Delayed Choice Quantum Eraser with altered beam splitter

[Athraxin]

Hi People,

In Kim's real setup for interference patterns we use a beam splitter as 50:50 (reflectance,transmittance rate), lets assume we use 20:80 (%20 reflectance, %80 transmittance) for beam splitter what could we observe in this situtation on detectors (except D3,D4)? Dont forget we use only a beam splitter for the erasing part.

Simply the question: Are we going to see constructive and destructive patterns on (D1,D2) dedectors? What would be different for these results in this case? (20:80) beam splitter(BS)

[1] Kim, Y.-H., Yu, R., Kulik, S. P. and Shih, Y. Delayed “choice” quantum eraser. Phys. Rev. Lett. 84, 1-5 (2000).
[2] Fankhauser, Johannes (2019). "Taming the Delayed Choice Quantum Eraser". Quanta. 8: 44–56.

 

[vanhees71]

I think the outcome for the subensembles/coincidence measurements, labelled $R_{01}$ and $R_{02}$, will be the same. Only the relative abundance will go from 50:50 to 20:80.

 

[Athraxin]

I think the outcome for the subensembles/coincidence measurements, labelled $R_{01}$ and $R_{02}$, will be the same. Only the relative abundance will go from 50:50 to 20:80.

So for instance we will see a pattern %80 photon of all on R01 and %20 photon of all on R02 also they will show same (destructive&constructive) patterns. In this case, one of them will be more crowded with photon counts, other one will be less crowded with photon counts. Thanks for you answer sir.

 

[vanhees71]

But there'll also be the $R_{03}$ and $R_{04}$ events.

 

[Athraxin]

But there'll also be the $R_{03}$ and $R_{04}$ events.

But I've removed them from the setup as you see in the picture. How can I see them even they are absent in this setup?

 

[vanhees71]

True, I've overlooked this. Then you are right: You split the full ensemble to two partial ensembles by the coincidence measurements and get $R_{01}$ and $R_{02}$ both showing two-slit interference patterns but one comes out in 20% and the other in 80% of all cases rather than the 50:50 of the original setup.

 

[Athraxin]

True, I've overlooked this. Then you are right: You split the full ensemble to two partial ensembles by the coincidence measurements and get $R_{01}$ and $R_{02}$ both showing two-slit interference patterns but one comes out in 20% and the other in 80% of all cases rather than the 50:50 of the original setup.

I'm right? That usually doesn't happen. :) Thank you btw.

 

[vanhees71]

I must however, correct myself concerning the explanation for the no-interference pattern at $D_0$ in the other thread. That was wrong. It's not due to the photons coming from the upper or lower slit being downconverted at different places but due to orthogonality of the two entangled states.

Now in the setup drawn the photons moving in direction of detector $D_0$ (at location $\vec{x}_1$), defining the "signal photons" are sent through the lens such that all rays in parallel direction are refracted to one point, i.e., there's no which-way information somehow stored in these photons, i.e., you can never say from which slit the original laser photon came and at which spot of the BBO it was down-converted.

In order for the eraser experiment to work, however, the idler photons must be separated depending on the location they came from, and that's what the prism is good for, i.e., for the idler photon, after the prism but before the mirrors and the beam splitter by putting detectors at the two well-separated places you know from which slit the original photon came.

So this situation is described by
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|\psi_1^{(S)},\psi_1^{(I)} \rangle + |\psi_2^{(S)},\psi_2^{(I)} \rangle).$$
Here the superscripts $(S)$ and $(I)$ refer to the idler photons and the subscripts 1 and 2 the source (slit 1 or slit 2 of the original then downconverted photons). These states are such that
$$\langle \psi_1^{(I)}|\psi_2^{(I)} \rangle=0$$
when the idler photons are registered at places after the prism but before the mirrors and the beam splitter.

Then you get
$$\langle \Psi|\Psi \rangle=\frac{1}{2} (\langle \psi_1^{(S)}|\psi_1^{(S)} \rangle \langle \psi_1^{(I)}|\psi_1^{(I)} \rangle + \langle \psi_2^{(S)}|\psi_2^{(S)} \rangle \langle \psi_2^{(I)}|\psi_2^{(I)} \rangle.$$
So you get no double-slit interference fringes at $D_0$ but rather an incoherent addition of the single-slit interference patterns for photons coming from either slit 1 or slit 2. These are slightly shifted against each other due to the distance between the sources, and you get a pretty broad pattern. As seen in the original paper (the PRL published version), in the range $D_0$ was placed this is simply a flat line.

Now take the situation after the mirrors and the beam splitter (for simplicity assuming a symmetric beam splitter). Then the idler photon states get modified in the following way (in your simplified experiment depicted in #1)
$$|\psi_1^{(I)} \rangle \rightarrow \frac{1}{\sqrt{2}} (\mathrm{i} \psi_{D_1}^{(I)} - \psi_{D_2}^{(I)})=| \psi_1^{\prime (I)}\rangle$$
and
$$|\psi_2^{(I)} \rangle \rightarrow \frac{1}{\sqrt{2}} (- \psi_{D_1}^{(I)} + \mathrm{i} \psi_{D_2}^{(i)})=\langle \psi_2^{\prime (I)} \rangle,$$
where I took into account the $\pi/2$ phase shift for each reflection either on the mirror or the beam splitter. That's a unitary transformation (assuming ideal, lossless mirrors and beam splitter).

Now we have for the two-photon state (after some simple algebra)
$$|\Psi ' \rangle=\frac{1}{\sqrt{2}} \left ( |\mathrm{i} \psi_1^{(S)}-\psi_2^{(S)},\psi_{D_1}^{(I)} \rangle + |-\psi_1^{(S)} + \mathrm{i} \psi_2^{(S)},\psi_{D_2}^{(I)} \rangle \right).$$
If you take all the photons again you don't get an interference pattern at $D_0$, and indeed the mirrors and beam splitters only interacted with the idler photon and doesn't affect the signal photon.

Of course, if you now only look at the cases, where $D_1$ registered an idler photon, then you get for the probability to register the signal photon
$$R_{01}=\frac{1}{2} \left [\langle \psi_1^{(S)}|\psi_1^{(S)} \rangle + \langle \psi_2^{(S)}|\psi_2^{(S)} \rangle + \mathrm{i} (\langle \psi_1^{(S)}|\psi_2^{(S)} \rangle-\langle \psi_2^{(S)}|\psi_1^{(S)} \rangle) \right],$$
and if you project to outcomes, when $D_2$ registered an idler photon,
$$R_{02}=\frac{1}{2} \left [\langle \psi_1^{(S)}|\psi_1^{(S)} \rangle - \langle \psi_2^{(S)}|\psi_2^{(S)} \rangle+ \mathrm{i} (\langle \psi_1^{(S)}|\psi_2^{(S)} \rangle-\langle \psi_2^{(S)}|\psi_1^{(S)} \rangle) \right].$$
In both cases you get a two-slit interference pattern due to the interference terms.

 

[Athraxin]

I must however, correct myself concerning the explanation for the no-interference pattern at $D_0$ in the other thread. That was wrong. It's not due to the photons coming from the upper or lower slit being downconverted at different places but due to orthogonality of the two entangled states.

Now in the setup drawn the photons moving in direction of detector $D_0$ (at location $\vec{x}_1$), defining the "signal photons" are sent through the lens such that all rays in parallel direction are refracted to one point, i.e., there's no which-way information somehow stored in these photons, i.e., you can never say from which slit the original laser photon came and at which spot of the BBO it was down-converted.

In order for the eraser experiment to work, however, the idler photons must be separated depending on the location they came from, and that's what the prism is good for, i.e., for the idler photon, after the prism but before the mirrors and the beam splitter by putting detectors at the two well-separated places you know from which slit the original photon came.

So this situation is described by
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|\psi_1^{(S)},\psi_1^{(I)} \rangle + |\psi_2^{(S)},\psi_2^{(I)} \rangle).$$
Here the superscripts $(S)$ and $(I)$ refer to the idler photons and the subscripts 1 and 2 the source (slit 1 or slit 2 of the original then downconverted photons). These states are such that
$$\langle \psi_1^{(I)}|\psi_2^{(I)} \rangle=0$$
when the idler photons are registered at places after the prism but before the mirrors and the beam splitter.

Then you get
$$\langle \Psi|\Psi \rangle=\frac{1}{2} (\langle \psi_1^{(S)}|\psi_1^{(S)} \rangle \langle \psi_1^{(I)}|\psi_1^{(I)} \rangle + \langle \psi_2^{(S)}|\psi_2^{(S)} \rangle \langle \psi_2^{(I)}|\psi_2^{(I)} \rangle.$$
So you get no double-slit interference fringes at $D_0$ but rather an incoherent addition of the single-slit interference patterns for photons coming from either slit 1 or slit 2. These are slightly shifted against each other due to the distance between the sources, and you get a pretty broad pattern. As seen in the original paper (the PRL published version), in the range $D_0$ was placed this is simply a flat line.

Now take the situation after the mirrors and the beam splitter (for simplicity assuming a symmetric beam splitter). Then the idler photon states get modified in the following way (in your simplified experiment depicted in #1)
$$|\psi_1^{(I)} \rangle \rightarrow \frac{1}{\sqrt{2}} (\mathrm{i} \psi_{D_1}^{(I)} - \psi_{D_2}^{(I)})=| \psi_1^{\prime (I)}\rangle$$
and
$$|\psi_2^{(I)} \rangle \rightarrow \frac{1}{\sqrt{2}} (- \psi_{D_1}^{(I)} + \mathrm{i} \psi_{D_2}^{(i)})=\langle \psi_2^{\prime (I)} \rangle,$$
where I took into account the $\pi/2$ phase shift for each reflection either on the mirror or the beam splitter. That's a unitary transformation (assuming ideal, lossless mirrors and beam splitter).

Now we have for the two-photon state (after some simple algebra)
$$|\Psi ' \rangle=\frac{1}{\sqrt{2}} \left ( |\mathrm{i} \psi_1^{(S)}-\psi_2^{(S)},\psi_{D_1}^{(I)} \rangle + |-\psi_1^{(S)} + \mathrm{i} \psi_2^{(S)},\psi_{D_2}^{(I)} \rangle \right).$$
If you take all the photons again you don't get an interference pattern at $D_0$, and indeed the mirrors and beam splitters only interacted with the idler photon and doesn't affect the signal photon.

Of course, if you now only look at the cases, where $D_1$ registered an idler photon, then you get for the probability to register the signal photon
$$R_{01}=\frac{1}{2} \left [\langle \psi_1^{(S)}|\psi_1^{(S)} \rangle + \langle \psi_2^{(S)}|\psi_2^{(S)} \rangle + \mathrm{i} (\langle \psi_1^{(S)}|\psi_2^{(S)} \rangle-\langle \psi_2^{(S)}|\psi_1^{(S)} \rangle) \right],$$
and if you project to outcomes, when $D_2$ registered an idler photon,
$$R_{02}=\frac{1}{2} \left [\langle \psi_1^{(S)}|\psi_1^{(S)} \rangle - \langle \psi_2^{(S)}|\psi_2^{(S)} \rangle+ \mathrm{i} (\langle \psi_1^{(S)}|\psi_2^{(S)} \rangle-\langle \psi_2^{(S)}|\psi_1^{(S)} \rangle) \right].$$
In both cases you get a two-slit interference pattern due to the interference terms.

I thought the similar process I guess. I thought that if we use that (20:80) beam splitter it will change possibilities of photons in one detector, for example in this case R01: will %80 photons from upper slit %20 photons from lower slit, R02: %80 photons from lower slit, %20 photons from upper slit. In this case we won't see anything in D0 because it will still be blurred but we will see two slit interference patterns on D1 And D2. So it will be like the old results. No mystery again.