# I Why must force-carrying particles be virtual?

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1. Aug 16, 2016

### referframe

Why must photons, for example, that carry the EM force be virtual? Is it because they are tied to gauge invariance and that is not observable?

2. Aug 16, 2016

### dextercioby

"Carriers of a certain force" are internal lines in Feynman diagrams, they are not on the mass sheet, because at each vertex the 4-momentum must be conserved. Gauge invariance produces observable effects, see the https://en.wikipedia.org/wiki/Aharonov–Bohm_effect

3. Aug 18, 2016

### haael

Carriers of force are not particles. They are fields.

There are configurations of fields that have changing potential. They do not obey equations of motion thus are not particles. But the gradient of potential means they extert forces.

And there are configurations of fields that have zero (mean) potential, but at the same time they obey wave equation = equations of motion. They are particles, but they don't carry forces.

A very useful analogy: you have a guitar string, attached to some point. You can take the string into your hand and pull it constantly. It will extert force. On the other hand, you can pluck the string. It will make a sound.
Pulling the string is analogous to force-carrying field. Sound (waves traveling on the string) is analogous to particles, excitations of the field.

TL;DR: Forces arise from non-wave non-constant solutions of fields, particles arise from wave solutions.

Remember: force carriers are not particles.

What's the fuss about virtual particles then? Well, they are a mathematical trick. You can express a non-wave configuration as a weighted sum of wave configurations. This is a procedure somewhat analogous to Taylor expansion or better Fourier expansion. You take a subset of configurations (wave solutions only) and use it to construct any other configuration.
The waves used in this construction are technically particle-like solutions, that's why we call them virtual particles. But they are not particles. They don't obey equations of motion. They are simply a mathematical re-expression of a non-particle field configuration.

Why are we doing this? Well, particles have a simple commutator relation. We know how to compute a commutator between two pure waves. We don't know instantly how to compute commutator between everything else. Until we express everything as a weighted sum (or integral) of particle configurations.

This is how Feynmann's quantization works.

4. Aug 18, 2016

### referframe

Photons are the gauge bosons for the EM field. I understand that in QFT, all particles, including photons, are characterized as excitations of a field. But, are they still not particles in the QFT sense? Isn't that what the Second Quantization was all about?

5. Aug 18, 2016

### haael

Real photons are excitations of a field (particles) - yes.
Virtual photons are excitations of a field - no.
Force-carrying EM field around an electron is an excitation of a field - no.

Only real particles are excitations.

6. Aug 19, 2016

### naima

Suppose that the carrier is a on shell particle. we would need another theory to explain how it attracts other particles!

7. Aug 19, 2016

### vanhees71

Yes, they are particles in the QFT sense and only in the QFT sense. Note that particularly for photons any idea to think of them as particles in the classical sense is doomed to misunderstanding. Note that you cannot even define a position observable for photons!

8. Aug 19, 2016

### referframe

OK, but is a force-carrying photon's "virtualness" due to local gauge invariance basically being unobservable (forgetting for the moment about the AB effect) ?

9. Aug 19, 2016

### haael

I would say no.

10. Aug 20, 2016

### vanhees71

Of course, the "virtual photons" are observable in the sense that you can observe the Coulomb force between static charges. This example shows you that "photons" (particularly "virtual photons") are very far from what you'd consider a "particle" in everyday life. The only way to understand what a photon is, is QED, and the only correct intuition (as far as we know today) is the underlying mathematics of relativistic quantum field theory.