Question about Bells tests?

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I have read a decent amount about bells test experiments and some of that recent thread about it. What I don't get is when they talk about rotating the detector an measuring the electrons spin. And lets use pi mesons that decay into an electron and positron. So if I have both of my detectors at 0 degrees I will measure one spin up and the other spin down. But if I rotate one of my detectors 45 degrees I don't see how these spin measurements can correlate because my axis has changed. For example if i measure spin up on the zero degree detector will I measure spin up or down on the 45 degree detector?
Any help will be much appreciated.
 

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  • #2
DrChinese
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The idea is that if you measure ONLY on fully non-commuting bases - say x and y - you don't see anything unusual. Bell realized that on ANY other basis, you were observing a mixture of x and y. What is 45 degrees other than a mixture? As long as you stay within QM, there is nothing new to see though.

But clearly there is if you want to assert realism, because you have to stay "true" to the mixture in a statistical sense. That won't be possible.
 
  • #3
JesseM
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I have read a decent amount about bells test experiments and some of that recent thread about it. What I don't get is when they talk about rotating the detector an measuring the electrons spin. And lets use pi mesons that decay into an electron and positron. So if I have both of my detectors at 0 degrees I will measure one spin up and the other spin down. But if I rotate one of my detectors 45 degrees I don't see how these spin measurements can correlate because my axis has changed. For example if i measure spin up on the zero degree detector will I measure spin up or down on the 45 degree detector?
Any help will be much appreciated.
According to QM, if you always get opposite spins when you pick the same angle, then the probability you'll measure opposite spins when you pick different angles A and B is cos^2(A-B). So If one detector is at 0 and the other is at 45, the probability of getting opposite results (spin-up at 0 and spin-down at 45, or spin-down at 0 and spin-up at 45) is cos^2(45-0) = 0.5.
 
  • #4
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I guess I will have to re-read a paper explaining bells theorem . Because I still don't understand the basic logic behind the experiment. Thank-you very much for your guys response.
 
  • #5
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So Jesse M I read your analogy with lemons and cherrys and it was good by the way.
So If Alice and bob pick different boxes to check they will get the same result 1/3 of the time. When they both get a lemon is this equivalent to them both measuring their particle spin up . And the fact that experimental evidence can show that this only happens 1/4 of the time, Is this enough to prove that the particles could not have been created with a preexisting spin?
 
  • #6
JesseM
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So Jesse M I read your analogy with lemons and cherrys and it was good by the way.
So If Alice and bob pick different boxes to check they will get the same result 1/3 of the time. When they both get a lemon is this equivalent to them both measuring their particle spin up . And the fact that experimental evidence can show that this only happens 1/4 of the time, Is this enough to prove that the particles could not have been created with a preexisting spin?
Well, you need some other conditions, for example it must be true that the choice of one particle's detector setting was at a space-like separation from the measurement of the other particle, so that in a local realist universe the first event can't have had a causal influence on the second (in terms of my analogy, if this wasn't true then each lotto card could have a radio receiver which picked up a signal about what box was scratched on the other card, and alter its hidden fruit so that if the same box was scratched on it it would show the same fruit, but if a different box was scratched it would always show the opposite fruit). You also need the assumption that all the emitted particles are measured (if this is violated then you could have a situation where some lotto cards are "defective" and burn up when you try to scratch a particular box so those trials don't get included in your data, see billschnieder's explanation [post=2767632]here[/post] and [post=2767828]here[/post]), in a situation where you don't have perfect detector efficiency you need a different Bell inequality that takes into account the level of efficiency of the detector, see here. But if you can close off these Bell test loopholes (and if you also theoretically rule out some more exotic possibilities like that the source printing the lotto cards could somehow know in advance what boxes the experimenters were going to scratch, which would require either backwards-in-time causal influences or superdeterminism), then you can rule out the possibility that the cards/particles just had predetermined answers for each possible measurement.
 
  • #7
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ok thanks for your response, And one more thing with your analogy of either a cherry or lemon. Could this be equivalent to light getting through the polarizer or not, If the light got through it is a cherry and if it doesn't get through it is a lemon?
 
  • #8
JesseM
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ok thanks for your response, And one more thing with your analogy of either a cherry or lemon. Could this be equivalent to light getting through the polarizer or not, If the light got through it is a cherry and if it doesn't get through it is a lemon?
Right, and the three boxes the experimenter has a choice of scratching would be equivalent to the three angles the experimenter can set the polarizer to.
 
  • #9
edguy99
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I have read a decent amount about bells test experiments and some of that recent thread about it. What I don't get is when they talk about rotating the detector an measuring the electrons spin. And lets use pi mesons that decay into an electron and positron. So if I have both of my detectors at 0 degrees I will measure one spin up and the other spin down. But if I rotate one of my detectors 45 degrees I don't see how these spin measurements can correlate because my axis has changed. For example if i measure spin up on the zero degree detector will I measure spin up or down on the 45 degree detector?
Any help will be much appreciated.
If modeled as a bloch sphere, you will have a 50% chance of measuring down and a 50% chance of not getting a measurement.
 
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  • #10
DrChinese
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If modeled as a bloch sphere, you will have a 50% chance of measuring down and a 50% chance of not getting a measurement.
Within the context of the OP's question, I don't think this will be very helpful. The likelihood of detection has nothing to do with a bloch sphere. It is dependent on the experimental setup.
 
  • #11
JesseM
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If modeled as a bloch sphere, you will have a 50% chance of measuring down and a 50% chance of not getting a measurement.
Here I think you are talking about the specific loophole I mentioned earlier:
You also need the assumption that all the emitted particles are measured (if this is violated then you could have a situation where some lotto cards are "defective" and burn up when you try to scratch a particular box so those trials don't get included in your data, see billschnieder's explanation [post=2767632]here[/post] and [post=2767828]here[/post]), in a situation where you don't have perfect detector efficiency you need a different Bell inequality that takes into account the level of efficiency of the detector, see here
The detector efficiency loophole has been closed in some experiments with entangled ions incidentally, see here (pdf file) and here.
 
  • #12
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If measuring one particle affects the other entangled particle. Then when we measure both at the same time it seems like we would have problems. If we measured both at the same time you wouldn't know which one influenced the other. Or maybe i'm missing something.
 
  • #13
DrChinese
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If measuring one particle affects the other entangled particle. Then when we measure both at the same time it seems like we would have problems. If we measured both at the same time you wouldn't know which one influenced the other. Or maybe i'm missing something.
The statistics are the same whether Alice influenced Bob or vice versa! Surprisingly, time ordering is not significant.
 
  • #14
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ok , And for jeeseM's analogy with the lotto cards . If we assumed they were predetermined fruits behind the boxes and they each scratched 3 different boxes on 3 different cards they would agree 1/3 of the time. But Is this assuming the lotto machine keeps making the same cards and sending it to them. Why couldn't it make a different card each time just make sure the cards going out had the same things on them.
 
  • #15
DrChinese
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But Is this assuming the lotto machine keeps making the same cards and sending it to them. Why couldn't it make a different card each time just make sure the cards going out had the same things on them.
Not sure I follow. How are they different cards if they have the same contents?
 
  • #16
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Ok the machine first spits out 2 cards with the same elements in each box. Then it spits out 2 more cards each with the same elements but different elements than the first set of cards. In Jesse M's analogy it seems like the mahine creates the same cards all the time?
 
  • #17
JesseM
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Ok the machine first spits out 2 cards with the same elements in each box. Then it spits out 2 more cards each with the same elements but different elements than the first set of cards. In Jesse M's analogy it seems like the mahine creates the same cards all the time?
No, in my [post=3237782]post[/post] I first examined a specific example where they both get cards with hidden fruits A+,B-,C+ (i.e. cherry under box A, lemon under box B, cherry under box C) and showed that in that case if they randomly picked different boxes there was a 1/3 chance they'd get the same fruit, but then I generalized to other combinations:
In this case, you can see that in 1/3 of trials where they pick different boxes, they should get the same results. 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.
 
  • #18
edguy99
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I have read a decent amount about bells test experiments and some of that recent thread about it. What I don't get is when they talk about rotating the detector an measuring the electrons spin. And lets use pi mesons that decay into an electron and positron. So if I have both of my detectors at 0 degrees I will measure one spin up and the other spin down. But if I rotate one of my detectors 45 degrees I don't see how these spin measurements can correlate because my axis has changed. For example if i measure spin up on the zero degree detector will I measure spin up or down on the 45 degree detector?
Any help will be much appreciated.
I wonder if everyone agrees with these statements, (flipping one spin for the moment as in photons to keep it simple). I choose the words "not measurable" carefully, but would be open to suggestions or changes. TIA:

At 0° - 100% measured up, 0% down, 0% not measurable
At 22.5° - 85% measured up, 0% down, 15% not measurable
At 45° - 50% measured up, 0% down, 50% not measurable
At 67.5° - 15% measured up, 0% down, 85% not measurable
At 90° - 0% measured up, 0% down, 100% not measurable
At 112.5° - 0% measured up, 15% down, 85% not measurable
At 135° - 0% measured up, 50% down, 50% not measurable
...
At 180° - 0% measured up, 100% down, 0% not measurable

and specifically:

At 120° - 0% measured up, 25% down, 75% not measurable
At 240° - 0% measured up, 25% down, 75% not measurable
 
  • #19
JesseM
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I wonder if everyone agrees with these statements, (flipping one spin for the moment as in photons to keep it simple). I choose the words "not measurable" carefully, but would be open to suggestions or changes. TIA:

At 0° - 100% measured up, 0% down, 0% not measurable
At 22.5° - 85% measured up, 0% down, 15% not measurable
At 45° - 50% measured up, 0% down, 50% not measurable
At 67.5° - 15% measured up, 0% down, 85% not measurable
At 90° - 0% measured up, 0% down, 100% not measurable
At 112.5° - 0% measured up, 15% down, 85% not measurable
At 135° - 0% measured up, 50% down, 50% not measurable
...
At 180° - 0% measured up, 100% down, 0% not measurable

and specifically:

At 120° - 0% measured up, 25% down, 75% not measurable
At 240° - 0% measured up, 25% down, 75% not measurable
What do you mean by "agree"? Are you asking whether your statements would actually be predicted by QM (in which case no, QM does not predict any reason that there must necessarily be a fraction that are "not measurable" beyond just inefficiencies in the design of your detector, which could always be improved upon), or are you asking whether we agree that there could be a self-consistent model where the above is true, or something else?
 
  • #20
edguy99
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What do you mean by "agree"? Are you asking whether your statements would actually be predicted by QM (in which case no, QM does not predict any reason that there must necessarily be a fraction that are "not measurable" beyond just inefficiencies in the design of your detector, which could always be improved upon), or are you asking whether we agree that there could be a self-consistent model where the above is true, or something else?
"agree" would mean both I guess, what do you think QM predicts and do you think it is correct.

I am guessing that you "agree" (both meanings) with the first 2 columns (probability of up and probability of down) but do not agree on the last column, but I am not sure, hence the question?
 
  • #21
DrChinese
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I wonder if everyone agrees with these statements, (flipping one spin for the moment as in photons to keep it simple). I choose the words "not measurable" carefully, but would be open to suggestions or changes. TIA:

At 0° - 100% measured up, 0% down, 0% not measurable
At 22.5° - 85% measured up, 0% down, 15% not measurable
At 45° - 50% measured up, 0% down, 50% not measurable
At 67.5° - 15% measured up, 0% down, 85% not measurable
At 90° - 0% measured up, 0% down, 100% not measurable
At 112.5° - 0% measured up, 15% down, 85% not measurable
At 135° - 0% measured up, 50% down, 50% not measurable
...
At 180° - 0% measured up, 100% down, 0% not measurable

and specifically:

At 120° - 0% measured up, 25% down, 75% not measurable
At 240° - 0% measured up, 25% down, 75% not measurable
They are 100% measurable in the ideal case. Just use a beam splitter, you see almost 100% at any angle.
 
  • #22
JesseM
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"agree" would mean both I guess, what do you think QM predicts and do you think it is correct.

I am guessing that you "agree" (both meanings) with the first 2 columns (probability of up and probability of down) but do not agree on the last column, but I am not sure, hence the question?
No, in QM with an ideally efficient detector, probability of up and probability of down should always add up to 1. Also, can we assume you are talking about photons that have already passed through a previous polarizer at angle 0°, or a photon that's a member of an entangled pair where the other member was found to pass through a polarizer at angle 0°? If not I don't understand why you would say there's a 100% chance of passing through at 0° (and this is assuming that by "up" you mean it passes through a polarizer at that angle, while by "down" you mean it's reflected...normally "up" and "down" are used when referring to spin measurements on particles like electrons)
 
  • #23
edguy99
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According to QM, if you always get opposite spins when you pick the same angle, then the probability you'll measure opposite spins when you pick different angles A and B is cos^2(A-B). So If one detector is at 0 and the other is at 45, the probability of getting opposite results (spin-up at 0 and spin-down at 45, or spin-down at 0 and spin-up at 45) is cos^2(45-0) = 0.5.
The OP asked the probability of an up and a probability of a down. From this post, I assume you think there is a 50% chance he will measure opposite results (Ie. a "down" as he has worded it). I assumed from the calculation that you have used, that the probability of the same results for this seup is 0% (Ie. a "up" as he has worded the experiment). Just wondering if I am reading you correctly?
 
  • #24
edguy99
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They are 100% measurable in the ideal case. Just use a beam splitter, you see almost 100% at any angle.
The OP asked how many up and how many down. Do you agree with Jesse (as I do) that you would measure 50% down (as he has worded the experiment)? What do you feel the percentage of "ups" would be measured as he has worded the experiment?
 
  • #25
JesseM
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The OP asked the probability of an up and a probability of a down. From this post, I assume you think there is a 50% chance he will measure opposite results (Ie. a "down" as he has worded it). I assumed from the calculation that you have used, that the probability of the same results for this seup is 0%
No, down does not refer to "opposite results", it's a specific result at one detector. If you have two detectors A and B, then "opposite results" would mean either A gets up and B gets down, or A gets down and B gets up (likewise "same results" would mean either they both get up or they both get down). And the OP was talking about spin of an electron and a positron (as measured by something like a Stern Gerlach magnet), not about photons passing through polarizers.

Anyway, the idea here would be that if B measured the positron to be spin-down at b degrees, then if A uses angle a, the probability that A will measure the electron to be spin-up at a degrees would be cos^2[(a-b)/2]. And that means that in the same scenario, the probability that A will get the result spin-down would be 1 - cos^2[(a-b)/2] = sin^2[(a-b)/2].

If you prefer to think about experiments with photon polarization, in this case I believe you won't have that factor of 1/2 inside the cos^2 (though I'm not sure if this is true for all types of entangled photon polarization experiments or just some). Then if B measured the photon to pass through a polarizer at b degrees, and A uses angle a for his polarizer, the probability the photon will pass through A's polarizer is cos^2[a-b]. This means that the probability the photon is reflected by A's polarizer is 1 - cos^2[a-b] = sin^2[a-b].
 
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