How Does Detector Rotation Affect Spin Correlation in Bell's Test Experiments?

[JesseM]

Well, his table shows that the prediction for a "match" (+ and -, or - and +) for B and C is the sum of cases [2] and [3]. That would be 37.5% which is experimentally testable. Of course the QM prediction is 25%.

Well, I was talking about a situation where the positron was measured at B and the electron at A rather than one where the two angles were B and C, but I agree, this would be another way of showing his model differs from the predictions of QM.

 

[SeventhSigma]

DrChinese: What exactly would the QM math be behind the 25% outcome?

 

[DrChinese]

DrChinese: What exactly would the QM math be behind the 25% outcome?

The general formula for spin 1/2 correlation is:

cos^2[(theta)/2]

where theta is the angle setting difference between the 2 entangled particles. (Note that theta is the only variable in the equation. And theta is a quantum non-local variable at that, because its value is determined at 2 different spacetime coordinates.)

In edguy99's example, because of his (nice) choice of 0/120/240 for a/b/c, we end up with the following for BC:

cos^2[(240-120)/2] = 0.25

 

[DrChinese]

Well, I was talking about a situation where the positron was measured at B and the electron at A rather than one where the two angles were B and C, but I agree, this would be another way of showing his model differs from the predictions of QM.

Yup, so he ends up "classically" with the following predictions:

AB=25%
AC=25%
BC=37.5%

Since this obviously won't match the results, the next step may be to explain (inappropriately) why BC "doesn't count". But who knows?

 

[SeventhSigma]

So that function describes the probability of having *different* spins? (e.g. probability of two electrons having opposite spins is 1 if I measure the same angle, making theta 0)?

What exactly would an "electron spin" designate? Is spin a property that is associated with a given axis (x, y, and z)? What does it mean for a particle to have "spin up" or "spin down" and how does this relate to polarizers?

 

[DrChinese]

So that function describes the probability of having *different* spins? (e.g. probability of two electrons having opposite spins is 1 if I measure the same angle, making theta 0)?

What exactly would an "electron spin" designate? Is spin a property that is associated with a given axis (x, y, and z)? What does it mean for a particle to have "spin up" or "spin down" and how does this relate to polarizers?

Polarizers are used for examining the "spin" (polarization) of photons, i.e. light.

Electrons have non-commuting spin components in 3 mutually orthogonal (perpendicular) directions: x, y and z. The labeling of these is arbitrary. Ditto with whether you call it up or down, or +1/2 or -1/2, or just + or -. It is just a convention. Any suitable measurement of ANY spin component of an electron always returns an up or a down result. There is no such thing as "no" spin as a result of a measurement.

Because x, y and z do not commute: If you measure and determine the electron is x+, then you have NO idea what its y and z components are. If later you measure and determine the electron is y-, then you have NO idea what its x and z components are. This is true for any electron any time. And note that although you knew it to be x+ at one time, now it could be x+ or x- (equally likely always).

 

[JesseM]

What exactly would an "electron spin" designate? Is spin a property that is associated with a given axis (x, y, and z)? What does it mean for a particle to have "spin up" or "spin down" and how does this relate to polarizers?

For an electron you would define spin-up and spin-down at a certain angle based on which direction the electron was deflected when passing through a Stern-Gerlach magnet oriented at that angle, see the explanation with diagrams here.

 

[SeventhSigma]

Why is the probability function a cosine wave? (as opposed to sin or tan or some other function, etc)

 

[edguy99]

OK, so you interpret edguy99 to be talking about hidden variables, so for example the probability of A+B+C- is the probability that this electron has hidden variables that ensure it's predetermined to give "up" if measured on A, "up" if measured on B and "down" if measured on C? While the positron must in this case have A-B-C+ to ensure that it always gives opposite results if they're both measured at the same angle? That does seem like the most likely interpretation. But then as you said this won't actually violate any Bell inequality--perhaps edguy99 would like to pick one to check? And it's easy to see that this probability distribution won't give the correct QM probabilities if the positron was measured at a different angle other than 0 degrees, say B=120 degrees. Then if the positron gives "down" at angle B we know the electron's hidden state must be one that has a B+, so it must be A+ B+ C+ or A+ B+ C-, which means if the electron is measured at angle A=0 it must give result "up" with probability 1. But according to QM, if the positron is measured at B=120 and gives "down", then if the electron is measured at A=0 the probability of the electron giving "up" should be cos^2[(120-0)/2] = 0.25.

I have re-drawn the picture to reflect that the observer of the positron starts to measure at B=120 degrees. Notice how 75% of the former "ups" are now measuring "down" and 25% still show "up"

So I would disagree with the statement:

"Then if the positron gives "down" at angle B we know the electron's hidden state must be one that has a B+"

In this simulation, the B+ is only 75% sure. The other 25% of the time, it is B-.

 

[JesseM]

I have re-drawn the picture to reflect that the observer of the positron starts to measure at B=120 degrees. Notice how 75% of the former "ups" are now measuring "down" and 25% still show "up"

I don't understand what the diagram is supposed to tell us about the hidden variables and how they are correlated for the two particles. Was the following statement an accurate description of your idea?

for example the probability of A+B+C- is the probability that this electron has hidden variables that ensure it's predetermined to give "up" if measured on A, "up" if measured on B and "down" if measured on C?

If A+B+C- does indeed refer to predetermined results for a given electron, I had assumed that anytime the source emitted an electron with predetermined results A+B+C-, then the positron that was paired with it would automatically have predetermined results A-B-C+, since that's the only way to get agreement with the quantum-mechanical prediction that they always are found to have opposite spins when measured on the same axis. If that's not the case, I don't really see what the point of this model is, since it clearly disagrees with quantum mechanics.

 

[SeventhSigma]

I think, long story short, if your results are not exactly consistent with cos^2((a-b)/2) = the probability that two entangled particles will have opposite spins when you observe them at angles a and b, your model is not consistent with what experiments show

 

[edguy99]

... clearly disagrees with quantum mechanics.

I disagree. It predicts "if the positron is measured at B=120 and gives "down", then if the electron is measured at A=0 the probability of the electron giving "up" should be cos^2[(120-0)/2] = 0.25" exactly as QM does.

 

[JesseM]

I disagree. It predicts "if the positron is measured at B=120 and gives "down", then if the electron is measured at A=0 the probability of the electron giving "up" should be cos^2[(120-0)/2] = 0.25" exactly as QM does.

But when I said 'if the positron gives "down" at angle B we know the electron's hidden state must be one that has a B+' I meant that if the electron was also measured at B in this case, it would have to give spin-up with probability 1. Do you disagree? If you don't disagree with that, then assuming you'd agree the hidden variables can't "anticipate" in advance what angle the particles will be measured at, it seems the only way to explain this is to say that any trial where the positron was measured "down" at B, the electron's hidden state must have included a B+ so it would be guaranteed to give "up" if it were measured at B. Yes?

 

[edguy99]

But when I said 'if the positron gives "down" at angle B we know the electron's hidden state must be one that has a B+' I meant that if the electron was also measured at B in this case, it would have to give spin-up with probability 1. Do you disagree? If you don't disagree with that, then assuming you'd agree the hidden variables can't "anticipate" in advance what angle the particles will be measured at, it seems the only way to explain this is to say that any trial where the positron was measured "down" at B, the electron's hidden state must have included a B+ so it would be guaranteed to give "up" if it were measured at B. Yes?

That particular electron - sure. But we are talking probabilities of many electrons and over time this would go to 75%/25%.

 

[JesseM]

That particular electron - sure. But we are talking probabilities of many electrons and over time this would go to 75%/25%.

Yes, but the point is that the source creating the particles doesn't know in advance what detector settings will be used on each trial, so there's always a chance this will be a trial where both are measured at B, so that means on every trial where the positron was measured along B and found to be "down", the electron must have the predetermined value of B+ "just in case" it's measured at angle B (so according to your table the electron must be either A+B+C+ or A+B+C- on any such trial). That's all I meant when I said 'if the positron gives "down" at angle B we know the electron's hidden state must be one that has a B+'. Do you still disagree with that statement now that I've elaborated?

Are you familiar with the idea of (and the notation for) conditional probability? If so I could state my point like this:

P(electron is either in state A+B+C+ or A+B+C- | positron was measured "down" on B) = 1

And therefore:

P(electron is measured "up" on A | positron was measured "down" on B and electron measured on A) = 1

 

[edguy99]

Yes, but the point is that the source creating the particles doesn't know in advance what detector settings will be used on each trial, so there's always a chance this will be a trial where both are measured at B, so that means on every trial where the positron was measured along B and found to be "down", the electron must have the predetermined value of B+ "just in case" it's measured at angle B (so according to your table the electron must be either A+B+C+ or A+B+C- on any such trial). That's all I meant when I said 'if the positron gives "down" at angle B we know the electron's hidden state must be one that has a B+'. Do you still disagree with that statement now that I've elaborated?

Are you familiar with the idea of (and the notation for) conditional probability? If so I could state my point like this:

P(electron is either in state A+B+C+ or A+B+C- | positron was measured "down" on B) = 1

And therefore:

P(electron is measured "up" on A | positron was measured "down" on B and electron measured on A) = 1

The positron, if measured at 0% is detected "down" 100% of the time. If the positron is measured at 120 degrees, it will measure "up" 75% of the time and "down" 25% of the time as you agreed earlier. But yet you are claiming here that it will be seen "up" 100% of the time? I am not sure I follow your reasoning?

 

[SeventhSigma]

The idea is that if observation didn't have any effect on the observable (e.g. if the states of the two cards were A+B-C+ and A-B+C-, respectively -- opposite charges on each letter), then we would of course see that when both individuals observed the same letter, they get opposite signs 100% of the site.

But then you would also expect that *if this were true*, then we should see certain probabilities emerge if we choose *different* letters (say I choose A and you choose B). If the charges of the letters are predetermined to be fully opposite prior to observation, then choosing *different* letters should yield us a probability of, at least, something like 33% -- but in practice we see 25%, so this way of setting up the problem *must* be false.

(Although, annoyingly, when I do cos((120-0)/2)^2, I get 0.907090485 and not .25)

 

[edguy99]

The idea is that if observation didn't have any effect on the observable (e.g. if the states of the two cards were A+B-C+ and A-B+C-, respectively -- opposite charges on each letter), then we would of course see that when both individuals observed the same letter, they get opposite signs 100% of the site.

But then you would also expect that *if this were true*, then we should see certain probabilities emerge if we choose *different* letters (say I choose A and you choose B). If the charges of the letters are predetermined to be fully opposite prior to observation, then choosing *different* letters should yield us a probability of, at least, something like 33% -- but in practice we see 25%, so this way of setting up the problem *must* be false.

(Although, annoyingly, when I do cos((120-0)/2)^2, I get 0.907090485 and not .25)

You say "if the states of the two cards were A+B-C+ and A-B+C-" and I agree.

The state we are dealing with has a 75% B+ and 25% B- on one end, and a 25% B+ and 75% B- on the other end.

 

[SeventhSigma]

Are you talking about those magnet systems where we have different orientations? Where particles enter one polarizer and then leave/enter the next, etc?

And so you are saying you can arrange these polarizers in a way where you have 75% B+ coming out, say, the left set of polarizers and 75% B- coming out of the right set?

 

[edguy99]

Are you talking about those magnet systems where we have different orientations? Where particles enter one polarizer and then leave/enter the next, etc?

And so you are saying you can arrange these polarizers in a way where you have 75% B+ coming out, say, the left set of polarizers and 75% B- coming out of the right set?

The experiment by the OP where he has turned the detector of the positron by 120 degrees and is taking the measurement where the angle of the electron detector matches the positron detector.

Note: the experimenter has not changed how the electrons/positrons are generated, they would still measure "up" if measured at 0 degrees.

 

[SeventhSigma]

Yeah, so if you have one detector measuring at 0 and the other measuring at 120, you'll see different charges cos^2((120-0)/2) = 25% of the time

 

[edguy99]

Yeah, so if you have one detector measuring at 0 and the other measuring at 120, you'll see different charges cos^2((120-0)/2) = 25% of the time

The sequence of post is maybe a little unclear. We start with our standard electron/positron setup and measure 100% up at one end (the electron) and 100% down at the other end (the positron).

We then turn the positron end by 120 degrees and restart the measurement.

Without touching the electron end, we still see 100% up at the electron end, but we now see 75% up at the positron end.

We then start the experiment at the electron end: The first set of measurements with the 2 detectors matched will result in 75% up at the positron end and 75% down at the electron end.

 

[SeventhSigma]

AFAIK, you can't say "We measure 100% up at one end and 100% down on the other." You will get an even split of up and down at both ends, but they'll always be opposite of one another assuming you're using the same angle. You could have 10 ups followed by 10 downs coming out the left, but then you'd also have the entangled particles coming out the other end to the tune of 10 downs followed by 10 ups.

 

[edguy99]

AFAIK, you can't say "We measure 100% up at one end and 100% down on the other." You will get an even split of up and down at both ends, but they'll always be opposite of one another assuming you're using the same angle. You could have 10 ups followed by 10 downs coming out the left, but then you'd also have the entangled particles coming out the other end to the tune of 10 downs followed by 10 ups.

The original poster has assumed that the 2 particles start out with opposite spin and that the positron is being detected down 100% of the time, hence the electron will be detected up 100% of the time (this defines the 0 measurement angle for both detectors).

 

[SeventhSigma]

I would then say that you can't make this assumption and be consistent with QM. I don't see how you can assume that you're going to get 100% up or 100% down coming out of a particular end unless you use filters (such as the lead blocks in that magnet page example), but this has nothing to say about the nature of the actual particle entanglement with respect to observations.

 

[edguy99]

I would then say that you can't make this assumption and be consistent with QM. I don't see how you can assume that you're going to get 100% up or 100% down coming out of a particular end unless you use filters (such as the lead blocks in that magnet page example), but this has nothing to say about the nature of the actual particle entanglement with respect to observations.

To quote the OP "So if I have both of my detectors at 0 degrees I will measure one spin up and the other spin down."

This is what QM predicts 100% of the time.

 

[SeventhSigma]

Yes -- if you have both detectors set to 0, you'll get opposite spins every single time. But that means you have either up/down or down/up. That doesn't mean you get all up at one detector and all down at the other. You have (afaik) a 50% chance of getting up/down and a 50% chance of getting down/up, but altogether this is 100% chance of getting one up and one down in combination.

 

[DrChinese]

AFAIK, you can't say "We measure 100% up at one end and 100% down on the other." You will get an even split of up and down at both ends, but they'll always be opposite of one another assuming you're using the same angle. You could have 10 ups followed by 10 downs coming out the left, but then you'd also have the entangled particles coming out the other end to the tune of 10 downs followed by 10 ups.

Yup.

 

[edguy99]

Yup.

So you would not agree with this for the OP setup?

At 0° - 100% measured up, 0% down
At 45° - 85% measured up, 15% down
At 90° - 50% measured up, 50% down
At 135° - 15% measured up, 85% down
At 180° - 0% measured up, 100% down
At 225° - 15% measured up, 85% down
At 270° - 50% measured up, 50% down
...
At 360° - 100% measured up, 0% down

and specifically:

At 120° - 25% measured up, 75% down
At 240° - 25% measured up, 75% down

 

[SeventhSigma]

I'd change the wording to "100% measured opposite-spin, 0% measured same-spin"

So:

At 120° - 25% measured opposite-spin, 75% measured same-spin
At 240° - 25% measured opposite-spin, 75% measured same-spin

 

[edguy99]

I'd change the wording to "100% measured opposite-spin, 0% measured same-spin"

So:

At 120° - 25% measured opposite-spin, 75% measured same-spin
At 240° - 25% measured opposite-spin, 75% measured same-spin

For context, please check the earlier posts (#36). We are talking about measurements at one end, not comparison measurments.

 

[DrChinese]

So you would not agree with this for the OP setup?

At 0° - 100% measured up, 0% down
At 45° - 85% measured up, 15% down
At 90° - 50% measured up, 50% down
At 135° - 15% measured up, 85% down
At 180° - 0% measured up, 100% down
At 225° - 15% measured up, 85% down
At 270° - 50% measured up, 50% down
...
At 360° - 100% measured up, 0% down

and specifically:

At 120° - 25% measured up, 75% down
At 240° - 25% measured up, 75% down

That the entangled stats for the subset (half) where you have A=up . Otherwise, you won't have an entangled pair.

 

[SeventhSigma]

For context, please check the earlier posts (#36). We are talking about measurements at one end, not comparison measurments.

Okay, if you're looking only at one end, then you'll see the following:

At x° (any angle) - 50% measured up, 50% down

 

[DrChinese]

For context, please check the earlier posts (#36). We are talking about measurements at one end, not comparison measurments.

Are we talking about entangled electrons? I thought that was the discussion topic, since we are on Bell tests.

 

[edguy99]

That the entangled stats for the subset (half) where you have A=up . Otherwise, you won't have an entangled pair.

The discussion assumes the particles are prepared in such as way so the positron always measures down at 0 degrees and the preparation method is never changed.

 

[SeventhSigma]

You can't control what spin a particle assumes afaik (edit: I have no idea if you actually can do this). It comes out either spin up or spin down with equal probability and you have no way of controlling this. You can control what kinds of spins are allowed to exit a detector, but this doesn't change the fact that the particle assumed that particular spin in the first place.

Even if you could change the spin by forcing it, you'd be breaking the entanglement anyway.

 

[DrChinese]

The discussion assumes the particles are prepared in such as way so the positron always measures down at 0 degrees and the preparation method is never changed.

This is not a spin entangled setup. The only way to get that is to look at the subset where your criteria is met. You cannot otherwise "force" it to be up AND entangled as to spin.

 

[DrChinese]

Entangled state, by definition, is not pure. It is a superposition of pure states.

 

[SeventhSigma]

Entangled state, by definition, is not pure. It is a superposition of pure states.

What does "a superposition of pure states" mean?

 

[DrChinese]

What does "a superposition of pure states" mean?

It can get confusing because of terminology.

Just means a combination of possible pure states: i.e. H + V. H is pure because it is known. There are 2 ways the state is called unknown: when it is in a superposition and when it is mixed. Mixed is actually an ensemble of items in pure states but is usually randomly distributed from a source.

If it is mixed or pure, it cannot be entangled on that same basis (spin for example).

A ensemble of entangled particles will be experimentally indistinguishable from mixed state UNLESS you perform matching on the entangled partner.

 

[edguy99]

You can't control what spin a particle assumes afaik (edit: I have no idea if you actually can do this). It comes out either spin up or spin down with equal probability and you have no way of controlling this. You can control what kinds of spins are allowed to exit a detector, but this doesn't change the fact that the particle assumed that particular spin in the first place.

Even if you could change the spin by forcing it, you'd be breaking the entanglement anyway.

These types of experiments are often conducted with the spin controlled and the measuring devices rotated to see the effects, especially the effects when 2 matching particles are detected. See http://arxiv.org/PS_cache/quant-ph/pdf/0205/0205171v1.pdf" as an example of preparing entangled spin up particles (photons in this case).

 

[DrChinese]

These types of experiments are often conducted with the spin controlled and the measuring devices rotated to see the effects, especially the effects when 2 matching particles are detected. See http://arxiv.org/PS_cache/quant-ph/pdf/0205/0205171v1.pdf" as an example of preparing entangled spin up particles (photons in this case).

Maybe you can quote something. Because it doesn't say that.

 

[edguy99]

Maybe you can quote something. Because it doesn't say that.

"To create the state | EPRi or something close to it, we adjust the parameters which determine the laser polarization. First we adjust l to equalize the coincidence counts N(0◦, 0◦) and N(90◦, 90◦). Next we set l by rotating the quartz plate about a vertical axis to maximize N(45◦, 45◦). When performing these optimizations, we typically collect a few hundred photons per point which requires an acquisition window of a few seconds."

 

[DrChinese]

"To create the state | EPRi or something close to it, we adjust the parameters which determine the laser polarization. First we adjust l to equalize the coincidence counts N(0◦, 0◦) and N(90◦, 90◦). Next we set l by rotating the quartz plate about a vertical axis to maximize N(45◦, 45◦). When performing these optimizations, we typically collect a few hundred photons per point which requires an acquisition window of a few seconds."

Yes, an EPR state is entangled. Notice the discussion of coincidences?

 

[edguy99]

Yes, an EPR state is entangled. Notice the discussion of coincidences?

Yes, they have "fixed" the orientation of the photon stream, and are rotating the two measuring devices to see the effect on the number of coincident (entangled) hits. An entangled hit is essentially when they see both detectors go off at "almost exactly" the same time.

 

[SeventhSigma]

That paragraph just talks about how they calibrated the equipment as far as I can tell -- it's not saying that they're somehow "fixing" the streams to produce certain types of spins that break the entanglements. They adjust the first parameter (the theta-l) so that N(0,0) and N(90,90) produce roughly the same output counts, and then they determine where the proper 45 degree mark is by going between those two thresholds (hence the "maximize N(45,45)" part).

Someone else can correct me if I'm wrong, here.

Nowhere in here is it implying that it's breaking entanglement by "fixing" the spins in any way. The conclusion of that paper is also consistent with what's been put forth in this thread (local realistic variables contradicted, Bell's Inequality violated, etc).

 

[DrChinese]

Yes, they have "fixed" the orientation of the photon stream, and are rotating the two measuring devices to see the effect on the number of coincident (entangled) hits. An entangled hit is essentially when they see both detectors go off at "almost exactly" the same time.

Yes, and we are looking at the subset where Alice's side is all up, which is 50% of the total stream from the PDC crystal.

The point is that the stream going to Bob consists of ones in which Alice is up, and where Alice is down. You must coincidence match to find the desired subset of Alice=up.

 

[DrChinese]

The sequence of post is maybe a little unclear. We start with our standard electron/positron setup and measure 100% up at one end (the electron) and 100% down at the other end (the positron).

We then turn the positron end by 120 degrees and restart the measurement.

Without touching the electron end, we still see 100% up at the electron end, but we now see 75% up at the positron end.

We then start the experiment at the electron end: The first set of measurements with the 2 detectors matched will result in 75% up at the positron end and 75% down at the electron end.

OK, I think I see what you are wanting to do. Let's just refer to this as the subset so it is clear. Then everything you are saying is fine. We will simply ignore the other subgroup for our discussion purposes, knowing that the "true" universe does not have this attribute.

 

[edguy99]

That paragraph just talks about how they calibrated the equipment as far as I can tell -- it's not saying that they're somehow "fixing" the streams to produce certain types of spins that break the entanglements. They adjust the first parameter (the theta-l) so that N(0,0) and N(90,90) produce roughly the same output counts, and then they determine where the proper 45 degree mark is by going between those two thresholds (hence the "maximize N(45,45)" part).

Someone else can correct me if I'm wrong, here.

Nowhere in here is it implying that it's breaking entanglement by "fixing" the spins in any way. The conclusion of that paper is also consistent with what's been put forth in this thread (local realistic variables contradicted, Bell's Inequality violated, etc).

If the photon is modeled as http://en.wikipedia.org/wiki/Polarization_(waves)" , he is setting the properties of the photons to keep the photon properties consistent throughout the experiment. This allows him to continue the experiment where the only change being made is the rotation of the measuring device. My use of the word "fix" means the properties of the particle stream are known and unchanging.

 

[edguy99]

OK, I think I see what you are wanting to do. Let's just refer to this as the subset so it is clear. Then everything you are saying is fine. We will simply ignore the other subgroup for our discussion purposes, knowing that the "true" universe does not have this attribute.

I am not sure I understand if you are agreeing or not (is there a 50% chance of us agreeing?). What probabilities would you assign to the OP question?