# A Massive gauge bosons in QFT in/out states

#### 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:

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.

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#### DarMM

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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.

#### PeterDonis

Mentor
Thread closed for moderation.

Edit: the thread will remain closed

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