Entanglement and Bell’s inequality Question

[DrChinese]

I was really hoping I could find out how to calculate different QM outcomes for testing the spin of entangled particles when using different angles other than 0, 120 and 240. Say, 10, 60 and 200. I did make another post but so far no luck. So was hoping someone could point me in the right direction?

The normal way you would do that would be to weight each of the options as 1/3 and sum to get an average. So:

a) Match(10,60)= .4132
b) Match(10,200)= .9698
c) Match(60,200)= .5868

However, there is no guarantee that you will get a scenario that leads to a Bell-like inequality in the exact same manner as before. The other one "happens" to work out nicely for the job. Most combinations of angles that lead to nonsensical results but it is sometimes quite complicated to see the problem. For example, there is a specific combo for the triple 10, 60 and 200 degree that is impossible to obtain classically. I know because I build a generator that graphs the permutations.

The quantum prediction for case (reference above a/b/c):

a) + ~b) - c is -.1434 (you can easily obtain that from above) whereas the classical predication is >=0; note that ~b in this equation is Mismatch(10,200) or .0302.

This probably makes no sense as I have described it, but the formula above works out to be a couple of permutations. Ergo, the classical sum must always be zero or above (for any set of outcome permutations) where experiment yields a different value. And yes, it's negative.

 

[rede96]

The normal way you would do that would be to weight each of the options as 1/3 and sum to get an average. So:

a) Match(10,60)= .4132
b) Match(10,200)= .9698
c) Match(60,200)= .5868

However, there is no guarantee that you will get a scenario that leads to a Bell-like inequality in the exact same manner as before.

Thank you for that. But at the moment I wasn't looking for a Bell-like inequality I was just trying to find the right way to calculate the quantum result for the experiments if it was done in the same way as 0, 120 and 240 but using the angles above. (or any set of angles) Which if I understood you would be 0.6566?

So if I did 100,000 tests where Alice and Bob selected at random from detectors at 10, 60 and 200 the ratio of valid matches would be 0.6566. Is that correct?

The other one "happens" to work out nicely for the job. Most combinations of angles that lead to nonsensical results but it is sometimes quite complicated to see the problem. For example, there is a specific combo for the triple 10, 60 and 200 degree that is impossible to obtain classically. I know because I build a generator that graphs the permutations.

The quantum prediction for case (reference above a/b/c):

a) + ~b) - c is -.1434 (you can easily obtain that from above) whereas the classical predication is >=0

Ah ok, so that is just one of the 8 possible permutations. e.g. (A+, B-, C-)? But if I look at the original example of 0,120,240, wouldn't I still get a negative result for this permutation? E.g. 0.25 + (-0.25) + (-0.25) = -0.25?

note that ~b in this equation is Mismatch(10,200) or .0302.

I didn't quite understand that. Could you explain why that is a mismatch and where 0.0302 comes from please? That seems to be around 80 degrees working backwards.

This probably makes no sense as I have described it, but the formula above wo255rks out to be a couple of permutations. Ergo, the classical sum must always be zero or above (for any set of outcome permutations) where experiment yields a different value. And yes, it's negative.

Ok, that really confused me! So are you saying that it is possible that the total results from an experiment (ie multiple testing of the three angles) will give a negative ratio of matches? I think I have misunderstood that as you can only get one of two results from each test, either Alice and Bob's detectors match or the don't. So impossible to get a negative number of matches as the ratio is total matches / total number of valid tests.

As I said at the moment, I just wanted to know if there was a way I could calculate what the experimental result would be using different angles. Which if I understood was just the average of the 3 probability options you mentioned.

 

[DrChinese]

1. Thank you for that. But at the moment I wasn't looking for a Bell-like inequality I was just trying to find the right way to calculate the quantum result for the experiments if it was done in the same way as 0, 120 and 240 but using the angles above. (or any set of angles) Which if I understood you would be 0.6566? ... So if I did 100,000 tests where Alice and Bob selected at random from detectors at 10, 60 and 200 the ratio of valid matches would be 0.6566. Is that correct?

2. Ah ok, so that is just one of the 8 possible permutations. e.g. (A+, B-, C-)? But if I look at the original example of 0,120,240, wouldn't I still get a negative result for this permutation? E.g. 0.25 + (-0.25) + (-0.25) = -0.25?

3. Could you explain why that is a mismatch and where 0.0302 comes from please? That seems to be around 80 degrees working backwards.

1. Assuming you selected 2 of the 3 angles to observe randomly, yes.2. 8 permutations, yes that is very good. +++, ++-, ... , ---. It turns out that 2 of the 8 cases can be represented in the manner I provided. Those cases are: ++- and --+. You can best see the logic chain in detail at a page of mine. That page basically shows you how to arrange the terms so this result is obtained.

http://www.drchinese.com/David/Bell_Theorem_Negative_Probabilities.htm

Basically, you go back to this:

a) Match(10,60)= .4132
b) MisMatch(10,200)= .0302
c) Match(60,200)= .5868

If you think of the 8 permutations, then each term above represents 4 of them. The terms can be rearranged* in such a way ( a + b - c) = 2 * P(++- , --+). And that equals -.1434 (obtained from: .4132 + .0302 - .5868) as I said previously. Obviously a classical prediction for cases 2 * P(++- , --+) is greater than or equal to 0. In actual experiments, you never see a negative result directly. You see experimental support for the individual a, b and c terms above. But assuming you also believe that Alice's choice of angle does not affect Bob's outcome, and vice versa, then the classical prediction and the experimental prediction (which is the same as: a + b - c or -.1434) are inconsistent.

So the conclusion from all of this manipulation is: any 3 angles you pick (with a few exceptions) will ultimately lead to a Bell Inequality. But the 0/120/240 case is easier to visualize.

3. Mismatch % is of course nothing but (1 - Match %). So 1 - .9698 = .0302* The rearrangement can be seen if you study the link. Basically: a b and c each represent 4 of the 8 permutations, but each a different 4. When you combine them via a + b - c you get cancellation of some terms, leaving you with 2 sets of terms for ++- , --+. This is equal to 2 * P(++- , --+). QED.

 

[DrChinese]

Yes, the above is a bit convoluted. But this is simply a way to prepare a Bell Inequality. The reason there are examples like hats/scarves/gloves is that it is easier to follow than the detail above. But it is really not that hard once you jump all the way in.

 

[rede96]

Yes, the above is a bit convoluted. But this is simply a way to prepare a Bell Inequality. The reason there are examples like hats/scarves/gloves is that it is easier to follow than the detail above. But it is really not that hard once you jump all the way in.

Thanks for your help with this. I think I understand the EPR argument a lot better now and can see how Bell's inequality and experimental results lead to the conclusion that there is something very non-classical going on. But my mind still struggles accepting that this is spooky action at a distance, even though it is obvious one is forced to accept that conclusion.

I was also curious to know is entanglement something that persists? So once the wave function has collapsed then I assume that anything further I do to particle A, such as run it through a magnetic field and change the spin direction for example, would have no effect on partial B? Or are they forever entangled?

 

[DrChinese]

Thanks for your help with this. I think I understand the EPR argument a lot better now and can see how Bell's inequality and experimental results lead to the conclusion that there is something very non-classical going on. But my mind still struggles accepting that this is spooky action at a distance, even though it is obvious one is forced to accept that conclusion.

I was also curious to know is entanglement something that persists? So once the wave function has collapsed then I assume that anything further I do to particle A, such as run it through a magnetic field and change the spin direction for example, would have no effect on partial B? Or are they forever entangled?

The general rule is that once you measure either particle, entanglement ceases.

However, there are multiple degrees of entanglement. It is also possible to collapse one degree and leave others intact. For example, you collapse the spin degree and leave position/momentum entanglement intact for the particle pair.