Observable particles as asymptotic states....

[asimov42]

Oi, very complicate stuff. So the constituents in the bound state don't have well defined energies and momentums - but the hydrogen atom certainly does. In the bound state, do the constituents then really maintain any identify of their own? They do continue to exist in the bound state...

For the on-shell condition above, I was thinking of the situation where you start with free particles as you described and then bring them together to form the atom - something has to happen to go from a collection of on-shell free particles to an off-shell bound state...

 

[PeterDonis]

In the bound state, do the constituents then really maintain any identify of their own?

What does "identity" mean?

something has to happen to go from a collection of on-shell free particles to an off-shell bound state...

You shouldn't call the bound state "off shell" for the reasons I've already given.

In any case, the "shell" condition, where it applies, is part of the model, not part of reality. So it's wrong to think that "something has to happen" for the "shell state" to change. What happens in the idealized experiment I described is that a free proton and a free electron are brought together and allowed to interact to form a hydrogen atom, and energy is emitted during the process.

 

[asimov42]

What does "identity" mean?

Here, I mean, for example, that the hydrogen atom is composed (in most cases) of one proton and one electron, and they're distinct. In the bound state, the electron can absorb and then emit a photon, and alter the quantum state of the atom ... there's still an electron there. Hence a bound state includes a number of particles that continue to persist in the bound state...

Also, you mentioned that constituent particles may not have well-defined energy or momentum - does the electron in the ground state of a hydrogen atom have no well-defined energy?

Lastly, when I think about identity, I think about a metal, again for example, where it's easy 'to 'scrape' electrons from the surface (.e.g., electrostatic charge on one plate). Then you've got an excess of electrons, that were once bound, and now (via electrostatic discharge) go off and do whatever.

The quantum mechanical or QFT description of the bound state makes clear the number and types of of constituent particles in the state, correct?

You shouldn't call the bound state "off shell" for the reasons I've already given.

In any case, the "shell" condition, where it applies, is part of the model, not part of reality. So it's wrong to think that "something has to happen" for the "shell state" to change. What happens in the idealized experiment I described is that a free proton and a free electron are brought together and allowed to interact to form a hydrogen atom, and energy is emitted during the process.

Sorry to ask @PeterDonis, but could you briefly recap this - I've gotten lost in the discussion (or don't worry if too tedious).. Ultimately. particles in bound states are considered "real" (whatever that means) ?

 

[DrDu]

It doesn't make sense to say a particle is "on the complex mass shell".

Is this really so? At least in non-relativistic potential scattering you can treat resonances as "Siegert states" with complex energy and momentum.

 

[vanhees71]

Do you have a reference about this (review article/textbook)?

 

[DrDu]

https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.79.2026

 

[asimov42]

@vanhees71 - if you are referring to Siegert states, there is some information in @A. Neumaier's response here: https://physics.stackexchange.com/questions/4349/are-w-z-bosons-virtual-or-not/22064 with references to both Siegert states and Gamov states.

 

[vanhees71]

Thanks!

 

[asimov42]

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!

@vanhees71 thanks for the above - the problem I've been searching for an answer to, is the following: yes, bound states are "real" states that are on-shell - but are the constituent particles forming the state also considered to be "real"? Since they do not necessarily satisfy the energy-momentum relationship, I can't tell.

Virtual particles are (supposed to be) a totally different thing - terms in a series expansion, but I'm still not clear on the the 'state' of the individual constituents in a bound state...

 

[asimov42]

Hi all,

Sorry for beating a dead horse here - I've read through the entire thread, and although there are answers that approach the question I've asked above, I still don't have a clear indication of how this works (and yes, a much more thorough look at the math would certainly help, but I feel that there should be some intuition aside from the math).

We have that the constituents of bound states are 'off-shell' (even though as @PeterDonis pointed out, this is likely the incorrect terminology) - but as @vanhees71 has noted, physical states are related to the DOF of the system.

So, a) what is the specific reason constituents of the bound state are off shell (or don't satisfy the energy momentum relationship)? b) can we still treat the components as physical, despite not satisfying the relationship? (this cannot be done for virtual particles).

If anyone could provide a direct answer to the above, it would be exceedingly helpful in my studying of the theory. Thanks all in advance, and apologies for so many posts.

 

[PeterDonis]

Here, I mean, for example, that the hydrogen atom is composed (in most cases) of one proton and one electron, and they're distinct.

Ok, but then I suggest that you go back and read the previous post of mine where I talk about degrees of freedom.

does the electron in the ground state of a hydrogen atom have no well-defined energy?

Strictly speaking, no. The atom as a whole has a well-defined energy in the ground state, but neither of its constituents do. However, most treatments of the hydrogen atom aren't that strict; they ignore the proton's quantum state altogether, treating it only as a source of the Coulomb potential that appears in the Hamiltonian for the electron, and then treat the electron, in the atom's ground state, as being in an energy eigenstate of that Hamiltonian. That turns out to be an extremely good approximation, but it's still an approximation.

The quantum mechanical or QFT description of the bound state makes clear the number and types of of constituent particles in the state, correct?

No. They make clear the number and types of the degrees of freedom of the system. Again, go back and read the previous post of mine where I talk about that.

Ultimately. particles in bound states are considered "real" (whatever that means) ?

"Real" isn't a useful term. If we have a proton and an electron and nothing else, then we have a quantum system with a certain number of degrees of freedom (I counted them up in the previous post I referred to). Those degrees of freedom can be entangled; if the system as a whole is in what you are calling a bound state, then the degrees of freedom are entangled. If the system starts out in a non-bound state, and we do something to it to put it into a bound state, then the degrees of freedom can start out not entangled, and then we entangle them by doing what we do to the system.

Notice that nowhere in what I just said did I talk about "on shell" or "off shell". Those are just bookkeeping devices, which might or might not even be meaningful, depending on the state of the system. But they certainly aren't needed to understand or predict what will happen in an experiment like the one I described.

 

[PeterDonis]

We have that the constituents of bound states are 'off-shell' (even though as @PeterDonis pointed out, this is likely the incorrect terminology)

If you know it's incorrect terminology, why do you keep on using it? What you should be doing is forgetting the terms on-shell and off-shell altogether.

what is the specific reason constituents of the bound state are off shell

This question is meaningless and cannot be answered. Do you understand why? If not, read what I wrote just above again and again until you get it.

can we still treat the components as physical

What does "physical" mean?

Part of your problem here seems to be that you keep on trying to ask questions about ordinary language terms like "real" and "physical" that don't have precise meanings (and technical terms like "off shell" that don't have well-defined meanings at all in the context in which you're asking about them), instead of forgetting about all that stuff and trying to understand the actual models.

 

[PeterDonis]

At this point the OP question has been answered, and the discussion is going in circles. Thread closed.