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]

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.

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]

It' like apples and oranges.

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]

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

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]

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.

See my post
https://www.physicsforums.com/showpost.php?p=4267048&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]

According to another (more meaningful, in my opinion), it is any internal line in a Feynman diagram. The two definitions are not equivalent.

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]

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

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]

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

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 $-i\frac{g^{\mu\nu}}{k^2+i\epsilon}$ doesn't have the right ingredients to allow me to do this.

 

[tom.stoer]

)... For example, for a photon, I'd like to be able to prepare/measure its momentum/polarization, but the photon propagator $-i\frac{g^{\mu\nu}}{k^2+i\epsilon}$ doesn't have the right ingredients to allow me to do this.

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]

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?

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?

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.

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.

 

[tom.stoer]

Could you solve he following puzzles: let's say virtual particles (internal lines in Feynman diagrams) are 'real'; then
1) why are there gauge bosons propagators with unphysical polarizations whereas in observations only two physical polarizations are present?
2) why is it not possible to attribute polarization and 4-momentum to gauge boson propagators whereas in observations we always have a certain polarization and 4-momentum?
3) why are there Fadeev-Popov ghost particles in certain gauges whereas they do appear in any observation? and why are they absent in other gauges? is reality gauge-dependent?

To answer your question: real particles do appear in observations. External lines in Feynman diagrams as representations of asymptotic states are an idealization for real particles. But what we observe in reality is rather close to these idealized asymptotic states, so it's OK to try to interpret them as a mathematical model of reality.

In contrast, internal lines in Feynman diagrams do not have attributes like 4-momentum, spin or polarization as we observe them. In addition the attributes of internal lines are gauge dependent whereas our observations aren't. So we can neither relate our observation of reality to these 'virtual particles', nor can we relate all 'virtual particles' we use in our calculations to our observation of reality. So the fundamental difference between internal and external lines is not only that internal lines are 'off-shell'. And b/c there is no working relation between attributes of observations of reality and attributes of internal lines, it is NOT OK to construct an interpretation of reality in terms of these virtual particles.

 

[mfb]

1-3: Your observations are too slow. You observe just long-living particles, which act like long-living particles.

But what we observe in reality is rather close to these idealized asymptotic states, so it's OK to try to interpret them as a mathematical model of reality.

But "rather close" is the main point. There is no fundamental line separating particles we observe from the virtual W in a weak decay.

If you try to "catch" a photon in the near field of an emitter, you get polarizations of the field which are impossible for real photons. If you go away, the radiative part gets more and more dominant, but there is no line after which you observe just radiation and nothing else.

 

[tom.stoer]

You ignore most of what I am saying.

 

[mfb]

No, I reduce it to the main point, and adress that.
Anyway, I don't think further discussion will produce anything new here.

 

[tom.stoer]

No, the main point are unphysical properties of virtual particles; ignoring these facts does not solve any problem. There are not even short-lived ghosts in reality; and reality is not gauge dependent. So focussing on livetime and off-shell is not the main point. It misses nearly everything which characterizes virtual particles.

 

[JK423]

Can i ask both of you a question?
The interaction time between two particles is finite. If we could make measurements during this finite period while interactions are still on, what would we see? What are the observables? Is, for example, the number operator of virtual particles (if such thing exists) an observable? Any idea?

 

[tom.stoer]

I don't think that the question regarding interaction time makes sense.

Regarding the number operator it's trivial: this operator acts on Hilbert space stars. But virtual particles (internal lines in Feynman diagrams) are propagators, not Hilbert space states. Therefore there is not operator 'counting' virtual particles.

 

[JK423]

I haven't seen anywhere a discussion of what happens during an interaction (not just in/out states). Why doesn't it make sense (theoreticaly at least)?

 

[mfb]

How would those measurements look like? They would be interactions with the particles, and not separable from the process you consider.

No, the main point are unphysical properties of virtual particles

If you call everything "unphysical" which cannot be "observed" in non-interacting particles, sure. That is just playing with words. The near field of an antenna would be unphysical then, as electric and magnetic fields do not follow the rules for radiation (=light particles).

 

[sheaf]

I haven't seen anywhere a discussion of what happens during an interaction (not just in/out states). Why doesn't it make sense (theoreticaly at least)?

Arnold Neumaier mentions this issue on his FAQ. Some results are known in <3 spatial dimensions, but it's a hard problem, and much of the formalism he discusses is beyond my knowledge.

 

[tom.stoer]

If you call everything "unphysical" which cannot be "observed" in non-interacting particles, sure. That is just playing with words.).

It seems that you are not familiar with the meaning of "unphysical" in the context of gauge theories. The 0-component of the gauge field A is an unphysical d.o.f. b/c it has no associated canonical momentum; states in the kinematical Hilbert space are unphysical if they are not annihilated by the Gauß constraint (= if they are not in the kernel of the Gauß Law operator = if they are not gauge singulets); Fadeev-Popov ghosts are unphysical d.o.f. b/c they are artificial d.o.f. Introduced to eliminate other unphysical d.o.f. (polarizations) of the gauge fied.

All these entities are unphysical simply b/c strictly speaking they are not required; you can find formulations avoiding them, so there is no fundamental reason to introduce them into the formalism (they are artifacts of the formalism) and therefore there is no reason to interpret them as physical entities.

 

[Maui]

I was under the impression that "physical" means exchange and interaction of virtual particles(at least according to qft) between...er... objects/real particles or whatever you want to label it. If they are not real(don't exist), is anything real? How would physicalness arise?

 

[mattt]

Imagine I develop a new mathematical formalism, that is a good enough mathematical approximation to, say, Newtonian Mechanics, based on a given mathematical Serie.

Imagine I call each element of the Serie, "a little green dwarf", because I like it.

Would you say that those "little green dwarfs" are "real" or "physical"?

Would you say that gravity exists because of the actions of those "little green dwarfs"?

 

[Maui]

Imagine I develop a new mathematical formalism, that is a good enough mathematical approximation to, say, Newtonian Mechanics, based on a given mathematical Serie.

Imagine I call each element of the Serie, "a little green dwarf", because I like it.

Would you say that those "little green dwarfs" are "real" or "physical"?

If i see a multitude of little green dwarfs all around me for a life time, i might be inclined to believe they are real and exist. The mathematics wouldn't work if it had no resemblance to reality. Why would it work otherwise? Just a happy coincidence?

Would you say that gravity exists because of the actions of those "little green dwarfs"

If little green dwarfs are the curvature of spacetime, yes.

 

[tom.stoer]

I guess we should come back to the question

I was recently told virtual particles don't cause decoherence. Why not?

My first answer was

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".

The discussion over the last couple of days did not change anything; the first answer is still correct.

Let me summarize some additional ideas

[one] can formulate QFT [and non-rel. QM] non-perturbatively w/o Feynman diagrams;

... there is no need to introduce perturbation theory and propagators when studying density operators.

... attributes of internal lines are gauge dependent whereas our observations aren't.

But all this is not directly relevant for the original question b/c virtual particles are completely irrelevant in the context of decoherence: they are not present in the full theory; they do neither introduce the above mentioned factorization of H nor the partial trace; and they are not states in any Hilbert space H_system, H_pointer or H_environment .

Last but not least: nobody would assume that any approximation like a Taylor series (or green dwarfs) do introduce additional effects which are not already present in the full theory w/o the approximation (w/o green dwarfs). So if the theory w/o virtual partices green dwarfs) already contains decoherence (gravity) it would be silly to say that decoherence (gravity) is due to virtual particles (green dwarfs). This changes if the theory cannot be formulated w/o virtual particles (w/o green dwarfs), or if the formulation is conceptally simpler (in the sense of Ockham's razor) using virtual particles (green dwarfs).

I am not an expert regarding green dwarfs, but I know that perturbation theory is incomplete and misses relevant non-perturbative effects. So I can't see any reason to rely on the interpretation of partially unphysical artifacts due to an incomplete approximation instead of using the full theory.

 

[Demystifier]

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.

There are internal lines which are exactly on-shell.
There are external lines which are slightly off-shell (see e.g. http://en.wikipedia.org/wiki/Scharnhorst_effect ).

 

[TrickyDicky]

I guess we should come back to the question

Hi Tom, I tend to sympathize with the way you present the answer to this FAQ, but I have a doubt about the "virtual particles are just artifacts of perturbation theory" issue, you often use the example of QCD where this methodology is not necessary, but if we consider only QED ("the jewel of the physics crown") for a moment it does seem that perturbation is needed to obtain reasonable results, so in that sense at least for QED VP seem like something you can't get rid of so easily.

 

[tom.stoer]

Hi Tom, I tend to sympathize with the way you present the answer to this FAQ, but I have a doubt about the "virtual particles are just artifacts of perturbation theory" issue, you often use the example of QCD where this methodology is not necessary, but if we consider only QED ("the jewel of the physics crown") for a moment it does seem that perturbation is needed to obtain reasonable results, so in that sense at least for QED VP seem like something you can't get rid of so easily.

Neither QED nor QCD require perturbation theory or virtual particles. In QED non-perturbative aspects are either discused using rel. QM + radiative corrections a la Lamb shift, or they are not relevant at all (due to small alpha 1/137 and abelian gauge symmetry). So virtual particles are common standard and mostly sufficient in QED, but not required. One can quantize QED non-perturbatively w/o using virtual particles.

In QCD nearly everything requires non-perturbative methods (even in DIS - using perturbation theory - one probes non-perturbative structure functions)

There is one problem, namely that QED is ill-defined in the UV (Landau pole), in contrast to QCD which is UV complete.

Anyway, most perturbation series (QED, QCD, phi^4 theory, ...) are ill-defined and divergent, so perturbation theory does not make sense to arbitrary high order; its radius of convergence is zero.