"True" quantum-mechanical description of the hydrogen atom

Question

I have recently read about the quantization of the energy levels that the electron in a neutral hydrogen atom can be in, and I noticed that all available treatments seem to treat the nucleus as a point charge $+e$ which determines the shape of the electronic wavefunction. However, are there any treatments which consider the nucleus as a quantum-mechanical particle as well, and determine the two-particle wavefunction of both the proton and electron? I know that a spherically symmetric charge distribution behaves as if all its charge were concentrated in the middle by Gauss's Law, but what about the portion of the ($1s$, for instance) wavefunction that lies inside the classically approximated proton radius?

Answer 1

Summary: as for your first question - what about the motion of the proton - the answer is: after a quick transformation you get the same potential, and the mass is replaced by the "reduced mass" (the same trick that works for two orbiting bodies in Newtonian mechanics). As for the second question, the proton radius (exactly as you suggest) gives a small shift to S-wave functions that can be found using perturbation theory.

To some extent you have asked two very different questions - what about the motion of the proton, and what about the radius of the proton - and they should be treated completely separately. For the first question, we have a two body wavefunction in a Hamiltonian:

$$ H=\frac{p_e^2}{2m_e}+\frac{p_p^2}{2m_p}-\frac{1}{4\pi\varepsilon_0}\frac{e^2}{|\mathbf{x}_e-\mathbf{x}_p|} $$ To make sure we're all on the same page, this is a wavefunction of six variables (the three components of the position of each particle), and it has a potential which depends on the difference between the two particles' positions. So now we use the same trick used in orbital mechanics between two bodies. Define the difference in position and the center of mass position: $$ \mathbf{x}_d=\mathbf{x}_e-\mathbf{x}_p $$ $$ \mathbf{x}_c=\frac{m_e\mathbf{x}_e+m_p\mathbf{x}_p}{m_e+m_p} $$ The only annoying thing to do is to convert the momentum operators. To show you how it goes, I'll convert this into the new coordinates $(1/m_e)d^2/d\mathbf{x}_{e,1}^2+(1/m_p)d^2/d\mathbf{x}_{p,1}^2$ (the part of the total momentum energy in the first direction): $$ \frac{1}{m_e}\left(\frac{d}{d\mathbf{x}_{e,1}}\right)^2+ \frac{1}{m_p}\left(\frac{d}{d\mathbf{x}_{p,1}}\right)^2= \frac{1}{m_e}\left(\frac{d}{d\mathbf{x}_{d,1}}+\frac{m_e}{m_p+m_e}\frac{d}{d\mathbf{x}_{c,1}}\right)^2 + \frac{1}{m_p}\left(-\frac{d}{d\mathbf{x}_{d,1}}+\frac{m_p}{m_p+m_e}\frac{d}{d\mathbf{x}_{c,1}}\right)^2 $$ The thing that's great about these coordinates is that the cross terms cancel (that was the whole point of them). The Hamiltonian turns into: $$ \frac{p_c^2}{2(m_e+m_p)}+\frac{p_d^2}{2\mu}-\frac{1}{4\pi\varepsilon_0}\frac{e^2}{|x_d|} $$ Where $\mu$ is called the "reduced mass" $m_em_p/[m_p+m_e]$. So basically, you have the normal hydrogen wavefunction solutions, but instead of $x$ being the electron position, it's the difference between the two positions, and instead of $m_e$, you have $\mu$ (which is real close to the electron mass). As for the center of mass position - there's no potential! So we arrive at the obvious fact that the hydrogen atom as a whole behaves as a quantum mechanical free particle with mass $m_e+m_p$.

So now the proton radius. You may have heard that the proton has a "charge radius." It's worth emphasizing that this is not the size of the proton's wavefunction. Typically, the size of a proton's wavefunction, or it's $\Delta_x$ in the formula $\Delta_x\Delta_p\geq \hbar/2$, is much bigger than "the proton radius." Instead, the proton radius is, in the sense of the previous derivation, the size of the "relative part" of the wavefunction of the quarks. This is all very handwavy, but in a sense, think of a wavefunction that initially has four particles: two up quarks, a down quark, and the electron, and for the quarks, we add this to the hamiltonian: $$ H=\frac{p_{u1}^2}{2m_u}+\frac{p_{u2}^2}{2m_u}+\frac{p_{d}^2}{2m_d}+V_{QCD}(u1,u2,d) $$ Remember how the electromagnetic interaction ensures that we have a wavefunction in terms of the relative positions of the proton and the electron, which localizes them to be within about a Bohr radius of each other. Like that, the QCD potential ensures that the quarks are within a proton radius of eachother (much smaller than a Bohr radius). So how do we deal with the effect this has on the Hydrogen wavefunction? An approximation we make is to say that instead of a "point charge-like" potential: $$ V_0=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{|x_d|} $$ We use a potential that would arise from the proton being a uniformly charged sphere, which looks like this (but scaled to the right proton radius and total charge):

The difference between the two potentials we treat using perturbation theory. Seeing as you're in a quantum mechanics course, you're soon to learn how perturbation theory works, and you might even have this as a homework question.

Answer 2

See Volume Effects on the isotope shift for atomic spectra. Two isotopes of the same element will have slightly different transition frequencies. The frequency difference is referred to as the isotope shift and it arises from two effects. The first is the different effective mass due to the change in nuclear mass and the second effect is the volume shift. The second effect is the volume effect which is essentially what you are asking about. Basically the nucleus is not exactly a point mass. To first order it can be modeled as an arbitrary charge distribution with some radius $R = \sqrt{\langle r^2\rangle}$. Such a charge distribution results in a slightly different electric potential felt by the electron which results in slightly different energy levels. This latter effect is small but detectable [citation needed].

I'm not an expert in this field, but I can copy the formula from Wikipedia for the volume shift: $$ \delta E \approx \frac{e^2}{4\pi \epsilon_0} \frac{4}{5} R^2 \frac{Z^4}{a_{\mu}^3 n^3} \frac{\delta R}{R} $$ Here $e$ is the electron charge, $\epsilon_0$ is the permittivity of free space, $R$ is the effective radius of the nucleus, $Z$ is the number of portons, $n$ is the principle quantum number for the electronic state, I think $a_\mu = (m_e/\mu) a_0$ is the effective-mass-corrected Bohr radius, an $\delta R$ is of course the difference in the effective nuclear radius due to different number of nuetrons.

In fact, since nuclear structure modifies atomic spectra (which we can measure extremely precisely) this means that we can use atomic spectroscopy to probe nuclear structure. See for example proposals (and experimental efforts) for a nuclear clock which uses two nuclear states to keep time the same way an atomic clock uses two (spin or orbital) electronic states to keep time.

• Nuclear clock wiki
• Thorium nuclear clock project

This is pure speculation: The treatments above all treat the nucleus as spherically symmetric and described by a simple effective radius. However, we could model the nucleus as having some charge distribution described be a series of spherical harmonics. It's possible that the spectroscopic shifts may depend on the orbital angular momentum differences between atomic states. Such effects would allow us to probe not only the spherically symmetric part of nuclear structure but also the angular parts! One issue I see with this, however, is that it is likely that in the "lab frame" we can control the orientation of the electronic states using lasers with well controlled polarizations. I don't know if it is similarly possible to control the spatial orientation of the nucleus. That is, maybe the nucleus is rolling and tumbling around with respect to the orientation of the electronic state. If this were the case, the spectroscopic shifts of any angular momentum structure of the nucleus would be time-averaged to zero.

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Another related effect is the Darwin term which results in a fine-structure shift to s-orbitals due to the overlap of the s orbital with the nucleus. But this shift is not due to the finite extent of the nucleus. Rather, I think it due to some higher order QFT dynamics of the electron which I frankly don't understand. In the words of Wikipedia:

The Darwin term changes potential energy of the electron. It can be interpreted as a smearing out of the electrostatic interaction between the electron and nucleus due to zitterbewegung, or rapid quantum oscillations, of the electron. This can be demonstrated by a short calculation."

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The Lamb Shift seems to also be related to the finite size (or at least position fluctuations?) of the nucleus but it is, like, the Darwin shift, a QFT/QED shift that I do not personally understand very well.

Answer 3

Adding to the answer from Jagerber48, I will add that the inverse is also true: due to this overlap in wave functions, nuclear energy levels are shifted by what the inner shell electrons are doing. This is commonly measured in Mössbauer spectroscopy, which measures the nuclear energy levels via X-ray scattering, and the spectra can be compared with theoretical predictions such as density functional theory calculations to try to pin down what the inner shell electronic structure of a material is like.