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I'm working through trying to understand this, but the math of the original paper, and subsequent actual academic papers are a bit beyond me. So I've turned to simplified explanations, all of which are more or less the same, but for reference i'll stick with this one

However, it seems to me that the assumptions about a locally deterministic theory are trivially easy to challenge simply by actually adding an extra "hidden" variable.

As a corollary, lets say we create a pseudo random number generator, that outputs a single number 1-6. And then we create a second one with the same exact seed, such that they would output the same number, except we add 2, so if the first rolls a 6 the second will be a 2.

Then lets say the program only reports to us if its either 1-3(A) or 4-6(B). Now, individually each is perfectly unpredictable happening 50% of the time, and thus we might assume, that the results should be AA,AB,BA,BB, and we'd have agreement 50% of the time. However, because only a first roll of 1 or 4 results in an agreement, the actual result is 1/3.

So here is an example of a system that's perfectly locally deterministic, and unpredictable, while still producing a disagreement, between the "random" outcome and a correlated one.

And thus my suggestion would be that Bell's Theorem fails, in assuming that the hidden variables have to adhere to the top level random odds.

First question, why is my explanation wrong?

Second Question, After a photon goes through a polarized filter it takes on the polarization of the filter. What if we setup two filters with one rotated to always have perpendicular polarization to the first. Then, set it up so the exact orientation is unknown to us and random(excepting the relation between the two) And then fired photons through this setup, and did the Bell Theorem tests, would the predictions of QM for some reason be different, then in the case of the "entangled" photons?

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  • $\begingroup$ I think your reference is misleading. Have a look at the image here en.wikipedia.org/wiki/Spin_angular_momentum_of_light which showd how circular light polarization is built up by the individual photons, which can only have +1 or -1 spin to their direction of motion. Light is emergent from zillions of photons in a quantum mechanical superposition ( not interaction). The probabilities are not random but weighted , they have to follow the solutions of the quantum mechanical equations which are axiomatic, in the sense that the postulates of qm are like axioms. $\endgroup$ – anna v Feb 1 '18 at 12:05
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Suppose we have two dice. You hold one, I hold the other. And:

Whenever we both roll with our right hands, our dice come up identically.

Whenever you roll with your left and I roll with my right, our dice come up identically.

Whenever you roll with your right and I roll with my left, our dice come up identically.

But whenever we both roll with our left hands, our dice come up different.

Now try constructing a locally deterministic explanation of that. Bell's Theorem says you can't. If you think you can, you are mistaken.

So:

  • Bell's Theorem can be applied to rule out locally deterministic explanations of some things but not others.
  • You've given a perfectly good example of a case where you can't apply Bell's Theorem.
  • That doesn't mean you can never apply Bell's theorem. In fact I've just given you an example where you can apply it.
  • What happens in quantum mechanics is neither exactly like your example nor exactly like mine, but it's a lot more like mine than it is like yours. In particular, Bell's Theorem applies.
  • In short: The fact that you can find a locally deterministic explanation for some made-up dice problem tells you nothing about whether you can find a locally deterministic explanation for the actual phenomena that we observe in quantum mechanics.
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  • $\begingroup$ each left hand imparts half of whats necessary for coherence? lol, but yah I'm perfectly willing to accept that the scenario ive outlined doesn't apply to QM, it just hasn't been made totally clear to me why yet. $\endgroup$ – Dekeita Feb 1 '18 at 6:50
  • $\begingroup$ @Dekeita quantum mechanical probabilities are not random, they are weighted by solutions of specific differential equations which follow specific postulates and principles to give the probability distributions. $\endgroup$ – anna v Feb 1 '18 at 11:55
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You've missed the point. It has nothing to do with producing a disagreement between a random outcome and a correlated one.

The hidden variables assumption is that the particle is created with definite polarization. So it already "knows" if it will go through a filter at any given angle.

If you make this assumption, and then imagine measuring the same photon along two different random axes (separated by $120^\circ$) many times (really we measure two separated, entangled photons), the link you've given shows that you can never get a result of less than $\frac{1}{3}$, no matter what state the photon is actually created in. Note that the randomness has nothing to do with the photon- the whole point is that in a hidden variable theory, there is no randomness. The only randomness is in the experiment itself.

Specifically, in a hidden variable theory like the one listed on that page, there are 8 types of photons. If you randomly choose axes and check if they agree, depending on what photons are made the probability can be anywhere from $1\over 3$ to $1$. It's only less than $1\over 3$ that can't happen, regardless of how the photons are picked.

Now to the main point. Quantum mechanics predicts (and experiments don't disagree) that if you actually do this, the probability is actually $\frac{1}{4}$. Since $\frac{1}{3} > \frac{1}{4}$, this demonstrates that hidden local variable theories are inadequate for explaining quantum mechanics.

If you want to play around with this, I've made a python program to do so. It simulates the experiment described on the page you linked to.

#bell_inequality.py
import random

random.seed()

def agree(state):
    x=random.randint(0,2)
    return state[x]*state[x-1] > 0

def make_photon(x):
    extra_hidden_variables={ 'dog' : 'no' , 'cat' : False , '1-3' : 'A' , '2-4' : B' } #The experiment can't measure these.
    #Whatever extra hidden variables I have, at the end I have to pick my behavior for each axis
    return (1,1,1) #All plus, replace with -1 for negative along specified axis

agreement=0
total=1000000
for x in xrange(total):
    state=make_photon(x)
    if agree(state): agreement+=1

print "%d/%d=%f" % (agreement,total,agreement/float(total))

The function make_photon(x) makes a photon and then it is measured along 2 axes at random. Put whatever you like in make_photon(x) and see if you can get the probability below $\frac{1}{3}$, or to the $\frac{1}{4}$ predicted by quantum mechanics!

The point here is, however the photons "decide" to act when they are measured along an axis, there is no possible set of decisions that leads to a probability of $\frac{1}{4}$ in the experiment. So, whatever you put in make_photon(x), you won't be able to get that probability.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Feb 2 '18 at 17:56

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