# Entanglement Swapping

by StevieTNZ
Tags: entanglement, swapping
 PF Gold P: 802 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?
PF Gold
P: 5,374
 Quote by 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?
I think the requirement is that the identities of B and Y are indistinguishable in the sense you don't know which is which.
P: 1,583
 Quote by 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.
Hmm, I wonder how entanglement swapping is dealt with in Fock space, where not only individual particle identity but also particle number is lost.

 PF Gold P: 802 Entanglement Swapping 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?
PF Gold
P: 5,374
 Quote by 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?
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!
 PF Gold P: 802 Even if the entanglement is broken between A and B, because you've measured A?
 PF Gold P: 802 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.
PF Gold
P: 5,374
 Quote by 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.
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.
PF Gold
P: 1,376
 Quote by 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.
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.
 PF Gold P: 802 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?
PF Gold
P: 1,376
 Quote by 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?
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.
PF Gold
P: 802
 Quote by 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.
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?
PF Gold
P: 1,376
 Quote by 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?
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.
PF Gold
P: 802
 Quote by 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.
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?
PF Gold
P: 1,376
 Quote by StevieTNZ When we detect HV and VH, would that when four-photon entanglement has NOT been created successfully?
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?

 Quote by StevieTNZ 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)?
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.
PF Gold
P: 802
 Quote by 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.
So in this case, there'd be no entanglement between 1 and 4?

 Quote by zonde And what do you mean by four-photon entanglement?
Four photons entangled.

 Quote by zonde No. Besides in this experiment it is explained as creation of entanglement between photons 1 and 4.
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.
PF Gold
P: 1,376
 Quote by StevieTNZ So in this case, there'd be no entanglement between 1 and 4?
Right

 Quote by StevieTNZ 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.
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.