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I was recently told virtual particles don't cause decoherence. Why not? Do they just never interact with their environment (apart from transferring energy/force) so they can never collapse a wavefunction?
Or in slightly oversimplified terms, virtual particles don't cause decoherence simply because virtual particles don't exist.Decohence is due to factorizing the full Hilbert space H in H_{system}, H_{pointer} and H_{environment } and then "tracing out" the environment degrees of freedom. The remaining "subsystem" can be described by an "effective density matrix" which is nearly diagonal in the pointer basis, so it seems as if it collapsed to the a pointer state with some classical probability.
Virtual particles are artifacts of perturbation theory, i.e they are not present in the full theory w/o using this approximation. Using virtual particles does not introduce the above mentioned factorization of H. And last but not least virtual particles are not states in any Hilbert space H_{system}, H_{pointer} or H_{environment }, but they are "integrals over propagators".
It' like apples and oranges.
I have a better analogy. If you have one apple, then in the equationIt' like apples and oranges.
If they do not exist, please provide a more appropriate way to describe all particles ever detected. They are all virtual, see Bill_K's post (or this one from me) for an explanation.Or in slightly oversimplified terms, virtual particles don't cause decoherence simply because virtual particles don't exist.
See my postIf they do not exist, please provide a more appropriate way to describe all particles ever detected. They are all virtual, see Bill_K's post (or this one from me) for an explanation.
Where is the difference? An internal line in a Feynman diagram is not exactly on-shell, and particles not exactly on-shell are internal lines in Feynman diagrams.According to another (more meaningful, in my opinion), it is any internal line in a Feynman diagram. The two definitions are not equivalent.
Are you aware of the fact that you can formulate QFT non-perturbatively w/o Feynman diagrams? Do you see any relevance for propagators in non-rel. QM and density operators?A mathematical artifact like our world?
In the QFT sense of real particles, do you see* any real particles in the world?
*actually, you must not be able to see it, as it must not interact with anything
But isn't that the essence of the difficulty here? Namely:A mathematical artifact like our world?
In the QFT sense of real particles, do you see* any real particles in the world?
*actually, you must not be able to see it, as it must not interact with anything
Very good point, the photon propagator does not carry momentum in the sense we measure it.)... For example, for a photon, I'd like to be able to prepare/measure its momentum/polarization, but the photon propagator [itex]-i\frac{g^{\mu\nu}}{k^2+i\epsilon}[/itex] doesn't have the right ingredients to allow me to do this.
I am aware of that. Could you answer my question, please? It would help me to understand where our views differ:Are you aware of the fact that you can formulate QFT non-perturbatively w/o Feynman diagrams? Do you see any relevance for propagators in non-rel. QM and density operators?
You got my point. I don't think there is a fundamental difference between an electron measured in a detector and a W boson mediating a weak decay.Yes, in reality, they may be slightly off-shell. They must be since they haven't been around for an infinite time. In this sense the model is an idealization.
But "rather close" is the main point. There is no fundamental line separating particles we observe from the virtual W in a weak decay.But what we observe in reality is rather close to these idealized asymptotic states, so it's OK to try to interpret them as a mathematical model of reality.