# I Are these photons entangled?

1. Feb 4, 2017

### edguy99

Photons are prepared in a vertical state (90º), split in a crystal and sent to Bob and Alice. Both Bob and Alice will measure 100% vertical and 0% horizontal.

Photons are prepared in a 45º state, split in a crystal and sent to Bob and Alice. Both Bob and Alice will measure 50% vertical and 50% horizontal. This can be calculated by applying a matrix to the Jones vector. Are these photons entangled and are Bob and Alice guaranteed the same reading for each photon? Is simply splitting the photon enough to cause entanglement or is a more complex process needed to ensure entanglement?

2. Feb 4, 2017

### DrChinese

Splitting photons using parametric down conversion does not automatically entangle them on the polarization basis. There are several Types of crystal (usually called Type I or Type II and there are some variations of these), each with somewhat different properties in this respect.

Also, there are things that must be done during the collection process following splitting. Again, these vary somewhat by Type. Generally, there are specific exit angles (regions) for harvesting entangled pairs - and if you collect outside that, the efficiency drops precipitously.

If you haven't seen these before, I would recommend them to provide additional insight:

https://arxiv.org/pdf/quant-ph/0103168v1.pdf
http://xqp.physik.uni-muenchen.de/publications/files/articles_2001/journmodopt_48_1997.pdf

3. Feb 4, 2017

### Strilanc

If you're using the same crystal in both cases, then it can't be true that both cases are fully entangled. That would imply the crystal was able to copy both vertically and diagonally, but measuring/controlling-based-on those two directions doesn't commute so in effect the crystal would be a polarization cloning machine. It would violate the no-cloning theorem.

Furthermore, entangled states never give 100% results on any measurement done on half of the system. So the first case can't be entangled. You need to some combined measurement of the entire system to find a property that's always the same (e.g. Bob=H could always match Alice=V).

The second case could be entangled or it could just be that you're sending separate diagonally-polarized photons to each party. Alice and Bob would have to compare notes and see if there was correlation between their H-vs-V results or not to tell the difference. And to eliminate the case where the sender is simply alternating between "H to Alice, V to Bob" and "V to Bob, H to Alice", Alice and Bob should also try checking for correlations when measuring in different directions. To be really sure there's entanglement present, they should try to use the photons win the CHSH game more than 75% of the time.

Last edited: Feb 4, 2017
4. Feb 7, 2017

### edguy99

A very important point. From your first reference: https://arxiv.org/pdf/quant-ph/0103168v1.pdf
There is, however, a common problem: the entangled photon pairs have 50% chances of leaving at the same output ports of the beamsplitter. Therefore, the state prepared after the beamsplitter may not be considered as a Bell state without amplitude post-selection as pointed out by De Caro and Garuccio [19]. Only when one considers the coincidence contributing terms by throwing away two out of four amplitudes (post-selection of 50% of the amplitudes), the state is then considered to be a Bell state

My concern is that many (all?) experiments use some kind of post selection algorithm, where they only consider entangled photons to be the ones that arrive with the same polarization at the same time. This presupposes the notion (required for a Bell test) that entangled photons will "always" measure the same, even off of their basis vectors.

In the Loophole-free Bell inequality violation using electron spins separated by 1.3 kilometres experiment http://www.nature.com/nature/journal/v526/n7575/full/nature15759.html, all mismatched photons (when α = β) are not considered entangled and are not used. Consider their selection criteria in this image at D).

Last edited by a moderator: May 8, 2017
5. Feb 7, 2017

### Strilanc

I think that sometimes people do misuse post-selection in their experiments, but I don't think of the Hensen et al. experiment as an example of that. They're picking out a subset of the games to keep, but the event that determines keep-vs-toss is spacelike separated from the "player" events. Strictly speaking it's not a CHSH game, but the changed game still has a Bell inequality to violate and they violated it.

The reason they expected to violate the inequality involved entanglement and post-selection, but the actual measured violations didn't depend on those assumptions. The classical limits on how often you can win don't depend on anything about entanglement and can't be perturbed by a spacelike separated choice to keep-or-toss.

Last edited: Feb 7, 2017
6. Feb 7, 2017

### DrChinese

You are mixing 2 different concepts. Theory says one thing, and that is used to tune the experimental model. Nothing wrong with that. And what they tune is both the apparatus and a time window (if pairs are desired).

Once they have that, they do NOT assume the polarization is the same. They measure it and compare to theoretical predictions. So basically I do not see where there is anything to object to.

7. Feb 7, 2017

### edguy99

I disagree. If two photons with with different polarization's arrived within the same timeslot, that event would not be used.

From the article:
If the photons are indistinguishable in all degrees of freedom, then the observation of one early and one late photon in different output ports projects the spins at A and B into the maximally entangled state , where ms = 0 ≡ |↑ and ms = −1 ≡ |↓ . These detections herald the successful preparation and play the role of the event-ready signal in Bell’s proposed set-up.

8. Feb 7, 2017

### Strilanc

I meant I didn't think it was an example of misuse of post-selection. Obviously they do use post-selection.

9. Feb 7, 2017

### edguy99

It looks like they are assuming the polarization is the same since they are excluding all polarizations that are mismatches (no matter how caused) from the experiment.

10. Feb 7, 2017

### DrChinese

This is in the swapping part of the unit. This prepares the desired state. They can include or exclude whatever they like in determining event ready pairs elsewhere. (I.e. these are determined outside the relevant time cone.) They could pick them by hand as long as that decision is independent of the results elsewhere.

11. Feb 7, 2017

### DrChinese

edguy99: We need to make sure we are talking about the same thing. There are many different ways of generating entanglement. Your original example used PDC and the later one uses swapping. So let's pick one and discuss that. Swapping uses a different post-selection technique than PDC. Post selection is fine, just needs to be clear what the parameters are so that it is "fair".

12. Feb 7, 2017

### edguy99

I will start a new topic for the Hensen et al. experiment.

WRT post selection, good question: are the parameters "fair"? If the post selection process is weeding out split pairs that do not match over split pairs that do match?

13. Feb 7, 2017

### DrChinese

The answer depends. Post selection of Bell states via swapping requires that. That produces entangled pairs elsewhere that can then be used for experimental purposes.

The Hensen experiemental setup is quite complex, a very nice experiment indeed!

14. Feb 7, 2017

### Strilanc

Note that entanglement swapping doesn't require post-selection. It can be done with fixup operations instead. In effect, entanglement swapping is just quantum teleportation of an EPR half. It's just easier to do the fixing when analyzing the results rather than as a step in the experiment.

(One of my pet peeves is quantum teleportation experiments that use post-selection instead of fixup operations. No one would trust their qubits to a teleportation process that failed 75% of the time! It's like... instead of proving they could move the information reliably, they proved that if they had done the fixup operations then they could have moved the information reliably. )

15. Feb 7, 2017

### DrChinese

Always ready to learn something new... what's the secret to skipping that other 75%?

16. Feb 8, 2017

### Strilanc

An entanglement swap is just a quantum teleportation of an EPR half. So basically you just do the fixup operations you'd do in a quantum teleportation. Here's the circuit:

When Charlie gets a "bad" measurement result, it just means the X or Z parity of the entanglement is backwards, and he fixes it by telling Alice or Bob to flip their qubit over correspondingly. You can very freely split the fixup operations between them. Both fixup operations can go to Alice, or both to Bob, or one to each, or even combinations where one rotates by R and the other by 180-R.

17. Feb 8, 2017

### DrChinese

Thanks!

As you have so generously explained with your diagram, it makes perfect sense. Never thought of it like that.