- #1

cragar

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Any help will be much appreciated.

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- Thread starter cragar
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- #1

cragar

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Any help will be much appreciated.

- #2

DrChinese

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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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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.

Any help will be much appreciated.

- #4

cragar

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- #5

cragar

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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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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.

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?

- #7

cragar

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- #8

JesseM

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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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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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Here I think you are talking about the specific loophole I mentioned earlier:If modeled as a bloch sphere, you will have a 50% chance of measuring down and a 50% chance of not getting a measurement.

The detector efficiency loophole has been closed in some experiments with entangled ions incidentally, see here (pdf file) and here.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

- #12

cragar

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- #13

DrChinese

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The statistics are the same whether Alice influenced Bob or vice versa! Surprisingly, time ordering is not significant.

- #14

cragar

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- #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

cragar

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- #17

JesseM

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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 answerif 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 assumea 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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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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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?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

- #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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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)"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?

- #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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No, down does not refer to "opposite results", it's a specific result atThe 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%

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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- #26

edguy99

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No, down does not refer to "opposite results", it's a specific result atonedetector. 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].

Sorry, you are losing me. You calculated 50% for "down", what do you feel the percentage of "ups" would be measured as he has worded the experiment? 0%, 10%, 20%, 50%?

- #27

JesseM

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My calculation there was slightly inaccurate because I didn't include the (1/2) factor in the cos^2--when measuring an electron spin it would actually be at 90 degrees (not 45 as I said originally) that you'd have a 50% chance of "down" when the other particle was measured "up" at 0 degrees. Anyway, as I said before the probabilities always add up to 1, so naturally in this example there would be a 50% chance of "up".Sorry, you are losing me. You calculated 50% for "down", what do you feel the percentage of "ups" would be measured as he has worded the experiment? 0%, 10%, 20%, 50%?

- #28

edguy99

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My calculation there was slightly inaccurate because I didn't include the (1/2) factor in the cos^2--when measuring an electron spin it would actually be at 90 degrees (not 45 as I said originally) that you'd have a 50% chance of "down" when the other particle was measured "up" at 0 degrees. Anyway, as I said before the probabilities always add up to 1, so naturally in this example there would be a 50% chance of "up".

So you don't think its 50% at 45 degrees?

- #29

JesseM

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For photon polarization I think it would be, but not for electron spin. That's what I was saying in the last paragraph of post #25.So you don't think its 50% at 45 degrees?

- #30

edguy99

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For photon polarization I think it would be, but not for electron spin. That's what I was saying in the last paragraph of post #25.

What do you think it is for the OP?

- #31

JesseM

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The OP said it was about spin of an electron/positron pair, so as I said in post #25, 50% at 90 degrees, not at 45 degrees (at 45 it should be 88.85355339% that one electron is spin-up if the positron was spin-down at 0 degrees)What do you think it is for the OP?

- #32

edguy99

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The OP said it was about spin of an electron/positron pair, so as I said in post #25, 50% at 90 degrees, not at 45 degrees (at 45 it should be 88.85355339% that one electron is spin-up if the positron was spin-down at 0 degrees)

No problem, the difference between particles (photons/electrons) and ways of measuring (beam splitters/filters) can certainly cause confusion. Just to double check the 88.85%, as I get 85.35% using the cos^2 with the 1/2? What did you use to get 88.85% or did I do it different then you?

- #33

JesseM

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You're right, I messed up when copying and pasting, I got that the answer was 0.85355339 but then somehow I thought I had just copied the part after the first two digits rather than the whole thing, so I thought I needed to type the first two digits and I misremembered them as "88". Anyway, it should be 85.355339%.No problem, the difference between particles (photons/electrons) and ways of measuring (beam splitters/filters) can certainly cause confusion. Just to double check the 88.85%, as I get 85.35% using the cos^2 with the 1/2? What did you use to get 88.85% or did I do it different then you?

- #34

DrChinese

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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?

In all cases, assuming either a mixed or entangled state, you see 50% up and 50% down. The only points I am making really are:

a) that Alice and Bob see only random patterns until they correlate their results (done at light speed or less, of course).

b) other than signal losses due to a less than ideal setup, there is nothing tangible about some particles being somehow detectable and others not. They are all detected. If they cannot be matched later (into a pair, one Alice and one Bob), that is simply an indication they were not entangled.

- #35

JesseM

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I think edguy99 is talking about the conditional probability that we will see "up" at one detector at angle a,In all cases, assuming either a mixed or entangled state, you see 50% up and 50% down.

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