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We have virtual particles in quantum field theory (QFT). In general, they don't have the need to obey causality.

My question is:

Do the processes in QFT (electron self-energy, photon self-energy, electron-photon vertex, etc.) have to obey causality?
For example, can some parts of the electron self-energy diagram fall out of the light cone?
Or can some parts of the electron-positron annihilation process fall out of the light cone?

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No, to my knowledge causality is a crucial part of quantum field theory. Everything that can be measured has to obey causality. For bosons, this corresponds to the fact that $\langle 0|[\phi(x), \phi^*(y)]|0 \rangle$ is always zero at space-like separation. For fermions, we know that $\langle 0| \lbrace \psi(x), \bar{\psi}(y) \rbrace |0 \rangle$ is zero outside the light cone. Since operators such as charge, energy, and momentum always involve an even number of spinor fields, this is enough to ensure that $[\mathcal{O}_1(x), \mathcal{O}_2(y)]$ is also zero outside the light-cone. Which says no two measurements can effect each other at space-like separation.

By the way, I don't have a mathmatical proof of how the anticommutation relation is related to the commutator of the operators. I would love to see one.

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  • $\begingroup$ physics.stackexchange.com/questions/159308/…: check this link virtual particles can violate causality and amplitude never dies off @tomtom1-4 $\endgroup$ – QFT addict. Oct 3 '20 at 17:05
  • $\begingroup$ Amplitudes can be non-zero outside the light-cone. But two real measurements can not effect each other at space-like separation. So in a sense, you could argue that virtual particles can interact somewhat outside the light-cone but since you can not measure them it does not really matter. $\endgroup$ – tomtom1-4 Oct 3 '20 at 17:10

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