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Confusion arised out of reading this answer, it says

You can immediately describe the 2 particles by their center of mass description (an atom) plus their individual attributes (i.e. what the particles are doing within the atom). Assuming the 2 particles start off at rest, then they are in a bound state already because they can't escape each other (go off to infinite separation) due to lack of energy.

It would seem to me that, that we could, for instance, we could pair any proton in the universe with an electorn anywhere else and call that as an atom. This is at least consistent with Bohr Theory as we can take radius as large as we want.

So, would a question asking "if a proton and electron can combine" to form an atom be non-sensical? or, is my understanding of what an "atom" means wrong?

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    $\begingroup$ "It would seem to me that, that we could, for instance, we could pair any proton in the universe with an electorn anywhere else and call that as an atom." How is this specific to atoms? You could attempt to do the same with any two celestial bodies in classical orbital mechanics and claim they form a two-body system. Do you understand why that makes no sense there? $\endgroup$
    – ACuriousMind
    Commented May 16, 2023 at 8:29
  • $\begingroup$ Hmm, yes, I had a hunch this was not specific to this case @ACuriousMind.. but I amnot sure what the abstract is $\endgroup$
    – Brian
    Commented May 16, 2023 at 8:33

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You've missed a key phrase "assuming the two particles start off at rest" most electrons are not moving at the same velocity as most protons in the universe.

You cannot make two particles be at rest by choosing a particular reference frame.

Take a very slight approximation where we assume the center of mass is determined by the proton. Then we choose the center of mass frame, and the proton is not moving. So we now consider the electron's velocity with respect to the proton. If $$\frac{m_ev^2}{2}-\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}>0$$ Then the electron has posive energy and is not bound to the proton.

If you consider also quantum mechanics, it becomes more dire. If the electron is localized so it has some uncertainty in its velocity this is another way for the electron to be unbound. If it's position uncertainty is $\delta_x$, then $\hbar/2\delta_x$ is roughly its momentum uncertainty, and $(\hbar/2\delta_x)^2/(2m)$ is roughly its kinetic energy expectation value. So even if $\langle v\rangle=0$, the electron may not be bound by the proton if it is reasonably localized.

Not to mention the issue that if there is lots of other stuff between the electron and the proton, one cannot approximate the system as only a force between an electron and a proton. The proton may not even be relevant for determining the motion of the electron.

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  • $\begingroup$ Valid points but what does that the velocity at which electrons and protons have to do with defining an atom? $\endgroup$
    – Brian
    Commented May 16, 2023 at 8:44
  • $\begingroup$ @TrystwithFreedom I added more detail $\endgroup$
    – AXensen
    Commented May 16, 2023 at 8:57
  • $\begingroup$ Okay, but what are you taking to be definition of the atom? Something with bound orbits? $\endgroup$
    – Brian
    Commented May 16, 2023 at 9:01
  • $\begingroup$ @TrystwithFreedom why are we identifying an electron and a proton as a hydrogen atom? Typically it is because the proton is important for determining the motion of the electron. If the electron is miles away and moving way too fast to be bound, then the proton is entirely irrelevant for determining the motion of the electron. $\endgroup$
    – AXensen
    Commented May 16, 2023 at 9:11
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    $\begingroup$ @TrystwithFreedom sure. If the electron is far away, but nothing else is nearby to mess up this approximation, you can describe the electron and the proton in terms of hydrogen atom wavefunctions. Again - don't forget about unbound vs bound states. If the electron is very far away it's wavefunction is probably composed mostly of unbound states (not typically referred to as "orbitals", which are all bound states with negative total energy). But why would you do that? What would it help you calculate? $\endgroup$
    – AXensen
    Commented May 16, 2023 at 9:21

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