Entanglement Swapping

StevieTNZ

Hi there,

If we set-up an experiment for entanglement swapping, the two entangled pairs of photons are:
Group 1: A, B
Group 2: X, Y

Take B and Y and perform a Bell-state measurement on them.
Obviously A and X are now entangled. But for this to occur, must B and Y have no definite polarisation?

DrChinese

I think the requirement is that the identities of B and Y are indistinguishable in the sense you don't know which is which.

lugita15

Hmm, I wonder how entanglement swapping is dealt with in Fock space, where not only individual particle identity but also particle number is lost.

StevieTNZ

Okay:

Because if we measure particles A and X, the entanglement with B and Y is lost. So when you go to entangle B and Y, because no entanglement exists between A and B, and X and Y, A and X won't become entangled?

When we talk of teleportation, is it when we measure A, Y takes on the same polarisation?

DrChinese

Just remember that the *ordering* does not matter in a swapping setup. You can measure A and X before you swap the entanglement via B and Y and the results are the same. Strange but true!

StevieTNZ

Even if the entanglement is broken between A and B, because you've measured A?

StevieTNZ

Okay, I've read this paper: http://www.univie.ac.at/qfp/publicat...es/2001-06.pdf

And I am lost!!

So what I understand: photons 1 and 4 show entanglement correlations; i.e. both are HH or VV, never seeing HV or VH.

"This is the so-called entanglement swapping , which can also be seen as teleportation either of the state of photon 2 over to photon 4 or of the state of photon 3 over to photon 1."
This is where I am confused, especially if photons 1 and 4 already have polarisations. Teleporting the state of photon 3 to 1, etc. in this case makes my brain fry.

DrChinese

Keep in mnd that they 1 and 4 photons did not have a well defined polarization when the swapping occurred. Now keep these 2 additional points in your head: forget the ordering, it doesn't matter. And remember that swapping only occurs sometimes, just when a certain results occurs with 2 and 3. I.e. a subset.

zonde

Correlations for photons detected in detectors 1 and 4 show up only when you condition them on detections in detectors 2 and 3. It's very similar to correlations in delayed choice quantum eraser.
I wouldn't say that teleportation is very good word for entanglement swapping.

StevieTNZ

If we implement the same experimental set-up, and measure photons 1 and 4 before measurement occurs on photons 2 and 3 - I assume when the polarising beam splitter combines particles 2 and 3 this is when the entanglement switch occurs. So if we detect photons 1 and 4 in both H (or V polarisations), prior to reaching the filters and detectors photons 2 and 3 take on V (or H polarisations) - the opposite of what the others take?

zonde

Ok, we should detect only HH or VV in detectors 1 and 4 for particular type of correlations observed in that experiment. But we detect HV and VH as well and at the same rate as HH and VV.

So how they disappear when we perform Bell state measurement? We throw them out because their pair photons reach the same detector (either both reach detector 2 or detector 3) and so we don't have 4-fold coincidence in 1,2,3 and 4.

StevieTNZ

I see.

If we analyse photon 3 in the 45 basis, according to QM it gets entangled with the filter so it both passes and fails at the same time. If we then measure photons 1 and 4, so they show VV or HH, will photon 4 take the opposite polarisation of photons 1 and 4, or will it written in the 45 basis so take on V or H, without dependence on what photon 1 takes?

zonde

I am not sure I understand what measurement settings has polarization analyzer before detector 2. It seems you have mixed up detector numbering somewhere.

Anyways you get entanglement for photons 1 and 4 only (in 4-fold coincidences) when you measure photons 2 and 3 in +/- (+45/-45) basis after PBS.

StevieTNZ

When we detect HV and VH, would that when four-photon entanglement has NOT been created successfully?


For entanglement swapping: is entanglement created between photons 2 and 3 when they hit the beam splitter (in this experiment scheme: http://www.univie.ac.at/qfp/publicat.../2001-06.pdf)?

Is entanglement swapping the same thing as creating four-photon entanglement?

zonde

When we detect HV in 1 and 4 we have two photons in 3 and no photons in 2.
When we detect VH in 1 and 4 we have two photons in 2 and no photons in 3.

And what do you mean by four-photon entanglement?

No. Besides in this experiment it is explained as creation of entanglement between photons 1 and 4. But basically you can take any two detectors, perform measurements in +/- basis and claim that 4-fold coincidence reveals entanglement between the other two.

So entanglement swapping happens when you register 4-fold coincidence in coincidence counter i.e. when filtering is performed.

StevieTNZ

So in this case, there'd be no entanglement between 1 and 4?

Four photons entangled.


Entanglement swapping is NOT a creation of entanglement between photons 1 and 4?

If I measured photon 2 in 45 basis and 3 in 135 basis, wouldn't that indicate whether the photons were described by the GHZ state? If one passes, the other fails. From the GHZ state, you can't have photon 2 in 45 and photon 3 in 135.

zonde

Right

Wrong

If you measure photon 2 in 45 basis and 3 in 135 basis then photons 1 and 4 will have positive correlation for +45 and +135 measurement and negative correlation (minimal rate of four-fold coincidences) for +45 and +45 measurement.