Indistinguishability and entanglement

[Dorilian]

I have read many papers about polarization entangled photons, but in all of them, the photons need to be indistinguishable in time. But, what about the other degrees of freedom? They need to have the same frequency, energy, angular momentum? So, do they need to be indistinguishable in all degree of freedom to be entangled?

 

[A. Neumaier]

I have read many papers about polarization entangled photons, but in all of them, the photons need to be indistinguishable in time. But, what about the other degrees of freedom? They need to have the same frequency, energy, angular momentum? So, do they need to be indistinguishable in all degree of freedom to be entangled?

One averages over all degrees of freedom that are deemed irrelevant for the experiments.

Particles cannot be distinguishable in some respects and indistinguishable in others. Once the are known to differ in at least one respect they are distinguishable.

 

[Dorilian]

One averages over all degrees of freedom that are deemed irrelevant for the experiments.

Particles cannot be distinguishable in some respects and indistinguishable in others. Once the are known to differ in at least one respect they are distinguishable.

Ok, but for example, in the Aspect experiments, the two photons emitted by the atomic radiation have different wavelength but they are entangled in polarization. What happen in that case?

 

[DrChinese]

Hmmm, are we talking about the same things?

Momentum & polarization entangled photons can be measured as to their polarization, I believe, without necessarily collapsing their momentum entanglement. And polarization entangled photons are not required to have the same wavelength.

 

[DrChinese]

I have read many papers about polarization entangled photons, ...

By the way, welcome to PhysicsForums, Dorilian!

 

[A. Neumaier]

Ok, but for example, in the Aspect experiments, the two photons emitted by the atomic radiation have different wavelength but they are entangled in polarization. What happen in that case?

How do you know that they have different wavelengths?

And don't you confuse being indistinguishable and being entangled?

 

[Dorilian]

How do you know that they have different wavelengths?

And don't you confuse being indistinguishable and being entangled?

In Aspect paper (http://prl.aps.org/abstract/PRL/v47/i7/p460_1 ) they mention that one photon is emitted at 551.3 nm and the second photon at 422.7nm.

My problem is that I don´t understand why we need temporal indistinguishability. In other type of experiments using PDC processes, they have to compensate the temporal walk-off of one of the photons to avoid the temporal distinguishability.

 

[A. Neumaier]

In Aspect paper (http://prl.aps.org/abstract/PRL/v47/i7/p460_1 ) they mention that one photon is emitted at 551.3 nm and the second photon at 422.7nm.

This means that the photons are distinguishable. But this has nothing to do with entanglement. Entangled particles are essentially always distinguishable. If you only measure polarization, the labels distinguishing the particles don't matter.

My problem is that I don´t understand why we need temporal indistinguishability. In other type of experiments using PDC processes, they have to compensate the temporal walk-off of one of the photons to avoid the temporal distinguishability.

Please put this into context by citing a corresponding source, so that I can see what you mean.

 

[Dorilian]

This means that the photons are distinguishable. But this has nothing to do with entanglement. Entangled particles are essentially always distinguishable. If you only measure polarization, the labels distinguishing the particles don't matter.

Ok, that makes sense. Thanks, I only need to understand why weed need the temporal indistinguishability.

Please put this into context by citing a corresponding source, so that I can see what you mean.

For example here in Kwiat paper (http://prl.aps.org/abstract/PRL/v75/i24/p4337_1) in Figure 2 they need to compensate the walk-off effect due to the birefringence of the crystal.

 

[A. Neumaier]

For example here in Kwiat paper (http://prl.aps.org/abstract/PRL/v75/i24/p4337_1) in Figure 2 they need to compensate the walk-off effect due to the birefringence of the crystal.

They compensate for the different velocities, so that when the signals are measured there is no retardation of one photon relative to the other.

They call this ''temporal indistinguishability'' - but it has nothing to do with the kind of indistinguishability that results in symmetrization or antisymmetrization of the wave function. A less confusing term would have been ''temporal synchronization''.

 

[Dorilian]

They compensate for the different velocities, so that when the signals are measured there is no retardation of one photon relative to the other.

They call this ''temporal indistinguishability'' - but it has nothing to do with the kind of indistinguishability that results in symmetrization or antisymmetrization of the wave function. A less confusing term would have been ''temporal synchronization''.

Ok, that's better. Well, thank you for your help. I have another question but I would like to put it in another thread.

 

[zonde]

I have read many papers about polarization entangled photons, but in all of them, the photons need to be indistinguishable in time. But, what about the other degrees of freedom? They need to have the same frequency, energy, angular momentum? So, do they need to be indistinguishable in all degree of freedom to be entangled?

To understand better this requirement for indistinguishability it helps to understand the difference between non commuting polarization measurements.
When you use PDC source for polarization entangled photons you know in what base photons are created. So you can separately look at polarization measurements in H/V basis and +45°/-45° basis.
Now the indistinguishability is requirement only for measurements in +45°/-45° base but is not needed for measurements in H/V basis. That means that when photons are "distinguishable" correlation reduces to classical product state (interference term disappears).

So in more details this measurement in +45°/-45° base looks like that:
You have two superpositions - photon pair can be H(Alice)/V(Bob) polarized or V(Alice)/H(Bob) polarized (I assume type II PDC source).
So when Alice makes measurement in +45°/-45° base H and V superpositions can interfere as both of them can pass with 50% probability through polarizer at +-45°. This measurement should correlate with similar measurement performed by Bob. That will happen if we adjust relative phase between H and V superpositions for maximum interference after polarizer. This phase adjustment is done with walk-off compensators.

As to your question if they need the same frequency - no, as long as H/V interference at Alice is symmetrical to V/H interference at Bob and we can adjust relative phase between two modes for maximum interference.

And if you are interested in experiments - I find this experiment rather fascinating for better understanding of this indistinguishability condition. However it is a bit more complicated than in photon Bell tests.
http://physics.nist.gov/Divisions/Div844/publications/migdall/psm96_twophoton_interference.pdf"

 

[Dorilian]

To understand better this requirement for indistinguishability it helps to understand the difference between non commuting polarization measurements.
When you use PDC source for polarization entangled photons you know in what base photons are created. So you can separately look at polarization measurements in H/V basis and +45°/-45° basis.
Now the indistinguishability is requirement only for measurements in +45°/-45° base but is not needed for measurements in H/V basis. That means that when photons are "distinguishable" correlation reduces to classical product state (interference term disappears).

So in more details this measurement in +45°/-45° base looks like that:
You have two superpositions - photon pair can be H(Alice)/V(Bob) polarized or V(Alice)/H(Bob) polarized (I assume type II PDC source).
So when Alice makes measurement in +45°/-45° base H and V superpositions can interfere as both of them can pass with 50% probability through polarizer at +-45°. This measurement should correlate with similar measurement performed by Bob. That will happen if we adjust relative phase between H and V superpositions for maximum interference after polarizer. This phase adjustment is done with walk-off compensators.

As to your question if they need the same frequency - no, as long as H/V interference at Alice is symmetrical to V/H interference at Bob and we can adjust relative phase between two modes for maximum interference.

And if you are interested in experiments - I find this experiment rather fascinating for better understanding of this indistinguishability condition. However it is a bit more complicated than in photon Bell tests.
http://physics.nist.gov/Divisions/Div844/publications/migdall/psm96_twophoton_interference.pdf"

Ok, I am going to take a look at that paper. Quick question... all the pair of photons created by PDC are also entangled in time-energy?

 

[Dorilian]

They compensate for the different velocities, so that when the signals are measured there is no retardation of one photon relative to the other.

They call this ''temporal indistinguishability'' - but it has nothing to do with the kind of indistinguishability that results in symmetrization or antisymmetrization of the wave function. A less confusing term would have been ''temporal synchronization''.

Ok, but, why is necessary that temporal synchronization?? Is it important for the entanglement? I understand that, I can measure the polarization on one photon and in that moment I know the polarization of the other photon, which I can measure in a later time, so I don´t understand why we need the temporal synchronization.

 

[A. Neumaier]

Ok, but, why is necessary that temporal synchronization?? Is it important for the entanglement? I understand that, I can measure the polarization on one photon and in that moment I know the polarization of the other photon, which I can measure in a later time, so I don´t understand why we need the temporal synchronization.

Entangled photon pair statistics depends on coincidence measurements, which is ruined by delays.

 

[zonde]

Ok, but, why is necessary that temporal synchronization?? Is it important for the entanglement? I understand that, I can measure the polarization on one photon and in that moment I know the polarization of the other photon, which I can measure in a later time, so I don´t understand why we need the temporal synchronization.

Temporal synchronization between two sites is not necessary. You can write in file time tags for detector clicks and find pairs by comparing tags from two files. There are experiments that where performed that way. For such a scenario you just have to establish correspondence between time tags from two files.

Temporal indistinguishability means something else. Different modes (or superpositions) should overlap (in time) in one arm of experimental setup i.e. localy the same way as corresponding modes overlap in other arm.

 

[kof9595995]

One averages over all degrees of freedom that are deemed irrelevant for the experiments.

Particles cannot be distinguishable in some respects and indistinguishable in others. Once the are known to differ in at least one respect they are distinguishable.

I thought if the particles species are the same then they are indistinguishable. For example, if I have a electron spin singlet, consisting of a spin up and a spin down, are they distinguishable or not? The notation |+->-|-+> seems to suggest indistinguishable, but according to you they'd be distinguishable?

 

[DrChinese]

Entangled photon pair statistics depends on coincidence measurements, which is ruined by delays.

I think the issue here is that you must be able to pair up the Alice and Bob streams. You can in fact observe them at different times as long as you can match them up. To be clear: if you sent Alice through a 1 meter optical fiber and Bob through 1000 meters of fiber, they can still be matched and they will be entangled.

 

[Dorilian]

I think the issue here is that you must be able to pair up the Alice and Bob streams. You can in fact observe them at different times as long as you can match them up. To be clear: if you sent Alice through a 1 meter optical fiber and Bob through 1000 meters of fiber, they can still be matched and they will be entangled.

So, why do we need to make them coincidence?

 

[DrChinese]

So, why do we need to make them coincidence?

In what sense do you use the word "coincidence" ?

Clearly, we want to match up Alice1 with Bob1, Alice2 with Bob2, etc. Based on estimated travel time from the source to the detector, fine grained time stamps are used to record each particle as + or -. There are roughly about 1000 per second for these setups. If you match up the wrong pair, you get meaningless results.

 

[Dorilian]

In what sense do you use the word "coincidence" ?

Clearly, we want to match up Alice1 with Bob1, Alice2 with Bob2, etc. Based on estimated travel time from the source to the detector, fine grained time stamps are used to record each particle as + or -. There are roughly about 1000 per second for these setups. If you match up the wrong pair, you get meaningless results.

I mean that the two photons arrive at the same time at each detector. Ok, I think that I got it, so, the coincidence time is useful to avoid a mismatch in the correlations.

 

[DrChinese]

I mean that the two photons arrive at the same time at each detector. Ok, I think that I got it, so, the coincidence time is useful to avoid a mismatch in the correlations.

Yes, and you can synchronize this a variety of ways as long as you match properly. What you really look for is time delta between the various apparati.