If you drop a coin it's affected by the air drag, bounces and tumbles on floor before it settles and you can read whether it was heads and tails.

If I understood it right, the Bell's theorem says that if the value what we'll read from a pair of coins is determined from the start, then the correlations between the two data will have limits given by the Bell's and other inequalities.

But in the coin tossing example there are lot of parameters that determine the result: air flows, the location it first hits the floor, etc. And these parameters can change in the meantime. So it's practically unpredictable to say at the time of tossing that it will be heads or tails. Pretty much similar situation what say about the particles. It's in a magical superposition state which then collapses and we can read heads and tails.

That's a bit similar to what we do in those particle experiment, we toss particles and measure whether they pass the polarizer or not or they'll be spin up or spin down.

So could we say the same about the experiments that aimed to show entanglement? That photon entering the polarizing beam splitter is passed back and forth millions times among rapidly thermally moving atoms of the crystal, which is quite chaotic. But at the end there is some order in the chaos since the two polarizations exit in two directions.

Can this chaos in that crystal cause correlations that violate the Bell's theorem?

  • $\begingroup$ No, the violation of Bell's theorem is a phenomenon of quantum physics, at the exclusion of any explanation by classical phenomena. $\endgroup$
    – Moonraker
    Jan 22, 2015 at 9:31
  • $\begingroup$ Someone asked a fairly similar question: Does Bell's inequalities also rule out non-computable local hidden variable theories?. If you substitute "non-chaotic" for "computable" and "chaotic" for "non-computable" in my answer, I think it covers your question, but read it over and let me know if you think it's missing something. $\endgroup$
    – Hypnosifl
    Jan 23, 2015 at 0:10

2 Answers 2


Suppose you have two electrons correlated in a singlet state: $$ |\psi\rangle=|ud\rangle+|du\rangle, $$ where $u$ and $d$ stand for up and down spins in the z direction.

What happens in the experiment is that regardless of which axis you measure along the probability of each result is always 50%/ but you find correlations when you compare the results of measurements. If you measure the spins of the particles along the same axis and compare them you find the results are anti-correlated (i.e. - opposite). If you measure one along the x-axis and the other z-axis then you find the results match 50% of the time and do not match the other 50% of the time. So whether the spins match depends on whether you measure them in the same direction. But this contradicts your assumption that there is a single value of the spin decided before you do the measurement. Bell's theorem quantifies the magnitude of this effect for all angles between the axes, explaining that it is larger than could be produced by any local explanation in which the value of the measured quantity is a stochastic classical variable (i.e. - a single number picked at random with some probability).

Your suggested explanation is that the measurement devices somehow produce this effect. But this doesn't get you out of the problem. For you can look on the state of the measurement devices as being the result of another measurement, in which case you still have measurement results that have larger correlations than allowed by classical physics.

The EPR experiment has been explained. The particles can't be described by a single spin because they don't have a single spin. Rather, the spin is described by Heisenberg picture observables, which are nothing like a single number representing the spin. The electrons each exist in multiple versions that can interact with one another in interference experiments. Each particle's observables describe quantum information about the relations between the different versions of each particle, but this information can't be revealed by measurements on either particle alone. For discussion of relevant issues see





The violation of Bell's inequalities is not exactly chaotic. Let me put the things simpler, for a singlet state of photons whose polarizations are measured along the same pair of axes, $x, y$, if one photon responds $x$ the other responds also $x$. This is not chaos, this is perfect order.

There is a big difference from tossing coins. If the two coins are identical, and you toss one independently of the other, the results they produce are independent: one produce had and tail in proportion of 1/2 and 1/2, the other produces the same, so in all you get results according to the law of independent events, i.e. 1/4 both tails, and 1/4 both heads, 1/4 one gives had and one gives tail, and 1/4 vice-versa.

With the entanglement the events are not independent, as said above.

What make the two photons to give sometimes $x$, and other times $y$, we don't know, if there is the chaos that you suggest ... maybe yes, maybe no. But in each single experiment, if one particle responded $x$, there is no more chaos.

Now, in Bell-type experiments we don't measure both particles along the same axis, one is measured in one system of axes, the other in a system of axes rotated by some angle. Here also, there are laws that appears directly from the singlet correlation $(|x>|x> + |y>|y>)/\sqrt{2} \ \ $ where the axes for one particle are rotated. Why in each single trial we get a certain result, we don't know, maybe is the chaos that you suggest. But this chaos turns out to be discipline. Returning many times on an experiment with one photon measured in the axes $x, y$ and the other photon along axes e.g. $x', y'$ one gets the response $x, x'$ not 1/4, but $1/2 \ cos^2(\theta)$.

The chaos is disciplined. Once a particle gave a result, the other one is constrained by the correlation.


Your Answer

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

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