Scalar QED atoms - will they pass through each other?

Question

Atoms generally do not pass through each other. This is usually attributed to the Pauli exclusion principal between the electrons (see links below).

If the electrons and nucleons were switched with scalar particles, would these scalar-atoms pass through each other, or would we experience some normal force (would we still have solids)?

More concretely, how does the potential between these atoms look like as a function of distance, at short distances? does it diverge like the London force between actual atoms? (figure below)

My point is to figure out if the divergence of the potential between atoms at close range is due to the Pauli exclusion principal.

Discussions in which the Pauli exclusion principal was deemed responsible for atomic repulsion:

How can I stand on the ground? EM or/and Pauli?

Pauli exclusion principle and van der Waals force

The Lennard-Jones potential and the Pauli exclusion principle

Answer 1

There are a few layers to this question. I'll try to address as many as I can.

To your concrete question about close-range atomic potentials, the divergence at short distances is entirely due to electrostatic repulsion between nuclei. Or at least this statement is unambiguously true for $\text{H}^2$. These atomic potentials can be computed using the Born-Oppenheimer approximation, in which nuclei are treated classically, and then we solve for the electronic wave functions and energies as a function of internuclear distance. In the case of hydrogen, Pauli exclusion plays no role in the calculation, because the bound state $\text{H}^2$ shows up when its two electrons are paired in the same orbital, one spin up and one spin down. The spatial wavefunctions are identical, and thus the Pauli exclusion principle does nothing to keep them apart.

However the story is more complicated for larger systems.

Turning to bosonic matter, it turns out that it is unstable. Wikipedia has the following equation, which makes this statement precise; for fermionic matter, the lowest energy of a collection of $N$ electronics and $K$ nucleons satisfies

$$E_{N,K} \geq -(N+K) C$$

where $C$ is some constant. This means that binding energy is extensive, like we are used to. However if electrons were bosons, then it was shown that the right-hand side would scale like $N^{7/5}$, and therefore binding energy becomes superextensive. Thus if you had two chunks of bosonic matter and you brought them together, they would collapse into a much lower energy state and release a ton of energy. So while there are no normal forces, it's not like particles would just pass through each other.

So back to interactions between pairs of atoms; I suppose interpreting forces between atoms larger than hydrogen becomes a bit subtle. If you compute the energy of $\text{He}^2$ in the Born-Oppenheimer approximation, then the Pauli exclusion principle plays no role when the helium atoms are far apart, but certainly plays a role at small separations. If you turned off the Pauli exclusion principle and repeated the calculation, which you can absolutely do in non-relativistic problems like this, then you would get a different Born-Oppenheimer potential and it would almost certainly have a minimum at smaller separation; Pauli exclusion and the notion of full orbitals is what makes helium inert, after all. But the potential would still diverge at small internuclear distances, again due to Coulomb repulsion.

OK so we get a different answer when we replace electrons with bosons here; does this mean that the repulsion is due to Pauli exclusion? That's a question of interpretation. My take is that at the end of the day, all of the energy in the problem is derived from Coulomb interactions. There is no 'Pauli force' showing up to keep the atoms apart. Pauli exclusion only eliminates certain low-energy states from the problem.

Previous answer before I learned that bosonic matter is definitely unstable

To the question "can we construct a scalar model in which we have stable solids? Could those solids be rigid and not pass through each other?" those are hard questions and I don't know the answers. Here are two potential almost-answers though:

• Superconductors are still rigid objects, even though their conduction electrons condense into Cooper pairs that almost obey Bose-Einstein statistics.
• Superfluid helium becomes a crystalline solid at high pressures, despite its nearby superfluid phase behaving like a Bose-Einstein condensate.
• Related to the above, helium 4 may or may not have a supersolid phase, which is a sort of blend of between a superfluid and a solid. I don't know much about it, but it could be an interesting source of further, potentially related reading.

Answer 2

This question is unanswerable and too hypothetical. Nevertheless, trying and failing can elucidate some principles.

1. The title "scalar QED". Once you bring in QED, as decidedly not scalar theory (spinors and vector boson), there is nothing that can be done, and if we did something: it would not be renormalizable, so our results would be meaning less.

So let's just say charged particles.

2. but, you mentioned nucleons. Peruse http://www.scholarpedia.org/article/Nuclear_Forces and you should convinced to avoid nucleon--I mean the interaction isn't even a central force.

so let's chose an exotic atom, maybe a mix between anti-muonium and positronium: bind an $e^-$ to a $\mu^+$.

Ofc muons are unstable, but their lifetime is long on the some of the atomic timescale. Or we can just "turn-off" the weak interaction, in which case they are stable and only interact with electrons via EM.

And we make them spin zero. If we make the photon scalar too, we can go in with Coulomb atom plus corrections.

Since hyper-fine structure is cause by nuclear spin, we can chuck that.

Fine structure is tricky. Walking through https://en.wikipedia.org/wiki/Fine_structure, we have:

Kinetic energy relativistic correction

We're still going to have this. If we turn off relativity, we turn off the Spin-Statistics Theorem, which cancels the Pauli exclusion term.

So we pick up an:

$$ H' = -\frac{p^4}{8m_e^3c^2} $$

Probably $m_{\mu}$ is large enough that it doesn't need a correction. If it does, make it a tau.

Spin–orbit coupling

Well, there's no spin, and we made EM a scalar theory: no magnetism, none of this stuff.

Darwin term

This is tricky. It's non-relativistic, but is derived from decidedly spinor Dirac equation. It looks like:

$$ H_{Darwin} = \frac{\hbar^2}{8m_e^2c^2} 4\pi \big(\frac{Ze^2}{4\pi\epsilon_0}\big)\delta^3(\vec r) $$

So a delta function at the origin. I think this is a result of the Pauli Exclusion principle via the Dirac eq, but maybe not. Let's chuck it.

So we have two scalar boson hydrogen-like atoms, with the $S$ shell holding infinity electrons.

How do they interact? Is there a Van der Waals force, maybe?

But if we have a bunch of them, they can all be in the same state, so the answer to your question is "yes".

But what does that mean? An atom is a Bohr radius in size, about

$$ a_0 \approx 50,000\,{\rm fermi}$$

Meanwhile, a nucleus is

$$ R_p \approx 1\, {\rm fermi} $$

(ofc, we have a point nucleus here), but an electron has a Compton wavelength

$$ \lambda_C = 2\pi\alpha a_0 \approx 2,400\, {\rm fermi}$$

so you can't even localize two electrons to the same position on the scale of a nucleus.

Edit: thanks to Rd Basha for the comment. I'm adding the reply here b/c I want my fitty:

So we got scalar structure less charged colorless particles? The localized nuclei is where the strong field it, and they would scatter in the standard manner with zero phase space for =0, but not for $<_$--so what does pass through even mean? If they are identical would they be in a symmetric wave function? Yes--then they wouldn't even have identity, so and channel are indistinguishable, there is no "atom 1" and "atom 2". Like I said: too hypothetical, but this is PSE, where it's OK to entertain silliness, since it can elucidate seriousness.

PS: I don't care about scalar QED's renormalizability. It's irrelevant to the question, but is relevant to the class of questions, which is, "non-physical hypotheticals".