0
$\begingroup$

I wonder if double-slit experiment can be considered a proof of non-existence of local hidden variables in quantum mechanics?

Consider this: probability $P(A \: \textrm{or} \: B)$ that either one of two events A and B or both happen is $$ P(A \: \textrm{or} \: B) = P(A) + P(B) - P(A \: \textrm{and} \: B) $$ Additional term here accounts for probability of two events being non-exclusive (happening at the same time), in which case we would have counted them twice.

In double-slit experiment, we can consider $A$ and $B$ to be events of going through one of the slits. According to QM the interference of amplitudes $\mathcal{A}$ and $\mathcal{B}$ gives $$ P(A \: \textrm{or} \: B) = |\mathcal{A} + \mathcal{B}|^2 = P(A) + P(B) + 2\Re \left(\mathcal{A}^*\mathcal{B} \right) $$ The interference term $2\Re \left(\mathcal{A}^*\mathcal{B} \right)$ can be both positive and negative, depending on the point on the screen. But in the first equation additional term can only be negative. Thus, it would seem, we can't describe quantum interference as a result of combining certain isolated events.

I'm sure it's not that simple, but can't really see what's I'm missing in this reasoning.

Improve this question
$\endgroup$
6
  • $\begingroup$ Are you assuming that the double slit experiment measures which slit the particle goes through? The whole point of the double slit experiment is that you don't know which slit the particle went through. $\endgroup$
    – BioPhysicist
    Commented Sep 21, 2020 at 13:01
  • $\begingroup$ @BioPhysicist No, I know it doesn't measure which slit the particle goes through. What I mean is, if there were some hidden variables that predict which slit a particle goes through (maybe in a very complicated way), the probabilities would still obey the first equation. But since they don't, we can deduce that no such variable exist. $\endgroup$
    – xaxa
    Commented Sep 21, 2020 at 14:42
  • $\begingroup$ @BioPhysicist see also my comments on Emilio Pisanty's answer below $\endgroup$
    – xaxa
    Commented Sep 21, 2020 at 14:44
  • $\begingroup$ I don’t think the bells-inequality tag is applicable here. $\endgroup$
    – Superfast Jellyfish
    Commented Sep 21, 2020 at 15:16
  • $\begingroup$ @SuperfastJellyfish I'd tag it "hidden variables" but there isn't one, so I chose the closest one found $\endgroup$
    – xaxa
    Commented Sep 21, 2020 at 15:20

1 Answer 1

Reset to default
1
$\begingroup$

Thus, it would seem, we can't describe quantum interference as a result of combining certain isolated events.

Yes, this is correct. However, nothing in your argument has any real bearing on the existence of hidden variables, or indeed on the double-slit experiment.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I mean... if there were any hidden variables that predict which of the slits a particle would actually go through... then the probabilities would obey the first equation, right? $\endgroup$
    – xaxa
    Commented Sep 21, 2020 at 14:38
  • $\begingroup$ I'm probably confusing non-existence of trajectories with hidden variables. But now that I think of it, I don't see what the difference is between the two concepts. $\endgroup$
    – xaxa
    Commented Sep 21, 2020 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.