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You can't. This is just covered by classical electrodynamics. Of course in a sense you can say anything is quantum, but then the stability of a table is quantum
by andi999 5y ago
You can't. This is just covered by classical electrodynamics. Of course in a sense you can say anything is quantum, but then the stability of a table is quantum (which it is). If you were to have a single photon source and avalanche detector, ok now you are talking quantum.
- admax88q 5y agoI'm curious, can you detail how classic electrodynamics covers this?
- kgwgk 5y agohttps://aip.scitation.org/doi/pdf/10.1063/1.4943751 https://aip.scitation.org/doi/pdf/10.1063/1.4943751
- infogulch 5y agoThe first 2 minutes of that video make it clear that they think it is "real" quantum effects (e.g. at 1:34 entanglement > EPR paradox > Bell's Theorem > local realism). Those are pretty quantum-y imo.
- tofof 5y agoYeah. At 9:34 they're dealing with entangled photons passing through 2 filters at different points in space, simultaneously. They affect each other, which is impossible without faster-than-light communication between them. This is absolutely quantum and cannot be explained at all by classical EM.
- andi999 5y agoIf you have entanglement then it is quantum. Yes. But the comment was you 'just' need polarized glasses. So if 'just' polarized glasses for a few $ means: "and of course a biphoton source for a couple of thousand $ and some coincidence counters for another couple of thousand $", then you only 'just' need polarizers.