Hidden Variables and Quantum Mechanics

[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-]