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Ecl1psed t1_izatrna wrote

I am having trouble wrapping my head around how it's possible that there are no local hidden variables in quantum mechanics. I came up with this argument that seems to me to be solid, but it would contradict the nonexistence of local hidden variables. Could someone point out where the argument goes wrong?

Suppose Alice and Bob take two entangled particles very far from each other. For simplicity, assume they are 1 ly apart, and with a relative velocity of zero. Suppose the particles become entangled at year 0.0, and Alice and Bob take 5.0 years to reach the 1 ly mark.
When I talk about "before" and "after" in this argument, I am referring to the reference frame of a point exactly in between Alice and Bob, also with a relative velocity of zero.
Suppose that the particles are entangled in the Bell state, so that A's and B's measurements will always produce the same binary result of either 1 or -1.
Without loss of generality, let's say Alice gets a result of 1. Now, Bob is guaranteed to measure 1 as well. This is true matter how long he waits. He could wait until year 50.0 and he would still get a result of 1. Consider two scenarios:
(1) Alice mesaures at year 10.0, and Bob measures at year 10.5
(2) Alice mesaures at year 10.0, and Bob measures at year 9.5
From Alice's perspective, there is absolutely nothing different between the two scenarios. In either case the no-communication theorem prevents any information from being transmitted, since they measure only 0.5 years apart. Thus, she will measure 1 in either scenario.
Now, consider a third scenario:
(3) Alice mesaures at year 9.0, and Bob measures at year 9.5
From Bob's perspective, this scenario is identical to the second scenario (due to the no-communication theorem), so he (and by extension Alice) is still guaranteed to measure a 1 in this case.
Now, consider a fourth scenario:
(4) Alice mesaures at year 9.0, and Bob measures at year 8.5
Again, from Alice's perspective, this is completely identical to the third scenario, so she (and by extension Bob) will end up measuring 1.
You can probably see where I'm going with this. We can follow this logic all the way back to the creation of the entangled pair. The conclusion of the argument is that if Alice measures a 1, then it must have been the case that measuring the particles at any previous point (after they were entangled) would also have resulted in a 1.
This seems to me to force the existence of a local hidden variable. Apparently this isn't right, because we have recently ruled out local hidden variables from being consistent with observations, but I am at a loss to see what is wrong with this argument. Or is what I said correct, but there is another explanation for it without involving a local hidden variable?
Could someone explain what is the issue with this argument? And, could someone explain Bell's theorem/Bell's inequality in layman's terms, and how it's possible that QM can violate it?

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SonOfOnett t1_j22snxg wrote

The problem with your argument is you are saying Alice is going to measure a 1: that’s a hidden variable you are introducing into the thought experiment! In reality and experiment, regardless of who measures first, we don’t know what the outcome of that first measurement will be.

You need to stop after stating that Bob and Alice’s particles are entangled to have the same measurement. Given that, neither observer, no matter how close in time to the start of the entanglement will know what they are going to measure, just that their measurements will match.

Veritasium has a decent video on Bells Inequality that may be useful to you as well. It explains how we can tell the difference between a local hidden variable or not

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Ecl1psed t1_j24gd5s wrote

I assumed that Alice measures a 1 because the argument would be the same no matter what she measures. Just replace the 1 with -1 and the logic still works. (Basically, "without loss of generality"). Or, I could have some variable X that represents Alice's measurement. We don't know X until Alice actually performs the measurement, but I think the argument should still work if we just replace all instances of 1 with X. I'll have a look at that Veritasium video, thanks for mentioning it.

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