with enough photons, you can send much longer messages.

I know it is far likelier that I've made a mistake here than that I've disproved the no-communication theorem, so what's wrong? If anyone doesn't understand the experiment, ask and I'll try to clarify.

  • $\begingroup$ I posted a comment on that article with the same proposal. $\endgroup$
    – ike
    Commented Apr 24, 2014 at 19:06
  • $\begingroup$ And yes, I saw the other five versions of this question here, but none seemed like my experiment, nor did the answers explain why mine wouldn't work. $\endgroup$
    – ike
    Commented Apr 24, 2014 at 19:12
  • 2
    $\begingroup$ Why the downvote? $\endgroup$
    – ike
    Commented Apr 24, 2014 at 20:00
  • 2
    $\begingroup$ I didn't downvote, but here is a suggestion why someone might have done. This is without a doubt the most frequently asked question on this Q&A site. For good reason, because it's a confusing subject. Have you thoroughly searched the previous proposals of this type on the site, and read the answers there? $\endgroup$ Commented Apr 24, 2014 at 20:44
  • $\begingroup$ There is an entirely local explanation for how the Bell correlations are created, see arxiv.org/abs/quant-ph/9906007 arxiv.org/abs/1109.6223 The confusion about whether the Bell correlations involve FTL signalling is due to lack of clarity about what quantum mechanics implies about the way the world works, or even about whether it should be used to understand how the world works. $\endgroup$
    – alanf
    Commented Apr 25, 2014 at 9:26

1 Answer 1


I'm assuming you're talking about plane-polarized photons, where a photon that passes a 0º analyzer is horizontally polarized, a photon that passes a 90º analyzer is vertically polarized, and there's an orthogonal polarization basis at ±45º.

Here's the trouble:

Now whoever is at A measures A first at 0°, then (if it goes through), at 40°.

Like a river, you can never step in the same photon twice. Once you have analyzed the photon at 0º, you have irrevocably changed that photon to be either horizontally or vertically polarized. The probability that it passes a second polarizer is independent of whatever state it had when it was "born."

You can see this for yourself if you can find someone selling polarized sunglasses and borrow three pairs. If you take two linear polarizers and set them perpendicular to each other, the light transmission is zero. But if you take a third polarizer and put it in the middle, you can tune the total transmission of the system. With the middle polarizer at 45º you get 1/4 of the photons through: any information about their state in the horizontal-vertical basis gets destroyed when the pass through a diagonal polarizer.

Here's a quantitative analysis. Your polarization analyzers tell you whether a photon's linear polarization is parallel or perpendicular to a particular direction. If I have two polarizers whose axis are different by $\phi$, and they analyze your entangled photons, they'll see a correlation $$ C = \frac{N_\parallel - N_\perp}{N_\parallel + N_\perp} = \cos 2\phi. $$ This has the correct limiting behaviors: perfect correlation for $\phi=0$, perfect anticorrelation for $\phi=90º$, zero correlation for $\phi=45º$. If the $N$ are fractions of the whole, $C$ is equivalent to $$ N_\parallel = \cos^2 \phi, \quad\quad N_\perp = \sin^2 \phi. $$ If A's primary analyzer is at 0º, and B's analyzer is at 20º, then A and B will compare notes after the experiment and find (neglecting experimental uncertainties) that their populations of photons fall into these four groups: $$ \begin{array} & & \parallel\text{ at B} & \perp\text{ at B} \\ \parallel\text{ at A} & 44.2\% & 5.8\% \\ \perp \text{ at A} & 5.8\% & 44.2\% \end{array} $$ Now add A's second analyzer. The key insight is that A's first analyzer acts as a polarizer: once the photon is analyzed, its correlation to B is destroyed. The transmission through A's second analyzer is given by Malus's Law, $$ I = I_0 \cos^2 \theta, $$ where $\theta$ is the angle between the two polarizers. For 40º, the transmission through the second polarizer is 58.7%. So the new results become $$ \begin{array} & & & \parallel\text{ at B} & \perp\text{ at B} \\ \parallel\text{ at A and } & \parallel\text{ at A}' & 25.9\% & 3.4\% \\ \parallel\text{ at A and } & \perp\text{ at A}' & 18.3\% & 2.4\% \\ \perp \text{ at A} & & 5.8\% & 44.2\% \end{array} $$ If B declines to make a measurement, or if B makes a measurement but his dog eats his notes, there is no effect on the transmission or polarization at A. Only when all the photon polarization measurements are compared can this pattern emerge.

  • $\begingroup$ So what would actually happen? A measures 20.7% regardless of what happens at B? $\endgroup$
    – ike
    Commented Apr 25, 2014 at 18:31
  • $\begingroup$ The results of measurements at A's second polarizer would show no correlations to the conditions at B. $\endgroup$
    – rob
    Commented Apr 25, 2014 at 18:38
  • $\begingroup$ No correlation to the measurement (choice of angle) at B, or no correlation to the result at B? $\endgroup$
    – ike
    Commented Apr 25, 2014 at 18:41
  • $\begingroup$ Your model doesn't describe a choice of angle at B: you say "measure B at 20º, or do nothing." But any entanglement correlation will be seen at the first polarizer at A; the behavior of light at A's second polarizer has nothing to do with the conditions at B $\endgroup$
    – rob
    Commented Apr 25, 2014 at 19:05
  • 1
    $\begingroup$ But if the first measurement of A is correlated with B, and is also related to the second measurement of A, doesn't that mean second(A) correlates to B? And can you give exact percents for my experiment? $\endgroup$
    – ike
    Commented Apr 25, 2014 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.