- #1
vintch
- 5
- 2
Hello.
This topic seems to be discussed a lot of times before. But I still have some questions which was not clarified earlier and I will be grateful if someone will give me explanations :)
First of all, to define a context, original experiment setup:
(source: https://upload.wikimedia.org/wikipedia/commons/2/21/Kim_EtAl_Quantum_Eraser.svg)
My understanding of what is happening:
Laser emits a powerful ray which meets an obstacle with a double slit. At this point, each of photons in that ray propagates through both slit simultaneously (i.e. superposition; or through one exact slit, depending on detectors) and meets BBO crystal, where SPDC process taking place which creates entangled photon pairs. Next, first photon ("signal") travels to the detector D0, and the second photon ("idler") travels to one of the "determined" detectors (D3 or D4), or to the quantum eraser circuitry (randomly chosen detector D1 or D2). Detector D0 all the time sees just a blur. But when we filter all the photons reached D0 by corresponding D1-D4 signals, we will get four images: a "blur" (diffraction pattern) for D3 and D4 (where the idler photon has which-way information), and an interference patterns for D1 and D2 (where the idler photon has no which-way information). Interference patterns for D1 and D2 will be different (in counterphase).
Questions:
1. As was noted in few sources, difference between interference patters for D1 and D2 caused by phase shift, which takes place in BSc half-silvered mirror. Is it so? If yes, please explain, how exactly it happens if the system is fully symmetrical?
2. Consider next image:
I removed detectors D3-D4 and corresponding half-silvered mirrors BSa-BSb. Detector D0 still sees kinda "blur". But if we filter photons which idler pair hit exactly detector D1 or D2, we will get two interference patterns as in previous case. But this patterns still be different. Am I right?
3. And finally, most interesting one. Consider this:
Following the same logic, here we should see an interference patterns on the both screens. Literally "see on the screen" - there is no detectors anymore. There is no which-way information exist. But if so, it suggests violation of causality/faster-than-light interaction! (e.g. if we simply place two polarizers before lens ("mark" photons), interference pattern should disappear)
So, my question is: what will actually happen and why?
Thank you very much if you read so far :)
Best regards.
This topic seems to be discussed a lot of times before. But I still have some questions which was not clarified earlier and I will be grateful if someone will give me explanations :)
First of all, to define a context, original experiment setup:
(source: https://upload.wikimedia.org/wikipedia/commons/2/21/Kim_EtAl_Quantum_Eraser.svg)
My understanding of what is happening:
Laser emits a powerful ray which meets an obstacle with a double slit. At this point, each of photons in that ray propagates through both slit simultaneously (i.e. superposition; or through one exact slit, depending on detectors) and meets BBO crystal, where SPDC process taking place which creates entangled photon pairs. Next, first photon ("signal") travels to the detector D0, and the second photon ("idler") travels to one of the "determined" detectors (D3 or D4), or to the quantum eraser circuitry (randomly chosen detector D1 or D2). Detector D0 all the time sees just a blur. But when we filter all the photons reached D0 by corresponding D1-D4 signals, we will get four images: a "blur" (diffraction pattern) for D3 and D4 (where the idler photon has which-way information), and an interference patterns for D1 and D2 (where the idler photon has no which-way information). Interference patterns for D1 and D2 will be different (in counterphase).
Questions:
1. As was noted in few sources, difference between interference patters for D1 and D2 caused by phase shift, which takes place in BSc half-silvered mirror. Is it so? If yes, please explain, how exactly it happens if the system is fully symmetrical?
2. Consider next image:
I removed detectors D3-D4 and corresponding half-silvered mirrors BSa-BSb. Detector D0 still sees kinda "blur". But if we filter photons which idler pair hit exactly detector D1 or D2, we will get two interference patterns as in previous case. But this patterns still be different. Am I right?
3. And finally, most interesting one. Consider this:
Following the same logic, here we should see an interference patterns on the both screens. Literally "see on the screen" - there is no detectors anymore. There is no which-way information exist. But if so, it suggests violation of causality/faster-than-light interaction! (e.g. if we simply place two polarizers before lens ("mark" photons), interference pattern should disappear)
So, my question is: what will actually happen and why?
Thank you very much if you read so far :)
Best regards.