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

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Detector rotation in Bell's test experiments affects spin correlation by changing the measurement basis, which influences the probability of obtaining opposite spin results. When detectors are set at 0 degrees, one measures spin up and the other spin down, but rotating one detector to 45 degrees introduces a probability of 50% for measuring opposite spins. This correlation is derived from quantum mechanics, where the probability of measuring opposite spins at different angles is given by cos²(A-B). The discussion also touches on the implications of realism, suggesting that if particles had predetermined spins, the observed correlations would differ from experimental results. Understanding these correlations is crucial for interpreting the outcomes of Bell's tests and the nature of quantum entanglement.
  • #61
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
 
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  • #62
JesseM said:
... 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.
 
  • #63
edguy99 said:
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?
 
  • #64
JesseM said:
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%.
 
  • #65
edguy99 said:
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
 
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  • #66
JesseM said:
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?
 
  • #67
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)
 
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  • #68
SeventhSigma said:
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.
 
  • #69
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?
 
  • #70
SeventhSigma said:
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.
 
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  • #71
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
 
  • #72
SeventhSigma said:
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.
 
  • #73
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.
 
  • #74
SeventhSigma said:
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).
 
  • #75
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.
 
  • #76
SeventhSigma said:
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.
 
  • #77
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.
 
  • #78
SeventhSigma said:
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.

:smile:
 
  • #79
DrChinese said:
Yup.

:smile:

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
 
  • #80
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
 
  • #81
SeventhSigma said:
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.
 
  • #82
edguy99 said:
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.
 
  • #83
edguy99 said:
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
 
  • #84
edguy99 said:
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.
 
  • #85
DrChinese said:
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.
 
  • #86
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.
 
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  • #87
edguy99 said:
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.
 
  • #88
Entangled state, by definition, is not pure. It is a superposition of pure states.
 
  • #89
DrChinese said:
Entangled state, by definition, is not pure. It is a superposition of pure states.

What does "a superposition of pure states" mean?
 
  • #90
SeventhSigma said:
What does "a superposition of pure states" mean?

It can get confusing because of terminology. :smile:

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.
 

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