Exploring Spin Correlation in Bells Test Experiments

[edguy99]

I think edguy99 is talking about the conditional probability that we will see "up" at one detector at angle a, given that we have already observed "down" with the entangled particle at another detector at angle b. In this case I think you would agree the probability of "up" is cos^2[(a-b)/2]

Yes, and I hope to extend it to other angles. Would you agree with these figures:

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

 

[DrChinese]

I think edguy99 is talking about the conditional probability that we will see "up" at one detector at angle a, given that we have already observed "down" with the entangled particle at another detector at angle b. In this case I think you would agree the probability of "up" is cos^2[(a-b)/2]

I thought he was switching back to the OP's scenario without correlation. But of course if there is matching to be done, we will have one of the cos formulae depending on whether we have spin 1 or spin 1/2 particles. The examples keep changing from post to post, so I have a hard time keeping it straight what the poster is aiming at. (Which is why I always recommend a pair of Type I polarization entangled photons as a clean example. And they feature identical spins rather than opposing.)

 

[JesseM]

Yes, and I hope to extend it to other angles. Would you agree with these figures:

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

Yes, I agree with all of these for an electron.

 

[JesseM]

I thought he was switching back to the OP's scenario without correlation. But of course if there is matching to be done, we will have one of the cos formulae depending on whether we have spin 1 or spin 1/2 particles. The examples keep changing from post to post, so I have a hard time keeping it straight what the poster is aiming at. (Which is why I always recommend a pair of Type I polarization entangled photons as a clean example. And they feature identical spins rather than opposing.)

I agree it's easier to deal with photons, but edguy99 said he specifically wanted to deal with the OP scenario involving an entangled electron/positron pair...and the OP was dealing with correlations as I understood it ("if i measure spin up on the zero degree detector will I measure spin up or down on the 45 degree detector"), in any case that seems to be what edguy99 wants to deal with.

 

[edguy99]

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.

One idea of "detectable" or "measureable" comes from thinking about what happens to the particles that entered the third SG machine as shown http://www.animatedphysics.com/sterngerlachmachine.gif" . Particles that have been measured oneway, are then measured at 90 degrees, then measured again in the original direction. These particles (apparently spin 1/2) do not measure as either an up or a down. I have seen it referred to as loss of kinetic energy of the particle.

 

[DrChinese]

One idea of "detectable" or "measureable" comes from thinking about what happens to the particles that entered the third SG machine as shown http://www.animatedphysics.com/sterngerlachmachine.gif" . Particles that have been measured oneway, are then measured at 90 degrees, then measured again in the original direction. These particles (apparently spin 1/2) do not measure as either an up or a down. I have seen it referred to as loss of kinetic energy of the particle.

As far as I know, all electrons will always be observed to have z spin of +1/2 or -1/2. So I don't think this is a good example. But I am ready to be corrected on this point.

 

[JesseM]

One idea of "detectable" or "measureable" comes from thinking about what happens to the particles that entered the third SG machine as shown http://www.animatedphysics.com/sterngerlachmachine.gif" . Particles that have been measured oneway, are then measured at 90 degrees, then measured again in the original direction. These particles (apparently spin 1/2) do not measure as either an up or a down. I have seen it referred to as loss of kinetic energy of the particle.

I don't think the numbers in that diagram represent percentages, rather they represent actual numbers of electrons detected in each channel of a simulated experiment. The diagram says it's from http://www.kcvs.ca/martin/phys/phys243/labs/sglab/stern_gerlach.html, try playing with the applet and you see it gives you numbers, not percentages. It may be that a certain number of electrons are lost, but this should just be a matter of inefficiencies in the design of the experiment (the applet claims its numbers are based on data from real experimenters), losing particles is not an inevitable consequence of the basic laws of QM.

edit: I see in the "control" menu there is an option to "do 100" (100 particles emitted), and in this case a setup with multiple simulated magnets may collect less than 100 in total, so it does appear that the simulation is including some detector inefficiencies, though when I "do 100" with setups that duplicate the ones from the diagram you link to, I don't get zero for both channels for the third and fourth setup like they show. Anyway if you click "simple configurations" on the right hand side of the page, they say "Moore suggests that in an arrangement like the one shown in question 3 you will get 50% of the electrons through each channel. Is that exactly what you got? Comment on why you may not have gotten exactly 50% through each channel. Does it get closer to 50% as you analyze more electrons? (Hint: this is a pretty good simulation! The numbers you see at the detectors represent real measurements.)"

 

[DrChinese]

I don't think the numbers in that diagram represent percentages, rather they represent actual numbers of electrons detected in each channel of a simulated experiment. The diagram says it's from http://www.kcvs.ca/martin/phys/phys243/labs/sglab/stern_gerlach.html, try playing with the applet and you see it gives you numbers, not percentages. It may be that a certain number of electrons are lost, but this should just be a matter of inefficiencies in the design of the experiment (the applet claims its numbers are based on data from real experimenters), losing particles is not an inevitable consequence of the basic laws of QM.

Which is exactly why discussion of these scenarios end up getting bogged down by issues which are peripheral to the main points. Now edguy99 will want to analyze this setup, which makes no sense as an item to discuss. For normal quantum particles, there is no "loss of kinetic momentum" effect to discuss.

 

[DrChinese]

Just to be clear: measuring spin does NOT intrinsically cause a particle to be absorbed, disappear or otherwise lose energy. A particle ALWAYS possesses spin in any direction of its spin axes. There is no such thing as a spin 1/2 particle without spin in the x, y or z directions.

 

[edguy99]

Yes, I agree with all of these for an electron.

I have applied these probabilities to the picture below:

and a "mini bell test" with Case A=0 degrees B=120 degrees C=240

[1] A+ B+ C+ 1.00*0.25*0.25 = .0625
[2] A+ B+ C- 1.00*0.25*0.75 = .1875
[3] A+ B- C+ 1.00*0.75*0.25 = .1875
[4] A+ B- C- 1.00*0.75*0.75 = .5625

Assume we know the measurement at A will be "up" since we know its entangled particle is "down". This chart would give the probability of measuring the configuration specified, certainly the most common is that you measure "down" at B and C.

 

[edguy99]

Just to be clear: measuring spin does NOT intrinsically cause a particle to be absorbed, disappear or otherwise lose energy. A particle ALWAYS possesses spin in any direction of its spin axes. There is no such thing as a spin 1/2 particle without spin in the x, y or z directions.

I agree. The only thing I am adding is that the spinning axis may be precessing.

 

[JesseM]

I have applied these probabilities to the picture below:

and a "mini bell test" with Case A=0 degrees B=120 degrees C=240

[1] A+ B+ C+ 1.00*0.25*0.25 = .0625
[2] A+ B+ C- 1.00*0.25*0.75 = .1875
[3] A+ B- C+ 1.00*0.75*0.25 = .1875
[4] A+ B- C- 1.00*0.75*0.75 = .5625

I don't get it, why are you multiplying the three probabilities? What is the final probability at the end of each line supposed to be? Are you imagining we look at three successive measurements of different particles at angles A,B,C (and only on trials where the other particle was measured "down"), so for example [1] would calculate the probability that our first measurement at angle A would get "up" and the second measurement at angle B would get "up" and the third measurement at angle C would get "up"? If not, can you explain the physical meaning of the probability at the end of each line?

 

[DrChinese]

I have applied these probabilities to the picture below:

and a "mini bell test" with Case A=0 degrees B=120 degrees C=240

[1] A+ B+ C+ 1.00*0.25*0.25 = .0625
[2] A+ B+ C- 1.00*0.25*0.75 = .1875
[3] A+ B- C+ 1.00*0.75*0.25 = .1875
[4] A+ B- C- 1.00*0.75*0.75 = .5625

I'm always going to like these angle settings to discuss.

JesseM, I think this is a fair classical approach and it should lead to a conclusion that can be shown to be false, by using the ideas of Bell.

 

[JesseM]

JesseM, I think this is a fair classical approach and it should lead to a conclusion that can be shown to be false, by using the ideas of Bell.

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.

 

[DrChinese]

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.

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%.

 

[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.