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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.tom.stoer said:Decohence is due to factorizing the full Hilbert space H in Hsystem, Hpointer and Henvironment 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 Hsystem, Hpointer or Henvironment , but they are "integrals over propagators".
It' like apples and oranges.
I have a better analogy. If you have one apple, then in the equationtom.stoer said:It' 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.Demystifier said:Or in slightly oversimplified terms, virtual particles don't cause decoherence simply because virtual particles don't exist.
See my postmfb said: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.
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.Demystifier said: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?mfb said: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
mfb said: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.sheaf said:)... 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:tom.stoer said: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.sheaf said: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.
If you call everything "unphysical" which cannot be "observed" in non-interacting particles, sure. That is just playing with words. The near field of an antenna would be unphysical then, as electric and magnetic fields do not follow the rules for radiation (=light particles).tom.stoer said:No, the main point are unphysical properties of virtual particles
JK423 said:I haven't seen anywhere a discussion of what happens during an interaction (not just in/out states). Why doesn't it make sense (theoreticaly at least)?
It seems that you are not familiar with the meaning of "unphysical" in the context of gauge theories. The 0-component of the gauge field A is an unphysical d.o.f. b/c it has no associated canonical momentum; states in the kinematical Hilbert space are unphysical if they are not annihilated by the Gauß constraint (= if they are not in the kernel of the Gauß Law operator = if they are not gauge singulets); Fadeev-Popov ghosts are unphysical d.o.f. b/c they are artificial d.o.f. Introduced to eliminate other unphysical d.o.f. (polarizations) of the gauge fied.mfb said:If you call everything "unphysical" which cannot be "observed" in non-interacting particles, sure. That is just playing with words.).
If i see a multitude of little green dwarfs all around me for a life time, i might be inclined to believe they are real and exist. The mathematics wouldn't work if it had no resemblance to reality. Why would it work otherwise? Just a happy coincidence?mattt said:Imagine I develop a new mathematical formalism, that is a good enough mathematical approximation to, say, Newtonian Mechanics, based on a given mathematical Serie.
Imagine I call each element of the Serie, "a little green dwarf", because I like it.
Would you say that those "little green dwarfs" are "real" or "physical"?
Would you say that gravity exists because of the actions of those "little green dwarfs"
Mukilab said:I was recently told virtual particles don't cause decoherence. Why not?
tom.stoer said:Decohence is due to factorizing the full Hilbert space H in Hsystem, Hpointer and Henvironment 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 Hsystem, Hpointer or Henvironment , but they are "integrals over propagators".
tom.stoer said:[one] can formulate QFT [and non-rel. QM] non-perturbatively w/o Feynman diagrams;
... there is no need to introduce perturbation theory and propagators when studying density operators.
... attributes of internal lines are gauge dependent whereas our observations aren't.
There are internal lines which are exactly on-shell.mfb said: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.
Some particles are just more off-shell than others.
tom.stoer said:I guess we should come back to the question
Neither QED nor QCD require perturbation theory or virtual particles. In QED non-perturbative aspects are either discused using rel. QM + radiative corrections a la Lamb shift, or they are not relevant at all (due to small alpha 1/137 and abelian gauge symmetry). So virtual particles are common standard and mostly sufficient in QED, but not required. One can quantize QED non-perturbatively w/o using virtual particles.TrickyDicky said:Hi Tom, I tend to sympathize with the way you present the answer to this FAQ, but I have a doubt about the "virtual particles are just artifacts of perturbation theory" issue, you often use the example of QCD where this methodology is not necessary, but if we consider only QED ("the jewel of the physics crown") for a moment it does seem that perturbation is needed to obtain reasonable results, so in that sense at least for QED VP seem like something you can't get rid of so easily.