Register to reply

Why don't virtual particles cause decoherence?

by Mukilab
Tags: decoherence, particles, virtual
Share this thread:
mfb
#19
Feb16-13, 06:32 PM
Mentor
P: 11,601
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
#20
Feb16-13, 06:33 PM
Sci Advisor
P: 5,366
You ignore most of what I am saying.
mfb
#21
Feb16-13, 06:38 PM
Mentor
P: 11,601
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
#22
Feb16-13, 06:43 PM
Sci Advisor
P: 5,366
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
#23
Feb17-13, 10:53 AM
P: 381
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
#24
Feb17-13, 11:08 AM
Sci Advisor
P: 5,366
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
#25
Feb17-13, 11:17 AM
P: 381
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
#26
Feb17-13, 11:24 AM
Mentor
P: 11,601
How would those measurements look like? They would be interactions with the particles, and not separable from the process you consider.

Quote Quote by tom.stoer
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
#27
Feb17-13, 11:29 AM
P: 203
Quote Quote by JK423 View Post
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
#28
Feb17-13, 12:29 PM
Sci Advisor
P: 5,366
Quote Quote by mfb View Post
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
#29
Feb17-13, 01:20 PM
P: 724
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
#30
Feb17-13, 03:51 PM
P: 125
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
#31
Feb17-13, 04:38 PM
P: 724
Quote Quote by mattt View Post
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
#32
Feb18-13, 01:08 AM
Sci Advisor
P: 5,366
I guess we should come back to the question

Quote Quote by Mukilab View Post
I was recently told virtual particles don't cause decoherence. Why not?
My first answer was

Quote Quote by tom.stoer View Post
Decohence is due to factorizing the full Hilbert space H in Hsystem, Hpointer and Henvironment 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 Hsystem, Hpointer or Henvironment , 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

Quote Quote by tom.stoer View Post
[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 Hsystem, Hpointer or Henvironment .

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
#33
Feb18-13, 03:12 AM
Sci Advisor
Demystifier's Avatar
P: 4,575
Quote Quote by mfb View Post
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
#34
Feb18-13, 08:44 AM
P: 3,009
Quote Quote by tom.stoer View Post
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
#35
Feb18-13, 10:31 AM
Sci Advisor
P: 5,366
Quote Quote by TrickyDicky View Post
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.
Demystifier
#36
Feb19-13, 01:52 AM
Sci Advisor
Demystifier's Avatar
P: 4,575
Quote Quote by tom.stoer View Post
... so perturbation theory does not make sense to arbitrary high order; its radius of convergence is zero.
Yes, and this important fact is usually ignored by those who try to ascribe some reality to virtual particles. Insisting on incorporating this fact into a realistic interpretation of virtual particles would be like saying that virtual particles are real only when their number is sufficiently small (say less than 137 in the case of QED). Which, of course, would not make sense.


Register to reply

Related Discussions
Virtual particles vs. real particles Quantum Physics 12
Virtual particles colliding with other virtual particles Quantum Physics 7
If virtual particles can appear, can real particles disappear? Quantum Physics 3
Virtual particles Quantum Physics 6