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Bell's theorem, together with experiments confirming the ordinary quantum mechanical distant correlation predictions which violate its conclusion for local theories, show that any entirely local theory is incorrect. QFT's microcausality postulate forbids an experiment at one space-time point from influencing the probabilities of an experimental outcome at a distant spacelike-related point, but doesn't forbid those distant correlations.

Does the current QFT of the Standard Model actually predict those distant correlations, or if not, does it at least allow them? It is difficult for me to determine this, since I have never taken a QFT course, and QFT is usually stated in the Heisenberg picture rather than the Schrödinger picture. If QFT is entirely local, and forbids those distant correlations, is this viewed as a serious flaw in it, perhaps as serious as the failure, so far, to make any quantum theory fully compatible with general relativity?

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The pair of entangled photons violating Bell's inequality are perfectly described by a state of the EM field containing two particles. Therefore QFT gives rise to non-local correlations, just in view of the said violation.

There is no problem with micro causality which refers to observables: localized field operators (if they represent observables as is the case for boson fields). They must commute when they are associated to spatially separated regions of the spacetime. Bell's inequality instead concerns outocomes of measurements in peculiar non-local (entanlged) states. Micro causality does not forbid non-local correlations.

DETAILS

The meaning of locality aka micro causality in QFT can be illustrated as follows.

Consider two regions $A$, $B$ in Minkowski spacetime which are spacelike related and a pair of corresponding (quantum-field) observables $O_A$ and $O_B$ localized at those regions respectively.

Let $P_A$ and $P_B$ be the orthogonal projectors of the spectral measures of $O_A$ and $O_B$ respectively, corrsponding to the oucomes $E_A$ and $E_B$, respectively.

Let $\Psi$ be a normalized vector representing the state of the system.

The locality axiom requires that (*) $$P_AP_B \Psi = P_BP_A \Psi\tag{1}$$ which, in particular, implies that $$||P_AP_B \Psi||^2 = ||P_BP_A \Psi||^2\:.$$ In other words, the probability to measure the outcome $E_A$ first and the outcome $E_B$ next on the state $\Psi$, -- when measuring first $O_A$ and next $O_B$ -- equals the probability to measure the outcome $E_B$ first and the outcome $E_A$ next on the state $\Psi$ -- when measuring first $O_B$ and next $O_A$.

This requirment is very natural since the temporal order of $A$ and $B$ cannot be fixed because these regions are spacelike related.

This natural requirement is in perfect agreement with the setup of Bell experiment. There the quantum field is the EM one and we measure $4$ couples of photon-polarization observables localized in the regions $A$ and $B$ as above

$O_A,O_B\quad$, $O_A,O'_B\quad$, $O'_A,O_B\quad$, $O'_A,O'_B\quad$.

The state $\Psi$ is very peculiar:

(a) it includes two particles only,

(b) it is a special case of an entangled state.

The Bell (actually CHSH) inequality concerns a relation among the expectation values

$\langle \Psi|O_AO_B\Psi\rangle\quad$, $\langle \Psi|O'_AO_B\Psi\rangle\quad$, $\langle \Psi|O_AO'_B\Psi\rangle\quad$, $\langle \Psi|O'_AO'_B\Psi\rangle\quad$.

If there were a (realistic-local) hidden-variable theory capable to explain the results of the above measurment procedures of quantum theory, the expectation values above should satisfy a certain inequality (the CHSH inequality).

Instead, quantum (field) theory predicts -- through an elementary computation -- that this inequality is violated.

DISCUSSION

The crucial fact that, together with the choice of the state (see below), produces the violation is the choice of the pairs $O_A, O'_A$ and $O_B, O'_B$: these pairs are made of certain incompatible observables. Though the observables of each such pair are never simultaneously measured, the total statistics feels this quantum incompatibility and produces the violation.

I finally stress that locality in the hypotheses of Bell inequality is used in a different way with respect to its use in QFT: it means that the result of the mesurement of, e.g., $O_A$ does not depend on the choice of the observable $O_B$ or $O'_B$ simultaneously measured, since this choice is made in a causally separated region.

This requirment holds true in classical physics, but not in quantum physics and the (entangled) quantum states are responsible for that, not the observables.


(*) The postulate is more usually stated in terms of bosonic field operators as

$$[\phi(f),\phi(g)]=0\tag{2}$$ if $\mathrm{supp}(f) \subset A$ and $\mathrm{supp}(g) \subset B$ with $A$ and $B$ spacelike related and $f$, $g$ real.

Assuming the field operators selfadjoint (or taking the closures of the standard operators) and considering the von Neumann algebras generated by them when $f$ and $g$ are taken as above, the commutation relation above extends to all elements of these algebras. These algebras includes the spectral projectors of every observable constructed as said. Therefore the standard requirement imples (1). Conversely (1) immediately produces the standard requirement (2) (on a suitable dense invariant domain) when the considerd observables are the field operators themselves.

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  • $\begingroup$ To Valter Moretti- You have some nice mathematics in your comment, but seem to completely fail to understand my question. Most of your comment is devoted to lecturing me on the fact that microcausality does not forbid non-local correlations. If you will read the last sentence of my first paragraph with understanding, especially the last 6 words, you will see that I said just that. My first paragraph serves to establishes a few facts. My question is contained in the second paragraph, which you totally ignore. By the way, EM fields, described by Maxwell, don't contain any particles. $\endgroup$ – Michael Fox Aug 19 at 0:44
  • $\begingroup$ Sorry, I do not understand and actually I cannot see your question. I mean a question whose answer is not contained in my post. Essentially I tried to answer the title of your post. Maybe it depends on the fact that English is not my mother tongue. I hope my post is useful for other readers. $\endgroup$ – Valter Moretti Aug 19 at 6:41
  • $\begingroup$ To Valter Moretti- My question was essentially what was in its title, although in SE posts such is often not the case. I asked whether non-locality (in the Bell sense) was compatible with QFT, even though it is not ruled out by microcausality, because MC is not all of QFT. As far as I know, there might be, somewhere in QFT, some rule or equation, such as a propagation equation, that rules out non-locality even though MC doesn't. On the other hand, somewhere in QFT there might be something that implies non-locality. There might even be both, since QFT might be inconsistent. $\endgroup$ – Michael Fox Aug 20 at 19:38
  • $\begingroup$ It seems to me that I actually answered, maybe not as directly as you wanted. QFT of EM field implies the theory of couples of photons used in Bell's argument. Therefore the statement "there might be, somewhere in QFT, some rule or equation, such as a propagation equation, that rules out non-locality even though MC doesn't" is false when the notion of non-locality is the one of Bell's argument. $\endgroup$ – Valter Moretti Aug 20 at 19:41
  • $\begingroup$ Roughly speaking, and details are written in my aswer, the notion of locality in QFT concerns observables, whereas the one in Bell's argument concerns states. There is no inconsitence in QFT arising from these notions and facts. $\endgroup$ – Valter Moretti Aug 20 at 19:45

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