New quantum experiments and its implications

zonde

Will wait for it.

StevieTNZ

Do we expect to see photons #1 and #4 both in VV with photons #2 and #3 in HH (and vice versa)? And not #1 and #4 in HH, and #2 and #3 also in HH?

StevieTNZ

Received a PDF from Johannes Kofler re: the evolution of the separable state. But this is without the EOW's etc in action (which should produce the same result).

However if we have definite states prior to the first BS, even though after they're in superposition of travelling both paths, they reach the plates and are converted into L and R polarisation. When they reach the PBS's after the 2nd BS, wouldn't they have 1/2 probability of taking on V and H? *shrugs*

Attached Files

SSM.pdf (41.4 KB, 4 views)

StevieTNZ

Okay - so I don't understand a few things:

1) We start off with two pairs of entangled photons, #1 and #2, and #3 and #4 as |H>|V> - |V>|H>.
2) How do we get from that to describing #2 and #3 (before the 1st beam splitter) as |H>|H> + |V>|V> and |H>|H> - |V>|V>, when no entanglement swapping hasn't occured yet?

If we sent #2 and #3 through the interferometry, and they were definite H or V polarised, would we even end up with the results obtained? Shouldn't we describe #2 and #3 before the 1st BS as if they're still entangled originally, and calculate the result from that? But even then we wouldn't end up with |H>|V>(b") or |H>|V>(c") or |H>(b")|H>(c") or |V>(b")|V>(c").

mathal

post edit -this should have referenced San K post # 19. My apologies, Dr. Chinese.
Whereas the order of magnitude seems immense, all but these scant few down converted photons pass 'straight' through the mechanism while the 'entangled' down converted photons travel in a cone away from this straight line. Each pair of down converted photons split 180 degrees from each other. These are the entangled pairs of photons which are used in experiments. There is 'nearly' no messy noise that can enter an experiment in this process. IMO.


mathal