1
$\begingroup$

Could physics still be local? Here's what I mean:

The Schrodinger/Dirac equations allow for quantum entanglement, right? So in that sense they are non-local physically. But they are mathematically local in the sense of being point-wise partial differential equations. So if you imagine that the QM wavefunction corresponds to a true field, as yet to be identified, or at least a close approximation to it, and particles are due to a true collapse of that field, whatever "collapse" means in reality, then you could imagine that physics might still be local.

Whereas, if the QFT wavefunction has as its domain, instead of spacetime, the Hilbert space of generally delocalized field states, then what does that say about locality? Does that mean that, even in principle, there is no way to write the laws of physics in a point-wise mathematical form with respect to spacetime? And in that sense, is it the case that the non-locality of QFT is much more unavoidable, more essential than that of regular QM?

I saw there's also a topic called non-local Lagrangians, but I'm not trying to go there...I just want to understand the inherent local/non-local character of QFT in general.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ The word "local" is overloaded. In relativistic QFT, the dynamics is local (all interactions are local), but the state of course can still be nonlocal (entangled). Which type of (non)locality are you asking about? Depending on what you're asking, one answer could be: yes, QFT is more non-local than QM, because all finite-energy states in QFT are entangled with respect to location, including the vacuum state. (But all interactions are still local.) Is that what you're asking? $\endgroup$
    – Chiral Anomaly
    Commented Jun 16, 2021 at 0:51
  • 1
    $\begingroup$ You're right that the QFT wavefunction is a function of the field configuration, not of the spacetime coordinates. The wavefunction is a representation of the state, which is nonlocal in the sense that it's entangled with respect to location. But the dynamics is still local in the sense that the equations of motion that govern the time-dependence of the field operators (the operators that act on states) involve only local interactions. "Fundamental quantities" and "fundamental laws" are ambiguous words, just like "local" is ambiguous: all of those words can mean several different things. $\endgroup$
    – Chiral Anomaly
    Commented Jun 16, 2021 at 1:47
  • 1
    $\begingroup$ Here's an example: physics.stackexchange.com/a/542253. The Hamiltonian shown in that answer is local in the dynamic sense: it involves only products of field operators at the same point (or neighboring points, because space is discretized in that example, so derivatives become finite-differences). The state is a function of the field configuration. The Hamiltonian generates time-evolution (of the field operators, in the Heisenberg picture, or of the state, in the Schrödinger picture). The dynamics is local, but the state is not -- and those are two different meanings of "local." $\endgroup$
    – Chiral Anomaly
    Commented Jun 16, 2021 at 1:55
  • 1
    $\begingroup$ It depends on what you mean by "purely local." Since the word "local" can mean different things that are only loosely related to each other, I doubt that any equation in physics would qualify as local according to all of the common meanings of that word -- but that's a language issue, not a physics issue. Words are necessary for communication, but they're not sufficient for communication, and equivocation is just one of the many reasons. $\endgroup$
    – Chiral Anomaly
    Commented Jun 17, 2021 at 0:22
  • 1
    $\begingroup$ No, QFT doesn't break it. The equations of motion for QFT are differential equations that apply separately in an arbitrarily small neighborhood of each point of spacetime. (The neighborhood allows room for defining derivatives, just like in classical field theory.) The QFT equations of motion are local in this sense, and this is the way most QFT-ists normally use that word. $\endgroup$
    – Chiral Anomaly
    Commented Jun 17, 2021 at 3:51

1 Answer 1

Reset to default
6
$\begingroup$

Entanglement doesn't break locality. Entanglement must be generated locally. Wavefunction collapse a la Copenhagen interpretation breaks locality.

QM and QFT are equivalent with respect to to locality. If you don't allow collapse in your interpretation then they are both unitary and local theory. If you do allow collapse then they are local until an entangled state collapses, at which point they are [something complicated that might be non-local depending on the exact philosophical details of your interpretation].

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.