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In quantum field theory, causality is imposed by demanding that field operators at spacetime points separated by spacelike intervals commute. However, time-ordered field correlation functions between such points can be finite, and is commonly understood to arise due to entanglement. (I read this in a comment by Gerard 't Hooft on https://www.quantamagazine.org/20160517-pilot-wave-theory-gains-experimental-support/)

However, it is possible to imagine situations with classical fields producing finite correlations at spacelike points. Imagine a spherical wave from a classical electromagnetic monochromatic point source. The electric fields at spacelike separated points on a large spherical wavefront can be perfectly correlated.

The question then is, is there a quantitative relation between the strength of correlations at spacelike intervals possible in quantum field theory vs classical field theory? I believe this would be tantamount to a reformulation of Bell's theorem in the language of quantum field theory.

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  • $\begingroup$ I think this question could be rephrased as follows: What is the interpretation of correlation functions in QFT? $\endgroup$ – Blazej Aug 3 '16 at 14:01
  • $\begingroup$ IMHO, the quantitative relation is simply the common use of phase différences. Looking at the momentum décomposition of the Feynmann propagator in position space, for each light-like momentum contribution, you have a imaginary exponential of à phase différence. One may imagine, for each momentum, à kind of source at spatial infinity, both in the past and the future. $\endgroup$ – Trimok Aug 6 '16 at 13:29
  • $\begingroup$ "field correlators between such points can be finite, and is commonly understood to arise due to entanglement..." can you expand on that? Moreover "Imagine a spherical wave from a monochromatic point source. The fields at spacelike separated points on a large spherical wavefront can be perfectly correlated" what fields and what correlators? $\endgroup$ – gented Aug 10 '16 at 7:49
  • $\begingroup$ In QFT, correlation functions don't characterise actual correlations, in the statistical sense of the word. We say "correlation functions" even though they have nothing to do with what we mean when we speak of correlations. We may have a non-zero value for a correlation function between two points, and this doesn't mean that those points are correlated in any practical way (in what sense could those points be correlated after all?) $\endgroup$ – AccidentalFourierTransform Aug 10 '16 at 14:32
  • $\begingroup$ I think the confusion lies with the fact that entanglement is often in a certain sense referred to as quantum correlations. This is different from the two-correlation functions that the question seems to refer to. For entanglement as far as I know one needs a 4-point correlation function. Is this more or less what the question is about? $\endgroup$ – flippiefanus Aug 11 '16 at 18:38
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The type of correlation that is required to violate the Bell inequality is not so much a matter of the strength of the correlation. It requires a type of correlation that is qualitatively different from the correlations in classical theories.

Here one needs to be more specific by what one means by the term correlation. Normally the term correlation refers to a two-point function $\langle \phi_1^* \phi_2 \rangle$. In quantum field theory such two-point functions give rise to propagators or mass.

In classical theories such as stochastic optics two-point functions give the mutual coherence functions that describe the coherence properties in an optical field. In the latter case one can have correlations at two space-like separated points. This would indicate spatial coherence in the optical field. However, this does not violate any causality principles, because the correlation does not represent a causal link. Neither does it represent entanglement.

To investigate the entanglement in a state (which is a requirement to violate the Bell inequality) one needs a different kind of correlation. This is represented by a four-point function $\langle\phi_m^* \phi_n^* \phi_p \phi_q\rangle$, where $m,n,p,q$ label the different degrees of freedom in the field. This four-point function is often represented as a density matrix/operator and denoted by $\rho$. One can now used this density matrix to compute quantities such as the concurrence to quantify the amount of entanglement in the state.

There is a classical analog to quantum entanglement which, is called classical or local entanglement to distinguish it from quantum or nonlocal entanglement. In the classical case the correlation function needs to use two different degrees of freedom (e.g. polarization and spatial mode) from the same field to replace the two different fields used in the quantum case. As a consequence the classical entanglement is always local. Apart form this distinction the two concepts are formally exactly the same. One can use classical entanglement to violate a local version of the Bell inequality, but not the normal nonlocal Bell inequality.

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  • $\begingroup$ But the four-point correlation function contains information about the entanglement in a two-particle system. For instance, the visibility of the interference fringes in the coincidence count rate (four point function) of photon pairs produced from parametric downconversion is a measure of two-particle entanglement. Also, concurrence is a good measure of entanglement only for two-qubit states. For multipartite entanglement, there is no universally accepted robust measure. $\endgroup$ – Girish Aug 14 '16 at 5:58
  • $\begingroup$ So your point that four-point correlation functions quantify the entanglement is well-taken. But then one could argue that four-point correlation functions can be computed even for a completely classical field. The question then remains: are four-point correlation functions for classical fields allowed to take the entire range of values accessible for quantum fields? $\endgroup$ – Girish Aug 14 '16 at 6:06
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    $\begingroup$ @Girish (first comment): Yes that's correct. In my answer I only consider the bipartite qubit case. The situation becomes more complicated with more particles or higher dimensions. For entanglement among more than two particles one would need a corresponding higher order correlation function (n-point function). If you stay with two particles but consider higher dimensions then you would still need a 4-point function, but the dimensions of your density matrix would increase. $\endgroup$ – flippiefanus Aug 14 '16 at 6:22
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    $\begingroup$ @Girish (second comment): Unfortunately, the answer (as far as I understand) is no. For the classical entanglement, the correlation is actually still a two-point function, but which contains two degrees of freedom for each field: $\langle \phi_{m,p}^* \phi_{n,q} \rangle$. The four-point function in classical fields cannot give nonlocal entanglement information, even though it can show some correlations as in the Hanbury-Brown-Twist type experiments. $\endgroup$ – flippiefanus Aug 14 '16 at 6:27
  • $\begingroup$ So since nonlocal entanglement is impossible for classical fields, the classical four-point correlation functions can at best quantify HBT-type correlations. In fact, photon antibunching which leads to $|g_{2}(0)|<1$ is a quintessentially non-classical feature. In fact $|g_{2}(\tau)\geq 1\,\,\forall \tau$ for classical fields, while $g_{2}(\tau)\geq 0$ for a general quantum field. So that is one instance where the range of accessible values for the four-point correlation function is clearly distinct for classical and quantum fields. $\endgroup$ – Girish Aug 14 '16 at 6:44
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The classical equivalent of spacelike commuting operators is fields with spacelike vaishing Poisson brackets - not spacelike vanishing correlators.

Note that the Wightman 2-point functions, which are the quantum analogues of classical correlation functions, also do not vanish at spacelike arguments!

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