Probabilities of couples of impacts - Bob and Alice and Young experiment

[Heidi]

TL;DR Summary

Bob and Alice do the same Young experiment with the EPR particles they share

Let O be where pairs of EPR particles are emitted to Alice on the righr and Bob on the left. the pairs have a constant total energy, a null total momentun and a null total angular momentum.
Alice at a distance D on the right of O have a Young double slits device with a screen at D+ L . Bob has the symmetrical device a -D
there is a coincidence setup that indicates when a pair gives impacts on de two screens a z1 and z2. what are the probablitiess for (z1,z2)? on the two screens . we only consider thes cases with coincidernce.
Are there couples of impacts which are avoided and others very frequent?
thanks

 

[DrChinese]

Summary:: Bob and Alice do the same Young experiment with the EPR particles they share

Let O be where pairs of EPR particles are emitted to Alice on the righr and Bob on the left. the pairs have a constant total energy, a null total momentun and a null total angular momentum.
Alice at a distance D on the right of O have a Young double slits device with a screen at D+ L . Bob has the symmetrical device a -D
there is a coincidence setup that indicates when a pair gives impacts on de two screens a z1 and z2. what are the probablitiess for (z1,z2)? on the two screens . we only consider thes cases with coincidernce.
Are there couples of impacts which are avoided and others very frequent?
thanks

In general terms (and assuming I understand the question): Entangled particles do not produce an interference pattern when passed through a double slit apparatus. So while there are coincidences of arrival times for Alice and Bob, there is no coincidence regarding related to interference patterns. They might produce coincidences for momentum as to "which slit" but I am not completely sure about that.

If you have them go through a diffraction slit (on each side) prior to the double slit, they would be independent but would then show interference.

 

[vanhees71]

Well, the photons are entangled in both the polarization states and the momentum, but this entanglement indeed doesn't do anything with respect to correlations between photon detectionprobabilities at a given place at Alice's and Bob's screens.

To see this note that the photons before the double slits are described by
$$|\Psi \rangle=\frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k},1) \hat{a}^{\dagger}(-\vec{k},-1) - \hat{a}^{\dagger}(\vec{k},-1) \hat{a}^{\dagger}(-\vec{k},1)] |\Omega \rangle.$$
For the double-slit experiment the polarizations and its entanglement doesn't play much of a role as long as you don't use it to imprint "which-way information" on the photons using the polarization (as e.g., used in Walborn's quantum-eraser experiment).

Let $\vec{k}=k \vec{e}_z$ and have put the Alice's screen to observe the photons with momentum $+\vec{k}$ behind her double slit at distance $z=L_A$ from these slits and the position on her screen $x_A \vec{e}_x$ an Bob's screen to observe the photons with momentum $-\vec{k}$ at $z=-L_B$ and the position on his screen at $x_B \vec{e}_x$ (with the slits of width $b_A$ and distance $d_A$ and $b_B$ and $d_B$ respectively infinite in the $y$ direction, so that you only get an interference pattern along the $x$ direction).

As explained on slide 3 of

https://itp.uni-frankfurt.de/~hees/publ/habil-coll-talk-en.pdf

The state after the slits is

$$|\Psi' \rangle=N_{\text{ESA}}(x_A)[1+\cos(\phi_{\text{A}}(x_A))N_{\text{ESB}}(x_B)[1+\cos(\phi_{\text{B}}(x_B)) |\Psi \rangle.$$
The only thing that happens is that you have a common position dependent factor which is due to the phase difference between the partial waves running through the one or the other slit of the double slits, leading to the "(1+cos)-factor" and the corresponding factor from the finite width of the single slits leading to the single-slit $N_{\text{ES}}$ factor. The interference pattern is then given by these amplitude factors squared, and they simply factorize for A's and B's photon. So there is no correlation for the detection probability at given places $x_A$ and $x_B$ for the single photons.

This simple calculation shows that the polarization state doesn't play any role here. It's only important that the single photons are coherent to get an interference pattern (in the here given idealized calculation with sharp plane-wave states, the photons are of course ideally coherent). The polarization state doesn't play any role in forming the detection probability and thus also the entanglement doesn't have any effect, i.e., there are no correlations between the detection positions of A's and B's entangled photons, as @DrChinese says.

 

[Heidi]

I know that no interference can be seen by Alice on her screen (same for Bob) . But see the Birgit Dopfer experiment where Alice has a 2 slits device but where Bob has a lens and a captor on his screen.
https://www.physicsforums.com/threads/heisenberg-lens-and-probabilities.1003862/
we could extract interfence in using the coincidence setup.
cannot we do the same here with two Young devices?
A captor is at a given place on Bob's screen and we consider the couples , one captured to the left and the other anywher on Alice's screen.
Is it very different thand the Dopfer experiment?

 

[vanhees71]

Of course both A and B can see intereference patterns if they choose sufficiently coherent sources. Case (a) as described in #7 of the quoted thread on Dopfer's Heisenberg-lens experiment with entangled is just using one of the entangled photons to project to a certain momentum, i.e., to provide a coherent source.

What's described here is much simpler: We simply use photons with (sufficiently) determined momenta to have coherent light sources.

Of course in Dopfer's experiment you need a sample of entangled photons that are not too coherent in order to be able to choose either the "which-way" or the "interference" setup to (post-)select the photons.

 

[DrChinese]

I know that no interference can be seen by Alice on her screen (same for Bob) . But see the Birgit Dopfer experiment where Alice has a 2 slits device but where Bob has a lens and a captor on his screen.
https://www.physicsforums.com/threads/heisenberg-lens-and-probabilities.1003862/
we could extract interfence in using the coincidence setup.
cannot we do the same here with two Young devices?
A captor is at a given place on Bob's screen and we consider the couples , one captured to the left and the other anywher on Alice's screen.
Is it very different thand the Dopfer experiment?

In the Dopfer experiment, the interference pattern is a subset extracted from a larger pattern. What are you planning to use on the other side (which has an identical setup) to select your subset?

 

[Heidi]

i am planning to select the couples with impact on the left screen (behind the slits) always at the same place and the rigbt ones anywhere on the screen. will there be interference pattern on the right?

are there other ways than the Dopfer trick on the left side to extract an interference pattern from the impacts on the right?

 

[Heidi]

lood at the picture here
i wonder if the lense can be replaces by something else.
my idea was to replace it by a second two slits device.

 

[DrChinese]

i am planning to select the couples with impact on the left screen (behind the slits) always at the same place and the rigbt ones anywhere on the screen. will there be interference pattern on the right?

No. Assuming you see any coincidences: probably all you will see is a single thin bar, relative to the selected location on the left. Due to conservation.

 

[Heidi]

Conservation of what here?
And in the Dopfer experiment which selects a given impact on one side but where the impact is random behind the slits on the other side what is conserved?

Suppose now that O is not the center of symmetry of the two young's devices. if the left particles goes through one slit, the line from this slit to O does not go to a slit on the right. we cannot use that kind of symmetry.

 

[Heidi]

As Bob have to use a reduced density matrix , z1 is a sum of probabilities (up slit and down slit) same thing for Alice for her z2
If we consider the system created with an occupation number = 2 it is in a pure state and the couple (z1,z2) which can be plotted on a sheet of paper could show an interference pattern. No?

 

[DrChinese]

As Bob have to use a reduced density matrix , z1 is a sum of probabilities (up slit and down slit) same thing for Alice for her z2
If we consider the system created with an occupation number = 2 it is in a pure state and the couple (z1,z2) which can be plotted on a sheet of paper could show an interference pattern. No?

No. See the following reference by Zeilinger, who was Dopfer's thesis advisor, specifically Figure 2:

"Experiment and the Foundations of Quantum Physics" (1999)
https://zbook.org/read/b80f_experiment-and-the-foundations-of-quantum-physics.html

Also check out this reference, which I believe is essentially a realization of the above (although perhaps not with all of the parameters you seek).

"Quantum double-double-slit experiment with momentum entangled photons."
https://www.nature.com/articles/s41598-020-68181-1

 

[Heidi]

thank you for the double double experiment paper in the Nature review.
why did you say no?
i read this at the end:
For a given location of detection of a photon the other photon shows interference pattern which corresponds to the conditional interference pattern. this is exactly what i wanted to know.

 

[Heidi]

I think that the captor on the left has also to be at an equal distance from the slits to destroy the which path information which would reduce the fringe visibility on the right.

 

[vanhees71]

No. See the following reference by Zeilinger, who was Dopfer's thesis advisor, specifically Figure 2:

"Experiment and the Foundations of Quantum Physics" (1999)
https://zbook.org/read/b80f_experiment-and-the-foundations-of-quantum-physics.html

Also check out this reference, which I believe is essentially a realization of the above (although perhaps not with all of the parameters you seek).

"Quantum double-double-slit experiment with momentum entangled photons."
https://www.nature.com/articles/s41598-020-68181-1

That's different from the case I've discussed above. Using the double slits in the plane perpendicular to the pump beam direction what's realized is the original EPR experiment concernig momentum entangled equally polarized photon pairs. The full ensemle of each of the single photons is incoherent, but selecting subensemles where the idler photon is detected at a small area behind its double-slit the corresponding signal photons show a double-slit interference pattern.

The math in the paper is straight forward, although the text is very hard to read...

 

[Heidi]

thanks this is clear now.
suppose that the selected common impact on the left is at an equaldistance from the slits. the interference pattern on the left could be the product on a cosine and of (G1 + G2) , two overlapping gaussians. if select a set of impacts on the left around the first impac,
i think that the interferences will disappear and we will only see the gaussian envelopes. is this correct?
is there a way to select a set of impacts on the left to keep only one of the gaussian envelopes ? just as if all the right photons were passing through the same right slit?
I think it woud be the case if the left screen were close to the slits and the impacts neat one of it. can we have one screen near on the left and one screen a greater distance from the screen on the right?

 

[Heidi]

what about the fringe visibility in the formulas when the selected impact is near a slit?