A Massive gauge bosons in QFT in/out states

[Michael Price]

This is untenable to me because the asymptotic particles are what actually registers in our detectors and virtual particles don't even appear in a non-perturbative formalism and depending on how you perform the perturbative expansion you get virtual particles with different properties.

Particles also change with gauge transformations, but we have learned to get used to that. But to return to the asymptotic business. I can pick up a U238 atom and mail to someone on the other side of the world - clearly there is more to reality than asymptotic in/out states.

 

[DarMM]

Particles also change with gauge transformations, but we have learned to get used to that. But to return to the asymptotic business. I can pick up a U238 atom and mail to someone on the other side of the world - clearly there is more to reality than asymptotic in/out states.

I never said there wasn't, I said:

1. The asymptotic states are real, i.e. they're not fictions.
2. Virtual particles are not part of reality as they only appear in the perturbative formalism.

What particles change with gauge transformations?

 

[vanhees71]

Particles also change with gauge transformations, but we have learned to get used to that. But to return to the asymptotic business. I can pick up a U238 atom and mail to someone on the other side of the world - clearly there is more to reality than asymptotic in/out states.

No! Nothing changes with gauge transformations. To the contrary gauge transformations simply transform from one description of a situation to another complete equivalent one. That's why there's no spontaneous symmetry breaking possible for a local gauge symmetry, and that's why the ground state is not degenerate, and that's why in breaking a local gauge symmetry doesn't imply the existence of Goldstone modes.

What, in your opinion is a U238 atom else, in the quantum description, than an asymptotic free state?

 

[Michael Price]

No! Nothing changes with gauge transformations. To the contrary gauge transformations simply transform from one description of a situation to another complete equivalent one. That's why there's no spontaneous symmetry breaking possible for a local gauge symmetry, and that's why the ground state is not degenerate, and that's why in breaking a local gauge symmetry doesn't imply the existence of Goldstone modes.

What, in your opinion is a U238 atom else, in the quantum description, than an asymptotic free state?

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.

 

[vanhees71]

Of course it does. Why shouldn't it? It's well observable after all!

 

[DarMM]

Of course it does. Why shouldn't it? It's well observable after all!

Strictly speaking in eletroweak theory it is not an asymptotic state.

In other theories it is. That's why what is asymptotic is theory dependent.

 

[Michael Price]

Of course it does. Why shouldn't it? It's well observable after all!

I refer you back to post #21 which says otherwise.

 

[DarMM]

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.

Particle states on the physical Hilbert space (BRST cohomology) do not transform under gauge transformations. Fields acting on the entire Krein space do.

 

[Vanadium 50]

Well, that is exactly how I do think of a virtual particle, namely as something flying about.

Well, that's wrong and you should stop doing that.

If seems to me that it is the asymptotic in/out states that are fictions.

At worst, they are approximations. If you are going to complain about this, you should start with massless pulleys, frictionless planes, spherical earth, etc. Furthermore, it seems a little like cutting off one's nose to spite one's face to reject some of the calculational successes: e.g. the muon magnetic moment to 12 decimal places - because there might be an error in the 20th.

 

[Michael Price]

Well, that's wrong and you should stop doing that.

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.

 

[DarMM]

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.

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]

No state in the Hilbert space has energy-momentum relations like those of virtual particles. This can be proven directly from the Wightman axioms.

Which begs the question a bit, and thus irrelevant to my point.

 

[DarMM]

Which begs the question a bit, and thus irrelevant to my point.

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.

What question is it begging?

 

[Vanadium 50]

So you could equally argue that it is the virtual particles that exist and the real particles that are fiction.

Sorry. Thought you were asking a question. Didn't realize you were here to tell everyone they are doing it wrong.

Sure, you could look at it that way. Indeed, I just said you could. But it is, as I said, a failure to use a convenient approximation that cuts off one's nose to spite one's face.

 

[Michael Price]

Sorry. Thought you were asking a question. Didn't realize you were here to tell everyone they are doing it wrong.

No, just pointing out there are two sides to some interpretational issues. Have a good day.

 

[vanhees71]

These are no interpretational issues, but mathematical facts about QFT.

 

[Michael Price]

These are no interpretational issues, but mathematical facts about QFT.

Whether virtual particles exist is not mathematics but interpretation.

 

[vanhees71]

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.

 

[DarMM]

Whether virtual particles exist is not mathematics but interpretation.

No, I agree with @vanhees71 that this is not interpretational.

No states we detect are like virtual particles. No states in the Hilbert space are like virtual particles.

The latter fact is insurmountable if you want to consider them as real, because they're are not an element of QM's state space.

As @vanhees71 said they are condensed notations for perturbative calculations. If we consider parts of a step in a calculational method as real, then we'd equally think the updating in the MCMC algorithm of a lattice calculation was a real process.

 

[Michael Price]

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.

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.

 

[weirdoguy]

We can certainly feel their effects

What effects? And no, Casimir effect is not the answer since it can be derived without virtual particles. Anyways, this topic has been discussed here many times, and there are even two insight articles about that:
https://www.physicsforums.com/insights/misconceptions-virtual-particles/https://www.physicsforums.com/insights/physics-virtual-particles/

and see no reason to drop this viewpoint

There is no reason to think that they exist in the first place. Mathematics of QFT are quite straightforward when it comes to virtual particles - they are simply propagators appearing in perturbation series. No perturbation series - no virtual particles.

Can you conduct an experiment to say whether a virtual particle exists or not?

There is no need to if you know and understand what is the very definition of a virtual particle. It's not that this notion existed before QFT and then physicists came to model it mathematically. It came after the mathematics were introduced and it's only a name for some part of it.

 

[DarMM]

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.

Because:

1. There is no state in the Hilbert space that corresponds to them. In Quantum Theory all physical states of matter are elements of the Hilbert space. If they're not in it they don't correspond to a physical arrangement of matter.
2. They only appear in one type of computational method and even then they are simply a way of drawing a term.

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.

 

[Michael Price]

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.

But does appear in the asymptotic in/out states.

 

[DarMM]

But does appear in the asymptotic in/out states.

It doesn't. The asymptotic fields are not the ones that appear in the perturbative calculations.

 

[Michael Price]

It doesn't. The asymptotic fields are not the ones that appear in the perturbative calculations.

How is that relevant? The asymptotic states have the interaction switched off.

 

[DarMM]

How is that relevant? The asymptotic states have the interaction switched off.

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]

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.

You have a soiurce? Weinberg, preferably. Or explain how they differ.

 

[DarMM]

You have a soiurce? Weinberg, preferably. Or explain how they differ.

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]

Just look at QCD. The asymptotic states are associated with color neutral fields, but the fields in the perturbative series carry color charge.

Can you find and paste a Feynman diagram to illustrate this?

 

[DarMM]

Can you find and paste a Feynman diagram to illustrate this?

Due to infrared ìssues related to confinement the asymptotic states in QCD are fundamentally non-perturbative.