Beam splitter in an entanglement swapping experiment

[Mentz114]

I have a misunderstanding about the beam splitter in this experiment

High fidelity entanglement swapping with fully independent sources
Rainer Kaltenbaek, Robert Prevedel, Markus Aspelmeyer,Anton Zeilinger
http://arxiv.org/pdf/0809.3991.pdf

Part of the diagram is reproduced here

Two photons ( 2 and 3 ) arrive at BS and coincidences between the four counters ##D_{Q1H}## to ##D_{Q2V}## are detected.
The authors say "A two-fold coincidence detection event between either DQ1H and DQ2V or DQ1V and DQ2H indicates ...". But this requires that one photon should be present in each output beam. I don't see how this can happen if they are expecting HOM interference.

What am I missing ?

 

[kurt101]

The authors say "A two-fold coincidence detection event between either DQ1H and DQ2V or DQ1V and DQ2H indicates ...". But this requires that one photon should be present in each output beam. I don't see how this can happen if they are expecting HOM interference.

Why exactly do you say that "this requires that one photon be present in each output beam"? Are you referring to the use of the word "two-fold coincidence"? Perhaps they use the term "two-fold" to mean timing and state.

 

[Mentz114]

Why exactly do you say that "this requires that one photon be present in each output beam"? Are you referring to the use of the word "two-fold coincidence"? Perhaps they use the term "two-fold" to mean timing and state.

If a coincidence means that detectors DQ1H and DQ2V both register - that requires one photon in each arm, surely ?

 

[kurt101]

If a coincidence means that detectors DQ1H and DQ2V both register - that requires one photon in each arm, surely ?

In the picture you posted there is 4 outputs and there is 2 inputs. If there is entanglement then a coindence will be detected between 2 of the outputs (i.e. either between DQ1H and DQ2V or DQ1V and DQ2H). So given that there are 2 inputs, why exactly do you find it unexpected that the setup is looking for 2 coincidental photons in the 4 outputs?

 

[Mentz114]

In the picture you posted there is 4 outputs and there is 2 inputs. If there is entanglement then a coindence will be detected between 2 of the outputs (i.e. either between DQ1H and DQ2V or DQ1V and DQ2H). So given that there are 2 inputs, why exactly do you find it unexpected that the setup is looking for 2 coincidental photons in the 4 outputs?

If two photons (coherent ?) arrive in a small time interval at the inputs of a beam splitter then the outputs are either empty or contain both photons. Look up the theory and tell me if I'm wrong. See Gerry&Knight 'Introductory modern optics' chapter 6, for example.

 

[StevieTNZ]

If a coincidence means that detectors DQ1H and DQ2V both register - that requires one photon in each arm, surely ?

After the first beam splitter, yes.

But there will be cases where both input photons go the same path at the first beam splitter. That's typical of such experiments - if that occurs, then no entanglement swapping occurs.

 

[Mentz114]

The problem of how to treat the beam splitter only affects the size of the selected sample. The experiment is actually easy to explain.

The probablity of getting a detection coincidence at D1 and D4 is given by ##P(11|\alpha, \beta)=\cos(\theta_1-\alpha)^2\cos(\theta_2-\beta)^2##. if ##\theta_1## and ##\theta_2## are both projected into ##\alpha## or both into ##\beta## then ##P(11|\alpha, \beta)=\cos(\beta-\alpha)^2##. So
[tex]
\begin{align*}
P(11|\alpha, \beta) &= \cos(\theta_1-\alpha)^2\cos(\theta_2-\beta)^2\ \Rightarrow \cos(\alpha-\beta)^2\ \ needs\ (\alpha\lor \beta)\\
P(10|\alpha, \beta) &= \cos(\theta_1-\alpha)^2\sin(\theta_2-\beta)^2\ \Rightarrow \sin(\alpha-\beta)^2\ \ needs\ (\alpha\lor \beta+\pi/2)\\
P(01|\alpha, \beta) &= \sin(\theta_1-\alpha)^2\cos(\theta_2-\beta)^2\ \Rightarrow \sin(\alpha-\beta)^2\ \ needs\ (\beta\lor \alpha+\pi/2)\\
P(00|\alpha, \beta) &= \sin(\theta_1-\alpha)^2\sin(\theta_2-\beta)^2\ \Rightarrow \cos(\alpha-\beta)^2\ \ needs\ (\alpha+\pi/2\lor \beta+\pi/2)
\end{align*}
[/tex]
The values in parentheses at end of the lines are the projections of ##\theta_1## and ##\theta_2## that give the value after the arrow.
This shows that if the random variates ##\theta_1## and ##\alpha## and ##\theta_2## and ##\beta## could act exactly like each other then the coincidence probabilities at D1 and D4 would be those of the symmetrically entangled singlet state.

This can be arranged by giving PBS1 the settings a, a' at random, and PBS 2 the settings b, b' at random. These are the settings of Pol 1 and Pol 4 which are set independently at random. There is no correllation between any of the settings in each run. The entanglement is expressed by projecting the partners of photon 2 and photon 3 into the same state when photon 2 and photon 3 pass PBS1 or PBS2.

Now if we only select those runs where the (setting of PBS1=setting of Pol1) and (setting of PBS2 = setting of Pol4) and (theta1=theta2) the selected sample shows the singlet state statistics (!) therefore violating the CHSH limit at the same values as the experiment cited in post#1.

This can be emulated in code and as predicted gives an average CHSH value of ##2.83\ \pm\ 0.06## with 70 instances of 32000 repetitions.

@DrChinese @StevieTNZ @kurt101

 

[StevieTNZ]

You raise an interesting issue, with regards to the beam splitter and the Hong–Ou–Mandel effect. I'll let others take it from this point as I'm not quite the Mathematician so lack the ability to work through your Maths.

 

[zonde]

But this requires that one photon should be present in each output beam. I don't see how this can happen if they are expecting HOM interference.

HOM interference can be observed only when both photons are identical. When two photons have orthogonal polarization states they are different and can appear at the same output or different outputs with the same probability.
So if you look at this comment below HOM interference graph there is information missing.
"FIG. 2: Hong-Ou-Mandel (HOM) interference scan. The fourfold coincidences are measured between the detectors D1, D4 and two detectors placed directly behind the fiber beam splitter in the BSM (see Figure 1). These are plotted over the relative time delay of the interfering photons, which is determined via cross correlation of the laser pulses. Background has not been subtracted. Error bars indicate s.d. The solid line is a Gaussian fit to the data (χ 2 ∼ 0.85) with a visibility of 96 ± 1 %."
Analyzers of D1 and D4 should be set to the same polarization angle and only then one can observe HOM dip. This in turn explains how you can observe photons 2 and 3 at different BS outputs when D1 and D4 have different settings.

 

[Mentz114]

HOM interference can be observed only when both photons are identical. When two photons have orthogonal polarization states they are different and can appear at the same output or different outputs with the same probability.
So if you look at this comment below HOM interference graph there is information missing.
"FIG. 2: Hong-Ou-Mandel (HOM) interference scan. The fourfold coincidences are measured between the detectors D1, D4 and two detectors placed directly behind the fiber beam splitter in the BSM (see Figure 1). These are plotted over the relative time delay of the interfering photons, which is determined via cross correlation of the laser pulses. Background has not been subtracted. Error bars indicate s.d. The solid line is a Gaussian fit to the data (χ 2 ∼ 0.85) with a visibility of 96 ± 1 %."
Analyzers of D1 and D4 should be set to the same polarization angle and only then one can observe HOM dip. This in turn explains how you can observe photons 2 and 3 at different BS outputs when D1 and D4 have different settings.

Yes. I agree on all counts. I inferred that their HOM dip figures come from a separate experiment.
I also think the write-up is not very clear. They should at some point have written out the probabilities ( did I miss that ?)
However, a very edifiying experiment.

 

[zonde]

However, a very edifiying experiment.

Certainly. Entanglement swapping using two independent lasers is very interesting.

And I like your question. Sometimes it seems people just go over such things - sort of QM is weird so anything can happen, nothing to wonder about.

 

[StevieTNZ]

HOM interference can be observed only when both photons are identical. When two photons have orthogonal polarization states they are different and can appear at the same output or different outputs with the same probability.

Thank you for clarifying this!

 

[Mentz114]

This is a correction - I just noticed a missing factor of 1/2 in the probabilities in post#7. So
##\cos(\theta_1-\alpha)^2\cos(\theta_2-\beta)^2\ \Rightarrow \cos(\alpha-\beta)^2##
should read
##\tfrac{1}{2}\cos(\theta_1-\alpha)^2\cos(\theta_2-\beta)^2\ \Rightarrow \tfrac{1}{2}\cos(\alpha-\beta)^2##
and so on. Apologies for my carelessness.