# Can the Schrödinger equation be solved for deuterium?

Or does the added neutron complicate things too much?

• What do you mean by Schrödinger's equation for deuterium? The electronic part of the neutral atom with a nucleus of D? Dec 7, 2021 at 18:43
• If all you care about is the electron, you can just use this with $\mu$ obtained for a $Z=1,\,A=2$ nucleus.
– J.G.
Dec 7, 2021 at 18:44
• @J.G. So the electron is unaffected by the neutron's addition except for the negligible change in μ?
– Iddo
Dec 7, 2021 at 19:03
• Somewhat related: the two answers to What is the difference between the Balmer series of hydrogen and deuterium? Dec 9, 2021 at 12:20

There are at least five ways of taking this question, all of them equally natural.

1. From first principles, a deuterium atom is a system of six quarks and one electron. The motion of the quarks is relativistic, so the nonrelativistic Schrodinger equation is useless.

2. Oh, wait...the term "Schrodinger equation" is ambiguous, though. In some contexts, it means the nonrelativistic SE, in others it just means the basic equation for time evolution in QM. Even by the latter interpretation, the seven-body problem does not usually have an exact solution. There may be approximate solutions. The first approximations you would naturally make would be to neglect the electron and treat the quarks as two 3-quark bags of nucleons. The nucleons are somewhat relativistic (v~5% of c), but it's not unreasonable to look at a nonrelativistic picture. Then there are pretty decent approximate solutions that tell you certain things, e.g., that the ground state has spin 1 and there are no bound excited states. Such solutions are going to be approximate partly because we don't have any exact way of solving QCD. In fact, nuclear structure physicists don't normally even attempt to use QCD. They use phenomenological models.

3. Similar to 1, but not even considering the possibility of treating the quarks separately.

4. To an atomic physicist, your question is obviously a question about the eV-scale observable spectrum, which has to do mostly with the electron's motion. An excellent approximation is just to use the reduced mass and otherwise apply the known solution for the hydrogen atom. The higher-order effects would be things like the coupling between the electron's magnetic dipole moment and that of the deuteron. These would be treated via perturbation theory, and would be accurate to many decimal places. Because the charge Z of the nucleus is small compared to the inverse of the fine structure constant (137), the theoretical results will be embarrassingly good, and experiment will never be able to give us interesting tests of the standard model. For bigger Z, getting closer to 137, there are people who are working on using the nth decimal place to compare theory to experiment.

5. It would never occur to a nuclear structure physicist that your question could be about anything other than the nuclear structure physics of the deuteron at MeV-scale energies. See 2.

• I assume in 4 you're referring to hydrogenlike high-Z ions? Because you can also compare QED calculations to experimental results with helium fine structure Dec 8, 2021 at 17:37
• I don't know why how the quark structure of the proton and deuteron is relevant in atomic physics. Fortunately so, as the traditional uud and udd quark configurations are strongly simplified with respect to more realistic, extremely complex structures of proton and neutron. Dec 9, 2021 at 20:31
• Six valence quarks. If I have a year where I earn €50,000 and spend €49,000, a statement like "this year I dealt with €1,000" does not adequately describe my financial situation.
– rob
Dec 9, 2021 at 20:39

The hydrogen atom consists of a proton of mass $$m_p$$ together with an electron of mass $$m_e$$. The interaction between them is the Coulomb potential between two point charges $$V= -\frac{1}{4 \pi \epsilon_0} \frac{e^2}{r},$$ the minus sign because the electron and proton have opposite charges.

The energies of the bound states are given as $$E_n = - \frac{ m_e c^2 \alpha^2}{ 2 n^2,}$$ where $$\alpha$$ is the fine structure constant (the details are given in the link in one of the comments).

In this model three effects are ignored:

(i) The proton's mass is assumed to be infinite. The finite mass of the proton is easily taken into account by replacing $$m_e$$ by $$\mu = \frac{ m_e m_p}{m_e + m_p }$$ that is aclled the reduced mass of the electron and the proton.

(ii) the electron and proton act a little like magnetic dipoles and there is an interaction between them, this is much smaller that the coulomb potential.

(iii) The proton is not a point charge and has a charge distribution. This means that the solutions (eigen energies and eigen wave functions) obtained for point charges are invalid, but are a small effect.

(There are other effects, too many to include here, for example relativistic effects.)

For deuterium, the most important fact to take into account is that the nucleus consists of a proton and a neutron that is called a deuteron. The neutron is uncharged so that the nuclear charge is still $$+ e$$.

The largest effect is still the mass effect, now you replace $$m_e$$ by $$\mu = \frac{ m_e m_d}{m_e + m_d }$$ where $$m_d$$ is the mass of the deuteron (which is approximately twice the mass of the proton).

I would say it can be very good approximated with computational resources, but it can't be solved analytically. This is just due to the fact that you have a three-body problem and there isn't any analytic solution for that.

This is a two body problem for which exact solutions of the Schrödinger and the Dirac equation exist to the same accuracy as for protonic hydrogen atom. The difference between the proton and the deuteron are:

• mass. This leads to a slightly different reduced mass.