Friday, July 10, 2020

Locality Confusion or: What Entanglement Can and Cannot Do For You

I really enjoyed last week's Zoom edition of the annual Strings conference. Clifford has said many of the things about it that I support wholeheartedly, so I don't have to repeat them here. One of the things I really liked was the active participation in the chat channel that accompanied the talks.

But some of the things I read there gave me the impression that there is some confusion out there about locality and things that can happen in quantum theories that showed up in discussions related to black hole information loss (or the lack thereof). So I though, maybe it's a good idea to sort these out.

Let's start with some basic quantum information: Entanglement is strange, it allows you to do things that maybe at first you did not expect. You can already see this in the easy, finite dimensional situation. Assume our Hilbert space is a tensor product
\[H=H_h\otimes H_t\]
of stuff here and stuff there. Further, for simplicity, assume both factors have dimension d and we can pick a basis
\[(e_\alpha)_{1\le \alpha\le d}\]

for both. If we have a maximally entangled state like
\[\Omega = \frac 1{\sqrt d} \sum_\alpha e_\alpha\otimes e_\alpha\]
the first observation is that instead of acting with an operator A here, you can as well act with the transposed (with respect to our basis) operator there, as you can see when writing out what it means in component:
\[(A\otimes id)\Omega = (id\otimes A^T)\Omega = \frac 1{\sqrt d}\sum_{\alpha\beta} a_{\alpha\beta} e_\beta\otimes e_\alpha.\]
That is, with the entangled state, everything, I can do here creates a state that can also be gotten by doing stuff there. And the converse is true as well: Take any state $\psi \in H$. Then I can find an operator $A$ that acts only here that creates this state from the entangled state:
\[ \psi = (A\otimes id)\Omega.\]
How can we find $A$? First use Schmidt decomposition to write
\[\psi = \sum_j c_j f_j\otimes\tilde f_j\]
where the $c$'s are non-negative numbers and both the $f$'s and the $\tilde f$'s are an ortho-normal basis. Define $V$ to be the unitary matrix that does the change of basis from the $e$'s to the $f$'s. Then
\[ A = \sqrt{d\rho_h}V\]
where we used the density matrix $\rho_h$ that is obtained from $\psi$ as a partial trace over the Hilbert space there (i.e. the state that we see here):
\[\rho_h = tr_{H_t}|\Omega\rangle\langle \Omega| = \sum_j c_j |f_j\rangle\langle f_j|.\]
It's a simple calculation that shows that this $A$ does the job.

In other words, you can create any state of the combined here-there system from an entangled state just by acting locally here.

But what is important is that as still operators here and there commute
\[ [A\otimes id, id\otimes B] =0 \]
you cannot influence measurements there by acting here. If you only measure there you cannot tell if the global state is still $\Omega$ or if I decided to act here with a non-trivial unitary operator (which would be the time evolution for my local Hamiltonian $A$).

It is easy to see, that you don't really need a maximally entangled state $\Omega$ to start with, you just need enough entanglement such that $\rho_h$ is invertible (i..e that there are no 0 coefficients in the Schmidt decomposition of the state you start with).

And from this we can leave the finite dimensional realm and go to QFT, where you have the Reeh-Schlieder theorem which tells you essentially that the quantum vacuum of a QFT has this entanglement property: In that setting, here corrensponds to any local neighbourhood (some causal diamond for example) while there is everything space-like localised from here (for a nice introduction see Witten's lecture notes on quantum information).

But still, this does not mean that by acting locally here in your room you can suddenly make some particle appear on the moon that somebody there could measure (or not). QFT is still local, operators with space-like separation cannot influence each other. The observer on the moon cannot tell if the particle observed there is just a vacuum fluctuation or if you created it in your armchair even though RS holds and you can create state with a particle on the moon. If you don't believe it, go back to the finite dimensional explicit example above. RS is really the same thing pimped to infinite dimensions.

And there is another thing that complicates these matters (and which I learned only recently): Localization in gauge theories is more complicated that you might think at first: Take QED. Thanks to Gauß' law, you can write an expression for the total charge $Q$ as an integral over the field-strength over a sphere at infinity. This seems to suggest that $Q$ has to commute with every operator localised in a finite region as $Q$ is localised in a region space-like to your finite reason. But what if that localised operator is a field operator, for example the electron field $\psi(x)$? Does this mean $Q, \psi(x)]=0$? Of course not, since the electron is charge, it should have
\[ [Q,\psi(x)] = e \psi(x).\]
But does that mean that an observer at spatial infinity can know if I apply $\psi(x)$ right here right now? That would be acausal, I could use this to send messages faster than light.

How is this paradox resolved? You have to be careful about the gauge freedom. You can either say that a gauge fixing term you add in the process of quantisation destroys Gauß' law. Alternatively, you can see that acting with a naked $\psi(x)$ destroys the gauge you have chosen. You can repair this but the consequence is that the "dressed" operator is no longer localised at $x$ but in fact is smeared all over the place (as you have to repair the gauge everywhere). More details can be found in Wojciech Dybalski's lecture notes in the very end (who explained this solution to me).

The same holds true for arguments where you say that the total (ADM) mass/energy  in a space time can be measured at spatial infinity.

So the upshot is: Even though quantum theory is weird, you still have to be careful with locality and causality and when arguing about what you can do here and what that means for the rest of the universe. This also holds true when you try to resolve the black hole information loss paradox. Did I say islands?

PimEyes knows what you did last summer

You might have come across news about a search engine for faces: . You can upload your photo and it will tell you where in the interwebs it has seen you before. Of course, I had to try it. Here are my results:

OK, that was to be expected. This is the image I use whenever somebody asks me for a short bio with a picture or which I often use as avatar. This is also the first hit when you search for my name on Google Images. This is fine with me, this image is probably my pubic persona when it comes to the information super-highway. But still, if you meet me on the street, you can use PimEyes to figure out who I am. But that was to be expected. Then there come some variants of this picture and an older one that I used for similar purposes.

Next come pictures like these:

There seems to be an Armenian politician with some vague resemblance and the internet has a lot of pictures from him. Fine. I can hide behind him (pretty much like my wife whose name is so common in Germany that the mother of one of our daughter's classmates has the same when you only consider first and maiden name as well as a former federal minister).

But then there is this:

And yes, that's me. This is some open air concert one or two years ago. And it's not even full frontal like the sample I uploaded. And there probably 50 other people who are as much recognisable as myself in that picture. And even though this search engine seems not to know about them right now, there must be hundreds of pictures of similar Ali Mitgutsch Wimmelbuch quality that show at which mass activities I participated. I have to admit, I am a little bit scared.