Entanglement Is Not That Magic

One of the things that made me very leery of the whole Brian Cox electron business was the way that he seemed to be justifying dramatic claims through dramatic handwaving: “Moving an electron here changes the state of a very distant electron instantaneously because LOOK! THE WINGED VICTORY OF SAMOTHRACE EINSTEIN-PODOLSKY-ROSEN PAPER!” On closer inspection, it’s not quite that bad, though it takes very close inspection to work out just what they are claiming.

That said, though, it’s fairly common to hear claims of the form “when two particles are entangled, anything you do to one of them changes the state of the other.” This is not strictly true, though, and it’s worth going through in detail, if only so I have something to point to the next time somebody starts using that line. This will necessarily involve some math, but I’ll try to keep it as simple as I can.

So: the problematic claim is that doing things to one particle of an entangled pair of particles affects the state of the other particle in the pair. This is true only for a very small subset of “doing things” and “affects the state”– that is, it is absolutely and unequivocally true that measuring the state of one entangled particle in some basis determines the possible outcomes of measurements on the other particle in the pair. However, the vast majority of things you might do to one of the two particles do not produce corresponding changes in the state of the other. In fact, most of the things you might do will appear to destroy the entanglement altogether.

To explain what I’m talking about, we need to consider an example, and how you describe matters mathematically. My background is in the quantum optics sort of world, so let’s think about an entangled state of two polarized photons, produced in a way that guarantees the polarization of the two will be exactly the same. If we think about linear polarization, there are two possible states, which are determined by the axis of a polarizer used to measure the state of one of the photons– by convention, we’ll call them horizontal (represented by |H>) and vertical (represented by |V>>). An entangled state then looks like:

|Ψ> = A(|H>1|H>2+|V>1|V>2

This wavefunction is the sum of two terms, one corresponding to both particles having horizontal polarization, the other corresponding to both particles having vertical polarization. If you measure the state of one of them, then, you can predict with certainty the state of the other, no matter how far away it is at the time you make the measurement. If you and a friend each take one photon and measure its state, horizontal or vertical, then repeat the experiment many times, when you compare your lists of measurements, you will find that they are always the same: when you find vertical polarization, your friend also finds vertical polarization, and vice versa.

Of course, horizontal and vertical polarization aren’t the only options. You could, for example, take your measuring apparatus and rotate it by 45 degrees counter-clockwise. In this case, the two polarization states you would measure are diagonal, which we’ll call |D+> (45 degrees counter-clockwise from vertical) and |D-> (45 degree clockwise from vertical). What happens if you and your friend use these diagonal polarizers to measure the entangled photons from the same source?

Your first guess might be that it would screw up the entanglement completely, given that a polarizer at 45 degrees has a 50% chance of passing a vertically polarized photon, and a 50% chance of passing a horizontally polarized photon. But something really neat happens when you look at the details, because we can write horizontal and vertical polarization as a combination of |D+> and |D-> states:

|H>=a(|D+> + |D->)

|V>=a(|D+> – |D->)

Using these, we can re-write our initial entangled state in terms of the new state:

|Ψ> = B[(|D+>1+|D->1)(|D+>2+|D->2) + (|D+>1-|D->1)(|D+>2-|D->2)]

If you multiply out all those parentheses, you get:

|Ψ> = B[(|D+>1|D+>2 + |D+>1|D->2+|D->1|D+>2+|D->1|D->2) + (|D+>1|D+>2-|D+>1|D->2 – |D->1|D+>2+|D->1|D->2)]

Looking closely at this, you should notice that there are four terms with one photon in one state and the other in the opposite state, and they cancel each other out. Thus, when you do all the agebra out, the wavefunction in terms of |D+> and |D-> states is:

|Ψ> = C[|D+>1|D+>2 + |D->1|D->2]

Mathematically, this is identical to our initial entangled state, meaning you see exactly the same correlations with your detectors rotated by 45 degree as you did with the initial state. If you measure one photon to have D+ polarization, your friend with the other photon will also find D+ polarization, and so on. Entangled states are entangled no matter how you measure them, which is one of the beautiful things about quantum entanglement.

Now, you might reasonably ask what happens when you and your friend set your polarizers to different angles. Which is an interesting question, and leads you to Bell’s theorem, which is one of the very coolest results in physics. But that’s not the question we want to think about here. The question of interest for this post is “What happens when you rotate the state of one of your photons?”

That is, let’s imagine that you stick in a waveplate that takes the polarization of one of your photons, and rotates that by 45 degrees. An initial state of |H>> gets turned into |D+>, and an initial state of |V> gets turned into |D-&;. What does this do to the entanglement?

Given our experience with rotating the detectors, you might want to say that the entanglement should still be there, which is what leads to the claim that changing the state of one entangled article always changes the state of the other. But that’s not exactly right, as you can see if you do the math. Rotating the polarization of photon 2 changes our initial entangled state to:

|Ψ> = B[|H>1|D+>2 + |V>1|D->2]

So, what do you and your friend see when you measure this state with your D+/D- detectors? To see that, we need to write everything in terms of D+ and D-, so:

|Ψ> = C[(|D+>1+ |D->1)|D+>2 + (|D+>1-|D->1)|D+>2]

|Ψ> = C[|D+>1|D+>2+|D->1|D+>2+ |D+>1|D->2-|D->1|D->2]

Looking at this, you see there’s no longer a solid correlation between the two photons. If you get a result of D+ for photon 1, your friend is equally likely to get D+ or D- for photon 2, and vice versa. If you each set your detectors to D+/D-, you see no sign of entanglement at all. If you’d like to do a little algebra, you can easily convince yourself that the same thing happens when both detectors are set to H/V (D+ and D- can be written in terms of H and V in the same way that H and V are written in terms of D+ and D-).

So, the rotation of one photon does not rotate the state of the other, as a naive approach to the idea of entanglement would have you believe. When you do that, you no longer see the same correlation that you did before.

Does that mean that the entanglement is gone forever? No, the two are still entangled, you just have to be more careful in your experimental design. If, rather than setting your detectors to the same angles, you set one to measure H/V and the other to measure D+/D-, you will see the same perfect correlation as before. It doesn’t even matter which detector is which, as long as they’re at different angles. The entanglement is still there, but it’s not between the same states any more.

So, what’s the point of all this math and stuff? A couple of things:

1) The “changing the state of one of the particles changes the state of the other” claim doesn’t hold up. Entanglement is about correlations between measurements, not the absolute states. Rotating the polarization of one photon doesn’t rotate the polarization of its distant partner, it just changes where you have to look to see the correlation between measurements.

This holds for whatever sort of system you’re looking at. If you’re working with atoms whose internal quantum states are entangled, kicking one of them up to a higher energy state does not change the energy state of the other in a spooky-action-at-a-distance way. All it does is change the nature of the correlation between your eventual measurement results.

2) If you don’t know exactly what happened to each of the two particles, entanglement is really fragile. A lot of “quantum” woo-woo New Age nonsense invokes entanglement between particles as an excuse, and it’s true that, in principle, everything is entangled with everything else, since every particle now in existence was initially packed into a tiny volume just after the Big Bang. But unless you know all of the ways the state of an electron in your brain was rotated in the intervening 13.7 billion years, you’ll never be able to detect the entanglement between it an an electron in somebody else’s brain, in order to form a telepathic connection with them, or whatever wacky thing you think you might be able to do.

3) If you have knowledge or control of the history of two particles, entanglement is pretty robust. You can pass it from one particle to another, or even from one type of particle to another– people have demonstrated quantum teleportation using ytterbium atoms entangled with light, and it works great. It requires you to know exactly what corresponds to what, though, as you move things around, and great care to avoid accidentally changing one of the states in a way that would mess up your measurement.

So, to sum up: Quantum entanglement is a really cool phenomenon, one of the most magical-seeming effects in the theory. It’s got that great history, too– Einstein! Spukhafte fernwirkung! Non-locality!

It’s not magic, though. At least, it’s not that magic. It’s just a correlation between measurement results, not a knob that lets you manipulate the states of distant particles in a magical manner.

Comments

  1. #1 David Marjanović
    March 14, 2012

    fernwirkung

    All Nouns begin with a capital Letter in german, and pretty much nothing else does; there aren’t even special Rules for Headlines.

  2. #2 David Robert Grimes
    March 14, 2012

    This is easily the best explanation of entanglement states I have ever seen. Prof, you have outdone yourself. It’s nice to see the wavefunction being used as I don’t tend to trust anything unless I can follow the math!

    Small technical question though; It’s been a while since I’ve done QM, so I’m a little rusty – The wavefunction squared always sums to 1 as probabilty amplitude , but am I correct in assuming in this case |Ψ> = A(|H>1|H>2+|V>1|V>2 itself sums to unity and the all the H terms and V terms are binary (1 or 0) ? This makes intuitive sense to me if so, but just wanted to check I had understood correctly…

  3. #3 Peter Morgan
    March 14, 2012

    I remind you of the Reeh-Schlieder theorem, which, for a Wightman field, allows us to determine (both in the sense of measurement and in the sense of preparation) the state of the whole universe to arbitrary precision using local observables that are associated with a finite region of space-time. The proof depends on the analyticity properties of Wightman fields, and the proof does not specify what precision is required for the local observables to achieve a given precision for a given nonlocal observable, so it’s not clear that a physically interesting implementation is possible, but it somewhat supports Cox’s perspective.

  4. #4 Matt Leifer
    March 14, 2012

    Let me try to be a bit more precise about what you can and can’t say about what happens when you perform a local rotation on one part of an entangled system.

    Assume, for the sake of argument, that you are one of those people who views the pure state of the global system as something physically real, and that you think that quantum theory is complete, i.e. there are no hidden variables. Then, when you have an entangled state of the sort described in the post, the local states of the two subsystems are maximally mixed, i.e. their density operators are proportional to the identity operator. The maximally mixed state is invariant under unitary operations, so when you rotate one of the polarizations, both of these local states remain unchanged. Assuming that local properties have to be defined in terms of these local states, no property of either photon has changed. You haven’t even changed the state of the photon you have acted on, let alone the other one.

    What you have to say in this instance is that there are no interesting local properties of the two subsystems for this particular state. Instead, you have a property, described by the entangled pure state, that does not belong to either of the photons individually, but only both of them together. When you rotate the polarization of one photon, you do in fact change this global property, and this is a nonlocal change because the property does not belong to either party individually but only the two of them together. You do not change any property of the second photon, but this is only because that photon does not actually have any interesting properties in the first place.

    Of course, all of this assumes that you accept the premises of completeness and state realism, but there are plenty of physicists who do accept them, particularly anyone who believes in many-worlds. The implied holism of the above argument is one of the reasons that I reject this whole approach myself. It seems to block approaches to understanding entanglement that are a lot more appealing, particularly analogies to classical probability theory, i.e. imagine how crazy it would seem to apply the above argument to a classical joint probability distribution. Personally, I would like to be able to say that when you rotate the polarization of one photon, that really does only effect a local change to the photon you acted upon, and I would like to be able to say that without resorting to any neo-Copenhagen nonsense.

  5. #5 Ambitwistor
    March 14, 2012

    This blog post might make a nice little AJP article.

  6. #6 Kevin Reid
    March 14, 2012

    May I suggest a substitution of “⟩” for “>” to make the notation a bit less visually unbalanced?

  7. #7 Sascha Vongehr
    March 14, 2012

    It is not magical, of course not, but EPR/Bell tell us that the directly real world does “not exist out there”. For many, this is as magical and unacceptable as it gets, nevertheless, quantum modal realism is thus experimentally proven (if you want to describe conserving Einstein locality(micro causality)).

  8. #8 Thony C.
    March 15, 2012

    All Nouns begin with a capital Letter in german, and pretty much nothing else does…

    As in English the first word of a sentence also starts with a capital letter ;)

  9. #9 Pierre Hemmerlé
    March 15, 2012

    on entanglement : all actual physics are based on a dish falling from Pisa’tower and the Maxwell’s equation forcibly given a Universe’s scope going as far as to preposterously and erroneously demanding the Cosmos to be curved. Mathematics are great. It is greater to have the humility of recognizing that extensions to the Universe cannot follow that pass with the Nessus’s tunic of bending the experiments to the theory. It has been done with light bending to please Einstein, it is done today with a LHC to please Higgs.

  10. #10 CCPhysicist
    March 15, 2012

    One typo: You messed up the closing ket for D- in the paragraph “imagine that we stick in a waveplate”.

    I second the suggestion in comment 5.

    There is an element of elementary physics in your situation. One reason you have to have entanglement is conservation of angular momentum (if we are talking about spin states). QM makes this “spooky” because of what you cannot know before hand, but it remains true that an external torque will affect the total angular momentum and destroy the correlation.

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