Is coherence restored after erasure of which-way info?

[San K]

After erasure (of which-way info) in DCQE or similar experiments, is coherence (constant relative phase) restored?

 

[SpectraCat]

After erasure (of which-way info) in DCQE or similar experiments, is coherence (constant relative phase) restored?

What makes you think it was ever lost?

 

[San K]

What makes you think it was ever lost?

here's the logic, not sure if its right, also as i write I realize there is a lot (of factors) to think about and there are more kinks, than I thought, in my logic...;)

when we get which-way, we loose interference pattern

this is because: when we try which-way, we introduce de-coherence (the phase difference is not longer constant)

when we do erasure we get back interference pattern,

this is because:

either 1. we have got back coherence, somehow

2. coherence in idler-photon

3. sub-sampling...i.e. only those ones get filtered (via coincidence counter) that fit/match the story (i.e. interference pattern)

on a separate but related note: when signal (or idler) photon is registered on the detector (Ds or Dp), is not the entanglement broken?
if the entanglement is broken does it not mean de-coherence has happened, or wave function has collapsed?

 

[SpectraCat]

here's the logic, not sure if its right, also as i write I realize there is a lot (of factors) to think about...;)

when we get which-way, we loose interference pattern

this is because: when we try which-way, we introduce de-coherence (the phase difference is not longer constant)

when we do erasure we get back interference pattern,

this is because:

either 1. we have got back coherence, somehow

2. coherence in idler-photon

3. sub-sampling...i.e. only those ones get filtered (via coincidence counter) that fit/match the story (i.e. interference pattern)

on a separate but related note: when signal (or idler) photon is registered on the detector (Ds or Dp), is not the entanglement broken?
if the entanglement is broken does it not mean de-coherence has happened, or wave function has collapsed?

That analysis is not consistent with Cthugha's explanation of the DCQE that we have discussed previously. I don't believe that decoherence is ever discussed in the context of that analysis. The issue there is with the sub-sampling of the coincidence measurements .. that is what causes the interference pattern to disappear/reappear. Remember that the single-photon measurements for the s-photon NEVER show an interference pattern in that experiment.

 

[San K]

Remember that the single-photon measurements for the s-photon NEVER show an interference pattern in that experiment.

would not the s-photon show interference if we did without QWPs? (in the DCQE), the paper shows the diagram/figure.

That analysis is not consistent with Cthugha's explanation of the DCQE that we have discussed previously. I don't believe that decoherence is ever discussed in the context of that analysis. The issue there is with the sub-sampling of the coincidence measurements .. that is what causes the interference pattern to disappear/reappear.

you would need de-coherence in addition to sub-sampling to explain the DCQE.

when s-photon is detected, the entanglement is broken, de-coherence between s and idler photon happen

and

then after that sub-sampling (filtering) comes into the picture...when we compare in the co-incidence counter (however the de-coherence has already happened when s-photon is detected)...

 

[SpectraCat]

would not the s-photon show interference if we did without QWPs? (in the DCQE), the paper shows the diagram/figure.

Sure, but that isn't really what the DCQE is about .. it is about how to recover the interference pattern after it has been destroyed by polarization-tagging the photon paths.

you would need de-coherence in addition to sub-sampling to explain the DCQE.

when s-photon is detected, the entanglement is broken, de-coherence between s and idler photon happen

and

then after that sub-sampling (filtering) comes into the picture...when we compare in the co-incidence counter (however the de-coherence has already happened when s-photon is detected)...

Well, I suppose that is correct strictly speaking, but it is only the decoherence that is associated with any kind of measurement. It is the same whether or not a polarizer is in place in the p-photon branch, so I don't see what it has to do with the interference pattern, or the DCQE results in particular. Adding the polarizer doesn't somehow remove decoherence from the results ... it just allows you to sub-select an ensemble that shows interference because of the well-defined relationship between the phases of the entangled photons.

Also, strictly speaking, it is the QWP's and not the detection that breaks the entanglement between the s- and p- photons.

 

[San K]

Sure, but that isn't really what the DCQE is about .. it is about how to recover the interference pattern after it has been destroyed by polarization-tagging the photon paths.

ok...i just wanted to validate that I understood the DCQE correctly...by doing tweaks (running various scenarios/modifications) to the DCQE...

Well, I suppose that is correct strictly speaking, but it is only the decoherence that is associated with any kind of measurement. It is the same whether or not a polarizer is in place in the p-photon branch, so I don't see what it has to do with the interference pattern, or the DCQE results in particular. Adding the polarizer doesn't somehow remove decoherence from the results ... it just allows you to sub-select an ensemble that shows interference because of the well-defined relationship between the phases of the entangled photons.

Also, strictly speaking, it is the QWP's and not the detection that breaks the entanglement between the s- and p- photons.

i meant that only...the entanglement breaks at the first instance of detection (whether by QWPs or the detector Ds or Dp)

QWPs are also doing detection (detection of which slit the photon went thought)...in absence of QWPs...the first instance of detection would happen at the detector...

It is the same whether or not a polarizer is in place in the p-photon branch, so I don't see what it has to do with the interference pattern, or the DCQE results in particular.

when the s-photon is detected (the entanglement is broken) ...and the behavior of p is fixed/determinable... probabilistically...?

 

[SpectraCat]

i meant that only...the entanglement breaks at the first instance of detection (whether by QWPs or the detector Ds or Dp)

QWPs are also doing detection (detection of which slit the photon went thought)...in absence of QWPs...the first instance of detection would happen at the detector...

Not sure if you are asking a question, but yes, that is correct.

when the s-photon is detected (the entanglement is broken) ...and the behavior of p is fixed/determinable... probabilistically...?

Not exactly sure what you are asking, but once the entanglement is broken, both photons have well-defined polarizations. In other words, if in some measurement basis you find the s-photon has right-handed circular polarization, then the p-photon will have left-handed circular polarization (assuming you started out with a Bell-state like |R>₁|L>₂ + |L>₁|R>₂), and vice-versa. The same is true if the "detection" projects the polarization state of one of the photons into a linear-polarization basis: Then if one photon is |H>, the other will be |V> (|H> and |V> stand for horizontal and vertical in the polarization basis).

 

[San K]

Not sure if you are asking a question, but yes, that is correct.

thanks for validating.

Not exactly sure what you are asking, but once the entanglement is broken, both photons have well-defined polarizations. In other words, if in some measurement basis you find the s-photon has right-handed circular polarization, then the p-photon will have left-handed circular polarization (assuming you started out with a Bell-state like |R>₁|L>₂ + |L>₁|R>₂), and vice-versa. The same is true if the "detection" projects the polarization state of one of the photons into a linear-polarization basis: Then if one photon is |H>, the other will be |V> (|H> and |V> stand for horizontal and vertical in the polarization basis).

because the p-photon now has well defined polarization (after s-photon detection, and hence collapse of the wave function) we can predict its behavior probabilistically?

x% probability of going through left slit

y% probability of going through right slit

i am not getting this correctly...but it's somewhere there...i guess

 

[SpectraCat]

thanks for validating.



because the p-photon now has well defined polarization (after s-photon detection, and hence collapse of the wave function) we can predict its behavior probabilistically?

x% probability of going through left slit

y% probability of going through right slit

i am not getting this correctly...but it's somewhere there...i guess

It's not even probabilistic .. we can tell with certainty (assuming we don't put the polarizer in place). The Walborn DCQE paper goes through this in full detail .. I would recommend reading through that section if you haven't already.

 

[San K]

It's not even probabilistic .. we can tell with certainty (assuming we don't put the polarizer in place). The Walborn DCQE paper goes through this in full detail .. I would recommend reading through that section if you haven't already.

i think i got it...its not probabilistic about the slit...

but probabilistic where it would fall on the screen... where it would fall on the (interference or non-interference) pattern...?

the (inherent) randomness persists...?