What does a Bell measurement mean in quantum teleportation?

  • #1

Main Question or Discussion Point

Hi,

I have a question regarding quantum teleportation. I understand that Alice starts of with a photon A whose state she wants to teleport. Alice and Bob share an entangled pair of photons, B and C.

Then Alice does something called a Bell measurement which entangles A and B and the result of this Bell measurement for Alice is one of the Bell states.

After this, she sends the results of which Bell state to Bob who can use the appropriate unitary transform on C to complete the teleportation.

What exactly does Alice do in the Bell measurement step? I can follow the mathematics but I don't understand what one would do in a lab if I was Alice and I had the two photons. Most searches for how to entangle photons bring me to SPDC which is obviously not what is going on here. Thank you!
 

Answers and Replies

  • #2
DrChinese
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Hi,

I have a question regarding quantum teleportation. I understand that Alice starts of with a photon A whose state she wants to teleport. Alice and Bob share an entangled pair of photons, B and C.

Then Alice does something called a Bell measurement which entangles A and B and the result of this Bell measurement for Alice is one of the Bell states.

After this, she sends the results of which Bell state to Bob who can use the appropriate unitary transform on C to complete the teleportation.

What exactly does Alice do in the Bell measurement step? I can follow the mathematics but I don't understand what one would do in a lab if I was Alice and I had the two photons. Most searches for how to entangle photons bring me to SPDC which is obviously not what is going on here. Thank you!
The Bell State Measurement collapses the wave function for 2 (sometimes called B & C) of the 4 particles, and has the potential to place the remaining 2 (sometimes called A & D) into an entangled (via swapping) state. This of course assumes that A & B started out entangled, and C & D started out entangled.

Keep in mind that A & D do NOT end up entangled in every case. Only some outcomes of the BSM lead to such swapped entanglement. This occurs randomly.
 
  • #3
Hi DrChinese,

My question is what actual operation constitutes a Bell State Measurement i.e. I want to know what is going on from an experimental point of view. What does Alice do in a lab with her pair of photons (B and C) that results in a possible entanglement swapping?

Thank you.
 
  • #4
DrChinese
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Hi DrChinese,

My question is what actual operation constitutes a Bell State Measurement i.e. I want to know what is going on from an experimental point of view. What does Alice do in a lab with her pair of photons (B and C) that results in a possible entanglement swapping?

Thank you.
Ah. Have you checked out this paper?

http://arxiv.org/abs/quant-ph/0201134

Experimental Nonlocality Proof of Quantum Teleportation and Entanglement Swapping
Thomas Jennewein, Gregor Weihs, Jian-Wei Pan, Anton Zeilinger (2002)

"Quantum teleportation strikingly underlines the peculiar features of the quantum world. We present an experimental proof of its quantum nature, teleporting an entangled photon with such high quality that the nonlocal quantum correlations with its original partner photon are preserved. This procedure is also known as entanglement swapping. The nonlocality is confirmed by observing a violation of Bell's inequality by 4.5 standard deviations. Thus, by demonstrating quantum nonlocality for photons that never interacted our results directly confirm the quantum nature of teleportation."

There is a diagram on page 9 (somewhat confusingly called Alice). As best I recall: the Psi- state is identified as being clicks in both Alice's D1 and D2 detectors. That is after they go through the same beam splitter, meaning they are measured at the same relative angle (I don't believe it matters what angle it is at). There is a polarization control prior to that for the 2 photon, I believe that simply rotates the polarization in a fashion similar to a wave plate - it does not collapse the polarization wave function.

If Alice see Psi-, this means that the other 2 photons (here identified as 0 and 3) are also in the Psi- state. That means they will be oppositely entangled (assuming I have all the signs correct). Thus they will violate a Bell Inequality.

There must be 4-fold coincidence for each trial that is to be considered (2 for Alice and 2 for Bob).
 
  • #5
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Ah. Have you checked out this paper?

http://arxiv.org/abs/quant-ph/0201134

Experimental Nonlocality Proof of Quantum Teleportation and Entanglement Swapping
Thomas Jennewein, Gregor Weihs, Jian-Wei Pan, Anton Zeilinger (2002)[..]
That's a very interesting paper - thanks!
 
  • #6
DrChinese
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Here is another which has a diagram, also a breakthrough paper (featuring independent pair sources):

http://arxiv.org/abs/0809.3991

High-fidelity entanglement swapping with fully independent sources
Rainer Kaltenbaek, Robert Prevedel, Markus Aspelmeyer, Anton Zeilinger (2008)

"Entanglement swapping allows to establish entanglement between independent particles that never interacted nor share any common past. This feature makes it an integral constituent of quantum repeaters. Here, we demonstrate entanglement swapping with time-synchronized independent sources with a fidelity high enough to violate a Clauser-Horne-Shimony-Holt inequality by more than four standard deviations. The fact that both entangled pairs are created by fully independent, only electronically connected sources ensures that this technique is suitable for future long-distance quantum communication experiments as well as for novel tests on the foundations of quantum physics. "
 
  • #7
Hi DrChinese,

That second paper was excellent! I think I understand how it works but I am still a little unclear on a couple of details.

I know that the arrangement of a beam splitter followed by two polarizing beam splitters can distinguish between the [itex]\psi^{-}[/itex] and [itex]\psi^{+}[/itex] states. That is, if we send in photons and they are already entangled in either [itex]\psi^{-}[/itex] or [itex]\psi^{+}[/itex], we can distinguish between the two Bell states by checking which detectors gave us coincident clicks.

Now, in the experiment, what we are claiming is the converse right? Let's assume that on one of the runs, we got coincident clicks on detectors [itex]D_{Q1H}[/itex] and [itex]D_{Q2V}[/itex]. But photons 2 and 3 are not even entangled when they go into the BSM box. Are we then claiming that because we got coincident clicks on [itex]D_{Q1H}[/itex] and [itex]D_{Q2V}[/itex], we have somehow entangled them into a [itex]\psi^{-}[/itex] state? I can't see why that is true.

How are we going from the results of a single measurement to conclude that we have collapsed the state of photons 2 and 3 into an entangled Bell state?

Thank you for your help! I'm very grateful for your efforts in finding easy to read papers for a relative newbie like me.
 
  • #8
Sorry about that last post, I think I get it now. I should decompose the state of photons 2 and 3 into the Bell basis since that is an orthonormal basis for any two qubit state. Coincident clicks on [itex]D_{Q1H}[/itex] and [itex]D_{Q2V}[/itex] cannot happen with [itex]\phi^{±}[/itex] or [itex]\psi^{+}[/itex]. So the state must be [itex]\psi^{-}[/itex].

I think that's right. And it is really cool now that I know how teleportation works. If I am wrong, do let me know! Thank you :)
 
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