A QFT & Entanglement: NRQT vs. PCT Transform

[stevendaryl]

Said someone who uses philosophical arguments himself. ;-)) SCNR.

In physics, "philosophy" means "a different philosophy than the one I believe".

 

[akvadrako]

Yeah, that's the possibility I referred to way back in

Yes, I know you got it right away; I was just replying to posts from you and Dr. Chinese and now I think the idea is more clear. I'm not sure what to believe — if you take your complete initial state and you trace out Erik's part, it seems like it leaves a mixed state of the two entangled states. I'm just curious, do you think this scenario needs some non-classical feature (besides the Bell test part), like retrocausality or "unitary collapse" ?

 

[Mentz114]

Yes, I know you got it right away; I was just replying to posts from you and Dr. Chinese and now I think the idea is more clear. I'm not sure what to believe — if you take your complete initial state and you trace out Erik's part, it seems like it leaves a mixed state of the two entangled states. I'm just curious, do you think this scenario needs some non-classical feature (besides the Bell test part), like retrocausality or "unitary collapse" ?

This is how it is done.

From

High-fidelity entanglement swapping with fully independent sources
arXiv:0809.3991v3

I'm hoping that someone can explain how this works. From my understanding, a 50:50 meam splitter with a photon in each input will have both photons in one output IFF the photons are completely in phase. Otherwise they behave independently. The explanation in the paper uses coincidences between detectors at possibly empty outputs.

 

[zonde]

I'm hoping that someone can explain how this works.

If both photons have the same polarization they will end up in the same output of BS (because of HOM interference) and in the same output of PBS (because they have the same polarization). So all these you can throw out as there will be no detections in two separate detectors.
If both photons have different polarization they will be either in the same or different outputs of BS subject to phase difference between $H_2/H_3$ modes vs phase difference between $V_2/V_3$. If phase difference between horizontal modes is the same as phase difference between vertical modes they will end up in the same port (I guess ... because one mode has it's phase flipped), and if relative phases differ by $\pi$ then in different ports. And because they have different polarization they end up in different ports of PBS. So they can detect both photons separately.
That way they single out two states:
$(|H_1 V_4\rangle + |V_1 H_4\rangle) \otimes (|H_2 V_3\rangle + |V_2 H_3\rangle)$
and
$(|H_1 V_4\rangle - |V_1 H_4\rangle) \otimes (|H_2 V_3\rangle - |V_2 H_3\rangle)$

 

[Mentz114]

If both photons have the same polarization they will end up in the same output of BS (because of HOM interference)

Thanks. The beam splitter equation does not include polarization. It is written in terms of photon creation and anihilation operators in the channels.
The HOM effect measures the relative phase - not polarization angle.

Unless the beam-splitter is different. Apart from that your explanation looks OK.

 

[zonde]

Thanks. The beam splitter equation does not include polarization. It is written in terms of photon creation and anihilation operators in the channels.
The HOM effect measures the relative phase - not polarization angle.

HOM interference happens for identical photons (whatever that means, but it certainly means the same polarization). Relative phase is certainly not a factor. Look at Fig.2 in the same paper. There are no fluctuations as relative delay is changed. This is common mistake to expect some phase dependent fluctuations in HOM interference, but there are none.

 

[Mentz114]

HOM interference happens for identical photons (whatever that means, but it certainly means the same polarization). Relative phase is certainly not a factor. Look at Fig.2 in the same paper. There are no fluctuations as relative delay is changed. This is common mistake to expect some phase dependent fluctuations in HOM interference, but there are none.

I think I've worked it out. Normally in the entangled scenario we can write $P(xy|\alpha\beta)=\tfrac{1}{2}P(x|\alpha)P(y|\alpha\beta) + \tfrac{1}{2}P(y|\beta)P(x|\alpha\beta)$ and applying Malus' law we get $P(11\ or\ 00) = \cos(\alpha-\beta)^2$. In the swapping case the information about the $\alpha (\beta)$ setting does not get transferred across from A to B. Instead the polarizations of photons 1 and 4 are projected in the same, or perpendicular polarizations ( ie V or H).

We can write $P(xy|\theta_1\theta_4)=P(x|\theta_1)P(y|\theta_1\theta_4)$

Applying Malus' gives $P(11|\theta_1\theta_4)=\cos(\theta_1-\alpha)^2\cos(\theta_1-\beta)^2$ and $P(00|\theta_1\theta_4)=\sin(\theta_1-\alpha)^2\sin(\theta_1-\beta)^2$ and summing the last two is the probability of a coincidence between A and B. The sum reduces to $\frac{\cos\left( 4\,\theta_1-2\,\alpha-2\,\beta\right) +\cos\left( 2\,\beta-2\,\alpha\right) }{4}+\frac{1}{2}$. For $\alpha=(0, \pi/4),\ \beta=(\pi/8,\ 3\pi/8)$ it turns out that $a+b=\pm(a-b)$ and also for the pairings $(a,b'),\ (a',b),\ (a',b')$ so setting $\theta_1=0$ we can write

$\frac{\cos\left( 4\,\theta_1-2\,\alpha-2\,\beta\right) +\cos\left( 2\,\beta-2\,\alpha\right) }{4}+\frac{1}{2}=\frac{\cos\left( 2\,\beta-2\,\alpha\right) }{2}+\frac{1}{2}=\cos(\alpha-\beta)^2$

So under certain assumptions it is possible for photons 1 and 4 to give the right statistics for total entanglement. This would be apparent if these events could be filtered out. I have not tried to work out what the Q1 and Q2 coincidences would be for this but they no doubt exist.

 

[akvadrako]

What I meant by "basis" is that A&B start out entangled either along the same bases or cross bases.

If it's same, $AB = |HH\rangle + |VV\rangle$, Erik will measure 1, $E = |HH\rangle + |VV\rangle$.
If it's cross, $AB = |HV\rangle + |VH\rangle$, Erik will measure 0, $E = |HV\rangle + |VH\rangle$.

Okay, I think I found a way to make sense of it. The AB system is normally written in the classical basis:

$|H_A + V_A\rangle \otimes |H_B + V_B\rangle$

but for the entanglement swapping experiment, the same/cross basis is more convenient:

$|H_A + V_A\rangle \otimes |same + cross\rangle$

This makes it look just like any standard measurement of entangled qubits. When Erik measures same/cross on his system, of course it collapses the AB system. And just like any fully entangled pair, they should also be able to violate a Bell inequality along that basis.

 

[zonde]

In entanglement swapping experiments polarization relationship is determined between two photons and relative phase relationship is determined between two pairs of polarization modes. Determining just one relationship is not enough to perform entanglement swapping.

One possible problem with a view that Bell state measurement (BSM) just reveals preexisting entanglement is that analysis of swapping is done in particular basis. Say maybe BSM forces photons from different entangled pairs to certain polarization and phase relationships and if it is not done we can't sort the quartets to four necessary bins before measurement of Alice's and Bob's photons.