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jgm340
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After reading about the
http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser" experiment,
I thought of two interesting questions.
Here I have taken http://en.wikipedia.org/wiki/File:Kim_EtAl_Quantum_Eraser.svg" and simplified it:
In the normal delayed choice quantum eraser experiment, the "BOX" will randomly do one of two things:
(i) gather information about which slit the photon has traveled through (either the red slit or the blue slit).
(ii) erase the information about which slit the photon has traveled through. It will erase the information one of two ways (call these methods of erasing "a" and "b"). It is detected which method of erasing was used.
There is a 50% chance of it doing (i) and a 50% chance of it doing (ii).
If we consider only photons detected to have passed through the red slit, we find no interference pattern. Likewise with those detected to have passed through the blue slit.
However, if we consider photons with path information erased by method "a", we DO see an interference pattern. Likewise with the path information erased by method "b".
The interesting thing is that the sum of these two interference patterns (formed through method "a" and method "b") is not an interference pattern!
Now here comes my first question:
Look at this diagram from the article:
What would happen if were were to simply allow light reflecting off Ma (blue) to bypass BSc. Would the photons striking D2 still form an interference pattern (albeit different than before) on D1?
What would happen if, in addition to the change above, I replaced D1 with a series of mirrors that diverted the light towards D4 instead, syncing it up so that whatever path a "red" photon takes, it arrives at D4 in the same amount of time. Would the photons registered by D4 still not show an interference pattern? And would there still be an interference pattern for the photons registered by D2 (as above)?
If the answer to all the above is "yes", then it seems to me that if we consider all the photons registered by D2, D3, and D4 all together, then there would be an interference pattern on D0.
Now here's my second question:
Suppose the answer to the first question is yes. Then suppose I were to replace BSc by a mechanical switch, which has the ability to change very quickly from a beam-splitter to a mirror.
Now, suppose I hook up this mechanical switch to D0, so that whenever a photon is detected at D0, the switch changes BSc to a mirror after a time delay just perfect to allow the switch to happen before the entangled photon twin reaches BSc. Suppose also I have the switch turn back into a beam-splitter very quickly afterwards.
In this case, it ought to be that all the photons that were registered before by D2, D3, and D4 all together will still be registered. The only difference, I would think, would be which of D2, D3, or D4 detects the photon. If this is true, then the photons registered by D2, D3, and D4 altogether ought to show the same interference pattern on D0 as before. (The photons have already struck D0, so assuming causality isn't violated, the pattern ought to be the same.)
Now, here's where it starts not making sense: now that BSc is always switched to a mirror when the "red" photons strike it, the photons registered by D2 ought to give the same interference pattern as that of the photons registered by D1 and D2 combined in the original (unmodified) experiment. But this should show no interference!
Which line of reasoning is correct? One shows that the photons registered by D2, D3, and D4 all together ought to show an interference pattern on D0, and the other line of reasoning shows that it should not.
Have any experiments been done testing similar ideas to this?
http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser" experiment,
I thought of two interesting questions.
Here I have taken http://en.wikipedia.org/wiki/File:Kim_EtAl_Quantum_Eraser.svg" and simplified it:
In the normal delayed choice quantum eraser experiment, the "BOX" will randomly do one of two things:
(i) gather information about which slit the photon has traveled through (either the red slit or the blue slit).
(ii) erase the information about which slit the photon has traveled through. It will erase the information one of two ways (call these methods of erasing "a" and "b"). It is detected which method of erasing was used.
There is a 50% chance of it doing (i) and a 50% chance of it doing (ii).
If we consider only photons detected to have passed through the red slit, we find no interference pattern. Likewise with those detected to have passed through the blue slit.
However, if we consider photons with path information erased by method "a", we DO see an interference pattern. Likewise with the path information erased by method "b".
The interesting thing is that the sum of these two interference patterns (formed through method "a" and method "b") is not an interference pattern!
Now here comes my first question:
Look at this diagram from the article:
What would happen if were were to simply allow light reflecting off Ma (blue) to bypass BSc. Would the photons striking D2 still form an interference pattern (albeit different than before) on D1?
What would happen if, in addition to the change above, I replaced D1 with a series of mirrors that diverted the light towards D4 instead, syncing it up so that whatever path a "red" photon takes, it arrives at D4 in the same amount of time. Would the photons registered by D4 still not show an interference pattern? And would there still be an interference pattern for the photons registered by D2 (as above)?
If the answer to all the above is "yes", then it seems to me that if we consider all the photons registered by D2, D3, and D4 all together, then there would be an interference pattern on D0.
Now here's my second question:
Suppose the answer to the first question is yes. Then suppose I were to replace BSc by a mechanical switch, which has the ability to change very quickly from a beam-splitter to a mirror.
Now, suppose I hook up this mechanical switch to D0, so that whenever a photon is detected at D0, the switch changes BSc to a mirror after a time delay just perfect to allow the switch to happen before the entangled photon twin reaches BSc. Suppose also I have the switch turn back into a beam-splitter very quickly afterwards.
In this case, it ought to be that all the photons that were registered before by D2, D3, and D4 all together will still be registered. The only difference, I would think, would be which of D2, D3, or D4 detects the photon. If this is true, then the photons registered by D2, D3, and D4 altogether ought to show the same interference pattern on D0 as before. (The photons have already struck D0, so assuming causality isn't violated, the pattern ought to be the same.)
Now, here's where it starts not making sense: now that BSc is always switched to a mirror when the "red" photons strike it, the photons registered by D2 ought to give the same interference pattern as that of the photons registered by D1 and D2 combined in the original (unmodified) experiment. But this should show no interference!
Which line of reasoning is correct? One shows that the photons registered by D2, D3, and D4 all together ought to show an interference pattern on D0, and the other line of reasoning shows that it should not.
Have any experiments been done testing similar ideas to this?
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