Strictly speaking in eletroweak theory it is not an asymptotic state.vanhees71 said:Of course it does. Why shouldn't it? It's well observable after all!
I refer you back to post #21 which says otherwise.vanhees71 said:Of course it does. Why shouldn't it? It's well observable after all!
Particle states on the physical Hilbert space (BRST cohomology) do not transform under gauge transformations. Fields acting on the entire Krein space do.Michael Price said:I agree, nothing measurable changes with a gauge transformation, but I don't want to get diverted off-topic here. To return to the U238 (or W boson) I got the distinct impression that some people are saying the U238 does not appear in an asymptotic state.
Michael Price said:Well, that is exactly how I do think of a virtual particle, namely as something flying about.
Michael Price said:If seems to me that it is the asymptotic in/out states that are fictions.
You should not be so dogmatic about interpretational issues. Virtual particles are off shell, and real particles are on shell. The on-shell relations are a sign of classicallity. Is any particle on the mass shell exactly? No. So you could equally argue that it is the virtual particles that exist and the real particles that are fiction.Vanadium 50 said:Well, that's wrong and you should stop doing that.
No state in the Hilbert space has energy-momentum relations like those of virtual particles. This can be proven directly from the Wightman axioms.Michael Price said:You should not be so dogmatic about interpretational issues. Virtual particles are off shell, and real particles are on shell. The on-shell relations are a sign of classicallity. Is any particle on the mass shell exactly? No. So you could equally argue that it is the virtual particles that exist and the real particles that are fiction.
Which begs the question a bit, and thus irrelevant to my point.DarMM said:No state in the Hilbert space has energy-momentum relations like those of virtual particles. This can be proven directly from the Wightman axioms.
I don't understand. All states of a quantum theory are elements of the Hilbert space. No elements of the Hilbert space have this relation.Michael Price said:Which begs the question a bit, and thus irrelevant to my point.
Michael Price said:So you could equally argue that it is the virtual particles that exist and the real particles that are fiction.
No, just pointing out there are two sides to some interpretational issues. Have a good day.Vanadium 50 said:Sorry. Thought you were asking a question. Didn't realize you were here to tell everyone they are doing it wrong.
Whether virtual particles exist is not mathematics but interpretation.vanhees71 said:These are no interpretational issues, but mathematical facts about QFT.
No, I agree with @vanhees71 that this is not interpretational.Michael Price said:Whether virtual particles exist is not mathematics but interpretation.
Can you conduct an experiment to say whether a virtual particle exists or not? We can certainly feel their effects; what more is needed? As I said, I think of them as something flying about, and see no reason to drop this viewpoint. YMMV.vanhees71 said:What do you mean by exist? "Virtual particles" are internal lines in Feynman diagrams with a clear mathematical meaning. Feynman diagrams are condensed notations for formulae to calculate perturbatively S-matrix elements and cross sections or decay rates. A propagator does not represent a state. So it's not clear what you think what should "exist" or being "real". For sure it's not representing a particle that is detectable in any way.
Michael Price said:We can certainly feel their effects
Michael Price said:and see no reason to drop this viewpoint
Michael Price said:Can you conduct an experiment to say whether a virtual particle exists or not?
Because:Michael Price said:Can you conduct an experiment to say whether a virtual particle exists or not? We can certainly feel their effects; what more is needed? As I said, I think of them as something flying about, and see no reason to drop this viewpoint. YMMV.
But does appear in the asymptotic in/out states.DarMM said:I mean look at what they are. In a QFT if ##\phi## represents a generic field in the Lagrangian, then a "virtual particle" is simply a drawing of:
$$\langle \phi_{0}(x)\phi_{0}(y) \rangle$$
i.e. a two-point correlator of the free field, a field that doesn't even appear in the Langrangian.
It doesn't. The asymptotic fields are not the ones that appear in the perturbative calculations.Michael Price said:But does appear in the asymptotic in/out states.
How is that relevant? The asymptotic states have the interaction switched off.DarMM said:It doesn't. The asymptotic fields are not the ones that appear in the perturbative calculations.
You said the free field in the perturbative calculations appeared in the asymptotic states. This is wrong. The free fields related to the asymptotic states are not the same free fields one uses in the perturbative series.Michael Price said:How is that relevant? The asymptotic states have the interaction switched off.
You have a soiurce? Weinberg, preferably. Or explain how they differ.DarMM said:You said the free field in the perturbative calculations appeared in the asymptotic states. This is wrong. The free fields related to the asymptotic states are not the same free fields one uses in the perturbative series.
Just look at QCD. The asymptotic states are associated with color neutral fields, but the fields in the perturbative series carry color charge.Michael Price said:You have a soiurce? Weinberg, preferably. Or explain how they differ.
Can you find and paste a Feynman diagram to illustrate this?DarMM said:Just look at QCD. The asymptotic states are associated with color neutral fields, but the fields in the perturbative series carry color charge.
Due to infrared ìssues related to confinement the asymptotic states in QCD are fundamentally non-perturbative.Michael Price said:Can you find and paste a Feynman diagram to illustrate this?