A Massive gauge bosons in QFT in/out states

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Summary
Are the massive gauge bosons of electroweak theory (W, Z) permitted in the external legs of Feynman diagrams?
Because massive gauge bosons have a finite half life, are they excluded from the (infinitely, asymptotically remote?) in and out states of QFT? Or, to put it another way, are they restricted to the internal legs of Feynman diagrams, i.e. to being virtual only? We can see W and Z tracks in bubble or cloud chambers, is this sufficient to guarantee their "real" existence?
 
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We can see W and Z tracks in bubble or cloud chambers, is this sufficient to guarantee their "real" existence?
For most people's definition of "real", this would seem to be sufficient, yes. :wink:
 

Vanadium 50

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vanhees71

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AFAIK they plotted "Lego diagrams" for electron-positron pairs from the decay of Z bosons using em. calorimeters. The Z-bosons are resonances and thus not asymptotic free states. You see them as peaks in invariant-mass plots of dileptons (electron-positron, or muon-antimuon pairs) in reactions like ##\text{hadrons} \rightarrow \ell^+ + \ell^- + X##.

For details on how the weak gauge bosons have been discovered, see Rubbia's Nobel lecture:

 
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AFAIK they plotted "Lego diagrams" for electron-positron pairs from the decay of Z bosons using em. calorimeters. The Z-bosons are resonances and thus not asymptotic free states. You see them as peaks in invariant-mass plots of dileptons (electron-positron, or muon-antimuon pairs) in reactions like ##\text{hadrons} \rightarrow \ell^+ + \ell^- + X##.

For details on how the weak gauge bosons have been discovered, see Rubbia's Nobel lecture:

I was thinking more of the charged W bosons. Being charged, should they not leave tracks?
 

vanhees71

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Hm, the W has a decay width of about ##\Gamma \simeq 2 \; \text{GeV}##. It's mean lifetime thus is ##1/\Gamma \simeq 0.2 \mathrm{GeV} \mathrm{fm}/(2 \mathrm{GeV}) =0.1 \text{fm}##. Thus it would leave a track of at most ##0.1 \; \text{fm} =10^{-16} \text{m}##, which is pretty hard to observe ;-)).
 
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Hm, the W has a decay width of about ##\Gamma \simeq 2 \; \text{GeV}##. It's mean lifetime thus is ##1/\Gamma \simeq 0.2 \mathrm{GeV} \mathrm{fm}/(2 \mathrm{GeV}) =0.1 \text{fm}##. Thus it would leave a track of at most ##0.1 \; \text{fm} =10^{-16} \text{m}##, which is pretty hard to observe ;-)).
Hah, I was about do the same calculation but you beat me to it. Yes, pretty hard to observe.... in principle observable, though.
 

Vanadium 50

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.... in principle observable, though.
Oh dear. This is going to be one of those threads.

Can we agree that a bubble or cloud chamber track can be no shorter than a single atom? What is the size of an atom compared to a track?
 
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Oh dear. This is going to be one of those threads.

Can we agree that a bubble or cloud chamber track can be no shorter than a single atom? What is the size of an atom compared to a track?
Well, let's put it back on track then; can W and Z states appear in the asymptotic in/out states in QFT? Seems to me they can. What do you think?
 

vanhees71

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No they can't since they are decaying! A resonance is a resonance is a resonance...
 

DarMM

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Summary: Are the massive gauge bosons of electroweak theory (W, Z) permitted in the external legs of Feynman diagrams?

i.e. to being virtual only?
As has been said they won't because they are resonances. However being a resonance is not the same thing as a virtual particle. Resonances correspond to states in the Hilbert space, just not ones that survive to asymptotic infinity. Virtual lines however don't correspond to any element of the Hilbert space, they're purely part of a term in the perturbative expansion.
 
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As has been said they won't because they are resonances. However being a resonance is not the same thing as a virtual particle. Resonances correspond to states in the Hilbert space, just not ones that survive to asymptotic infinity. Virtual lines however don't correspond to any element of the Hilbert space, they're purely part of a term in the perturbative expansion.
Would a U238 atom appear in an in/out state? I presume yes, but how does that differ from a W boson?
 
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vanhees71

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Strictly speaking not ;-)). Of course, an U238 atom lives "a bit" longer than the W boson though...
 

DarMM

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Would a U238 atom appear in an in/out state? I presume yes, but how does that differ from a W boson?
It wouldn't. Of course these things can be theory dependent. A particle could be stable in one theory, but a resonance in a more extensive theory that includes more of its interactions.
 
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So a U238 atom can't appear in an asymptotic in/out state?
Okay, I always suspected that the in/out states were mythical and that it is the virtual particles that comprise reality...
 

DarMM

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So a U238 atom can't appear in an asymptotic in/out state?
Okay, I always suspected that the in/out states were mythical and that it is the virtual particles that comprise reality...
The in/out states aren't mythical, they're the particles that survive to asymptotic infinity.

Also U238 and the W bosons aren't virtual but resonances, that's the point I was making in #12. There's a difference between resonances and internal virtual lines in graphs.
 

vanhees71

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Which difference? Do mean that resonances are discovered in s-channel contributions, when measuring invariant-mass spectra of "their decay products"?
 

DarMM

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Which difference? Do mean that resonances are discovered in s-channel contributions, when measuring invariant-mass spectra of "their decay products"?
Virtual lines are just terms in the perturbative expansion. Resonances are actual states in the Hilbert space. In other words resonances would appear in a non-perturbative formulation, virtual particles would not.
 

A. Neumaier

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Virtual particles do not correspond to states. See https://www.physicsforums.com/insights/physics-virtual-particles/, where the difference to unstable particles and resonances is explained.
Resonances correspond to states in the Hilbert space, just not ones that survive to asymptotic infinity.
Small correction: Resonances do not correspond to states in the Hilbert space but to unnormalizable states in the dual of a nuclear space whose completion is the Hlbert space.
 

A. Neumaier

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So a U238 atom can't appear in an asymptotic in/out state?
Not in the standard expansion. But one can analytically continue the standard expansion into asymptotic states (more precisely, the scalar product and the Möller operator). By deforming the integration contour in the scalar product (in a representation of the asymptotic space where the energy is diagonal) from the real line into the so-called unphysical sheet (of the Riemann surface associated with the resolvent), one gets an expansion in terms of long-living resonances.

If the half-life is very long (such as for U238), hardly any deformation is needed and one can approximate the contour by the real line, thus getting an approximate expansion in which the decay is neglected. But for short-lived resonances such as W or Z, the deformation of the contour is very substantial and the approximation bcomes ridiculously poor.

On the other hand, virtual particles can in no way be approximated in this sense.
 
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dextercioby

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Virtual particles do not correspond to states. See https://www.physicsforums.com/insights/physics-virtual-particles/, where the difference to unstable particles and resonances is explained.

Small correction: Resonances do not correspond to states in the Hilbert space but to unnormalizable states in the dual of a nuclear space whose completion is the Hlbert space.
As a reference for the second part, one just has to look for "Gamov vectors" (the Americans spell Gamow) and the work of Arno Böhm and his coworkers. The standard work on rigged Hilbert spaces (the PhD thesis of Rafael de la Madrid) has a chapter on Gamov vectors, as far as I remember.
 

DarMM

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Small correction: Resonances do not correspond to states in the Hilbert space but to unnormalizable states in the dual of a nuclear space whose completion is the Hlbert space.
I just want to understand this better. So if we take a reasonably long lived particle like a free neutron, ultimately it is still a resonance. What's going on during those ~15 minutes? Surely there is a state representing what is occurring in some manner. Or do you mean there is no state directly corresponding to the resonance pole on the second sheet of the S-matrix's analytic continuation.
 

vanhees71

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Virtual lines are just terms in the perturbative expansion. Resonances are actual states in the Hilbert space. In other words resonances would appear in a non-perturbative formulation, virtual particles would not.
Resonances are indeed states in the continuous-spectrum part of the Hamiltonian. The most simple example from QM 1 is a particle moving in a finite-depth attractive potential pot in 1 dimension:

https://itp.uni-frankfurt.de/~hees/qm1-ss09/wavepack/node19.html

In QFT it's an internal line in a Feynman diagram (in the case that the corresponding "particles" are described as an elementary field in the Lagrangian). As such it's as "virtual" as any internal line in a Feynman diagram. More generally, it's a complex pole in the energy argument of some Green's function. E.g., in the pion-nucleon elastic scattering amplitude you have some prominent resonances like the Δ(1232)Δ(1232).

There's a lot of confusion about what a resonance really is, leading sometimes to long debates. E.g., "what's a ρρ meson"? According to the particle data group, it's a strong resonance state in the process e++e−→π++π−e++e−→π++π−. Then it has a well-defined mass and width of around 770MeV770MeV and 150MeV150MeV respectively. So looking at it in the point of view of hadron physics, you'd say "it's a two-pion resonance". Of course you see the corresponding peak also in the corresponding spin-isospin channel in elastic ππππ scattering. Now you build a model of pions, ρρ mesons and couple some baryons too. Now you can easily populate "ρρ-meson states" below the two-pion threshold in reactions like πN→ρNπN→ρN. There's a phenomenolgically quite well working model, where all couplings of leptons to hadrons go only through ligth vector mesons (it's by Sakurai from the 60ies and known as the vector-meson dominance model). Now building such a model for dielectron production, you all of a sudden see a "modified ρρ-meson lineshape" just from coupling the baryons on top of the pions to the ρρ meson. In other words: What's a ρρ meson is in some sense model dependent. You can as well use another effective model describing the hadron physics and take care of some dielectron production channels through other "mechanisms", which maybe work as well. What you can at the end only measure is how some incoming hadrons react somehow and produce a dielectron pair. Whether or not you identify certain structures in the dielectron-invariant-mass spectrum as a "##\rho##-meson" is a question of your interpretation in terms of some effective model. That's why it's in general not consistent to just mix "resonances" from different models into a new model. You have to carefully fit all the parameters (couplings, masses, etc.) to the model you are using since you can shuffle strength of the same scattering processes to run through different resonances, and there are tons of hadron resonances around, whose branching ratios to various decay channels are not well known. Thus there's a lot of freedom in modeling, and with new data from experiments it may well be that you have to refit your model!
 
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DarMM

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In QFT it's an internal line in a Feynman diagram (in the case that the corresponding "particles" are described as an elementary field in the Lagrangian). As such it's as "virtual" as any internal line in a Feynman diagram
I agree that's how it appears in the perturbative formalism. My point to @Michael Price was more that this is simply a perturbative representation and you shouldn't think of virtual particles actually flying about.
 

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