Delayed choice quantum eraser scanning telescope

[0xDEAD BEEF]

Hi! :)

I have this question regarding delayed choice quantum eraser -
If my understanding is correct then interference pattern will form/not form based on whether there is erasing mirror in path of entangled particle - regardless of time when particle arrive at mirror.

Should not it be possible then to create a scanning telescope, that sends particles out in vast space of universe (focused at some very distant point) and get result of whether they hit "loose-path-information" material or "known-path-material" at the same instant measurement is done?

Can we build space scanning telescope and get instant results (no waiting for particles to actually reach far targets and return)?

Thank you,
0xDEAD BEEF

 

[DrChinese]

Are you imagining using entangled particle pairs as the source? Entangled photons do not produce an interference pattern (while still entangled on that basis).

 

[0xDEAD BEEF]

DrChinese - as my understanding is - entangled particles do produce interference pattern as long as both particles are not measured for '"which-path" information.

I am referring to this video (and sorry for my naive approach) -

As long as entangled particle hits BSc half-mirror (and it does not matter if that happens before or after other entangled particle reaches D0 screen) interference pattern is observed at D0 screen. And vice versa - when entangled particle hits D3 or D4 detector, there will never be interference pattern at D0 screen.

So let's call particles, that are heading for D0 screen particles A and particles, that go for double-eraser setup particles B (A and B are entangled).

My idea is - if we setup scanning telescope and scan universe using B particles, then at the same instant moment we should see particles A form or not form interference pattern (based on if particle B will hit D3/D4 setup in future or hit path information loosing mirror BSc in future). There should be crystals in universe that form D3/D4 setup and crystals that act as half-mirror (BSc). So - should we be able to scan universe (million of light years away) at the very instant for BSc mirrors?

Are my assumptions wrong?Beef

 

[MichPod]

If my understanding is correct then interference pattern will form/not form based on whether there is erasing mirror in path of entangled particle - regardless of time when particle arrive at mirror.

Briefly, no, this understanding is not exactly correct.

If it was correct, that would practically allow a faster than light communication or, in your case, sending some information from the future to the past by some civilization which may put or remove an erasing mirror in a far galaxy when your entangled particle arrives there.

There is no such thing as "erasing mirror" working with 100% guarantee. In the delayed quantum eraser experiment the path information may be erased if D1 or D2 detect a particle, but it is not erased when D3 or D4 detect a particle. The interference pattern does not form just because one puts all the construction of mirrors on the way of the entagled particle and the detector D0 does not, in fact, encounter any interference pattern in the experiment by itself. Yet, if you take a subset of D0 detection events for which D1 or D2 also detected the other entangled particle, then these detection events of D0 do from an interference pattern.

 

[DrChinese]

DrChinese - as my understanding is - entangled particles do produce interference pattern as long as both particles are not measured for '"which-path" information.
...
Are my assumptions wrong?

Entangled photons generally do not produce interference effects. Were this not the case, the mechanism could be exploited for FTL communication.

The quantum eraser produces the "interference" signature only when data from both sides are brought together. So your assumption is not correct.

 

[0xDEAD BEEF]

So if in "delayed quantum eraser experiment" we remove BSa and BSb will D0 always display interference pattern (even ignoring D1 and D2 data) or will there be some noise that now would be impossible to filter out due to lack of D3 and D4 input?

 

[MichPod]

Hmmm... indeed.

 

[DrChinese]

Entangled photons do not produce an interference pattern in the first place. The general explanation is that they are not coherent. See page S290 in the following reference, Figure 2, which makes this more clear. This paper is from Zeilinger, one of the premier researchers in this field.

http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

 

[DrChinese]

And yes, it is confusing as to why interference patterns show up when there is a comparison between the 2 sides in an eraser type experiment. But that involves an entirely different theoretical discussion. If you are only looking at one side, there is no pattern unless you first make the stream coherent - which collapses entanglement of momentum/position basis.

 

[DrChinese]

But since this thread is about entangled photons and a long range scanning scope which probes FUTURE celestial object positions:

I have previously asked about the possibility of a similar setup that does not rely on interference patterns. You can get some background from the following thread I started a few years back:

https://www.physicsforums.com/threads/does-cmbr-include-free-photons.558253/

The idea requires some assumptions, although they are mostly mainstream:

a) There are no free photons - photons require an emitter and an absorber. This is consistent with standard QFT.
b) The universe is expanding so fast that the observable universe is shrinking (the portion which can even receive light from us). This is fairly orthodox.
c) The universe (and the apparent horizon) is NOT a perfect absorber for otherwise free photons. I am not aware of any particular theory/evidence either way on this, but I am fairly ignorant in this area.

So assuming all of the above: if you pointed the output of one side (call it A) of an entangled photon stream towards deep space, no photon could be emitted in the direction(s) in which there is nothing to absorb the photon in the far future. If that could not be emitted, then the other of the pair (call that B) could not either. You could detect that now by noting the relative intensity of the B stream (here), and that would in turn give you information about the future in the A direction.

I cannot imagine this is actually feasible. So at least one of the 3 assumptions must be wrong.

 

[MichPod1]

So if in "delayed quantum eraser experiment" we remove BSa and BSb will D0 always display interference pattern (even ignoring D1 and D2 data) or will there be some noise that now would be impossible to filter out due to lack of D3 and D4 input?

All right, I now see that my answer was not quite correct. Now I much hope to have the right answer.
Contrary to what I said previously, there is no interference pattern on D0 even for events for which any of two detectors D1 or D2 detected a particle. But there is an interference pattern on D0 for events for which D1 detected a particle and there is (another) interference pattern on D0 for events for which D2 detected a particle. Hope this removes all the misunderstanding.

 

[Strilanc]

Should not it be possible then to create a scanning telescope, that sends particles out in vast space of universe (focused at some very distant point) and get result of whether they hit "loose-path-information" material or "known-path-material" at the same instant measurement is done?

Nope. The interference patterns are in the joint outcomes, not the local outcomes.

No matter what the sender Alice does, the receiver Bob sees the same thing. To find the interference patterns (or confirm they aren't there), Alice and Bob need to put their measurement results together. Bob can't do it on his own. Alice needs to fly over and help, which takes time, which puts a kink in the whole "have it happen instantaneously" idea.

 

[entropy1]

I think there is not an inerference pattern 'produced' by D0 only! The interference pattern in this setup requires, as @DrChinese pointed out, the comparing of the data coming from D0 with the other detectors, that is to say, it is a correlation.

 

[0xDEAD BEEF]

Thank you guys! This explain a lot of it. I have to think about all this for a while now! :)
Cheers!