Hidden Variables and Quantum Mechanics

[LikesIntuition]

You first thought will be to say that we don't have to give up locality to produce that effect; for all we know, the partner particle might have been created in the right state in the beginning. But that's not possible, because the states down-0, down-120, and down-240 are different states that can be experimentally distinguished - the partner particle cannot have been created in all of them at once, and if it were created in one of them then it would only be right for one of the three possible measurements of the first particle. Under those circumstances, where the second particle does not change its state in response to the measurement of the first, Bell's inequality cannot be violated.

That was absolutely my first thought. Do we have to give up realism and locality, or just one or the other? Aren't we stipulating that it's impossible for one of the particles not to "match" the other in both QM and using locality? It's never possible to have one particle be up-120 and the other be down-0, is it? And if that's the case, I still can't see why it matters when the particles decide what to do. But obviously, there's something about the non-locality that I am blatantly overlooking. Can anyone see what that is? Maybe I'm not fully understanding what non-locality is or something along those lines...?

Thanks again for the help. I'm learning a great deal, even if I'm still confused about this one thing.

 

[Nugatory]

It's never possible to have one particle be up-120 and the other be down-0, is it?

Of course it is possible. If I align the left-hand detector on the 0-degree axis and the right-hand detector on the 120-degree axis, there are exactly four possible outcomes:
1) left-hand particle measures up-0; right-hand particle measures up-120
2) left-hand particle measures up-0; right-hand particle measures down-120
3) left-hand particle measures down-0; right-hand particle measures up-120
4) left-hand particle measures down-0; right-hand particle measures down-120

I still can't see why it matters when the particles decide what to do. But obviously, there's something about the non-locality that I am blatantly overlooking.

Consider the ratio of case #1 above to case #2 above when we observe a large number of pairs. The quantum-mechanical prediction is that this ratio will be what we get when we pass a down-0 particle through a 120-degree detector, and this makes sense because our up-zero measurement of the left-hand particle tells us that the right-hand particle should be a down-zero.

We also have the four cases:
5) left-hand particle measures up-240; right-hand particle measures up-120
6) left-hand particle measures up-240; right-hand particle measures down-120
7) left-hand particle measures down-240; right-hand particle measures up-120
8) left-hand particle measures down-240; right-hand particle measures down-120
from when we have the left-hand detector set to the 240-degree axis. Again, quantum mechanics predicts and experiment confirms that the ratio of of #5 to #6 is what we get when we pass a down-240 particle through a 120-degree detector, and again this makes sense because our measurement of the left-hand particle tells us that the right should be a down-240.

But (and this is the key!) the two ratios #1 to #2 and #5 to #6 are different. The only way of explaining this result is if the right-hand particle, as it approaches its 120-degree detector, behaves differently when the left-hand detector is set to 0 degrees and when it is set to 240 degrees. Furthermore, we can wait until both particles are in flight before we decide whether to set the left-hand detector on the 0-degree or the 240-degree axis, so the behavior of the right-hand particle cannot have been determined when the pair was created - the right-hand particle has to change its behavior in flight as a result of the setting of the left-hand detector.

 

[Nugatory]

Do we have to give up realism and locality?

You have to give up at least one of the two.

 

[LikesIntuition]

But (and this is the key!) the two ratios #1 to #2 and #5 to #6 are different. The only way of explaining this result is if the right-hand particle, as it approaches its 120-degree detector, behaves differently when the left-hand detector is set to 0 degrees and when it is set to 240 degrees. Furthermore, we can wait until both particles are in flight before we decide whether to set the left-hand detector on the 0-degree or the 240-degree axis, so the behavior of the right-hand particle cannot have been determined when the pair was created - the right-hand particle has to change its behavior in flight as a result of the setting of the left-hand detector.

Alright. That is what I was missing. Thank you, that was extremely helpful!

 

[stevendaryl]

Of course it is possible. If I align the left-hand detector on the 0-degree axis and the right-hand detector on the 120-degree axis, there are exactly four possible outcomes:
1) left-hand particle measures up-0; right-hand particle measures up-120
2) left-hand particle measures up-0; right-hand particle measures down-120

...

5) left-hand particle measures up-240; right-hand particle measures up-120
6) left-hand particle measures up-240; right-hand particle measures down-120

...

But (and this is the key!) the two ratios #1 to #2 and #5 to #6 are different.

Hmm. What are the numbers for those cases? I thought that

Probability of both measurements resulting in spin-up = 1/2 sin²(Θ/2) = 1/2 sin²(60) = 3/8 (where Θ = angle between the two detector orientations).

 

[morrobay]

Of course it is possible. If I align the left-hand detector on the 0-degree axis and the right-hand detector on the 120-degree axis, there are exactly four possible outcomes:
1) left-hand particle measures up-0; right-hand particle measures up-120
2) left-hand particle measures up-0; right-hand particle measures down-120


We also have the four cases:
5) left-hand particle measures up-240; right-hand particle measures up-120
6) left-hand particle measures up-240; right-hand particle measures down-120

But (and this is the key!) the two ratios #1 to #2 and #5 to #6 are different.

I also see the ratios as being equal:
x...y...z......x...y...z
0^o120^o240^o......0^o120^o240^o
+...+...+......-...-...- (1)
+...+...-......-...-...+ (2)
+...-...-......-...+...+ (3)
-...-...-......+...+...+ (4)
-...-...+......+...+...- (5)
-...+...+......+...-...- (6)
-...+...-......+...-...+ (7)
+...-...+......-...+...- (8)

x+y+ lines 3,8 = #1
x+y- lines 1,2 = #2

z+y+ lines 5,8 = #5
z+y- lines 1,6 = #6

 

[morrobay]

I also see the ratios as being equal:
x...y...z......x...y...z
0^o120^o240^o......0^o120^o240^o
+...+...+......-...-...- (1)
+...+...-......-...-...+ (2)
+...-...-......-...+...+ (3)
-...-...-......+...+...+ (4)
-...-...+......+...+...- (5)
-...+...+......+...-...- (6)
-...+...-......+...-...+ (7)
+...-...+......-...+...- (8)

x+y+ lines 3,8 = #1
x+y- lines 1,2 = #2

z+y+ lines 5,8 = #5
z+y- lines 1,6 = #6

Is this table from measurements when both detectors are aligned for 0⁰,120⁰and 240⁰ producing perfect anti correlations , a valid example/reply to Nugatory's statement on ratios 2:3 and 5:6 ? in post # 32

 

[morrobay]

Yes.
If you give up locality, than you can have a theory in which if one particle of the entangled pair is measured be up on a 120-degree axis its partner will become down-120; if the first particle is measured to be up-240 its partner will become down-240; and if the first is measured to be up-0 its partner will become down-0. In such a theory, Bell's inequality will be violated. This theory is also necessarily non-local; there's no way for the partner particle to "know" what state it should switch into without some non-local communication from the site of the first measurement to the site of the second. (And, at the risk of repeating myself, this is pretty much what the traditional collase interpretation says happens - we measure one particle and an instantaneous non-local influence collapses the wave function of the other).

You first thought will be to say that we don't have to give up locality to produce that effect; for all we know, the partner particle might have been created in the right state in the beginning. But that's not possible, because the states down-0, down-120, and down-240 are different states that can be experimentally distinguished - the partner particle cannot have been created in all of them at once, and if it were created in one of them then it would only be right for one of the three possible measurements of the first particle. Under those circumstances, where the second particle does not change its state in response to the measurement of the first, Bell's inequality cannot be violated.

You can convince yourself of this by constructing a sample data set; there are some examples in the link I pointed you at earlier.

Re: Second paragraph. Here is a possible example where the second particle does change its state in response to the measurement of the first, resulting in Bell inequality violation:

[x+y+] + [z+y-] ≥ [x+y-] This is expected from table values when both detectors are aligned and holds if there are no changes during measurements
Suppose [x+y+] coverts during measurement to [x+y-] The inequality is violated.

Possible change during measurement
x...y...z......x...y...z
0⁰..120⁰..240⁰.....0⁰120⁰240⁰
+...-...-......-...+...+ = [x+y+] Converts during measurement non locally to:
+...+...-......-...-...+ = [x+y-]