# I Single photon entanglement

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1. Jan 5, 2017

I've recently been reading about entanglement between two spatial modes of a single photon. It's a little over my head and there is one aspect about it that I'm particularly unclear on, which I was hoping someone here might be willing to shed some light on it.

The basic setup is described with this picture from the paper (I hope I'm allowed to link this):

So simplified quite a bit, my understanding is that the procedure goes like this:
1. A single photon is sent to Bob and Alice with a $R=50\%$ probability distribution, and hence is in the state $| \psi \rangle = |0\rangle_A|1\rangle_B + |1\rangle_A|0\rangle_B$.
2. Next Alice rotates her setup by a chosen angle $\theta$.
3. Then Bob and Alice both measure their photon steams using "full quantum-state tomography using homodyne detection". I'm not very familiar with this but I assume you can't get more than one bit of information for each photon, so it must be an aggregate of many runs.
4. Then when their measurements are compared it can be shown that Alice's choice of measurement angle has affected Bob's results.
I'm okay with accepting the above story without grasping all the details, but my question is how there can ever be two simultaneous measurements to be compared. Binning Bob's results requires at least one bit from Alice, $s \in\{+,-\}$. Yet each "run" consists of only a single photon; if Alice can detect anything, then surely Bob will detect nothing, and vise-versa.

If there are not simultaneous measurements by both parties, I don't see how Bob's tomography results can be binned based on the binary result from Alice, unless there is some relationship between subsequent photons.

So what's wrong with my understanding?

2. Jan 5, 2017

I realise this may be an overly specific question, but I searched the forums and the internet and couldn't find much in the way of discussions on single particle entanglement, though I see the phenomena has been known for many years. I think if I could understand how a single particle is enough to produce entanglement it would provide an even simpler picture than the normal two-particle case.

3. Jan 5, 2017

### BvU

I found it a funny reference: eight feet of gedankenexperiment, then one foot of experiment description and no clarity (to my mind) wrt the actual experimental results.
I recall a later http://www.tudelft.nl/en/current/latest-news/article/detail/einsteins-ongelijk-delfts-experiment-beeindigt-80-jaar-oude-discussie/[/URL] from Delft University in [URL]http://www.nature.com/nature/journal/v526/n7575/full/nature15759.html[/URL] with results (article [URL='https://arxiv.org/abs/1508.05949']here[/URL]). I think it was discussed in PF as well. [URL='https://www.physicsforums.com/threads/another-loophole-free-test-of-bells-theorem.842620/page-3#post-5294454']Twice[/URL]. or more.

Last edited by a moderator: May 8, 2017
4. Jan 5, 2017

### zonde

They are using heralded photons. This means that first they produce two photons with downconversion, then they detect one photon and send other photon to Alice or Bob along with classical signal that heralding photon was detected.

And they are testing steering inequality. In order to demonstrate entanglement they would have to violate Bell inequality. These are different inequalities.
Steering inequality demonstrates that more than one bit of information is needed to produce observed correlations.

5. Jan 5, 2017

I had missed that word "heralded" and I've now got an idea what that entails, but I still understand that only one of that pair will be sent to Alice and/or Bob. Thus I'm still left wondering how there can be more than a single detection event (at Alice and/or Bob) for each heralded pair.

I understand it's not as loophole-free as a Bell test, but if Alice's choice of angle effects Bob's result FTL it still demonstrates Bell-style non-locality.

6. Jan 5, 2017

### secur

From the reference,

"If Alice were simply to detect the presence or absence of a photon, then Bob’s measurement of the same observable will be anticorrelated with Alice's, as in ref. 42. However, this does not prove that Alice’s measurement affected Bob’s local state because such perfect anticorrelations would also arise from a classical mixture of |0›A |1›B and |1›A |0›B, in which Bob’s measurement simply reveals a pre-existing local state for him, |1›B or |0›B."

So, detecting the photon's presence or absence wouldn't do any good. Instead A and B are both measuring phase information, "homodyne detection":

"However, there is a key difference, which enables demonstration of the nonlocal collapse experimentally: rather than simply detecting the presence or absence of the photon, homodyne detection is used."

With these measurements they never find out "which path" the photon took, that info is lost. Instead they both get phase info. And Alice's decision of how to measure the phase affects Bob's measurement.

BTW it's interesting to note that experimenters - the people who actually know what's going on - happily use the concept of "nonlocal collapse". Only theoretical physicists pretend there's no collapse.

Last edited: Jan 5, 2017
7. Jan 6, 2017

### zonde

They where making homodyne detections. We can say that beamsplitter is producing two outputs from two particular inputs. Now for homodyne detection you place some laser beam (local oscillator) at second input of beamsplitter so that you mix your input beam with that local beam and measure difference in amplitudes for two detector outputs (if my understanding about homodyne detection is correct).
It's not about loopholes (loopholes can be there for either test). It's that steering inequality does not rule out local hidden variable explanations. So you can't claim that "Alice's choice of angle effects Bob's result FTL". Local explanations are still possible for observed correlations (that violate steering inequality).
Well, it's true that the authors in abstract are claiming that they test "whether the choice of measurement in one laboratory really causes a change in the local quantum state in the other laboratory" but I don't see how they can claim that from violation of steering inequality. I say that they would have to violate Bell's inequality to claim that.

8. Jan 6, 2017

Um, I think I better read more about homodyne detection too. But I can picture how it would work if it functions as a kind of amplification.

I think that's just short-hand for "Bell non-locality" and "apparent collapse" or they are implicitly assuming the Copenhagen interpretation. It's hard to tell when clearly they aren't concerned with the foundational aspects.

9. Jan 6, 2017

I think they are all manifestations of entanglement. Quantum steering can be used to construct bell non-local states. Also, in my own words I would say to be able to use steering you need entanglement. And if you have entanglement you can also violate a bell inequality.

10. Jan 6, 2017

### zonde

The abstract indeed says that "Bell nonlocal states can be constructed from some steerable states." But if you look at equation (1) that gives condition for their claim than it looks exactly the other way around:
$\rho_{AB}=\mu \tau_{AB}+(1-\mu)\tau'_{AB}$
here $\rho_{AB}$ is steerable state and $\tau_{AB}$ is Bell nonlocal state.
Maybe. What arguments can you prose to back up your claim?

11. Jan 6, 2017

Well, I would define steering as Alice's ability to perform some operation, $f()$, on their shared state that somehow effects the Bob's measurement, $g()$. In the general case they both operate on the shared state: $$\psi_{final} = g(f( \psi_{original} ))$$ However if there is no entanglement, the state is separable: $$\psi_{original} = \psi_{alice} * \psi_{bob}$$ and $\psi_{bob}$ is inaccessible to Alice, while $\psi_{alice}$ is inaccessible to Bob, so this becomes: $$\psi_{final} = f(\psi_{alice}) * g(\psi_{bob})$$
Clearly the result of $g()$ doesn't depend on $f()$

12. Jan 6, 2017

### secur

Right. Steering definitely requires entanglement. For this paper it's defined as follows:

"From a quantum information perspective, EPR-steering is equivalent to the task of entanglement verification when Bob (and his detectors) can be trusted but Alice (or her detectors) cannot. This is strictly harder than verifying entanglement with both parties trusted, but strictly easier than violating a Bell inequality, where neither party is trusted."

The difference is just about who can be trusted.

BTW that's "affects" not "effects". This experiment does NOT verify FTL collapse:

"We note that, unlike ref. 12, we do not claim to have closed the separation loophole."

The "steering inequality" in this case simply shows that Bob's results are not equal to what he would get without Alice's actions. They take for granted that wavefunction collapse is nonlocal, since that's been proved by other experiments. BTW Bell's inequality can be viewed as a type of steering inequality, specially designed to rule out realist local HV explanations.

Sure, you could call it that. The key point is, it's a way to acquire info about a photonic wave function's phase. Tomography, BTW, means collecting that info at a full range of angles (0 to 2pi), from an ensemble of identically-prepared photons. Comparing that to the result you'd get without Alice's actions, verifies her ability to "steer" Bob's wavefunction.

No. From the experimenter's point of view there are no "interpretations". Experimentally, what we see is FTL collapse - that's all there is to it. As they correctly say, "the choice of measurement in one laboratory really causes a change in the local quantum state in the other laboratory". It's not an interpretation, but a fact. That's why those who deny it are ... well, let me quote Lubos Motl on this point. He says they're "idiots". It's hard to disagree. If there was a Nobel prize for this mental attribute identified by Motl, theoretical physicists would win it every year.

13. Jan 6, 2017

### Staff: Mentor

"FTL collapse" and "really causes a change in the local quantum state in the other laboratory" ARE an interpretation. You are certainly allowed to think that this interpretation is the only sensible one and that only an idiot would entertain any other, but this is not going to lead to any sort of productive discussion.

Do not continue making this argument here. Doing so is experimentally indistinguishable from trolling... And interpretation debates are already quite hard enough to moderate without that.

14. Jan 6, 2017

### Staff: Mentor

If a bunch of people who are as knowledgeable in a field as you are, and who appear sane in all other respects, have a viewpoint that appears "idiotic" to you, you might want to consider the strong possibility that there is something fundamental about their viewpoint that you are misunderstanding. That doesn't necessarily mean they're right and you're wrong; it just means that strong language like "idiotic" should be used with extreme caution, even over and above the general caution that Nugatory has given.

(That goes both ways, btw; it applies just as much to MWI proponents who are unable to understand how anyone could believe anything else.)

15. Jan 7, 2017

Yes, indeed. But loophole-free tests of Bell non-locality have been investigated by other experiments, so I doubt some loophole is manifesting itself here. What's interesting for me about this setup is that the non-local behavior involves only unrelated single qubits.

I see that even though CHSH tests are not too complex, they are still not intuitive enough to explain to most laymen. But maybe with a scheme based around only a single degree of freedom we can make a story that's more accurate and intuitive.

16. Jan 7, 2017

### zonde

17. Jan 7, 2017

18. Jan 7, 2017

### zonde

Well, Bell inequalities demonstrate what explanations do not work. It does not say what explanations should work.

But I am not sure if steering inequalities can help. As I see violation of steering inequalities is convenient starting point for experimental Bell tests as Bell's reasoning starts with perfect correlations for matching angles. And they are useful for quantum cryptography as they show what results you can't get with known classical laws (without inventing unknown classical laws) i.e. what you have to get to be sure that the other party is not cheating you using known classical methods.

19. Jan 7, 2017