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The electric field in an excited hydrogen atom is non-zero over a bigger volume than the electric field in a ground-state hydrogen atom. The energy associated with this extra electric field must be provided to excite the ground-state atom.

Intuitively, it is clear that in a system where the potential energy of the electron is lower, the overall mass is also lower. There's less energy here.

When two nucleons bind together. Their masses decrease. Is there a similar analog change in wave function between them akin to a reduction in the volume of the electric field of the electron? Is it possible to conceptualize this loss of mass through the change of component wave functions/behavior?

Does the exchange of pions/residual strong force alter the probability densities in a way where it is clear that the wave function of the bound systems possess less energy than unbound?

I would like to know if it's possible to explain mass defect without using the phrase "binding energy" or "Minus the energy it takes to separate the components"

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An under-appreciated fact about electromagnetism is that the Lorentz follows from Maxwell's equations (see user4552's answer to this question). You can derive the potential energy between two charges by simply computing the energy of the electromagnetic charges via

$$Energy = \int \left[\frac{\epsilon_0}{2}|\vec{E}|^2 + \frac{1}{2\mu_0}|\vec{B}|^2\right] dV$$

So there is not actually a fundamental distinction between binding energy and energy stored in the electromagnetic field, and you can think of it however you like.

The weak and strong nuclear forces are much more complicated than the electromagnetic force, but the same conclusion carries over. In fact it's common to understand the binding energy of the strong force by thinking about the energy stored in the field; the potential energy due to the strong nuclear force scales linearly with distance, as long as the distance is small. This arises because two particles with charge under the strong force (such as quarks) develop strings of gluons connecting them, called flux tubes. The length of the flux tube scales linearly with distance, and therefore the energy grows linearly as well. I assume the story is the same when you consider pion exchange at larger scales than individual quarks.

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  • $\begingroup$ Just to tie this together into a one-sentence summary "the total energy in the electromagnetic fields $\int E^2$ decreases, and that energy that used to be higher was part of the inertia/mass of the unbound particles" $\endgroup$
    – AXensen
    Commented Mar 22, 2023 at 20:23
  • $\begingroup$ I think this is getting closer to what I want to comprehend. Because there are now pion exchanges, is the "average distance" between quarks and other quarks decreased? So if we were to bring two protons together to the distances of fusion, but w/out pions. Then the average distances between the 6 quarks is kinda high. However, when you add in the pion exchange that distance or "strong energy density" decreases. So the sum of the energy in the fused system is less than isolated systems? $\endgroup$
    – TheJeran
    Commented Mar 23, 2023 at 8:04
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    $\begingroup$ I'm not totally sure I understand the question in your comment here; but the basic statement is that the total energy stored in the field (virtual pions and what-not) of 4 free nucleons is greater than when you bring them together and for a helium atom. The field produces an attractive force, which is equivalent to saying that there is less energy stored in the field when the attracted particles are brought closer together. $\endgroup$
    – user34722
    Commented Mar 26, 2023 at 18:04
  • $\begingroup$ @user34722 What I'm trying to figure out. Is that pions are like more quarks. And since quarks interacting can shrink fluxtubes. Does the increased quarks from pions increase the stability? $\endgroup$
    – TheJeran
    Commented Mar 28, 2023 at 9:34
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Does the exchange of pions/residual strong force alter the probability densities in a way where it is clear that the wave function of the bound systems possess less energy than unbound?

It is better to be consistent and use the same mental picture when comparing coupling in atoms and nuclei (Indeed, there are parallels.) Thus, if we speak of nuclear forces as an exchange of pions, we should also visualize the electromagnetic forces as an exchange of photons. The difference is that pions are massive, that is they are described by Klein-Gordon equation, whereas photons are described by wave equation (i.e., a Klein-Gordon equation with zero mass.) This means that nucleon potential exhibits a Yukawa-like behavior rather than a more simple Coulomb one. (This is further complicated by the Coulomb repulsion between protons.)

When two nucleons bind together. Their masses decrease. Is there a similar analog change in wave function between them akin to a reduction in the volume of the electric field of the electron? Is it possible to conceptualize this loss of mass through the change of component wave functions/behavior?

One we have agreed that the nucleon interaction can be approximately described by a potential, the same reasoning applies as in the case of a hydrogen atom, although with correction for the shape of the potential and the fact that the nucleons have approximately equal masses - whereas in an atom the center-of-mass can be safely assumed to coincide with the nucleus. Here is a figure taken from these notes:
enter image description here

What might be confusing is that realistic nuclear potentials are quite sharp, and often depicted as nearly square potential wells:
enter image description here
This is however only a convenient approximation.

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