Insights Misconceptions about Virtual Particles - Comments

[vanhees71]

Sure, it's also a question of time scales. On a time scale of a typical nuclear reaction a neutron or a pion is a very stable particle and thus can be treated almost always as such.

 

[A. Neumaier]

"Rizzo Phys. Rev. D22 (1980) 389"

I think the intended reference was p.722, http://journals.aps.org/prd/abstract/10.1103/PhysRevD.22.722

 

[A. Neumaier]

"Hey what does the star mean?" "It is off-shell".

If you like, you can draw one of the possible reactions as H --> Z l l, where the real Z is treated as an unstable particle that subsequently decays (in another diagram that you draw) and we reconstruct the usual peak (which we can interpret in the usual way as a pole of our S matrix process). The internal Z is a virtual particle as per your own definition!

So for me there is something more to be understood clearly enough that I can explain it in my own words.

I think I understand now in my own words what happens in the $H\to WW^*$ decay. I added the following two sections to the collection of definitions in the Insight article The Physics of Virtual Particles.

Branching fractions. If an unstable particle can decay in several different ways, the branching fraction of each single decay (or group of decays) is the relative frequency of this decay (or group of decays) compared to all decays. It can be computed from the S-matrix elements of all individual processes.

External lines and off-shell particles.
As a consequence of the description of S-matrix elements, the external lines usually correspond to on-shell particles. and then describe real particles before and after a collision or decay. However, there is the custom of using (generalized) Feynman diagrams also in certain cases where one or more out-particles are off-shell (typically denoted by a *). An example (see Figure 2 in http://arxiv.org/abs/hep-ph/9807536) is the Higgs decay $H\to WW^*$ in which one of the $W$ produced is off-shell, hence not a real particle but an unobservable label. Such a Feynman diagram is short-hand for a family of Feynman diagrams obtained by attaching to each off-shell particle another vertex and its admissble interaction partners, in case of the $W^*$ two leptons. Thus the single Feynman diagram visualizing the decay $H\to WW^*$ stands in fact for $H\to WX$, where $X$ are two external lepton lines attached to a vertex that turns $W^*$ into an internal line, as it should be for off-shell particles. The branching fractions for decays involving off-shell particles must be interpreted in the same way. For example, the branching factor for the decay $H\to WW^*$ is defined as the inclusive branching factor for all $H\to WX$, where $X$ are two observable leptons consistent with the standard model interactions.

 

[DarMM]

The same applies to unstable particles. Strictly speaking they are never particles but resonances and as such appear in the description of scattering matrix elements as internal lines.

A difference between resonances and virtual particles is that resonances are actual states in the Hilbert Space, i.e. they are present even non-perturbatively.

Also the debate about the "inner structure of protons" is funny. I think, the whole debate is much ado about nothing or say about sloppy language in the QFT community. Any practitioner of QFT,...

I agree that this debate is in a sense "about nothing", but many people, even professionals do not think as you seem to believe they do. They think virtual particles are really physical states, not just perturbative labels and that the proton is really a "sea" of quarks and gluons, rather than this being an approximation model.

That is why it is still worth pointing these things out, sloppy language can lead to sloppy thinking.

 

[DarMM]

I think the much more interesting issue is the series is asymptotically divergent. How can we get answers from a divergent series?

We can get answers because the series is an asymptotic series, i.e. a specific kind of divergent series where the early terms do give a good approximation. In such a series the series approaches the full result up until the point where the error is of order $e^{-\frac{1}{\lambda}}$, with $\lambda$ the expansion parameter. At this point it begins to diverge.

No surprise in QFT as $e^{-\frac{1}{\lambda}}$ is basically an instanton contribtuion.

 

[atyy]

A difference between resonances and virtual particles is that resonances are actual states in the Hilbert Space, i.e. they are present even non-perturbatively.

So this Physics FAQ is wrong?
http://www.desy.de/user/projects/Physics/Quantum/virtual_particles.html
"Then, the use of virtual particles as a communication channel is completely consistent with quantum mechanics and relativity. That's fortunate: since all particle interactions occur over a finite time interval, in a sense all particles are virtual to some extent."

Presumably the answer should be that virtual particles do not exist, whereas resonances are short lived states in Hilbert space which can be seen by a suitable measurement?

 

[RockyMarciano]

So this Physics FAQ is wrong?
http://www.desy.de/user/projects/Physics/Quantum/virtual_particles.html
"Then, the use of virtual particles as a communication channel is completely consistent with quantum mechanics and relativity. That's fortunate: since all particle interactions occur over a finite time interval, in a sense all particles are virtual to some extent."

Presumably the answer should be that virtual particles do not exist, whereas resonances are short lived states in Hilbert space which can be seen by a suitable measurement?

IMO this whole discussion is pointless. All that is ever measured are interactions(that's why the meaningful quantities come from the vertices in the diagrams), never "real particles" in the idealized infinite time of the free field theory nor "virtual particles". The detections, whether one refers to them as clicks, dots or tracks or scattering cross sections all refer to interactions and it is nonsensical to debate whether the particles exist as it will only depend on the kind of book-keeping one previously decides to use based on their particular prejudices.
Resonances are just inelastic scattering cross sections and again measure short lived interactions. If one decides to call them "particles" it is totally irrelevant if one classifies them as virtual or as idealized states in Hilbert space in the free theory. It doesn't affect the physics at all since the physics deals with the interacting fields.

 

[A. Neumaier]

So this Physics FAQ is wrong?

It tells (just as wikipedia) the ''popular science'' story based on the silent (but meaningless) identifications mentioned in post #58, not the ''research level'' one which has precise conventions. In addition, the Physics FAQ makes some other meaningless statements such as ''A virtual particle with momentum p corresponds to a plane wave filling all of space''. In fact it treats virtual particles as if they had wave functions (which would make them on-shell).

 

[mfb]

It tells (just as wikipedia) the ''popular science'' story based on the silent (but meaningless) identifications mentioned in post #58, not the ''research level'' one which has precise conventions. In addition, the Physics FAQ makes some other meaningless statements such as ''A virtual particle with momentum p corresponds to a plane wave filling all of space''. In fact it treats virtual particles as if they had wave functions (which would make them on-shell).

As discussed before, there is no clear line between a W* that is clearly off-shell and a muon that is treated as real particle, despite its finite lifetime and corresponding uncertainty in the invariant mass of its decay products. Yes, the muon is a much better approximation to a real particle with a proper mass, but it is still just an approximation.

 

[A. Neumaier]

there is no clear line between a W* that is clearly off-shell and a muon that is treated as real particle

What is regarded as existent (i.e., modeled by a state rather than symbolically by an internal line) in a particular experimental situation is in borderline cases a matter of modeling. But since it changes the predictions (factorized or unfactorized computations in the simulation!), one can in principle distinguish the two situation - though (due to limited data and/or simulation accuracy) perhaps not in practice.

In any case, it makes no sense (except in sloppy talk) to model a particle as off-shell and then claim it has a lifetime or another spatio-temporal property since the latter are properties of on-shell particles only. If one wants something to have a lifetime one must model it as on-shell (even when the sloppy talk says otherwise).

Saying (as the Physics FAQ does) ''A virtual particle with momentum p corresponds to a plane wave filling all of space'' is scientifically empty since this sentence cannot be translated to something more definite. The plane wave can formally be associated to a state only, in this case probably a bare state in a simplified model - but to associate on the formal level (not in one's fantasy) a plane wave with an internal line in a Feynman diagram has no function at all - it helps neither do nor to interpret calculations. One could as well associate plane waves to the momentum of a car, rocket or planet...

Thus the status of this FAQ entry is not better than a lie-to-children.

 

[mfb]

What is regarded as existent (i.e., modeled by a state rather than symbolically by an internal line) in a particular experimental situation is in borderline cases a matter of modeling. But since it changes the predictions
(factorized or unfactorized computations in the simulation!), one can in principle distinguish the two situation - though (due to limited data and/or simulation accuracy) perhaps not in practice.

Your model never influences physics. It would be like claiming 2 meters is a fundamental concept of nature if you use one model above 2 meters and a different one below that just because they are better approximations.

In any case, it makes no sense (except in sloppy talk) to model a particle as off-shell and then claim it has a lifetime or another spatio-temporal property since the latter are properties of on-shell particles only. If one wants something to have a lifetime one must model it as on-shell (even when the sloppy talk says otherwise).

The decay width is an observable - it does not care about the model you use. Every unstable particle has a decay width, if you measure the invariant mass of its decay products precise enough you will always get different values for different decays.

 

[A. Neumaier]

The decay width is an observable - it does not care about the model you use.

An observed decay width always needs for its interpretation a model involving a process observed in time, hence a state, hence a real particle (resonance). A virtual particle never has a decay width since it does not make formal sense to talk about its temporal properites. Thus one cannot model a decay width by a virtual particle.

But where you may have a choice (and hence a subjective aspect depending on the model used) is in the interpretation of a particular composite decay process for which an experimental decay width is available. You may model it as a single process with unresolved time, expressed by intermediate off-shell virtual particles. Or you may model it as a partially time-resolved process with an intermediate on-shell resonance and a factorizing computation. In borderline cases, the two models will produce approximately the same decay width. At a given finite resolution the models may agree, in which case one cannot decides whether the intermediate particle is on- or off-shell. But with sufficient computing time and experimental accuracy it should in principle be possible to distingush the two situations. This is what I was referring to in my post.

 

[mfb]

An observed decay width always needs for its interpretation a model involving a process observed in time, hence a state, hence a real particle (resonance). A virtual particle never has a decay width since it does not make formal sense to talk about its temporal properites. Thus one cannot model a decay width by a virtual particle.

I'm not talking about virtual particles now, I'm discussing the "on-shell"/"off-shell" difference which does not have a clear dividing line. The W* in a Higgs decay is clearly off-shell, while muons in decays are considered as on-shell. And there is no sharp line dividing those two cases. You can treat one as off-shell and one as on-shell, fine, but that is just your model used, it is not the physics behind it. You can also treat the muon as off-shell particle (with way too much computing effort) or the W* as "not so far away from on-shell particles" (with a huge error).

 

[A. Neumaier]

You can treat one as off-shell and one as on-shell, fine, but that is just your model used, it is not the physics behind it.

So what is the same physics behind the two different treatments?

 

[vanhees71]

I'm not talking about virtual particles now, I'm discussing the "on-shell"/"off-shell" difference which does not have a clear dividing line. The W* in a Higgs decay is clearly off-shell, while muons in decays are considered as on-shell. And there is no sharp line dividing those two cases. You can treat one as off-shell and one as on-shell, fine, but that is just your model used, it is not the physics behind it. You can also treat the muon as off-shell particle (with way too much computing effort) or the W* as "not so far away from on-shell particles" (with a huge error).

Well, and the good thing is that it doesn't matter, how you call the diagrammatic elements of a Feynman diagram symbolizing a mathematical expression to calculate an S-matrix element. You just calculate it, square it, multiply it with the appropriate kinematical factors, and you get (within a model, of course, because without a model, you'd have no Feynman diagram to begin with) a lifetime/decay width or cross section.

There's, of course, a well defined meaning when calculating, say at tree level, the life time or decay width of a muon, which you treat as an "on-shell" external line in the corresponding diagram, using the Feynman rules from QFD, as there is a well-defined meaning for the diagrams describing a Higgs-boson decay via W mesons, where you finally of course measure not W mesons directly but their decay products. How call them, on shell or off shell is completely irrelevant. What counts is the well-defined meaning of an (approximately calculated) S-matrix element, and this you can compare with counts in the detectors at the LHC and be excited about the discovery of the Higgs boson of the Standard Model checking all the details about its nature, maybe hoping for a deviation from the Standard Model to find new physics.

 

[mfb]

So what is the same physics behind the two different treatments?

What is different? After a long time, you get stable particles flying away. W* or muon are "just" our description how those particles got created.

How call them, on shell or off shell is completely irrelevant.

Someone in this thread argued that there was a difference between those categories I think.

 

[vanhees71]

Of course there is a difference, but it's totally irrelevant. I rarely use the words "on-shell" and "off-shell" in my daily conversation although we also use QFT and Feynman diagrams all the time. Sometimes you talk about "on-shell" and "off-shell" form factors or the like. There was never ever any problem understanding each other.

What's sometimes a problem is, however, the notion of resonances, and that often resonances are also a model-dependent notion, particularly if you have them involved (again in our case often in electromagnetic transition form factors of hadrons in vector-meson-dominance approaches) in "far-off-shell situations", i.e., where you probe the "far-off-shell tails" of broad spectral functions of such resonances as the $\rho$ meson. Then there's often confusion about "what's a $\rho$ meson" or "what's the $\rho$ meson's spectral shape" etc. Then it's good to have the arguments given in the Insights article at hand.

What I wanted to say is only, that now this discussion goes a bit out of sense, because it's now becoming a purely semantical fight about the meaning of a $W^*$ vs. a $W$ although the meaning is very clear in terms of (perturbative) Feynman diagrams and the underlying QFT (the GSW model of the electroweak interaction within the Standard Model in this case).

 

[A. Neumaier]

After a long time, you get stable particles flying away. W* or muon are "just" our description how those particles got created.

After a long time, we are all dead, and physics no longer matters.
Physics happens on every time scale, not only at arbitrarily long time.

W* or muon are "just" our description how those particles got created.

No. A $W^*$ is never responsible for the creation of these particles, while a $W$ can be. For getting created is a process in time, and this requires for its description a state - honce (on the level of the particle language) an on-shell particle.

 

[mfb]

No. A $W^*$ is never responsible for the creation of these particles, while a $W$ can be.

But we get leptons from both in a Higgs decay.

For getting created is a process in time, and this requires for its description a state - honce (on the level of the particle language) an on-shell particle.

And there is the "on-shell" again. No unstable particle is "exactly on-shell". Not the W, and not even the muon.

 

[A. Neumaier]

But we get leptons from both in a Higgs decay.

No. We get leptons from Higgs, and we calculate it either in partially resolved time as $H\to Wll$ and $W\to ll$ or unresolved in time as $H\to llll$. The $W^*$ is only a formal book-keeping device, not something existing in time, hence nothing from which we can gat anything - except figuratively.

No unstable particle is "exactly on-shell". Not the W, and not even the muon.

Of course they are on-shell, on a mass shell with a complex rest mass, as explained in the companion Insight article. That's what defines an unstable particle or resonance. They have an associated eigenstate for the analytically continued Hamiltonian in the second sheet of the Riemannian surface defined by the resolvent. In terms of the original Hilbert space they have a state formed by a superposition of states in the continuous spectrum of $H$ (the system Hamiltonain in the rest frame of the unstable particle), with energies corresponding to a continuous range of masses - which may be visible as a resonance peak. Thus their real effective mass (defined by $m=H/c^2$) is unsharp, in the same sense as the position of a particle in a beam is unsharp. But this doesn't make unstable particles off-shell.

 

[mfb]

Fine, then we use different definitions for "on-shell". Forget about the word. You can include the muon in the Higgs decay for $H \to e^-e^+ (e^- \nu_\mu \bar\nu_e) (e^+ \bar\nu_\mu \nu_e)$, brackets for clarity. Is the muon gone?

 

[A. Neumaier]

Fine, then we use different definitions for "on-shell". Forget about the word. You can include the muon in the Higgs decay for $H \to e^-e^+ (e^- \nu_\mu \bar\nu_e) (e^+ \bar\nu_\mu \nu_e)$, brackets for clarity. Is the muon gone?

In this process as written, time is not resolved at all, and the S-matrix elements (that only describe the in-out behavior, not the behavior at finite time) are computed in terms of Feynman integrals described by virtual particles in which no muon is present. Thus the muon is gone from the description - there is no occasion to talk about it. If you want to have a more detailed picture in time you either need to factor the S-matrix in terms of on-shell but unstable intermediate particle states such as a muon, or you need to work in terms of a hydrodynamic or kinetic description where states are abstracted from the CTP (Schwinger-Keldysh) formalism. In the latter case, the muon appears as an on-shell resonance state.

So the unstable muon exists once you sufficiently resolve the time, but it is invisible in the S-matrix elements of the total process when the latter are computed in unfactored form.

 

[vanhees71]

Be, however warned that "off-shell transport" is a very tricky business, much trickier than S-matrix theory, were you deal with asymptotic states and don't care about transient states, let alone the very ambigous if not impossible interpretation of Heisenberg field operators and their expectation values at finite times. In defining the S-matrix properly, by the way a lot is hidden by the ingenious use of $\mathrm{i} \epsilon$ prescriptions (aka. the Gell-Mann-Low description). In short, it is very difficult if not impossible to interpret Heisenberg-picture field operators in terms of particles. The particle interpretation is only unambiguously working for asymptotic free states. In any case a transport theory for broad resonances is very delicate and not fully solved (after more then 20 years of research, including practical implementations of some ideas concerning off-shell transport). Also the full quantum equations (Kadanoff-Baym equations)are not fully developed to the extent that you can use it for simulations of real-world problems (in heavy-ion collision physics), because they are simply to difficult and "CPU demanding" to bring them to the computer. There are, however some studies, with simpler toy models like $\phi^4$ in (1+2) dimensions using a $\Phi$-derivable approximation up to the two-point level (look for papers by Juchem & Cassing and also for Berges).

Of course, a muon is a good example for an "almost stable" particle, i.e., you can take it as a stable particle in QED and put the weak interaction perturbatively on top. The muon becomes a very narrow resonance, and it makes some sense to use the (tree-level) diagram to calculate its decay widht/lifetime although "strictly speaking" the muon is not a asymptotic free state anymore when taking into account its finite lifetime due to the weak interaction.

Of course, you must not take the formal $t \rightarrow \pm \infty$ limits in the Gell-Mann-Low prescription to define S-matrix elements as transition rates between asymptotic free states too literally. It's rather a matter of physical (finite) time scales, i.e., a separation between the distances involving the (approximately) free moving particles in the initial and final states compared to the usually very short duration of the collision itself. Usually you don't resolve the collision ("transient state") dynamics in time and are content with S-matrix elements as transition rates between asymptotic free states.

Of course, there's some tension between mathematical rigor and the physicists' intuition about the scattering process, having to do with Gell-Mann-Low and Lehmann-Symanzik-Zimmermann (LSZ) reduction and taking the limits. As far as I know, there's no mathematically rigorous definition of these limits and thus the so overwhelmingly successful S-matrix elements used to evaluate the Standard Model and comare it to the measured cross sections.

 

[Buzz Bloom]

Nothing virtual happens. The dry facts are that two real particles are created from gravitational energy (from two gravitons or from an external gravitational field), not from the vacuum. One particle escapes, the other is absorbed. A valid description is given on p.645 of the book
B.W. Carroll and D.A. Ostlie, An Introduction to Modern Astrophysics, 2nd. ed., Addison Wesley 2007.

Hi @A. Neumaier:

It has taken a month, but my local library has finally gotten for me a copy of the Carroll & Ostlie book you recommended. I think I now understand the phenomenon of Hawking Radiation more clearly than I had before, but some aspects I still find confusing.

This is what I now think I understand:

The total matter-energy M producing a black hole with an event horizon at radius R_s=2GM/c² includes the energy of the gravitational field that exists outside the event horizon. When this energy creates a pair of real particles, which is a common occurrence, and the tidal force moves one of the pair towards the event horizon while the other escapes towards infinity, which happens only a relatively small fraction of the time, the escaping particle is Hawking Radiation, and it's mass-energy is subtracted from the total mass energy M of the black hole.​

What are the particles that make up the energy of the static gravitational field? Are they virtual gravitons?

I have been thinking about the similar phenomenon about a sphere with a net charge distributed uniformly on its surface. What are the particles that make up the contents of the static electric field about this sphere? Are they virtual photons? If the field is strong enough, won't pairs of real particles also occasionally be formed, from the static electric field, (almost ?) all of which will then very soon afterwards annihilate each other? However, is it not possible for some of these pairs of particles to not annihilate each other? If these pairs are charged, then would this not also result in a weakening of the electric field? But, perhaps this phenomenon only creates pairs of uncharged particles. What would the effect be if a pair of non-charged particles did not annihilate each other? Would there then be some form of radiation emitted from the sphere?

Regards,
Buzz

 

[A. Neumaier]

the particles that make up the contents of the static electric field about this sphere? Are they virtual photons?

No. A static field is not made up of photons. It is made up of the fields of the particles in the sphere, which in turn is due to the fact that every charged physical particle carries with it an electromagnetic field - in a good approximation a Coulomb field.

In formal terms, physical electrons and nuclei are so-called infraparticles. If one describes these particles in perturbation theory based on the bare theory with a cutoff, they look like bare particles accompanied by a cloud of infinitely many arbitrarily soft photons. But this picture does not survive infrared renormalization via coherent states, hence has no physical meaning beyond pure analogy to condensed matter physics.

In a very strong classical electromagnetic field, real (not virtual) electron-positron pairs are created spontaneously. For example, http://arxiv.org/abs/hep-th/0005078 treats the corresponding case with scalar particles.

What are the particles that make up the energy of the static gravitational field? Are they virtual gravitons?

No. In a quantum field theory of gravity and matter, all massive physical particles also carry a gravitational field; everything is completely analogous. A static gravitational field behaves quantum mechanically in the same way, except that it can also produce two photons, because photons are their own antiparticles. Or two neutrinos, etc.. There is no energy barrier for photon production since photons are massless, but to achieve a noticeable effect, the field has to be extremely strong. (To date, Hawking radiation has not been experimentally demonstrated, primarily for this reason. The black holes with very strong fields are fortunately too far away to affect us significantly.)

 

[A. Neumaier]

Concerning the misconceptions in the interpretation of vacuum polarization in QED via virtual particles, see the discussion here.

 

[Buzz Bloom]

No. In a quantum field theory of gravity and matter, all massive physical particles also carry a gravitational field; everything is completely analogous. A static gravitational field behaves quantum mechanically in the same way, except that it can also produce two photons, because photons are their own antiparticles. Or two neutrinos, etc.. There is no energy barrier for photon production since photons are massless, but to achieve a noticeable effect, the field has to be extremely strong.

Hi @A. Neumaier:

I think I am gradually understanding more about the Hawking Radiation phenomenon, but I still have some confusion.

I get that the energy in the gravitational field outside the event horizon is able to create pairs of particles, but I have come to understand that the concept of this field is entirely represented by its space-time distortion represented by the Swartzchild metric. Is this correct? If so, does this mean that when a particle pair is created, and the particles do not annihilate each other, the corresponding loss of energy in the field is manifested by a corresponding change in the distortion of space-time? If this is so, then presumably after a brief time, the distortion becomes represented by a Swartzchild metric with a slightly smaller value for

R_s=2GM/c²,​

since a part of M corresponding to the energy in the field has been reduced by the creation of the particle pair.

If all of this is correct, then what is the mechanism that reduces the greater part of M represented by the mass inside the event horizon? In the absence of such a mechanism, then the net result of the Hawking radiation would seem to be the eventual elimination of the gravitational field outside the event horizon, while the mass inside the event horizon remains there forming a black hole that has no observable gravitational field.

Regards,
Buzz

 

[A. Neumaier]

does this mean that when a particle pair is created, and the particles do not annihilate each other, the corresponding loss of energy in the field is manifested by a corresponding change in the distortion of space-time?

No. The loss of energy is encoded in the distortion of the metric of the 3-dimensional space when moving the time of slicing spacetime into 3D spaces at fixed times. Note that in a space-time view, the latter corresponds to dynamics!

 

[Buzz Bloom]

No. The loss of energy is encoded in the distortion of the metric of the 3-dimensional space when moving the time of slicing spacetime into 3D spaces at fixed times. Note that in a space-time view, the latter corresponds to dynamics!

Hi @A. Neumaier:

I apologize for being sloppy with my use of language. I meant that the Swartzchild metric components for both dr² and dt² are distorted relative to flat space by the respective factors

(1 - R_s/r)-1 and c² (1 - R_s/r) .​

Is it correct to say that the creation of the particle pair (which survives rather than self annihilating) reduces the energy in the field, will ultimately manifest itself by changing the R_s factor in these metric components, and that this means that the energy in the entire field outside the EH has been reduced by the energy of the created particle pair?

My confusion is whether this change in the field metric has any effect on the amount of mass inside the EH? If it does, then what is the mechanism that makes this happen?

Regards,
Buzz

 

[A. Neumaier]

the energy in the entire field outside the EH has been reduced by the energy of the created particle pair?

Of course, if the mass of a black hole is reduced, the gravitational field everywhere also weakens.

But note that the Schwarzschild metric only describes a classical black hole in isolation, whereas the Hawking effect describes a quantum black hole interacting with QED. The quantum black hole can be described by a Schwarzschild metric with a variable mass only in some approximation