# Most general non-relativistic Hamiltonian for hydrogenic atom in quantised electromagnetic field

In many textbooks, the derivation of the energy levels in a hydrogen atom starts from the basic Hamiltonian $H = \frac{\mathbf{p}^2}{2m} + \frac{e^2}{4\pi\epsilon_0\mathbf{r}}$ and then adds relativistic fine structure or hyperfine structure as corrections to these basic energy levels. The effect of an electromagnetic field is then often included by modifying the Hamiltonian to $H = \frac{\left(\mathbf{p} - e \mathbf{A}\right)^2}{2m} + e \phi$ for electromagnetic potentials $\mathbf{A}$ and $\phi$.

What I find strange about this approach is that from the very beginning, the interaction between proton and electron is electrostatic (thus requiring the presence of an electromagnetic field), so why is the quantised electromagnetic field only added later and not right at the start? Also, the energy of the spins of the particles in an external magnetic field does not seem to be included in this treatment.

What I wonder is the following: If we do not start from the basic Hamiltonian and add all terms later, but set out to derive from first principles the energy levels for a proton and a (bound) electron in a quantised electromagnetic field (generated by proton and electron, but also due to external radiation that might be present), what is the Hamiltonian that describes this system? The proton and electron should be considered non-relativistic, so should be regarded as quantised in the sense that observables should be operators, but without the need for quantum field theory to describe proton and electron. How could relativistic effects then be included perturbatively in this general treatment?

EDIT: Just to clarify, the situation that I am interested in is the following: Two charged, spin-carrying non-relativistic particles (proton and electron) move under the influence of an internal electric field arising from their charges and an external field due to radiation. Both particles should be included in the Hamiltonian, so the proton will also have a coupling to any electromagnetic fields. Questions: How is the internal field treated as a quantised electromagnetic field, leading to a potential energy? Which terms in the Hamiltonian does the interaction with the external field lead to? What is therefore the most general Hamiltonian for a non-relativistic proton-electron system moving in a quantised electromagnetic field?

Any references to textbooks or journal articles with a detailed treatment of the most general and (within the limits of neglecting relativistic effects) most exact Hamiltonian describing such a system would be very useful.

• What do you mean: "so why is the quantised electromagnetic field only added later and not right at the start?"? If I got it right, the E.M. field added via minimal coupling is an external field perturbing the system. Interesting the question about spin. – Matteo Oct 10 '17 at 22:38
• As I understand it, the electromagnetic fields $\mathbf{E}$ and $\mathbf{B}$ (or corresponding potentials $\mathbf{A}$ and $\phi$) should be treated as quantised fields (that is, being represented by operators with appropriate commutation relations), regardless of what causes them. The interaction between proton and electron involves an electromagnetic field (to be precise, an electric field), so I would have expected to see a corresponding field operator in the Hamiltonian for the proton-electron interaction. – Quantum Oct 10 '17 at 22:51

What you want, I think, is starting from Dirac equation, find the non-relativistic limit, which is basically Pauli equation, plus spin-orbit coupling, plus the Darwin term. The classic way to do that is through the Foldy-Wouthuysen transformation [FW50]. The paper takes a bit of extra effort to read because they use old-fashioned notations for the Dirac machinery but manageable from memory. I did quote the Wikipedia pages mostly for the bibliography, as I don't find them particularly easy to learn from.

In any case, the final answer is a series in $v/c$, the classic terms of which being given by the following Hamiltonian, taken from the classic textbook by Cohen-Tannoudji, Diu and Laloë [CTDL77, chapter 12, appendix B, eqn (B-I)]:

$$H = m_ec^2 + \underbrace{\frac{P^2}{2m_e} + V}_\text{classic} \underbrace{- \frac{P^4}{8m_e^3c^2}}_\text{relativistic momentum} + \underbrace{\frac{1}{2m_e^2c^2}\frac{1}{R}\frac{\partial V}{\partial r}L\cdot S}_\text{spin orbit}+\underbrace{\frac{\hbar^2}{8m_e^2c^2}\Delta V}_\text{Darwin}$$

In the presence of an electromagnetic field $A$, the prescription is $P\to P-eA$, and then the extra term

$$-2\mu_B \frac{1}{\hbar}S\cdot B.$$

This correction has to be used to take into account the magnetic field from the magnetic moment of the proton, actually.

Finally, there is the contribution of the electromagnetic field itself. This is a very simple term actually, just the equivalent of a sum of harmonic oscillator, one per photon mode,

$$H_{em} = \sum_i\hbar\omega_i\left(a_i^+a_i\frac{1}{2}\right).$$

So here, $a_i^+$ is a creation operator and $a_i$ is a destruction operator, both for a photon of wave vector $k_i$, energy $\omega_i$, and helicity $\epsilon_i$, perpendicular to $k_i$: this is the point of using the Colomb's gauge, it completely decouples the time and longitudinal degrees of freedom of the photons, which can then be ignored.

[FW50] Leslie L. Foldy and Siegfried A. Wouthuysen. On the Dirac theory of spin 1/2 particles and its non-relativistic limit. Phys. Rev., 78:29–36, Apr 1950.

[CTDL77] Claude Cohen-Tannoudji, Bernard Diu, and Franck Laloë. Quantum mechanics. Wiley, New York, NY, 1977. Trans. of : M ́ecanique quantique. Paris : Hermann, 1973.

• This accounts for fine structure and hyperfine structure, but does not include a quantised electromagnetic field. To clarify, I guess my question is two-fold: 1) The electron and proton are modelled as charged, spin-carrying particles that move under the influence of an electromagnetic field. This field arises partly due to an internal field created by their charges. Should this internal field considered to be quantised (and how does the potential energy in the Hamiltonian arise from this)? 2) Another part of the field is due to external radiation - how is this rigorously included? – Quantum Oct 11 '17 at 0:32
• ah, ok, you did not mention that… well, in Coulomb gauge, the Hamiltonian will look almost what I have already written plus a term for the electromagnetic field but it's getting late in the part of the world where I live! So, tomorrow… – user154997 Oct 11 '17 at 0:36
• In any case, the Foldy-Wouthuysen is still an essential ingredient. – user154997 Oct 11 '17 at 0:37
• Sorry, it slipped from my mind to add the little bit about the radiation Hamiltonian. – user154997 Oct 20 '17 at 21:08

You'd need to solve Dirac's equation

$$(i\hbar\gamma^\mu\partial_\mu - mc)\phi = 0$$

Here's a pretty good reference on how to do it for the Hydrogen atom

• I am aware of the Dirac equation, but I mentioned in my post that the treatment should be non-relativistic as far as the proton and electron as particles are concerned and that it should consider the electromagnetic field (both internal due to electron-proton electrostatic potential and external due to radiation that may be present) as a quantum field. Thanks for the link! – Quantum Oct 10 '17 at 22:53