Why don't virtual particles cause decoherence?

Mukilab

I was recently told virtual particles don't cause decoherence. Why not? Do they just never interact with their environment (apart from transferring energy/force) so they can never collapse a wavefunction?

mfb

Interaction with real particles can be mediated via virtual particles, and cause decoherence.
I think it is misleading to distinguish between real and virtual particles here.

tom.stoer

Decohence is due to factorizing the full Hilbert space H in H_system, H_pointer and H_environment and then "tracing out" the environment degrees of freedom. The remaining "subsystem" can be described by an "effective density matrix" which is nearly diagonal in the pointer basis, so it seems as if it collapsed to the a pointer state with some classical probability.

Virtual particles are artifacts of perturbation theory, i.e they are not present in the full theory w/o using this approximation. Using virtual particles does not introduce the above mentioned factorization of H. And last but not least virtual particles are not states in any Hilbert space H_system, H_pointer or H_environment , but they are "integrals over propagators".

It' like apples and oranges.

Demystifier

Or in slightly oversimplified terms, virtual particles don't cause decoherence simply because virtual particles don't exist.

John86

they are not existing in the physical (space time) sense !.

Demystifier

I have a better analogy. If you have one apple, then in the equation
1 apple = (2 apples) + (-1 apple)
1 apple is a real apple, while 2 apples and -1 apple are virtual apples.

mfb

If they do not exist, please provide a more appropriate way to describe all particles ever detected. They are all virtual, see Bill_K's post (or this one from me) for an explanation.

Demystifier

See my post
http://www.physicsforums.com/showpos...8&postcount=12

The confusion stems from the unfortunate fact that physicists use two different DEFINITIONS of the word "virtual particle". According to one, it as any off-shell particle. According to another (more meaningful, in my opinion), it is any internal line in a Feynman diagram. The two definitions are not equivalent.

mfb

Where is the difference? An internal line in a Feynman diagram is not exactly on-shell, and particles not exactly on-shell are internal lines in Feynman diagrams.
Some particles are just more off-shell than others.

tom.stoer

Please have a look at the formal definition: an internal line is not a state but a propagator; and it's therefore not a particle

mfb

In that case, our universe has no particles.
There are no particles (!) which will not interact with anything else in the future.

tom.stoer

No, the only problem is that you try to interpret a mathematical artifact

mfb

A mathematical artifact like our world?
In the QFT sense of real particles, do you see* any real particles in the world?

*actually, you must not be able to see it, as it must not interact with anything

tom.stoer

Are you aware of the fact that you can formulate QFT non-perturbatively w/o Feynman diagrams? Do you see any relevance for propagators in non-rel. QM and density operators?

Mukilab asked why virtual particles do not cause decoherence.

The answers is simple: usually there is no need to introduce perturbation theory and propagators when studying density operators. The formalism is different. So there are no propagators in this context, and therefore they cannot cause anything.

sheaf

But isn't that the essence of the difficulty here? Namely:

Our world is a physical entity, including the ability to measure things. One of the things that can be measured is the state of incoming/outgoing particles in a scattering experiment. In the model, this corresponds to the in/out state containing noninteracting particles. Yes, in reality, they may be slightly off-shell. They must be since they haven't been around for an infinite time. In this sense the model is an idealization.

What happens whilst the particles are interacting, however, cannot be measured. In a model based on perturbation theory, this interaction includes the exchange of what we're calling virtual particles. If we could solve QED (say) exactly, presumably we wouldn't even need to chop up the interaction into these virtual particle contributions.

As soon as we wish to talk of an in state particle as something whose properties we can prepare or an out-state particle as something whose properties we can measure, it's no longer appropriate to model that particle by a propagator (and hence by our definition no longer appropriate to call it a virtual particle). For example, for a photon, I'd like to be able to prepare/measure its momentum/polarization, but the photon propagator [itex]-i\frac{g^{\mu\nu}}{k^2+i\epsilon}[/itex] doesn't have the right ingredients to allow me to do this.

tom.stoer

Very good point, the photon propagator does not carry momentum in the sense we measure it.

In addition gauge boson propagators are gauge-dependent objects and are therefore unphysical. So virtual particles DO depend on the specific gauge fixing. Temporal gauge, Lorentz gauge, Coulomb gauge, ... result in different propagators and 'potentials', so you can't interpret these entities directly. The difference becomes visible in QCD, where you have ghost propagators only in some gauges!

mfb

I am aware of that. Could you answer my question, please? It would help me to understand where our views differ:
Do you think there are any real particles? If yes, in which way?
Do they have a fundamental difference to, say, a short-living top quark at the LHC? Or an even shorter-living W boson in the weak decay of a neutron?
You got my point. I don't think there is a fundamental difference between an electron measured in a detector and a W boson mediating a weak decay.