Observable particles as asymptotic states....

[asimov42]

I've read Arnold Neumaier's excellent Insight article on virtual particles, but I'm very confused about one thing:

Observable particles are considered to be on-shell, and as 'asymptotic states' at time +- infinity. Now, in a scattering experiment, I may produce a new particle, which will travel off until it interacts with something else (another scattering process). This new particle did not exist at time -infinity, nor will its existence extend to +infintity, and yet it is observable.

The fact the newly-created particle is observable implies it is on-shell, but it definitely does not satisfy the asymptotic condition. So then how does this fit into the theory, since we're only supposed to talk about asympotic (free?) states? Is the new particle stricly on-shell? Is it 'real'? (apparently yes, but by the QFT definition this isn't clear to me at all).

Thanks.

 

[DrDu]

Maybe an example is the beta decay: A neutron decays into a proton and a W- boson, the latter decays shortly afterwards into a (anti-)neutrino and an electron.
Strictly speaking, neither the electron nor the W- boson are stable asymptotic particles, neither are they completely on shell as the time energy uncertainty leads to an uncertainty in energy for a particle with a finite lifetime. I think there are methods to cope with these situations although I am not very familiar to their application in QFT. Basically, a particle with a finite lifetime can be described as a pole of the scattering matrix at complex energy (the imaginary part of which is basically the inverse lifetime), and there are transformations, like e.g. "complex scaling" to rotate these complex poles back on shell.

 

[vanhees71]

Well, of course taking $t \rightarrow \pm \infty$ in the scattering processes (carefully using the adiabatic-switching prescription a la Gell-Mann and Low in the interaction picture as can be found in any good textbook!) is a mathematical abstraction. Physically what's meant is that you produce particles far away from each other so that you can neglect the interaction between them and then observe what's coming out of a scattering process a sufficiently long time after the collision, when the particles can be consdidered free again.

In fact, for (unscreened) long-range interactions (in HEP that's the electromagnetic interaction) the naive definition of asymptotic free states is not correct, and the usual sloppyness of course hits back in the IR problems (e.g., for bremsstrahlung diagrams), which are then resolved by doing soft-photon resummations (see Weinberg, QT of Fields, Vol. I for the best treatment).

Alternatively one can use alternative asymptotic states which take into account the long-range interaction part, leading to states which consist of what we'd roughly describe as a bare charge particle accompanied by coherent-photon states.

The most simple example is the analogous problem of Coulomb-potential scattering in non-relativistic QT. All this is nicely explained in

P. Kulish and L. Faddeev, Asymptotic conditions and infrared divergences in quantum electrodynamics, Theor. Math. Phys., 4 (1970), p. 745.
http://dx.doi.org/10.1007/BF01066485

 

[asimov42]

Thanks @vanhees71. I may not have sufficient background to read and fully understand the paper (trying) - but essentially what you're saying is that the definition of asymptotic states can be modified (to consider long-range interactions), such that we still end up with on-shell states (plus coherent photon states), and this is really the right way to do it?

 

[vanhees71]

Yes, and in the end you get the same result as with the conventional method. The soft-photon resummation technique takes the coherent em. fields into account.

 

[DrDu]

Does this also work with quarks? Can you do a soft gluon resummation?

 

[vanhees71]

Quarks and gluons are special. They don't exist as asymptotic states due to confinement.

 

[asimov42]

Quoting Dr. Du,

Strictly speaking, neither the electron nor the W- boson are stable asymptotic particles, neither are they completely on shell as the time energy uncertainty leads to an uncertainty in energy for a particle with a finite lifetime. I think there are methods to cope with these situations although I am not very familiar to their application in QFT. Basically, a particle with a finite lifetime can be described as a pole of the scattering matrix at complex energy (the imaginary part of which is basically the inverse lifetime), and there are transformations, like e.g. "complex scaling" to rotate these complex poles back on shell.

Reading Dr. Neumaier's notes about stable vs. unstable particles, unstable particles are still on-shell, even though the mass shell is real only for stable particles. So this would imply that the decay products are still on-shell - correct (and Dr. Du's answer would not be quite correct)? And the electron is in fact a stable decay product.

 

[asimov42]

Also, observable particles with finite lifetimes much still be on-shell, no? The time-energy uncertainty relation isn't the same as position momentum uncertainty, so even if the the particle energy is uncertain, doesn't QFT demand that the mass shell constraint holds?

 

[vanhees71]

It depends on the lifetime and the relevant times of observation. E.g., a neutron lives several minutes, and as long as you consider neutrons only in processes which are on a much smaller timescale you can consider them as stable. Of course, if you are interested in the decay of neutrons you have to treat them as a resonance with finite mass width. Formally everything that is an internal line in a Feynman diagram is not a particle but at best an unstable resonance.

 

[asimov42]

Hmm, interesting - so in a previous StackExchange post, @A. Neumaier noted that ''All observable particles are on-shell, though the mass shell is real only for stable particles.' But here you're saying that, since a free neutron would decay, it's actually not subject to the mass shell constraint? But doesn't QFT demand that observable particles are on-shell? I could certainly observe a free neutron moving through space before it decayed.

 

[vanhees71]

An unstable particle has a mass width given by $\hbar/(c^2 \tau)$, where $\tau$ is the (proper) lifetime in the particle (in Breit-Wigner approximation of the particle's spectral function).

 

[asimov42]

Thanks @vanhees71 - I'm just not familiar with mass width - isn't the mass width effectively the complex portion of the mass (hence maintaining the on-shell requirement, even if the mass is complex)?

 

[asimov42]

Ok @vanhees71 - got what you're saying. So here's one more question. Let's say a have a free neutron moving in space - so it's going to decay at some point. Now, I put a neutron detector in place and observe the neutron... meaning it can be treated as stable and on the (complex) mass shell. After detection, the neutron carries on it's merry way and eventually decays.

So first, I have an on-shell observable particle and then I have what I would assume is a virtual particle (internal line). How can this be? Or is it that the detector is effectively a scattering experiment?

 

[asimov42]

Ok @vanhees71 - got what you're saying. So here's one more question. Let's say a have a free neutron moving in space - so it's going to decay at some point. Now, I put a neutron detector in place and observe the neutron... meaning it can be treated as stable and on the (complex) mass shell. After detection, the neutron carries on it's merry way and eventually decays.

So first, I have an on-shell observable particle and then I have what I would assume is a virtual particle (internal line). How can this be? Or is it that the detector is effectively a scattering experiment?

@A. Neumaier any comments?

 

[asimov42]

Gah, more confusion! Quoting a (what I believe is reliable) StackExchange post:

Electrons and all leptons are real particles when on the mass shell, i.e. on external Feynman lines. The same is true for photons and Z and W bosons when on mass shell. Quarks and gluons can never be free and measured individually, because of the nature of the strong force. So quarks are always virtual and have to appear in pairs (mesons) or triplets (protons) in the external legs of the diagrams, and it is the pairs and triplets that can be on the mass shell. Gluons cannot be on their mass shell either, and their only externally measured evidence is in hadronic jets.

The nucleus is a more complicated grouping, it contains off the mass shell protons neutrons gluons photons, but all together, the nucleus is on the mass shell. when isolated. When surrounded by electrons it becomes an atom, the atom then is the one that is on the mass shell and contains electrons +nucleus off the mass shell. When observing a molecule, the molecule is on the mass shell and contains off mass shell atoms. When observing a crystal, the whole crystal is on the mass shell and the molecules are off mass shell.

Ok, so quarks are virtual because they can't be observed as free particles, fine. But why are the individual components of a nucleus (or atom) off shell? Shouldn't each component be on-shell? Why would the individual particles be off shell (and hence virtual?)? And how does this then relate to virtual particles in general (which are supposed to just be terms in perturbative series expansion)?

HELP! Thanks all.

 

[asimov42]

Sorry to keep asking so many questions - can't I describe the nucleus as a multiparticle state? And if so, doesn't each particle have to be on-shell? If not, what does this say about the 'reality' of the individual constituent particles?

 

[PeterDonis]

@asimov42 Your questions can't really be answered at the "B" level. Fundamentally this is an "A" level topic. I could bump the thread up a level or two, but I'm not sure if you would have the background to be able to get any value from the responses at those levels.

I'll try to give brief "B" level responses to some of your questions, but I don't think you're going to find them very satisfying. Unfortunately, the only real cure for that is to increase your background knowledge, which takes time.

first, I have an on-shell observable particle and then I have what I would assume is a virtual particle (internal line).

No, you don't. If you observe the neutron, it's a real particle. Basically you now have two experiments: first a trivial "scattering" experiment where the "in" state is a neutron (which you presumably prepared, for example by firing it out of a neutron source) and the "out" state is a neutron (which you observe); second a decay experiment where the "in" state is an unstable neutron and the "out" state is a proton, an electron, and an antineutrino.

why are the individual components of a nucleus (or atom) off shell?

Because you're not directly observing them; you're only observing the nucleus as a whole (or the atom as a whole).

how does this then relate to virtual particles in general (which are supposed to just be terms in perturbative series expansion)?

Bound states like nuclei or atoms are very different from scattering experiments. Perturbation theory was really intended to analyze scattering experiments. In a bound state you don't have "asymptotic" free particles the way you do in scattering experiments (except in the trivial sense that the bound state itself, the nucleus or the atom, could be treated as a "free" particle at some higher level of analysis).

can't I describe the nucleus as a multiparticle state?

Not the way you mean.

If not, what does this say about the 'reality' of the individual constituent particles?

"Reality" isn't a good physics term, and questions about what is "real" can't really be answered by physics. Physics can answer questions about what you will observe if you run a given experiment or make a given observation.

 

[asimov42]

@PeterDonis - this is actually very helpful - if you feel like bumping the questions (or thread) to 'A' level, feel free, and I'll eventually get there . Got the answer about the neutron, makes perfect sense.
'

@asimov42
Because you're not directly observing them; you're only observing the nucleus as a whole (or the atom as a whole).

Bound states like nuclei or atoms are very different from scattering experiments. Perturbation theory was really intended to analyze scattering experiments. In a bound state you don't have "asymptotic" free particles the way you do in scattering experiments (except in the trivial sense that the bound state itself, the nucleus or the atom, could be treated as a "free" particle at some higher level of analysis).

In the case of the nucleus (bound states), if constituents are off shell, what is presumed to be 'happening' internally? (I realize that this isn't a well formed question, and may not be able to be answered clearly.) Given constituent particles being off shell, are they then considered to be virtual? Is it that the nucleus must instead be analyzed as a whole? Is a perturbative treatment then the incorrect approach?

I guess fundamentally my question is what we can say about (bound) off shell particles in a nucleus, or at any higher level up the chain (an atom, etc.). We know the constituent particles, but when combined, it seems that their description as distinct elements vanishes (since QFT assigns states to on shell particles only, or groups, only).

Last q (sorry again): what's the right multiparticle description for bound states?

Thanks again @PeterDonis for taking the time.

p.s. I guess everything I've said above would apply equally to quarks bound in protons, neutrons, etc.

 

[DrDu]

Sorry to keep asking so many questions - can't I describe the nucleus as a multiparticle state? And if so, doesn't each particle have to be on-shell? If not, what does this say about the 'reality' of the individual constituent particles?

I always find it clarifying to look at solid state analogues. Take a metal, which can be described in terms of the Fermi Liquid theory. A one particle excitation with sharp momentum k will have a finite lifetime as the electron will quickly scatter off other electrons via the Coulomb interaction. Hence these excitations are at best virtual electrons. An exception occurs for the electrons very near the Fermi energy. Due to phase space restrictions, scattering is suppressed, so these electrons can be considered real particles. However, their effective mass will be quite different from that of free electrons, which makes them quasi-particles at best.
In principle, this picture also applies to the protons and neutrons making up a nucleus.

 

[vanhees71]

Ok @vanhees71 - got what you're saying. So here's one more question. Let's say a have a free neutron moving in space - so it's going to decay at some point. Now, I put a neutron detector in place and observe the neutron... meaning it can be treated as stable and on the (complex) mass shell. After detection, the neutron carries on it's merry way and eventually decays.

So first, I have an on-shell observable particle and then I have what I would assume is a virtual particle (internal line). How can this be? Or is it that the detector is effectively a scattering experiment?

It doesn't make sense to say a particle is "on the complex mass shell". What you do is rather to consider the neutron as a stable particle with its real mass. From a more theoretical point of view you can look at this as first neglecting the weak interaction. From the poin of view of the strong interaction the neutron is a stable particle and as such has well-defined asymptotic free states. Then you consider the electroweak interaction as a perturbation (e.g., using Fermi's theory of $\beta$ decay as an effective model). This let's you calculate in leading order the decay of the neutron with the stable neutron as an asymptotic free state (with the real mass) in the initial state and the (asymptotic free) electron, proton, and anti-electron-neutrino in the final state.

All this of course makes only sense if the imaginary part of the neutron self-energy (quantifying its width) is small compared to the real part (its mass) around the corresponding pole of the neutron propagator.

 

[asimov42]

Ok, a more direct question (which as @PeterDonis said) I may not have the background to understand - nevertheless I'll try (or at least know what to learn):

Bound states are considered to be physical ("real") states, and yet they are off shell - how is this possible? Does this have to do with nuclear binding (invariant mass is smaller than sum of constituent particle masses, hence off shell? If this is the case, then that makes sense to me)

 

[Vanadium 50]

Bound states are considered to be physical ("real") states, and yet they are off shell - how is this possible?

I don't think that word means what you think it means.

 

[vanhees71]

Ok, a more direct question (which as @PeterDonis said) I may not have the background to understand - nevertheless I'll try (or at least know what to learn):

Bound states are considered to be physical ("real") states, and yet they are off shell - how is this possible? Does this have to do with nuclear binding (invariant mass is smaller than sum of constituent particle masses, hence off shell? If this is the case, then that makes sense to me)

Stable bound states are indeed "real" states, and they are "onshell". Their mass squared is defined by the center-mass four-momentum squared. Note that the bound-state problem for relativistic particles is very difficult and not (yet) completely understood. It can be solved only approximately, where non-relativistic QM is a good approximation (e.g., positronium or heavy quarkonia, nuclei).

The strong interaction is particularly awful in this regard. There you have the extreme that the "elementary particles" of the theory (QCD) are not observable as (asymptotic free!) states, but they are confined in hadrons as prototons, pions, and all the many other baryons and mesons. Fortunately the bound-state problem can be numerically addressed with lattice-QCD calculations, leading to a pretty satisfactory calculation of the hadron masses, which is one indication that QCD is the correct theory of the strong interaction also in the non-perturbative limit, including confinement!

 

[asimov42]

Great, thanks @vanhees71 - just for my own clarity, you're saying that for stable bound states, it's the group of particles considered that are then on-shell (from center-mass four-momentum), even though this would/could be quite different from the on-shell four-momentums of the individual particles were they not bound?

Does this apply equally to quarks?

 

[vanhees71]

No, the constituents of the bound state are not onshell within a bound state since they are not asymptotic free states but bound. Even in the most simple example of a non-relativistic hydrogen atom the proton and electron are in a complicated entangled state. The hydrogen atom as a whole is of course an asymtptotic free state. The center-of-mass motion is that of a free particle (provided there are no fields disturbing it, as already good old Newton teaches us :-)).

 

[asimov42]

Ah, ok - so off shell to me implies virtual - but here I think you're saying that bound states, while having constituents that are off shell, do not necessarily involve virtual particles (although they may), rather a different approach is required? Because the bound state involves very complicated interactions, the off shell situation is a consequence?

p.s. and thanks - I'll stop asking and start reading more soon :)

 

[PeterDonis]

Moderator's note: Thread level bumped to "A" based on subject matter.

 

[PeterDonis]

Bound states are considered to be physical ("real") states, and yet they are off shell

No, they aren't. As @vanhees71 said, the full bound state--the nucleus, or the atom--is on shell.

Internally, the terms "on shell" or "off shell" aren't really good terms to use when looking at the constituents of a bound state, because, as I said before, a bound state is very different from a scattering experiment. The constituents don't have asymptotic free states in the past and future, with some kind of interaction in between. You can't view the bound state as a perturbation of the "no interaction" case of asymptotic free states, which is where the whole machinery of "on shell" vs. "off shell", "real particles" vs. "virtual particles" comes from.

 

[asimov42]

Thanks @PeterDonis. There is a StackExchange post here: https://physics.stackexchange.com/questions/337805/are-bounded-particles-off-shell that covers exactly what I'm now wondering (i hope it's not tabu to post outside links, if so, apologies).

The relevant quote is:

I think bound and virtual particles are not the same. Virtual particle is a particle in some intermediate state (for example, as an internal line of a Feynman diagram). Bound particle exists in a "real" state (it has definite energy), but it can be considered as a superposition of freely propagating states. The latter are virtual, because the particle spends only a fraction of time in each of them, then scattering on the external potential and going to a state with different momentum.

Presumably these are virtual states and not virtual particles (last part above).

My main concern is what physical description can be given to the constituents of bound states? Clearly, at some point, free particles went into the bound state - but if one uses the term 'off shell' (perhaps incorrectly), then it's not clear to me what remains of those free particles as a result of interaction.