Last week I saw an excellent talk by philosopher Wayne Myrvold.
(Download MP4 video here.)
The topic was well-defined, and of reasonable scope. The theorem is easily and commonly misunderstood. And Wayne’s talk served to dissolve the confusion around it, by unpacking the theorem into a handful of pieces so that you could quickly see where the rub was. I would that all philosophy of physics were so well done.
Here are the key points as I saw them:
- The vacuum state in QFTs, even non-interacting ones, is entangled over arbitrary distances (albeit by exponentially small amounts). You can think of this as every two space-like separated regions of spacetime sharing extremely diluted Bell pairs.
- Likewise, by virtue of its non-local nature, the vacuum contains non-zero (but stupendously tiny) overlap with all localized states. If you were able to perform a “Taj-Mahal measurement” on a region R, which ask the Yes-or-No question “Is there a Taj Mahal in R?”, you always have a non-zero (but stupendously tiny) chance of getting “Yes” and finding a Taj Mahal.
- This non-locality arises directly from requiring the exact spectral condition, i.e., that the Hamiltonian is bounded from below. This is because the spectral condition is a global statement about modes in spacetime. It asserts that allowed states have overlap only with the positive part of the mass shell.
- This is very analogous to the way that analytic functions are determined by their behavior in an arbitrarily small open patch of the complex plane.
- This theorem says that some local operator, when acting on the vacuum, can produce the Taj-Mahal in a distant, space-like separated region of space-time. This is often mistakenly interpreted as it being possible to create, in a causal sense, a Taj Mahal elsewhere. But in fact, the Taj Mahal is already “there”, deep in the belly of the vacuum, and the best you can do from your local region is make a measurement (essentially on the entangled Bell pairs) that has a (extraordinarily small likelihood) outcome that the Taj-Mahal will be found elsewhere.
- Basically all of the key moving parts can be seen by looking at non-interacting QFTs (e.g., just the Klein-Gordan equation), Rindler wedges, and other simple idealizations.
In other words: The vacuum of a QFT is non-local (entangled) over arbitrary distances because of the spectral condition.
Things I still don’t understand:
- I believe any two regions can formally access an arbitrarily number of shared Bell pairs, but the energetic costs are extraordinary. However, I don’t know how to make this “scarce resource” picture precise. In particular, I’d like a better way to explain what’s fishy about this sort of thing.
- All of this depends on requiring that there is precisely zero contribution from negative-energy modes, and then carefully analyzing exponentially tiny overlaps inherent in a non-local vacuum. But the justification I usually hear for the former are very weak, along the lines of “otherwise you’d have a Hamiltonian unbounded from below, which could set off a negative-energy cascade”. But a very tiny chance of this seems fine, for the same reason that we don’t worry about very tiny chances of spontaneous Taj Mahals. Is it obvious why we don’t go this route? Presumably there are deep considerations here with axiomatic QFT.
- What is the precise connection (or “point of closest approach”) between this vacuum non-locality and the non-locality of analytic functions?
Edit: “Yet More Ado About Nothing: The Remarkable Relativistic Vacuum State” (arXiv:0802.1854) by Stephen J. Summers is a good place for further reading.
Edit 2: In a more recent blog post I constructed a state with the Reeh-Schlieder property in the traditional non-relativistic context (i.e., a finite number of countable-basis Hilbert spaces tensored together). This avoids relying on things like analyticity.
Edit 3: This recent Ed Witten lecture (video, notes) has the simplest derivation I’ve seen, and he interprets the result sensibly. As you can see, the derivation relies heavily on analyticity. (H/t Scott Aaronson.)