In the non-relativistic limit, the Schrodinger equation for the hydrogen atom can be solved using reduced mass techniques to account for the motion of both the electron and proton.

I am wondering if a similar thing can be done with the Dirac equation. Wikipedia suggests so as the energy levels are written with the reduced mass, but I cannot find any discussion of this point, and most books just work with a single electron moving in a fixed Coulomb potential. I thought the Dirac equation was just for a single particle, so is it possible to add a Dirac Hamiltonian for the proton and solve the combined system?


1 Answer 1


Things become indeed more complicated in the relativistic case, since both the electron and the proton are spin $1/2$ particles, so the total wave function must be a spinor of $4^2=16$ components. While the spatial part of the wave function can be reduced to an effective one-body form (by the usual introduction of COM and relative coordinates), such a reduction cannot be done exactly for the spinor part. For this reason, the one-body Dirac equation cannot provide exact solutions in case of a finite-mass proton.

This problem was investigated by Salpeter[1] (following the seminal work of Bethe and Salpeter[2]), who derived a wave equation for the hydrogen atom, which was in principle exact on the level of QED (with relativistic and radiative/non-radiative QED effects taken into account). After a series of approximations, resorting to "pure relativistic" corrections (neglecting negative energy, retardation and QED contributions) gives the following two-body wave equation: $$ \left[\hat{h}_{\text{e}}\otimes I_{\text{P}}+I_{\text{e}}\otimes\hat{h}_\text{P}+V_{\text{CB}}^{++}\right]\Psi(\boldsymbol{r})=E\Psi(\boldsymbol{r}) \ , $$ where $\Psi(\boldsymbol{r})$ is a 16-component spinor depending on the relative coordinate $\boldsymbol{r}$, $$ \hat{h}_{\text{e}}=-i\boldsymbol{\alpha}_\text{e}\cdot\vec{\nabla}+\beta_{\text{e}}m_{\text{e}} \ , $$ and $$ \hat{h}_\text{P}=+i\boldsymbol{\alpha}_\text{P}\cdot\vec{\nabla}+\beta_{\text{P}}m_{\text{P}} $$ are one-body Dirac Hamiltonians for the electron and proton, respectively, and $$ V_{\text{CB}}^{++}=\Lambda_{++}\left\{-\frac{Z\alpha}{r}I_{\text{e}}\otimes I_{\text{P}}+\frac{Z\alpha}{2r}\left[\boldsymbol{\alpha}_\text{e}\otimes\boldsymbol{\alpha}_\text{p}+\frac{1}{r^2}(\boldsymbol{\alpha}_\text{e}\cdot\boldsymbol{r})\otimes(\boldsymbol{\alpha}_\text{P}\cdot\boldsymbol{r})\right]\right\}\Lambda_{++} $$ is the Coulomb-Breit interaction projected onto the positive-energy subspace (also, $\alpha=e^2/(4\pi)$ is the fine structure constant). Note that Salpeter worked in momentum space; the above formulae are transformed into coordinate space for convenience.

Salpeter treated this equation perturbatively, and obtained finite-mass corrections to the fine structure of hydrogen. The accurate numerical solution of the above equation could also be found with variational techniques, with the small caveat that the $m_{\text{P}}\rightarrow\infty$ limit would not reproduce exactly the one-body Dirac solution. This is due to the presence of the positive energy projection operators (or in the language of QFT, due to the absence of crossed-ladder type Feynman diagrams).

The formula obtained from the perturbative investigation of the Coulomb-Breit interaction reads[3] $$ E_{nlj}\approx M-\frac{\mu(Z\alpha)^2}{2n^2}-\frac{\mu(Z\alpha)^4}{2n^3}\left[\frac{1}{j+\frac{1}{2}}-\frac{3}{4n}+\frac{1}{4n}\frac{\mu}{M}-(1-\delta_{l0})\left(\frac{1}{j+\frac{1}{2}}-\frac{1}{l+\frac{1}{2}}\right)\frac{\mu^2}{m_{\text{P}}^2}\right] \ . $$ You can now see that the wikipedia formula is not exact, replacing the electron mass with $\mu$ only works approximately (note e.g. the new $\mu/M$ or $\mu^2/m_{\text{P}}^2$ terms). Such a replacement is more or less justified in "leading order", since the dominant contribution to the fine structure can be calculated by evaluating expectation values of relativistic operators with the non-relativistic wave function (a similar approximation can also be made when calculating finite-mass corrections to the Lamb shift).

The hyperfine structure arising from interactions with the proton spin was left out of $E_{nlj}$ on purpose, even though it is also an ${\cal{O}}(\mu\alpha(Z\alpha)^3\mu/M)$ finite mass contribution. The reason is that treating the proton as a spin $1/2$ elementary particle could only give an incomplete account of hyperfine structure; the naive correction would be calculated with the Dirac $g$-factor $g=2$ instead of the correct $g_{\text{P}}\approx5.586$ value. To correct for this, the excess part of the proton magnetic moment must be introduced via Pauli coupling[4,5].

As a curiosity, note that $E_{nlj}$ depends on the orbital angular momentum quantum number $l$. This means that proton recoil lifts the $j$-degeneracy, and predicts a Lamb-like splitting of the $2s_{1/2}$ and $2p_{1/2}$ levels already on the fine structure level! Of course, this effect is extremely tiny compared to the main QED contribution (for hydrogen, it is ${\cal{O}}(m_{\text{e}}\alpha^4m_{\text{e}}^2/m_{\text{P}}^2)$ against ${\cal{O}}(m_{\text{e}}\alpha^5\ln(\alpha)$).

Update (2023.10.16)

As pointed out by naturallyInconsistent (whom I thank again), there is actually an analitically solvable two-body Dirac Hamiltonian (with Coulomb interaction only) proposed by Marsch[6,7]. After a lengthy derivation, he found (in the rest frame) $$ {\cal{E}}_{nk}=M\sqrt{1-\frac{2\mu}{M}+\frac{2\mu}{M}f\left(n,k;Z\alpha,\frac{Z\alpha}{2}\right)} \ , $$ where $$ M=m_{\text{e}}+m_{\text{P}} \ \ \ , \ \ \ \mu=\frac{m_{\text{e}}m_{\text{P}}}{m_{\text{e}}+m_{\text{P}}} $$ are the total and reduced masses, respectively, and $$ f(p,q;x,y)=\left(1+\frac{x^2}{\left(\sqrt{q^2-y^2}+p-|q|\right)^2}\right)^{-\frac{1}{2}} \ . $$ We have $k=1$ when the electron and nuclear spin are coupled to a singlet, and either $k=l+1$ (for $J=l+1$) or $k=-l$ (for $J=l-1$) in case of spins coupled to a triplet. This energy formula describes recoil effects to all orders of $\mu/M$. It is useful to compare the results with those obtained from the "tinkered" one-body Dirac equation of wikipedia: $$ {\cal{E}}'_{nj}=\mu f\left(n,j+\frac{1}{2},Z\alpha,Z\alpha\right) \ , $$ where $m_{\text{e}}$ is manually replaced by $\mu$. Up to $\mathcal{O}(\alpha^4)$ accuracy, we find $$ \begin{aligned} {\cal{E}}_{nk}&\approx M-\frac{\mu(Z\alpha)^2}{2n^2}-\frac{\mu(Z\alpha)^4}{2n^3}\left[\frac{1}{4|k|}-\frac{3}{4n}+\frac{1}{4n}\frac{\mu}{M}\right] \ , \\ {\cal{E}}'_{nj}&\approx \mu-\frac{\mu(Z\alpha)^2}{2n^2}-\frac{\mu(Z\alpha)^4}{2n^3}\left[\frac{1}{j+\frac{1}{2}}-\frac{3}{4n}\right] \ , \end{aligned} $$ to be compared with the perturbative result shown above.

Up to $\mathcal{O}(\mu(Z\alpha)^4\mu/M)$, $E_{nlj}$ agrees perfectly with the formula of Brézin, Itzykson and Zinn-Justin[8]. Apart from the uninteresting rest mass terms, all three formulae produce the same non-relativistic energy (which is not surprising). The fact that $\mathcal{E}'_{nj}$ misses the higher recoil effects is also expected.

But interestingly, while the $\mathcal{O}(\mu(Z\alpha)^4\mu/M)$ effects match nicely between $\mathcal{E}_{nk}$ and ${E}_{nlj}$, the $\mathcal{O}(\mu(Z\alpha)^4)$ part of $\mathcal{E}_{nk}$ differs from $\mathcal{E}'_{nj}$ and $E_{nlj}$. Due to the effect of nuclear spin, 4$|k|$ seems to take rather different values than $j+1/2$. I will need to think about this. [To be continued...]


[1] Salpeter: Phys. Rev. 87 2 328 (1952)

[2] Salpeter, Bethe: Phys. Rev. 84 1232 (1951)

[3] Eides, Grotch, Shelyuto: Phys. Rep. 342 2 63 (2001)

[4] Arnowitt: Phys. Rev. 92 4 1002 (1953)

[5] Newcomb, Salpeter: Phys. Rev. 97 4 1246 (1955)

[3] Marsch: Ann. Phys. 14 5 324 (2005)

[4] Marsch: Ann. Phys. 15 6 434 (2006)

[6] Brézin, Itzykson, Zinn-Justin: Phys. Rev. D 1 2349 (1970)

  • 2
    $\begingroup$ Your answer is correct in spirit but not technically correct. In 2005 Marsch published a RQM solution of the proton-electron system using reduced mass. (But you might also want to check the 2006 addendum.) His crazy solution involves finding the spin+orbital angular momentum eigenstates, so as to factorise the 16-components to 4 spin and 4 space. He only treated the Coulomb problem, so hyperfine corrections and QED corrections do not appear, but it is technically possible to treat proton & electron thusly. $\endgroup$ Oct 16, 2023 at 7:35
  • $\begingroup$ @naturallyInconsistent Thanks for the feedback, this sounds interesting. I cannot access the original 2005 paper (only the 2006 follow-up which does not show the energy formula), but I will try to get hold of it and read it. In the meantime, please consider writing an answer about this. Just let me point out that the Breit term gives contributions to the fine structure too, not just the hfs, so one way or another, you need to treat that term too, even on the "pure relativistic" level of theory (but I understand that this was not OP's original question). $\endgroup$ Oct 16, 2023 at 8:29
  • $\begingroup$ I don't think I can do justice to this; that does not read like a paper, it reads like a textbook. He used a lot of operator algebra just to extract the entangled spin eigenstates. I think he explicitly says that the scheme he is doing is not the full Breit term; it has spin-orbit coupling but is missing spin-spin coupling. But that is scarcely a problem: it is still a wonderful initial analytically exact solution from which to compute QED perturbation corrections. $\endgroup$ Oct 16, 2023 at 9:09
  • $\begingroup$ @naturallyInconsistent I updated the answer, and tried to give an account of the results of Marsch. Strangely enough, there is an $\mathcal{O}(\mu(Z\alpha)^4)$ discrepancy between his and the standard perturbative results. I must look into this later... $\endgroup$ Oct 16, 2023 at 16:01
  • $\begingroup$ He openly said that his model would be neglecting quite a lot of effects, so it is unlikely to be surprising that something is missing. Thanks! $\endgroup$ Oct 17, 2023 at 0:22

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