Relativistic Corrections to Hydrogen

Three first-order matrix elements reproduce the leading fine structure of a one-electron Coulomb problem: the relativistic kinetic correction, spin–orbit coupling, and the Darwin contact. Their separate shifts depend on orbital angular momentum $\ell$, but their sum depends only on $n$ and total angular momentum $j$. The cancellation provides a stringent check on signs, factors of two, and the treatment of S states. Here the nucleus is an infinitely heavy point source and $Z\alpha\ll1$. Atomic Fine Structure owns the spectroscopic hierarchy and many-electron interpretation.

Required background. The Foldy–Wouthuysen Expansion supplies the operators, Radial Wavefunctions supplies the Coulomb moments, and Spin–Orbit Angular-Momentum Algebra supplies the coupled basis. Helpful background. The Darwin Term explains the contact limit; Degenerate Perturbation Theory explains why the basis within each shell must be chosen before taking diagonal matrix elements.

Coulomb reference problem and perturbations

Let $m$ be the electron mass and define

$$a=Z\alpha,\qquad M=mc^2,\qquad b=\frac{\hbar}{mc a}.$$

The reference Hamiltonian, binding energies, and potential are

$$H_0=\frac{\mathbf p^2}{2m}+V,\qquad V=-\frac{a\hbar c}{r},\qquad E_n^{(0)}=-\frac{Ma^2}{2n^2}.$$

The total energy will include the rest term $M$ in addition to these binding energies. The mass here is not the electron–nucleus reduced mass: the fixed-source limit is part of the model. Reduced-mass wavefunctions can incorporate part of recoil physics, but a mass substitution alone is not the complete relativistic recoil calculation.

From the electrostatic FW Hamiltonian, use

$$\begin{aligned} H_{\rm T}&=-\frac{\mathbf p^4}{8m^3c^2},\\ H_{\rm SO}&=\frac{a\hbar}{2m^2c\,r^3} \mathbf L\cdot\mathbf S,\\ H_{\rm D}&=\frac{\pi a\hbar^3}{2m^2c} \delta^{(3)}(\mathbf r). \end{aligned}$$

The last two coefficients are derived and interpreted on Spin–Orbit Coupling and Darwin Term. They are not independent additions to an exact Dirac eigenvalue.

At fixed $n$, the unperturbed shell is degenerate. Use $|n,\ell,j,m_j\rangle$ with spin $1/2$. The kinetic operator is an orbital scalar, the spin–orbit operator is radial times $\mathbf L\cdot\mathbf S$, and the contact acts only on $\ell=0$ states. Their matrices are diagonal in this basis. In particular, same-$j$ states of opposite parity do not mix under these parity-even operators. This explains why diagonal expectation values suffice despite the Coulomb degeneracy.

Radial moments used in the calculation

With the Coulomb length $b$ above, the normalized radial states give

$$\begin{aligned} \langle r^{-1}\rangle &=\frac{1}{bn^2},\\ \langle r^{-2}\rangle &=\frac{1}{b^2n^3(\ell+\tfrac12)}. \end{aligned}$$

For $\ell>0$,

$$\langle r^{-3}\rangle =\frac{1}{b^3n^3\ell(\ell+\tfrac12)(\ell+1)}.$$

For $\ell=0$, use the contact density instead:

$$|\psi_{n00}(0)|^2=\frac{1}{\pi b^3n^3}.$$

These are matrix elements of the unperturbed Schrödinger states. Their normalized Laguerre representation belongs to Radial Wavefunctions. The $r^{-3}$ formula deliberately excludes S states; its divergent continuation to $\ell=0$ is not a spin–orbit matrix element.

Relativistic kinetic shift

On an eigenstate of $H_0$,

$$\mathbf p^2\psi =2m(E_n^{(0)}-V)\psi.$$

Consequently the finite quadratic-form expectation of $\mathbf p^4$ is

$$\begin{aligned} \langle\mathbf p^4\rangle &=\|\mathbf p^2\psi\|^2\\ &=4m^2\left\langle(E_n^{(0)}-V)^2\right\rangle. \end{aligned}$$

This uses the norm of $\mathbf p^2\psi$. It is not an operator replacement of $\mathbf p^4$ by $4m^2(E_n^{(0)}-V)^2$ on arbitrary states; such a replacement would ignore derivatives of $V$.

The Coulomb moments imply

$$\langle V\rangle=2E_n^{(0)},\qquad \langle V^2\rangle =\frac{M^2a^4}{n^3(\ell+\tfrac12)}.$$

Thus

$$\begin{aligned} \left\langle(E_n^{(0)}-V)^2\right\rangle &=\langle V^2\rangle-3(E_n^{(0)})^2,\\ \Delta E_{\rm T} &=Ma^4\left[ \frac{3}{8n^4} -\frac{1}{2n^3(\ell+\tfrac12)} \right]. \end{aligned}$$

The shift is negative for every allowed $0\leq\ell\leq n-1$, as it must be from $H_{\rm T}=-\mathbf p^4/(8m^3c^2)$. Checking that sign is useful before combining the other terms.

Spin–orbit and Darwin shifts

For $\ell>0$, combine the inverse cubic moment with

$$\langle\mathbf L\cdot\mathbf S\rangle =\frac{\hbar^2}{2} \left[j(j+1)-\ell(\ell+1)-\frac34\right].$$

This gives

$$\Delta E_{\rm SO} =\frac{Ma^4}{4n^3} \frac{j(j+1)-\ell(\ell+1)-\tfrac34} {\ell(\ell+\tfrac12)(\ell+1)}.$$

It is often clearer to write the two branches separately:

$$\Delta E_{\rm SO}= \begin{cases} \dfrac{Ma^4}{4n^3(\ell+\tfrac12)(\ell+1)}, &j=\ell+\tfrac12,\\[5pt] -\dfrac{Ma^4}{4n^3\ell(\ell+\tfrac12)}, &j=\ell-\tfrac12. \end{cases}$$

For $\ell=0$ the spin–orbit operator is zero because $\mathbf L=0$. Set it to zero directly, rather than multiplying a divergent radial integral by a vanishing angular factor.

The Darwin contact is positive and gives

$$\Delta E_{\rm D} =\frac{Ma^4}{2n^3}\delta_{\ell0}.$$

The contact term therefore matters precisely where the spin–orbit contribution is absent. All three matrix elements have the common energy scale $Ma^4$.

Cancellation of separate orbital dependence

For $j=\ell+\tfrac12$ with $\ell>0$, the terms proportional to $1/n^3$ combine as

$$-\frac{1}{2(\ell+\tfrac12)} +\frac{1}{4(\ell+\tfrac12)(\ell+1)} =-\frac{1}{2(\ell+1)}.$$

For $j=\ell-\tfrac12$,

$$-\frac{1}{2(\ell+\tfrac12)} -\frac{1}{4\ell(\ell+\tfrac12)} =-\frac{1}{2\ell}.$$

Since $j+\tfrac12$ equals $\ell+1$ or $\ell$ respectively, both yield the same formula in terms of $j$. For $\ell=0$, the kinetic coefficient $-1/n^3$ plus the Darwin coefficient $+1/(2n^3)$ gives the required $-1/(2n^3)$ with $j=1/2$. The complete shift is

$$\Delta E_{nj} =Ma^4\left[ \frac{3}{8n^4} -\frac{1}{2n^3(j+\tfrac12)} \right].$$

The total energy through this order is $E_{nj}=M+E_n^{(0)}+\Delta E_{nj}$. The $m_j$ degeneracy remains, and different $\ell$ values with the same $n,j$ retain equal energies in this model. Rotational invariance alone does not force that latter equality; it follows from the specific Coulomb coefficients and their cancellation.

A complete n = 2 check

The individual shifts below are divided by $Ma^4$.

State	Kinetic	Spin–orbit	Darwin	Total
$2S_{1/2}$	$-13/128$	$0$	$1/16$	$-5/128$
$2P_{1/2}$	$-7/384$	$-1/48$	$0$	$-5/128$
$2P_{3/2}$	$-7/384$	$1/96$	$0$	$-1/128$

The $2P_{3/2}-2P_{1/2}$ interval is $Ma^4/32$ and is positive. The $2S_{1/2}$ and $2P_{1/2}$ totals agree despite completely different decompositions. Omitting the Darwin term would destroy that agreement.

The Lamb shift lifts the same-$n,j$ degeneracy once the appropriate radiative and other precision effects are included. It is not the missing fourth term in this minimal tree-level FW calculation. Atomic Fine Structure and the Lamb Shift Overview explain the spectroscopic separation.

Comparison and controlled use of the expansion

An independent expansion of the exact point-Coulomb Dirac energy gives

$$E_{nj}^{\rm Dirac} =M-\frac{Ma^2}{2n^2} +\Delta E_{nj}+O(Ma^6)$$

at fixed $n,j$ as $a\to0$. Thus the first-order effective-Hamiltonian calculation reproduces the exact answer through $a^4$. Littlejohn (2021) derives the Dirac spectrum by solving the radial equations; the agreement checks the perturbative operators independently.

The binding scale is $Ma^2$ and the fine-structure scale is $Ma^4$. Higher terms and second-order perturbative effects enter the fixed-source expansion at higher powers, beginning with $Ma^6$. This counting is not permission to iterate singular contact operators without a consistent regulator and matching prescription.

In particular, the truncated kinetic Hamiltonian $\mathbf p^2/(2m)-\mathbf p^4/(8m^3c^2)$ is unbounded below at large momentum. It must be used as a low-energy expansion, not minimized over arbitrary wavefunctions as though it were an exact stable Hamiltonian. For large $Z\alpha$, use a relativistic starting point; for precision atomic work, include recoil, finite source size, nuclear spin, and QED effects at the required accuracy.

Exercises

Ground-state cancellation. Compute all three $1S_{1/2}$ shifts and compare their sum with the expansion of $M\sqrt{1-a^2}$.

Solution

The kinetic shift is $-5Ma^4/8$, the spin–orbit shift is zero, and the Darwin shift is $+Ma^4/2$. The sum is $-Ma^4/8$. Indeed,

$$M\sqrt{1-a^2} =M-\frac{Ma^2}{2}-\frac{Ma^4}{8} +O(Ma^6).$$

A general doublet interval. For fixed $n,\ell\geq1$, find the energy of $j=\ell+1/2$ relative to $j=\ell-1/2$.

Solution

The kinetic terms coincide and both Darwin terms vanish. Equivalently, subtract the complete $j$ formulas:

$$E_{n,\ell+1/2}-E_{n,\ell-1/2} =\frac{Ma^4}{2n^3\ell(\ell+1)} +O(Ma^6).$$

The interval is positive for the attractive point Coulomb potential.

Why squaring an eigenvalue relation is dangerous. Explain why the kinetic expectation calculation is valid even though $\mathbf p^2$ and $V$ do not commute.

Solution

The eigenvalue equation determines the vector $\mathbf p^2\psi$. Taking its squared Hilbert-space norm requires no second differentiation: $\|\mathbf p^2\psi\|^2 =4m^2\|(E_n^{(0)}-V)\psi\|^2$. Acting again with $\mathbf p^2$ on the right-hand side would differentiate $V$ and is a different operation. For Coulomb states the norm identity is the appropriate quadratic-form statement.

References

• Bethe, Hans A., and Edwin E. Salpeter. Quantum Mechanics of One- and Two-Electron Atoms. Springer (1957).
• Foldy, Leslie L., and Siegfried A. Wouthuysen. “On the Dirac Theory of Spin 1/2 Particles and Its Non-Relativistic Limit.” Physical Review 78, 29–36 (1950). doi:10.1103/PhysRev.78.29.
• Littlejohn, Robert G. “Solutions of the Dirac Equation and Their Properties.” Physics 221B, Notes 50, University of California, Berkeley, academic year 2021–22 (2021). Lecture notes.