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January 6, 2020

Entanglement for Laymen

I’ve been asked, innumerable times, to explain quantum entanglement to some lay audience. Most of the elementary explanations that I have seen (heck, maybe all of them) fail to draw any meaningful distinction between “entanglement” and mere “(classical) correlation.”

This drives me up the wall, so each time I am asked, I strive to come up with an elementary explanation of the difference. Rather than keep reinventing the wheel, let me herewith record my latest attempt.

“Entanglement” is a bit tricky to explain, versus “correlation” — which has a perfectly classical interpretation.

Say I tear a page of paper in two, crumple up the two pieces into balls and (at random) hand one to Adam and the other to Betty. They then go their separate ways and — sometime later — Adam unfolds his piece of paper. There’s a 50% chance that he got the top half, and 50% that he got the bottom half. But if he got the top half, we know for certain that Betty got the bottom half (and vice versa).

That’s correlation.

In this regard, the entangled state behaves exactly the same way. What distinguishes the entangled state from the merely correlated is something that doesn’t have a classical analogue. So let me shift from pieces of paper to photons.

You’re probably familiar with the polaroid filters in good sunglasses. They absorb light polarized along the horizontal axis, but transmit light polarized along the vertical axis.

Say, instead of crumpled pieces of paper, I send Adam and Betty a pair of photons.

In the correlated state, one photon is polarized horizontally, and one photon is polarized vertically, and there’s a 50% chance that Adam got the first while Betty got the second and a 50% chance that it’s the other way around.

Adam and Betty send their photons through polaroid filters, both aligned vertically. If Adam’s photon makes it through the filter, we can be certain that Betty’s gets absorbed and vice versa. Same is true if they both align their filters horizontally.

Say Adam aligns his filter horizontally, while Betty aligns hers vertically. Then either both photons make it though (with 50% probability) or both get absorbed (also with 50% probability).

All of the above statements are also true in the entangled state.

The tricky thing, the thing that makes the entangled state different from the correlated state, is what happens if both Adam and Betty align their filters at a 45° angle. Now there’s a 50% chance that Adam’s photon makes it through his filter, and a 50% chance that Betty’s photon makes it through her filter.

(You can check this yourself, if you’re willing to sacrifice an old pair of sunglasses. Polarize a beam of light with one sunglass lens, and view it through the other sunglass lens. As you rotate the second lens, the intensity varies from 100% (when the lenses are aligned) to 0 (when they are at 90°). The intensity is 50% when the second lens is at 45°.)

So what is the probability that both Adam and Betty’s photons make it through? Well, if there’s a 50% chance that his made it through and a 50% chance that hers made it through, then you might surmise that there’s a 25% chance that both made it through.

That’s indeed the correct answer in the correlated state.

In fact, in the correlated state, each of the 4 possible outcomes (both photons made it through, Adam’s made it through but Betty’s got absorbed, Adam’s got absorbed but Betty’s made it through or both got absorbed) has a 25% chance of taking place.

But, in the entangled state, things are different.

In the entangled state, the probability that both photons made it through is 50% – the same as the probability that one made it through. In other words, if Adam’s photon made it through the 45° filter, then we can be certain that Betty’s made it through. And if Adam’s was absorbed, so was Betty’s. There’s zero chance that one of their photons made it through while the other got absorbed.

Unfortunately, while it’s fairly easy to create the correlated state with classical tools (polaroid filters, half-silvered mirrors, …), creating the entangled state requires some quantum mechanical ingredients. So you’ll just have to believe me that quantum mechanics allows for a state of two photons with all of the aforementioned properties.

Sorry if this explanation was a bit convoluted; I told you that entanglement is subtle…

Posted by distler at January 6, 2020 2:30 PM

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Re: Entanglement for Laymen

Thank you for this post. Can you please tell me how this last result comes from a state such as (|up>|down> - |down>|up>)/sqrt(2) ?
If Adam’s photon goes through, does the state collapse to something? How does it imply that Betty’s photon must also go through? I’m just trying to understand within the context of the above state and the axioms if quantum mechanics.

Posted by: Dave on January 7, 2020 8:27 PM | Permalink | Reply to this

Re: Entanglement for Laymen

We are indeed studying an entangled state in a product of two-state systems. But we are measuring the operators (0 1 1 0)𝟙\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\otimes\mathbb{1} and 𝟙(0 1 1 0)\mathbb{1}\otimes\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right) (in the basis where |up=(1 0)|\text{up}\rangle=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right) and |down=(0 1)|\text{down}\rangle=\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)).

I am sure you can construct the corresponding projection operators and compute the probabilities for the various outcomes of those measurements.

Posted by: Jacques Distler on January 9, 2020 5:56 PM | Permalink | PGP Sig | Reply to this

Re: Entanglement for Laymen

Why did my question get deleted?

Posted by: Dave on January 9, 2020 12:35 PM | Permalink | Reply to this

Re: Entanglement for Laymen

It was swept up with a bunch of spam. Restored from backup now.

Posted by: Jacques Distler on January 9, 2020 6:04 PM | Permalink | PGP Sig | Reply to this

Re: Entanglement for Laymen

A layman here, thank you for this very clear explanation (the less convoluted I came across, from what I remember) !

I have a question, is it possible then that the entanglement operation makes each time different particles (something that we are not able yet to measure) to explain why we can know the state of the second particle once we observed if the first one is filtered ?

Posted by: Sidney on May 10, 2020 9:16 AM | Permalink | Reply to this

Re: Entanglement for Laymen

I suppose I should have started with an exposition of the difference between classical probabilities (there’s a 50% chance that Schrödinger’s cat is alive or dead) and quantum-mechanical superposition.

The entangled state is a particular kind of quantum-mechanical superposition.

Distinguishing between the two requires “non-commuting observables” — a measurement of a second observable that can change the result of measuring the first observable. In the case at hand, the act of measuring the polarization of the photon at a 45 45^\circ axis changes what the result for measuring whether it is horizontally/vertically polarized.

Posted by: Jacques Distler on May 10, 2020 1:15 PM | Permalink | PGP Sig | Reply to this

Re: Entanglement for Laymen

This criteria makes sense indeed, is it correct to interpret it as an “action at distance” ?

So if I understood correctly, the second photon polarization probabilities can change when the first photon mirror’s orientation change ?

Posted by: Sidney on May 12, 2020 3:47 PM | Permalink | Reply to this

Re: Entanglement for Laymen

This criteria makes sense indeed, is it correct to interpret it as an “action at distance” ?

No more than the classical correlations that I talked about in my introduction constitute “action at a distance.”

So if I understood correctly, the second photon polarization probabilities can change when the first photon mirror’s orientation change ?

We know things about the first photon (and hence also about the second) that we didn’t before we made the measurement on the first photon. The funny thing about non-commuting observables is that they also wipe out things that we previously knew about the system.

Just to go back to the polaroid filters, remember that they work solely by absorbing photons (of one polarization, while letting the other polarization pass).

So imagine a setup with a pair of filters, one oriented horizontally and the other vertically. No photons pass through the pair: those which are not absorbed by the first are absorbed by the second.

But now put a third filter in between the two, oriented at 45 45^\circ. Even though all it does is absorb photons, now suddenly light can pass through the system, whereas none could before.

Posted by: Jacques Distler on May 12, 2020 4:01 PM | Permalink | PGP Sig | Reply to this

Re: Entanglement for Laymen

This example is very useful !

I still don’t fully grasp the distinction between what the filter does (absorbing photons) and what happens (the lost of information on the system allowing photons to pass through the filter).

Does it means that the system doesn’t really change, only what we know about it, so the filter’s influence cannot be interpreted as an action at a distance ?

I’m sorry for the delay in my answer, and thank you for your patience :)

Posted by: Sidney on May 15, 2020 3:06 PM | Permalink | Reply to this

Re: Entanglement for Laymen

I guess, I am a bit late for the party, but haven’t been here for several years, sorry.

In fact, this is also a pet peeve of mine (popularisations of entanglement that would not work for correlation). I will have to do this again in about two weeks and had already ordered Polaroid filters for exactly this demonstration (including the one where the third filter makes light come through) but am a bit disappointed by how low the contrast is that these filter actually achieve.

I think that your explanation has to be supplemented, though, by something that rules out a variant of hidden variables (along the lines of: the photons do not only have a single polarisation but actually one for each angle you probe them in. That could explain the total correlation the 45 degree rotated version).

So, for a full argument, I think one also has to invoke some Bell type inequality, for example of the form

| X1 (X2 + Y2) - Y2 (X2 - Y2) | < 2

as with hidden variables, either the first or the second parenthesis vanishes while the other is two (when all variables have +/- 1 outputs) but there are quantum states where the LHS has expectation value 2 sqrt(2).

To a lay audience, this is of course harder to explain as there are formulas involved but I think the argument is incomplete without anything like this.

And then there is this very nice phrasing of entanglement that I learned form Hajo Leschke: “In a composite quantum system, you can have complete information about the composite system while not having complete information about its parts”. Unfortunately, that doesn’t help a lot, since the non-trivial meaning of “complete information” requires explanation.

Posted by: Robert on March 18, 2022 3:47 AM | Permalink | Reply to this

Hidden variables

Fair enough. I wasn’t trying to rule out hidden variable theories (which, if you are willing to entertain sufficiently baroque ones, can be an arduous task), but rather to simply state (in the terms of conventional quantum mechanics) what “entanglement” is and how it differs from mere classical correlation.

Once we know what quantum physicists mean by entanglement, then we can discuss whether there are “alternative” theories that yield the same result.

Posted by: Jacques Distler on March 18, 2022 4:01 AM | Permalink | PGP Sig | Reply to this

Re: Hidden variables

I guess what I was trying to say was that when you allow hidden variables, entanglement can still look like correlation. So, to drive home that entanglement is something different you need to close that loophole.

My background here is that I want to explain quantum cryptography (at least the idea) and its safety requires that Eve cannot “copy” the correlation.

Posted by: Robert on March 19, 2022 10:07 AM | Permalink | Reply to this

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