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When solving the hydrogen atom in non relativistic quantum mechanics we use Coulomb's potential, $\sim1/r$, to describe the interaction between the proton and the electron.

I am worried about two conceptual points:

1-Wouldn't this potential be modified as r becomes smaller and smaller? (because now the electron will feel different potential at very short distances because then it can see 3 quarks with different charges, not a point like proton)

2-Also very short distances means very strong electric field, strong enough for particle anti particle creation, hence we do not have non relativistic quantum mechanics anymore, specially when the number of particles is not conserved.

Then what justifies our solution of the hydrogen atom without taking the previous two points into account?

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In QM the potential acts between "waves". These "waves" do not feel the potential differences in the regions much smaller than the characteristic wave-length. As explained in the answers of Misha's and Lubosh's, the numerical effect of the potential difference in question is rather small indeed. –  Vladimir Kalitvianski Jan 23 '12 at 12:20
    
Related: physics.stackexchange.com/q/9415/2451 –  Qmechanic Jan 23 '12 at 12:46
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2 Answers 2

Misha's answer is just fine but I would like to be more quantitative.

  1. The first correction from a non-point-like nature of the nucleus may be approximated by replacing the potential $V(r)$ inside the nucleus by a constant equal to the value of the Coulomb potential on the "surface" of the nucleus. In the first-order perturbation theory, the change of the energy eigenvalue is calculated as the expectation value of $\Delta H$, i.e. roughly $e^2\int dV |\psi(r)|^2/r$ over the nucleus (ignoring purely numerical constants); the constant part is at most of the same order. Now, $dV=4\pi r^2dr$ so $1/r$ is still beaten by the Jacobian and we have $e^2\int dr |\psi|^2 r$. This is completely negligible for $l>0$ when $\psi(r)=0$ for $r=0$. However, for the $l=0$ states, it's somewhat greater, but still small, suppressed by $r_{\rm nucleus}/r_{\rm atom}\sim 10^{-5}$ relatively to the normal binding energies of order 13.6 eV. These corrections are known pretty accurately and some really tiny, high-precision discrepancies were the focus on renewed discussions about a "wrongly measured radius of the proton" etc.

  2. One needs 2 times 511 keV of energy to create an electron-positron pair. Clearly, the interaction energies inside the atom aren't enough for that physically. Still, the virtual particles do affect the energy levels of the atoms, even though the amounts are even smaller than the finite-nucleus correction from the point 1. First of all, one has to include the "normal" relativistic corrections from the Dirac equation. Then, the effect of QED-like corrections from virtual particles may also be incorporated. The most famous one is called the Lamb shift, linked to virtual photons. It's even much smaller than the Dirac relativistic corrections as well as the finite-nucleus corrections, but it's still significant so that it was accurately measured and the prediction of QED works.

To summarize, there are various corrections to the simplest solvable textbook treatment of the Hydrogen atom but all of them are very small, making the textbook approximation extremely important. The corrections are proportional to positive powers of the fine-structure constant (the relativistic corrections) 1/137.036, to powers of the nuclear-to-atomic-radius ratio, and other tiny numbers.

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That's correct for calculation of atomic energies and wave functions. There is also a kinematic parameter valid only for an "external observer" (projectile) of sufficiently short wave-length: when probing short atomic distances, the fast projectile "sees" a positively charged cloud of the size $\frac{m_e}{M_A}a_0$, which is the negative charge cloud rescaled down by the ratio $m_e /M_A$. It's a result of a mutual electron-nucleus motion "available" for a fast projectile, but not for the atomic electrons. The latter interact with the nucleus via $1/r$, as it goes from the Schroedinger equation. –  Vladimir Kalitvianski Jan 23 '12 at 15:59
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  1. Large value of the potential takes place in small volume. The higher potential you are interested in, the smaller the volume. It turns out, that there is no problem as long as this volume decreases faster than potential increases.

  2. Try to calculate the volume where field is high enough to create electron-positron pairs. Obviously, in that region Coulomb potential is wrong. But you will hardly see any changes if you change potential there.

This approach is justified by the fact that it works. Other real life perturbations like surrounding of the atom, corrections due to fine and hyperfine interactions, etc. are orders of magnitude more important

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