Bound states and the energy-momentum relation....

[asimov42]

Hi all - forgive me, I'd asked a series of questions in a previous post that was deemed to be circular, but I still didn't obtain a satisfactory answer to the question I was asking. In this post, I'm going to try to be very careful to use terms that are at least less 'misplaced', per se.

Consider the hydrogen atom, a bound state involving a proton and an electron - the overall atom satisfies the relativistic energy-momentum relation (I've carefully avoided using the words 'on shell' here)... yet, the individual components do not, which would seem to violate conservation of energy, at least naively. So, the question is, essentially, why is this the case?

The previous thread suggested not paying attention to the energy-momentum relation for bound states, but this absolutely seems fundamental to QFT, and so hard to ignore. Only observable states satisfy the energy-momentum relation, so it seems very odd that you assemble a set of constituents that ultimately don't and somehow end up with a result that does... simply ignoring the idea of the energy-momentum relation for bound states (as suggested) doesn't make the problem of understanding go away.

 

[PeterDonis]

Consider the hydrogen atom, a bound state involving a proton and an electron - the overall atom satisfies the relativistic energy-momentum relation

Yes.

yet, the individual components do not

No; the individual components do not even have a well-defined energy-momentum relation, because they do not have a well-defined energy and momentum.

The previous thread suggested not paying attention to the energy-momentum relation for bound states

No; it said just what I said above, with regard to the constituents of bound states.

Only observable states satisfy the energy-momentum relation

And the individual constituents of a bound system are not observable.

it seems very odd that you assemble a set of constituents that ultimately don't and somehow end up with a result that does

Sorry, but "it seems very odd" is simply not a valid argument. We can tell you what the physics is, but we can't make it not seem odd to you. That's not a matter of physics.

 

[jtbell]

As was noted in the other thread, or maybe somewhere else, in the case of hydrogen it turns out to be a good approximation to consider the proton as stationary and deal only with the dynamics of the electron. Perhaps it would bring out your issues more sharply if you were to consider instead, positronium (a bound electron and positron) during its lifetime before it annihilates into photons.

 

[asimov42]

Thanks again @PeterDonis for your patience, and yes, "seems very odd" is quite a poor argument indeed. I think I'm starting to get it - from you comment below...

No; the individual components do not even have a well-defined energy-momentum relation, because they do not have a well-defined energy and momentum.

And also from @A. Neumaier's physics FAQ, which says:

Note that the mass shell loses its meaning in external fields or dense media, where instead a so-called 'gap equation' appears. In particular, in matter, particles can be off-shell in the different sense of ''not on any mass shell'' -- due to their interaction with the background matter; in fact, due to ''collisional broadening'', they no longer have a well-defined mass and energy but must be characterized by a space- and time-dependent spectral function.

Presumably this could be any field, such as that generated by a nearby particle?

Thanks all!

 

[asimov42]

So what, then, does the final solution to the Schrödinger equation for the hydrogen atom actually describe? An 'electron like' thing surrounding a 'proton like' thing? The state had 8-DOF, but we can only observe the atom as a whole - you can't say you have an electron orbiting a proton, because this isn't really what the state describes, is it?

 

[vanhees71]

Consider the hydrogen atom, a bound state involving a proton and an electron - the overall atom satisfies the relativistic energy-momentum relation (I've carefully avoided using the words 'on shell' here)... yet, the individual components do not, which would seem to violate conservation of energy, at least naively. So, the question is, essentially, why is this the case?

Take the hydrogen atom in an energy eigenstate. The mass is given by the usual relation $p_{\text{cm}}^2=m^2$, where $p_{\text{cm}}$, is the center-momentum four-momentum and the square means the Minkowski product of this four-vector with itself. I use natural units with $c=1$. This is so in Q(F)T as well as in classical physics. It's a definition of what's called (invariant) mass of a composite system in relativistic physics.

The constituents (proton and electron) are in an entangled state and thus their four-momenta do not take definite values when they are in such an energy eigenstate of a hydrogen atom.

 

[PeterDonis]

what, then, does the final solution to the Schrödinger equation for the hydrogen atom actually describe?

The usual solution, as I said before, ignores the quantum state of the proton and treats it as a source of Coulomb potential and nothing more. Then the Schrödinger equation is solved for the electron only, with a Hamiltonian that includes the potential from the proton. So this solution only describes the electron, not the atom as a whole. As I said before, this is only an approximation; it turns out to be a very good one (basically because the proton is so much more massive than the electron), but it's still an approximation, and won't help you with what you are trying to understand.

 

[asimov42]

Ok, but in the full case (treating the complete atom), you get an atom which is essentially not independently separable into parts (electron, proton, etc.) - instead the state describes the composition... a "blending" of sort? (absolutely the wrong terminology)

 

[asimov42]

Take the hydrogen atom in an energy eigenstate. The mass is given by the usual relation $p_{\text{cm}}^2=m^2$, where $p_{\text{cm}}$, is the center-momentum four-momentum and the square means the Minkowski product of this four-vector with itself. I use natural units with $c=1$. This is so in Q(F)T as well as in classical physics. It's a definition of what's called (invariant) mass of a composite system in relativistic physics.

The constituents (proton and electron) are in an entangled state and thus their four-momenta do not take definite values when they are in such an energy eigenstate of a hydrogen atom.

@vanhees71 thanks - sorry last question. When you say "do not take definite values" you don't mean that there is a (probabilistic) distribution over values, but that the values are actually not well-defined (I believe this what @PeterDonis said). However the hydrogen atom in the energy eigenstate does have a definite energy as already established.

 

[PeterDonis]

in the full case (treating the complete atom), you get an atom which is essentially not independently separable into parts

Yes, because the atom is in an entangled state, where its constituents do not have well-defined quantum states individually. So asking what state the individual constituents are in is simply meaningless.

 

[vanhees71]

@vanhees71 thanks - sorry last question. When you say "do not take definite values" you don't mean that there is a (probabilistic) distribution over values, but that the values are actually not well-defined (I believe this what @PeterDonis said). However the hydrogen atom in the energy eigenstate does have a definite energy as already established.

Of course, you can give a probability distribution for measuring, e.g., the proton's energy, but this will be a broad probability distribution. The energy of the proton within a hydrogen atom in one of its energy eigenstates will not have a sharp value. That's what I called "indetermined".

 

[asimov42]

All - thanks very much (esp. @PeterDonis and @vanhees71) for all your help with this - it's deeply appreciated.

One thing that occurred to me at perhaps a more philosophical level: if we take what we've been discussing up to the scale of, say, the wavefunction for the entire observable universe, would it satisfy the mass shell condition? It seems like general relativity would demand this? I've only been able to find one paper speculating on the topic (and literally using the words "off shell", sorry Peter :) but it doesn't seem very convincing- highly speculative).

 

[PeterDonis]

if we take what we've been discussing up to the scale of, say, the wavefunction for the entire observable universe, would it satisfy the mass shell condition?

You can't even apply the mass shell condition to the entire universe, because it doesn't have a well-defined total energy and momentum.

 

[asimov42]

Ok - but if we're looking at the universe as a whole, what's the reason for the lack of a well-defined energy? Sure, huge number of entangled states, but ultimately, just like the hydrogen atom (a microcosm example), don't you end up with a result that satisfies energy-momentum? I'm talking about something bounded here (observable universe) ... not an infinite space.

 

[john ambrose]

If an object is a certain size, at what point would adding weight to its perimeter have less effect on the objects total MOI?

 

[PeterDonis]

if we're looking at the universe as a whole, what's the reason for the lack of a well-defined energy?

Because there's no way of looking at the universe as a whole from the outside and measuring it.

I'm talking about something bounded here (observable universe)

The observable universe by itself does not have a well-defined quantum state, because it's entangled with the rest of the universe.

 

[asimov42]

Thanks @PeterDonis - ok, so let's them consider the entire universe (whatever its extent)...

We can't look at the universe from the outside and measure it, but measurement is not necessary to have a quantum state, is it? We may not know what that state is (or cannot know because no measurement of an observable has been performed), but that doesn't mean that a 'wave function for the universe' doesn't exist, does it?

Could one not, in theory, write down the Hamiltonian for the entire universe - and should this Hamiltonian not be time independent?

 

[PeterDonis]

measurement is not necessary to have a quantum state, is it?

It depends on which interpretation of QM you adopt. On some of them, it is--more precisely, you have to be able to prepare a system in a controlled manner in order to assign a quantum state to it, and "prepare" basically means make a particular measurement on it whose outcome you can control.

Could one not, in theory, write down the Hamiltonian for the entire universe - and should this Hamiltonian not be time independent?

The problem is that such a Hamiltonian might end up telling you nothing. For example, in one scheme for quantizing General Relativity, the Hamiltonian you end up with is just $H = 0$. Which gives no information whatever.

 

[asimov42]

Ah, yes, I see the problem. Is there an interpretation for which preparation is not necessary, and you end up with a definite energy (those are two separate questions) - or, I guess, are you often stuck with a Hamiltonian that basically leaves the energy undetermined? (at least as far as we know)

 

[PeterDonis]

Is there an interpretation for which preparation is not necessary, and you end up with a definite energy

Not that I'm aware of.

are you often stuck with a Hamiltonian that basically leaves the energy undetermined?

For the universe as a whole, yes, as far as I know.

 

[A. Neumaier]

So what, then, does the final solution to the Schrödinger equation for the hydrogen atom actually describe? An 'electron like' thing surrounding a 'proton like' thing? The state had 8-DOF, but we can only observe the atom as a whole - you can't say you have an electron orbiting a proton, because this isn't really what the state describes, is it?

The solution of the time-independent Schroedinger equation is a product of a planar wave depending on the center of mass coordinates (describing free motion of the center of mass of the atom) and a wave function for the coordinates of the electron relative to the center of mass. The latter is the equation usually solved in the textbook, it satisfied a Dirac equation with a radially symmetric external electromagnetic field (which in some approximation may be taken to be a Coulomb field).
The complete solution expresses the entanglement of the two constituents in a way analogous to how a wave function with two spin indices describes the entanglement of two coupled spins.