Build Your Own Quantum Entanglement Experiment?

[Steven Ellet]

Came across a pair of websites claiming to be DIY Quantum Entanglement Experiment. Problem is, I don't know how realistic it is. Essentially, is this real, or am I being taken for a fool?
Part 1: https://blogs.scientificamerican.co...-quantum-entanglement-experiment-part-1-of-2/

Part 2: https://blogs.scientificamerican.co...-quantum-entanglement-experiment-part-2-of-2/

 

[Vanadium 50]

I wouldn't call this entanglement.

 

[Steven Ellet]

I wouldn't call this entanglement.

What would you call it?

 

[Vanadium 50]

Not entanglement.

 

[stevendaryl]

Why isn't it entanglement? Isn't the article talking about essentially a home-made version of the EPR experiment with twin photons?

 

[Nugatory]

Why isn't it entanglement? Isn't the article talking about essentially a home-made version of the EPR experiment with twin photons?

It's not doing the EPR thing in which entanglement is used in attempt to determine the value of two non-commuting observables... but it is a valid demonstration of entanglement (with a very high value*accessibility figure of merit).

What it doesn't do is demonstrate that the observed correlations must be due to entanglement, as opposed to something equivalent to Bertlemann's socks.

 

[Futurestar33]

Do it yourself for just 20,000 dollars!

 

[zonde]

What it doesn't do is demonstrate that the observed correlations must be due to entanglement, as opposed to something equivalent to Bertlemann's socks.

Just to add.
To tell apart simplest Bertlemann's socks type explanation from entanglement, results would have to be significant enough to tell apart $\sin(\alpha-\beta)$ correlation from $\frac14+\frac12\sin(\alpha-\beta)$ correlation. For $\alpha-\beta=90^{\circ}$ this would be telling apart correlations 1 and 0.75.

 

[vanhees71]

The articles are, unfortunately a bit vague, indeed. E.g., you should always tell your readers which observables you are after that are supposed to be entangled. In the case of photons it cannot be position and momentum as in the original EPR, because there's no position observable of a photon. What's, entangled are the two momenta. The pair momentum is given by the total momentum of the electron and positron. In the center-momentum frame you have $\vec{q}_1=-\vec{q}_2$, if $\vec{q}_{1,2}$ are the momenta of the photons in the considered pair-annihilation reaction $e^+ + e^- \rightarrow \gamma+\gamma$. So there's an entanglement between the momenta of the photons, but the author doesn't demonstrate this entanglement but just doing coincidence measurements. As far as I can see, it's a correct measurement to demonstrate pair annihilation, but it's not a demonstration of entanglement.

 

[zonde]

Author of the article was talking about polarization measurement.

To check for entanglement, I measure the photons' polarization with a technique called Compton polarimetry.

 

[vanhees71]

Argh! I should have read the 2nd part too. That's of course a nice demonstration of polarization entanglement. This is referring to the annihilation of positronium in matter. You have two classes of positronium states with regard to spin: Either positron and electron couple with total spin 0 (odd in the spin states) or total apin 1 (even in the spin states). The parity is even and odd, respectively, and the annihilation process is electromagnetic and thus parity is conserved. Thus only in the first case you can have an annihilation to 2 photons (in the 2nd case it must be an odd number of photons, and for kinematic reasons the annihilation can only be to 3 (or more)). So, if the decay is from positronium into 2 photons it must be from the spin-odd state, i.e., total spin 0, and thus the photon helicities must be opposite, i.e., the photon polarization are entangled to total helicity 0 (or written in the linear-polarization basis $|HV \rangle-|VH \rangle$).

 

[Vanadium 50]

I have a bit more time now, so...

The article spends a good fraction of time on the angular correlation between the gammas, and sees an effect at 180 degrees - as expected. This is not entanglement any more than Bertlemann's socks. I hope we don't have to delve further into this.

I don't know exactly what he's supposed to be measuring with respect to polarization, but he's not doing what he thinks he's doing. Compton polarimetry correlates the electron's spin with the photon's spin (and direction). An aluminum cube doesn't have its electrons spinning in any particular direction. Also, for a spin-0 state, the photons have the same helicities, so he should see a defiict, not an excess. So his polarimetry shouldn't work, and by the evidence that he has presented, it doesn't.

Whatever it is, it's not a demonstration of entanglement.

 

[vanhees71]

I think, it in fact is a demonstration of entanglement, because my argument above is also valid for free electrons and positrons annihilating: In the center-mass frame (which we can use to argue without loss of generality due to Lorentz invariance) the angular momentum (i.e., spin) of the pair is either $S=0$ or $S=1$. Due to parity conservation under electromagnetic interactions, the exclusive channel $e^+ + e^- \rightarrow 2 \gamma$ (i.e., the annihilation into two photons and nothing else) can only occur for $S=0$, and thus no matter, whether you have polarized or unpolarized electrons ans positrons (in the here discussed setup, they are of course unpolarized) the initial state must have $S=0$ in order to end up with two (and not three) photons. But then, due to angular-momentum conservation the helicities of the photons must be opposite, and thus you have polarization entanglement.

 

[zonde]

Compton polarimetry correlates the electron's spin with the photon's spin (and direction).

Are you saying that there is no azimuthal modulation for linearly polarized gamma photons when they scatter from electrons with random spins?
I don't know anything about Compton polarimetry but quick google search turned up this document: Developing a Compton Polarimeter to Measure Polarization of Hard X-Rays in the 50-300 keV Energy Range
I couldn't however get access to the paper mentioned by author: Correlation between the States of Polarization of the Two Quanta of Annihilation Radiation

 

[zonde]

This another paper mentioned by the author (again behind the paywall): The Angular Correlation of Scattered Annihilation Radiation

 

[Vanadium 50]

Are you saying that there is no azimuthal modulation for linearly polarized gamma photons when they scatter from electrons with random spins?

The photons are not linearly polarized, but in any event, in what direction should it point? (Remember, the EM interaction is parity conserving, and that puts huge constraints on the answer)

Normally Compton polarimetry looks at the incoming photon, the outgoing photon, and the scattered electron. Here you lose all information about the scattered electron because it is absorbed.

 

[vanhees71]

The single photons are of course totally unpolarized but you have the entangled state $|HV \rangle-|VH \rangle$ (provided my parity argument above is right ;-)).

 

[Vanadium 50]

Sure, the photons in the decay of a spin-0 states are entangled. But he's not measuring the entanglement if his polarimetry doesn't work.

And the helicities are the same. The spins are opposite and the momenta are opposite.

 

[zonde]

But he's not measuring the entanglement if his polarimetry doesn't work.

There are references to experiments that show it works. There are theoretical calculations that say it should work (Klein - Nishina differential cross section for polarized photons).
What then is the basis for your doubt?

Normally Compton polarimetry looks at the incoming photon, the outgoing photon, and the scattered electron. Here you lose all information about the scattered electron because it is absorbed.

Why does it matter?

 

[Vanadium 50]

The JLAB Compton polarimeter has an analyzing power of about 6%. This is what you can do with the full resources of a national laboratory and a ton of money.

Musser doesn't give his exact number, but he says "one per minute", "runs all day" and "consistently greater". There are 1440 minutes in a day, so you need a difference of about 100 or 8% to determine that one orientation is better. But this 8% requires both polarimeters to give the correct answer, which means that the cheap aluminum cubes need an analyzing power of at least 28%. For this to be physics, the cheap aluminum cubes have to be 4-5x better than the state of the art.

With the colimator, he says the rate falls by a factor of 10, but gets a pair every 20 minutes. These numbers aren't completely consistent (perhaps by rounding), but you now get 72 counts per day, which means you need a difference of about 25, or a 30% effect. That requires an analyzing power of 54%.

Whatever he is seeing, I don't think it's photon polarization.

 

[zonde]

The JLAB Compton polarimeter has an analyzing power of about 6%.

What is this analyzing power? Do you mean that this is visibility of anisotropy of scattered linearly polarized 511 keV photons?

 

[Vanadium 50]

Analyzing power is a term used in polarimetry. If you have a 100% polarization, a polarimeter with an analyzing power of 20% will show a 20% asymmetry - i.e. the absolute polarization is 5x larger than what you measure.

 

[zonde]

I found this paper https://arxiv.org/abs/1507.02824. It calculates how this polarization asymmetry changes with photon energy.
To me it seems that the number you gave makes sense for higher energy photons (considerably more than 511 keV).

Another thing is that when we consider polarization entangled photons, polarization of first scattered photon determines polarization of second (incident) photon. So if photon has certain polarization after scattering we don't have to square probability of scattering in the "right" direction for coincidences.

 

[Vanadium 50]

Figure 2 shows a theoretical maximum analyzing power of 30%. Actual devices always do less well. The numbers from the collimator runs show he is doing almost twice as good as the theoretical maximum.

I stand by my comment: Whatever he is seeing, I don't think it's photon polarization.

 

[zonde]

Figure 2 shows a theoretical maximum analyzing power of 30%. Actual devices always do less well.

Do you mean that scattering asymmetry is reduced because of multiple scatterings in poorly chosen scattering block?

The numbers from the collimator runs show he is doing almost twice as good as the theoretical maximum.

Musser doesn't give his exact number, but he says "one per minute", "runs all day" and "consistently greater". There are 1440 minutes in a day, so you need a difference of about 100 or 8% to determine that one orientation is better. But this 8% requires both polarimeters to give the correct answer, which means that the cheap aluminum cubes need an analyzing power of at least 28%. For this to be physics, the cheap aluminum cubes have to be 4-5x better than the state of the art.

With the colimator, he says the rate falls by a factor of 10, but gets a pair every 20 minutes. These numbers aren't completely consistent (perhaps by rounding), but you now get 72 counts per day, which means you need a difference of about 25, or a 30% effect. That requires an analyzing power of 54%.

The number 25 out of 72 is the minimum to call result "consistently greater", right?
But I don't understand where you got that it requires an analyzing power of 54%. Do you mean that heralding photon does not have certain polarization (orthogonal to scattering direction) after scattering? Or you have on mind some estimates of double scattering of the same photon?

 

[zonde]

The number 25 out of 72 is the minimum to call result "consistently greater", right?

Musser says: "I continue cycling through different ways to align the detectors either parallel or perpendicular to each other."
So this suggests that he made more than one run and it should allow him to to claim "consistently greater" result with smaller difference in results.

heralding photon does not have certain polarization (orthogonal to scattering direction) after scattering

Forget about that. Of course after scattering, electron has to be considered as well.

 

[Vanadium 50]

Do you mean that scattering asymmetry is reduced because of multiple scatterings in poorly chosen scattering block?

There are many, many effects that go into analyzing power - indeed, the reason people use analyzing power is that it's a single number that covers everything. One reason is the energy needed to flip the spin of a particle is small (for a photon it's zero) so interactions with the environment - of many sorts - reduce the analyzing power. Ending up with a few percent is doing very well. And remember here that he's scattering off UN-polarized electrons. That will be a huge factor since you have to integrate over the unmeasured electron spin and the unmeasured electron recoil.

o this suggests that he made more than one run

Maybe you can shoehorn his report so that he's not doing better than the theoretical limit. But his claim is orders of magnitude better than what professionals with orders of magnitude more resources can do, all built out of odds and ends. I simply do not believe that he is measuring what he thinks he is measuring.

I'd also be more comfortable if he would explain why he thinks two photons with the same helicity give an excess in differently scattered photons.

 

[zonde]

But his claim is orders of magnitude better than what professionals with orders of magnitude more resources can do, all built out of odds and ends. I simply do not believe that he is measuring what he thinks he is measuring.

I'd also be more comfortable if he would explain why he thinks two photons with the same helicity give an excess in differently scattered photons.

He simply made a cruder version of 70 years old experiment, that was properly published in physics journal. That original experiment was not very sophisticated itself. Later there where other improved versions of that first experiment.
It is of course possible that Musser is cheating himself with wishful thinking. But I would say that your doubts have to be first examined in light of that 70 years old experiment (as they seem to apply to that version as well) before directing them against this DIY version.