Bell's Inequality Experiments Question

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Bell's Inequality Experiments Question

Postby lgw » Wed Sep 24, 2014 9:57 pm UTC

I've been following the mounting evidence for Bell's Theorem from experiments over the past couple of years, but I don't quite understand the experiment. When a pair of entangled photons are measured for polarization (or electrons for spin polarization, same deal), the predicted results are the same as for one photon passing through 2 polarizing filters, right? That is, the chance that the two (entangled) photons will be measured as "correlated" by two detectors at an angle θ is the same as the chance a single photon has of passing through two polarizing filters in sequence at that same angle: cos2(θ), right? (And similar for measuring the spin polarization of one electron twice vs two (entangled) once.)

Unless I'm missing something, this is basically the same simple QM result for a 2-state system being demonstrated in both cases. I'm struggling to see why measuring the entangled photon pair is any more interesting than the "two filters one photon" case as far as hidden variables, and especially why it's any more indicative of "no hidden variables". Why does the entangled case rule out hidden variables better than the simpler case? I feel like I must be missing something important here - like the entire point of the exercise!
"In no set of physics laws do you get two cats." - doogly

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Re: Bell's Inequality Experiments Question

Postby Hypnosifl » Wed Sep 24, 2014 10:56 pm UTC

The correlations between the two measurements would be easy to explain if the measurement of one particle could directly influence the hidden variables of the other. In the case of a single photon, there'd be nothing problematic about the idea that the first measurement alters the photon's hidden variables in a way that depends on the setting of the polarizer angle. But for two entangled photons, the spacetime separation between the two measurements can be such that no signal moving at the speed of light or slower could "update" one photon with information about what polarizer angle had been used in measuring its entangled twin.

A while ago I wrote up a little analogy to help explain why Bell's inequality follows naturally from any assumption of local hidden variables, and thus why it's a big deal to see it violated:

Suppose we have a machine that generates pairs of scratch lotto cards, each of which has three boxes that, when scratched, can reveal either a cherry or a lemon. We give one card to Alice and one to Bob, and each scratches only one of the three boxes. When we repeat this many times, we find that whenever they both pick the same box to scratch, they always get the same result--if Bob scratches box A and finds a cherry, and Alice scratches box A on her card, she's guaranteed to find a cherry too.

Classically, we might explain this by supposing that there is definitely either a cherry or a lemon in each box, even though we don't reveal it until we scratch it, and that the machine prints pairs of cards in such a way that the "hidden" fruit in a given box of one card always matches the hidden fruit in the same box of the other card. If we represent cherries as + and lemons as -, so that a B+ card would represent one where box B's hidden fruit is a cherry, then the classical assumption is that each card's +'s and -'s are the same as the other--if the first card was created with hidden fruits A+,B+,C-, then the other card must also have been created with the hidden fruits A+,B+,C-.

The problem is that if this were true, it would force you to the conclusion that if Alice and Bob are picking randomly which box to scratch on each trial (with a 1/3 chance of A, B, or C each time), then if they do this a large number of times, we should expect that in the subset of trials where Alice and Bob happened to pick different boxes to scratch, they should find the same fruit at least 1/3 of the time. For example, if we imagine Bob and Alice's cards each have the hidden fruits A+,B-,C+, then we can look at each possible way that Alice and Bob can randomly choose different boxes to scratch, and what the results would be:

Bob picks A, Alice picks B: opposite results (Bob gets a cherry, Alice gets a lemon)

Bob picks A, Alice picks C: same results (Bob gets a cherry, Alice gets a cherry)

Bob picks B, Alice picks A: opposite (Bob gets a lemon, Alice gets a cherry)

Bob picks B, Alice picks C: opposite results (Bob gets a lemon, Alice gets a cherry)

Bob picks C, Alice picks A: same results (Bob gets a cherry, Alice gets a cherry)

Bob picks C, Alice picks picks B: opposite results (Bob gets a cherry, Alice gets a lemon)

In this case, you can see that that if they are equally likely to pick each combination of boxes, then 2 times out of 6 when they choose different boxes, they will get the same fruit (i.e. a 1/3 chance of the same result). You'd get the same answer if you assumed any other preexisting state where there are two fruits of one type and one of the other, like A+,B+,C- or A+,B-,C-. On the other hand, if you assume a state where each card has the same fruit behind all three boxes, so either they're both getting A+,B+,C+ or they're both getting A-,B-,C-, then of course even if Alice and Bob pick different boxes to scratch they're guaranteed to get the same fruits with probability 1. So if you imagine that when multiple pairs of cards are generated by the machine, some fraction of pairs are created in inhomogoneous preexisting states like A+,B-,C- while other pairs are created in homogoneous preexisting states like A+,B+,C+, then the probability of getting the same fruits when you scratch different boxes should be somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if 100% of all the pairs were created in inhomogoneous preexisting states, it wouldn't make sense for you to get the same answers in less than 1/3 of trials where you scratch different boxes, provided you assume that each card has such a preexisting state with "hidden fruits" in each box.

But now suppose Alice and Bob look at all the trials where they picked different boxes, and found that they only got the same fruits 1/4 of the time! That would be the violation of Bell's inequality, and something equivalent actually can happen when you measure the spin of entangled photons along one of three different possible axes. So in this example, it seems we can't resolve the mystery by just assuming the machine creates two cards with definite "hidden fruits" behind each box, such that the two cards always have the same fruits in a given box.

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Re: Bell's Inequality Experiments Question

Postby BedderDanu » Thu Sep 25, 2014 12:51 am UTC

I recently encountered a crackpot who "disproved" Bell's Theorem, which forced me to actually try to pry it apart and figure it out. It's actually quite simple, but it takes more text to explain than a simple summary usually suffices.

Bell's theorem is actually from statistics, not physics, and has nothing to do with Quantum Mechanics... at first.
So let's see how far we can get without it.

Goal: Devise an experiment that produces different results whether "Quantum Mechanics" is true, or any other theory of local hidden variables is true.

Setup: Fire polarized photons at a polarized screen. Based on the angle between the screen and the photon, the photon has a chance of Passing through (1) or hitting the screen and being blocked (0). If we use 2 entangled photons, and fire them at two different screens, we can observe the effects of entanglement. Basically if one side is blocked the other side is also blocked, as long as the screens are angled the same way.

Test: We fire off the entangled photons, and while they are in flight we randomize the angle of the screens. So even though the photons are entangled, they can't send information between themselves before they get measured at the screens. Therefore, if there are hidden variables, they won't have time to be changed by the settings of the screens, and we shouldn't be able to find any entanglement effects. If we still find the effects, then we know that the only explanations are either 1) Hidden Variables can transmit information faster than light, or 2) QM description of entanglement exists.

Discussion: The problem with this is that the chance of a photon passing through the screen is inherently random. So how do you know if your experiment is actually working, or if you've simply "Lucked" your way into one answer or another. Enter Bell's Theorem, or more specifically Bell's inequality.

Admission: I'm not familiar with how to apply the inequalities. I've never had to use them. However, I've seen the following analogy to kind of explain how they work. Going back to our Experimental setup, lets say we collect the result of all the tests that had 75% chance of transmission on both sides (basically, we do this test thousands of times, and ignore all the results where the settings weren't ±30°) if we look at the results of each test, they probably would look something like this (idealized to prove a point)

Side A: 1110 1101 1011 0111 1110 1110 1011: 75% transmission rate
Side B: 0111 1011 0111 0111 1110 1011 1101: 75% transmission rate
Match? 0110 1001 0011 1111 1111 1010 1001: 64.3% match rate

Match simply states whether the two sides agreed with each other or not. However, if you think about it, there is a maximum amount of disagreement between the two sides that can be obtained:

Side A: 1110 1110 1110 1110 1110 1110 1110: 75% transmission rate
Side B: 0111 0111 0111 0111 0111 0111 0111: 75% transmission rate
Match? 0110 0110 0110 0110 0110 0110 0110: 50% match rate

It is impossible for randomness to produce less than a 50% match between the two results, as long as that transmission rate is correct.

So when we run the experiment for a known transmission rate, we know that if we get less than a 50% match, we know that true randomness cannot be the source of the errors. But our experimental design prevents local variables from mattering unless they transmit information faster than light.

When the above test is tried, we get a match rate of 25%. So, not a random answer, and no local variables.

This is the best I've been able to put together.

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Re: Bell's Inequality Experiments Question

Postby Hypnosifl » Thu Sep 25, 2014 2:28 am UTC

BedderDanu wrote:It is impossible for randomness to produce less than a 50% match between the two results, as long as that transmission rate is correct.

But why assume the match must be due to "randomness"? The main point is to rule out local hidden variables theories, and hidden variables theories try to explain matching results in terms of the two entangled particles having been created with correlated (hidden) properties which predetermine the results of different possible measurements that could be made on each. In my example, both scratch lotto cards had the same series of 3 fruits hidden under the 3 squares A-B-C (say, cherry-cherry-lemon), which obviously would determine what an experimenter would see if they scratch off a given square (for example, if they both scratch B they would both see a cherry). The 3 squares are meant to be analogous to 3 possible angles for the polarizer, say 0 degrees from vertical, 60 degrees, and 120 degrees. If both experimenters choose the same angle they are guaranteed to get the same result with probability 1 (both photons pass or both are blocked), but if they choose different settings the probability of getting the same result would be (cos(60))^2 or (cos(120))^2, which in both cases gives a probability of 1/4. But the Bell inequality whose derivation I sketched says that if you want to explain the perfect match when both experimenters make the same choice using local hidden variables, the probability of a match when they make different choices should be no lower than 1/3.
Last edited by Hypnosifl on Sat Oct 11, 2014 7:03 pm UTC, edited 1 time in total.

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Re: Bell's Inequality Experiments Question

Postby lightvector » Thu Sep 25, 2014 4:35 am UTC

Here's an interesting article that explains how quantum mechanics can allow two cooperating players to win a certain game ("CHSH game") with strictly greater probability than would be allowed by classical mechanics, assuming no FTL communication. Plus a wacky theoretical application of this game that's actually the more primary focus of the article. ... ndomness/1

The fact that you can win the game this frequently quantum-mechanically but not classically is sort of just Bell's inequality and its violation in a different guise, but I find it perhaps easier, or more accessible in some ways, to understand what's going on with Bell's inequality in this form.

In particular, if you accept as an axiom that the probability that the two players will measure and choose the same result is equal to cos^2(theta) where theta is the angle between their measurement bases, then it's easy to see how the stated quantum strategy does indeed achieve the desired winning percentage of about 85%. And deriving this result itself is not hard with some very basic quantum mechanics, and also is intuitively analogous and consistent with things like the percentages of light blocked by polarizing filters at different angles.

Additionally, it's also easy to show that the game cannot be won classically more than 75% on average. Here's a proof sketch: every classical strategy is simply an affine combination of one of 16 "pure" strategies, namely the deterministic strategies determined by the following 4 bits: What Alice does on blue, what Alice does on red, what Bob does on blue, what Bob does on red - that is any classical strategy is equivalent to the players on each round choosing among of these 16 pure strategies each with a certain probability and then playing that pure strategy. By linearity, the winning probability of the overall strategy is the corresponding affine combination of the winning probabilities of the 16 pure strategies. Each of these pure strategies wins at most 75% of the time on average, this is pretty obvious if you just start writing down what the strategies are and work through the rules of the game to see how often they will win. Therefore, any classical strategy wins at most 75% of the time on average.

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