# Observable particles as asymptotic states...

• A
vanhees71
Gold Member
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 :-)).

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

PeterDonis
Mentor
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.

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.

PeterDonis
Mentor
The relevant quote is
I'm not sure the heuristic description given in that quote is a good one. For example, the electron in a hydrogen atom in its ground state is not, I think, usefully thought of as "bouncing" between various freely propagating states, with each "bounce" being an interaction with the Coulomb potential of the nucleus. This looks to me like mistaking mathematical artifacts of a particular description (in this case, describing the electron's state as a superposition of momentum states from the free theory) for actual physical happenings.

My main concern is what physical description can be given to the constituents of bound states?
The best simple way of putting this that I can think of is that the constituents are in quantum states that do not have any useful description in terms of ordinary "classical" states that we're used to. Take the electron in the hydrogen atom in its ground state again as an example. We have a good description of its state mathematically: it's just the 1s orbital of the hydrogen atom. But this state is not like any state we're used to classically. And it is even less like any state we're used to classically when we remember that it's also entangled with the state of the nucleus, as @vanhees71 pointed out in an earlier post.

"Virtual particles" are basically a way of trying to describe certain quantum states that occur in perturbation theory in terms more like those we're used to--we say, oh, it's just like a real particle except it can be off shell, meaning its mass, momentum, and energy don't obey the usual relation $E^2 - p^2 = m^2$. But that's already an oversimplification, and can lead to lots of misconceptions if you take it too seriously. Trying to describe constituents of bound states is even more problematic because even the "virtual particle" description I just gave doesn't really work for them; the quote you gave is an attempt to come up with something similar, but it doesn't really work, not even as well as the virtual particle description works for scattering experiments.

At a more technical level, the answer to your question is that the best description we have is the math of QFT, which can be expressed in various ways. The answer by Eric Yang in the Stack Exchange thread, and some of the other posts in this thread, give examples of ways of expressing that math. We can use these mathematical descriptions to make accurate predictions about many properties of bound states--an early example was the Lamb shift in electrodynamics, more modern examples include the calculations of hadron masses that @vanhees71 referred to earlier. But whether those mathematical descriptions give you a "physical" description of the constituents of bound states is really a question about words, not physics.

@PeterDonis that's very helpful. So (casting aside all the on-shell off-shell stuff), each constituent has a quantum state, and so is at least somewhat physical (simply being representable by a state). I am of course familiar with the various orbitals in standard atomic descriptions. Presumably the off shell analogy comes from the interaction energy between bound particles?

I think my confusion arose because, in, e.g., @A. Neumaier 's insight article, he makes clear that virtual particles (which are 'off-shell') cannot be assigned states in QFT, at least from a scattering perspective (because there are no creation operators etc. for off-shell particles) - hence the virtual particle picture is a mathematical tool for series expansion, but should not be taken literally (being used in summing infinitely many Feynman diagrams).

PeterDonis
Mentor
each constituent has a quantum state
Strictly speaking, the overall bound system has a quantum state. If that state is entangled, which it will be in basically any case of interest, the constituents don't have well-defined quantum states by themselves.

and so is at least somewhat physical (simply being representable by a state)
For the constituent to be "physical", in quantum theory, really means that it corresponds to one or more degrees of freedom of the overall system. So an atom, for example, is better thought of as being made up of a certain number of degrees of freedom rather than a certain number of "particles". If we just consider the electron and the nucleus for hydrogen, then, at least in non-relativistic QM, each one has three configuration space degrees of freedom and one spin degree of freedom, for eight degrees of freedom total in the overall system.

Presumably the off shell analogy comes from the interaction energy between bound particles?
I've seen heuristic descriptions of this sort (for example in one of the responses in the Stack Exchange thread linked to earlier), but I don't think they're useful. Even in classical (non-quantum) relativity, invariant masses aren't additive, so the fact that, for example, the mass of a hydrogen atom is not the sum of the masses of a proton and electron is not in itself very useful. It can be given a physical meaning if we envision a particular process of "assembling" the atom by taking a proton and electron that are very, very far apart, so the interaction between them can be neglected, and bringing them together to form a hydrogen atom, allowing them to emit energy in the process. (And you also have to do this in a region that is completely empty of anything else, so the proton and electron don't interact with anything else.) Then you will find that the difference between the mass of the hydrogen atom, and the sum of the proton and electron masses, is equal to the energy emitted during the process. But that doesn't mean there is any "interaction energy" in the final hydrogen atom that somehow puts the proton and electron "off shell" compared to what their free states would be. At least, I don't think that's a useful way of viewing things.

Ok - then can I ask what puts them off shell? Everything else makes sense, except for the off shell part. Although you said the 'off shell' concept may not be very useful in the context of bound states, I continue to see it mentioned in many places.

There must be some reason why bound states don't satisfy the on-shell relation for constituent particles - my assumption would be that it has to do with binding energy.

PeterDonis
Mentor
There must be some reason why bound states don't satisfy the on-shell relation for constituent particles
I'm not even sure you can apply the on-shell condition in this case. The constituent particles don't even have well-defined energy and momentum.

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
Mentor
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.

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
Gold Member

vanhees71
Gold Member
Thanks!

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....

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