Timeline for Do Feynman path integrals satisfy Bell locality assumption?
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Feb 20, 2020 at 13:08 | comment | added | Jarek Duda | Indeed, but if measurement is in your t1 in the center, we have symmetrically kind of two "hidden variables": one from past, one from future, as in TSVF: en.wikipedia.org/wiki/Two-state_vector_formalism or Ising model: one amplitude from left, one from right - we get Born rule from symmetry ... instead of standard Kolmogorov probability theory in which we can derive Bell-like inequalities. | |
Feb 20, 2020 at 12:56 | comment | added | Eric David Kramer | The path integral has a slicing formula: $A(t_2,t_0)=\int\!dt_1\,A(t_2,t_1)A(t_1,t_0)$ so you can always insert your operators at $t_1$ and have a wave function there. | |
Feb 20, 2020 at 12:50 | comment | added | Jarek Duda | Ok, in this case you are right. But I have meant solving (CPT symmetric) physics in time-symmetric way: imagine t0 = -infinity, t = +infinity, measurement is between them. Is Bell locality assumption still satisfied in this case? While it is highly nonintuitive, Ising model is great to gain intuitions here - as it uses spatial instead of temporal direction in analogous way. | |
Feb 20, 2020 at 12:37 | comment | added | Eric David Kramer | All the measurements are at the time $t$. Just the order in which they are applied is important. | |
Feb 20, 2020 at 11:03 | comment | added | Jarek Duda | But what if measurement is between this time t and t0? Do Bell's "hidden variables" contain both information before (n0) and after measurement (n)? Both are required to solve what you have written, but I believe "hidden variables" use only information before (?) Think about Ising model (e.g. diagram before) - the math is nearly the same (Feynman -> Boltzmann path ensemble), but past/future become more intuitive left/right, still getting e.g. Pr(i)=(psi_i)^2 probability distribution inside. | |
Feb 20, 2020 at 9:02 | history | answered | Eric David Kramer | CC BY-SA 4.0 |