0
$\begingroup$

In the following video:

https://www.youtube.com/watch?v=qd-tKr0LJTM

DrPhysicsA describes the results of an experiment involving three polarized mirrors randomly selected by the perennial Alice and Bob.

enter image description here

The first three columns represent possible pre-determined polarizations (hidden variables) of entangled photons sent to Alice's and Bob's randomly selected mirrors, A, B, and C. The second three columns represent the possible results of observations by Alice & Bob for three different combinations of mirrors. The results tabulated show that considering only the middle six rows there will be six occurrences when Bob and Alice measure the same result, either no photon is detected by both, or they both detect a photon. According to the presenter this means that they will make the same 33% or more of the time if there are hidden variables, while Quantum Mechanical analysis predicts, and experimental results show 25%. Thus the possibility of local hidden variables is eliminated.

My question is this: On what basis are the top and bottom row of the hypothetical data eliminated from the calculation? Considering all possible results there would be 12 "Same" observations out of 24 possible results, or 50% instead 33%. Also, when running the experiment, it would not be possible to exclude a top or bottom row from the results. Any "Same" looks like any other. So why are only the six middle rows counted?

$\endgroup$
1
$\begingroup$

Independent of the question of which rows you count, the presenter's claim (at least in your paraphrase) does not make a whit of sense unless you add the observation that each row is equally likely --- and this isn't something you want to assume.

The important point is that no matter what probabilities you assign to the different rows, the implied probabilities for the experimental outcomes cannot match what we actually observe. That's a far far stronger statement than what you're quoting here, which is only about one particular set of probabilities (namely zero on the first and last rows and 1/6 for all the others). That's not something you can learn from naive counting (though you can learn it from some fairly simple algebra).

(If that's not clear, imagine the case where Alice and Bob get "same" every time in every experiment. Does that rule out hidden variables? No, it just means that if there are hidden variables, they are set so that we're always in either the first or last row---that is, the probabilties on all the other rows are zero. To rule out hidden variables, we need an outcome that can't be explained by any assumption about the probabilities of the rows.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't see that the situation is improved if each row is equally likely. The result will be 33% or greater no matter what rows are selected, even if it's consistently one row, ie, one set of hidden variables, including rows one and eight. My question, and perhaps it's just a detail, why are the first and last rows excluded? $\endgroup$ – John Fistere Aug 4 '16 at 1:11
  • $\begingroup$ @JohnFistere : I cannot tell you why someone else chose a particular probability distribution (the one with first and last rows having probability zero and the others equally likely.) I can tell you that by choosing any probability distribution at all, this person was entirely missing the point. The calculation with the first and last rows included would be as pointless as the calculation you quote. The only interesting calculation is the one that says that no matter what probabilities you assign to the rows, you can't match the observations. $\endgroup$ – WillO Aug 4 '16 at 1:27
  • $\begingroup$ Does the observation probe that for any given pair of entangled photons, if A and B use the same polarizer either both photons will go through or neither will do? $\endgroup$ – Natxo Aug 30 '16 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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