Hydrogen Fine Structure Revisited

The Dirac equation in an attractive point Coulomb field has an exact bound-state spectrum whose weak-coupling expansion reproduces hydrogen fine structure. Solving the coupled first-order radial equations fixes more than an energy formula: it determines the allowed angular labels, the relation between spinor components, and the boundary condition at the point source. Here the nucleus is infinitely heavy, the electron couples minimally, and nuclear size, recoil, and radiative effects are excluded.

Required background. Hamiltonian Form supplies the Dirac operator, Orbital Plus Spin supplies coupled angular functions, and The Hydrogen Atom supplies the nonrelativistic Coulomb comparison. Helpful background. Relativistic Corrections to Hydrogen derives the separate perturbative matrix elements.

Point-Coulomb Hamiltonian and angular labels

Let $m$ be the electron mass and write

$$\begin{aligned} a&=Z\alpha>0,\qquad M=mc^2,\\ V(r)&=-\frac{a\hbar c}{r}. \end{aligned}$$

The stationary equation in the Dirac basis is

$$\left(c\boldsymbol\alpha\cdot\mathbf p +\beta M+V\right)\Psi=E\Psi.$$

The energy $E$ includes the rest energy. For the main derivation take $0<a<\sqrt3/2$, where the standard point-source operator has a unique self-adjoint closure from its minimal domain away from the origin. The stronger point-Coulomb regime is qualified below. The eventual low-energy expansion additionally requires $a\ll1$.

Choose normalized spinor spherical harmonics $\Omega_{\kappa m_j}$ satisfying

$$(\boldsymbol\sigma\cdot\mathbf L+\hbar) \Omega_{\kappa m_j} =-\kappa\hbar\,\Omega_{\kappa m_j},$$

and fix their relative phase by

$$\boldsymbol\sigma\cdot\widehat{\mathbf r}\, \Omega_{\kappa m_j} =-\Omega_{-\kappa m_j}.$$

The nonzero integer $\kappa$ encodes the upper-component orbital angular momentum:

Angular branch	$\kappa$	Upper orbital label
$j=\ell+\tfrac12$	$-(\ell+1)$	$\ell=-\kappa-1$
$j=\ell-\tfrac12$	$\ell$	$\ell=\kappa$

In either case $j=|\kappa|-1/2$. The lower component has orbital label $\ell'=2j-\ell$. The complete spinor is a $J^2,J_z$ eigenstate and generally contains two different orbital $L^2$ eigenvalues.

If an angular Dirac operator is used, our dimensionless convention is $K_{\rm D}=-\beta(\boldsymbol\Sigma\cdot\mathbf L/\hbar+1)$. It has eigenvalue $\kappa$. Some references define the opposite sign; translating $\kappa$ is necessary before comparing radial equations.

Coupled radial equations

Separate the spinor as

$$\Psi_{\kappa m_j}(\mathbf r) =\frac1r \begin{pmatrix} G(r)\Omega_{\kappa m_j}\\ iF(r)\Omega_{-\kappa m_j} \end{pmatrix},$$

with radial normalization

$$\int_0^\infty \left(|G(r)|^2+|F(r)|^2\right)\,dr=1.$$

The angular phase above implies

$$\boldsymbol\sigma\cdot\mathbf p \left[\frac{G}{r}\Omega_\kappa\right] =\frac{i\hbar}{r} \left(G'+\frac{\kappa}{r}G\right)\Omega_{-\kappa}.$$

Substituting into the two Dirac blocks gives

$$\begin{aligned} G'+\frac{\kappa}{r}G &=\frac{E+M-V}{\hbar c}F,\\ F'-\frac{\kappa}{r}F &=\frac{M-E+V}{\hbar c}G. \end{aligned}$$

Both the explicit $i$ in the lower component and the phase of the harmonics matter for these signs. Littlejohn (2021) gives a compatible separation after translating his radial symbols and units.

For a bound state, $|E|<M$. Define

$$\lambda=\frac{\sqrt{M^2-E^2}}{\hbar c}, \qquad \tau=\sqrt{\frac{M-E}{M+E}}.$$

At large radius, normalizability selects the decaying exponential $e^{-\lambda r}$ and $F/G\to-\tau$. Near the origin, inserting $G,F\propto r^s$ gives

$$\begin{pmatrix} s+\kappa&-a\\ a&s-\kappa \end{pmatrix} \begin{pmatrix}G_0\\F_0\end{pmatrix}=0.$$

Thus $s=\pm\gamma$ with $\gamma=\sqrt{\kappa^2-a^2}$. The usual atomic boundary condition selects the $+\gamma$ branch. This refers to reduced radial functions: the full spinor behaves as $r^{\gamma-1}$, so “regular branch” need not mean a bounded four-spinor at $r=0$.

Polynomial termination and the energy sign

Set $\rho=2\lambda r$ and write

$$\begin{aligned} G&=\rho^\gamma e^{-\rho/2} \sum_{j=0}^\infty g_j\rho^j,\\ F&=\rho^\gamma e^{-\rho/2} \sum_{j=0}^\infty f_j\rho^j. \end{aligned}$$

Here the summation index $j$ is only a polynomial index, not the angular momentum label in the preceding sections. Let $x_j=(g_j,f_j)^{\mathsf T}$ and $x_{-1}=0$. The radial equations yield

$$A_jx_j=Bx_{j-1},$$

where

$$A_j= \begin{pmatrix} \gamma+j+\kappa&-a\\ a&\gamma+j-\kappa \end{pmatrix},$$ $$B=\frac12 \begin{pmatrix} 1&1/\tau\\ \tau&1 \end{pmatrix}.$$

The indicial equation is $\det A_0=0$. Up to normalization, take

$$g_0=1,\qquad f_0=\frac{\gamma+\kappa}{a}.$$

For $j>0$, $\det A_j=j(j+2\gamma)$, so the recurrence uniquely generates the regular solution. The rank-one matrix $B$ has image spanned by $w=(1,\tau)^{\mathsf T}$ and satisfies $Bw=w$. Define

$$\nu=\frac{aE}{\sqrt{M^2-E^2}}-\gamma.$$

A direct two-by-two inverse gives

$$BA_j^{-1}w =\frac{j-\nu}{j(j+2\gamma)}w.$$

For a generic nonterminating regular series, the large-order recurrence produces the growing confluent-hypergeometric exponential, which defeats the prefactor $e^{-\rho/2}$. Normalizable regular solutions terminate. For degree $N\geq1$, termination is $Bx_N=0$ and occurs at $\nu=N$. Degree zero needs the separate condition $Bx_0=0$, considered next.

The quantization condition is

$$N+\gamma=\frac{aE}{\sqrt{M^2-E^2}},$$

and therefore

$$E_{N\kappa} =M\left[ 1+\frac{a^2}{(N+\sqrt{\kappa^2-a^2})^2} \right]^{-1/2}.$$

The unsquared equation selects $E>0$: its left side is positive and $a>0$. Appending an independent minus sign after squaring would introduce spurious attractive Coulomb bound states. The negative continuum of the Dirac operator still exists; charge conjugation also changes the external-charge problem.

The lowest degree and allowed principal shells

For $N=0$, the candidate energy is $E=M\gamma/|\kappa|$, and $\tau=a/(|\kappa|+\gamma)$. If $\kappa<0$, then

$$f_0=-\tau,\qquad Bx_0=0,$$

so the constant polynomial is allowed. For $\kappa>0$, instead $f_0=1/\tau$, which does not lie in the kernel of $B$. Thus

$$N\geq0,\qquad N=0\ \Longrightarrow\ \kappa<0.$$

Define the principal quantum number by $n=N+|\kappa|$. Its allowed angular labels are

$$\kappa=-n,\ldots,-1,\quad +1,\ldots,n-1.$$

Both signs occur when $|\kappa|<n$. At $|\kappa|=n$, only the negative sign occurs. In particular, $n=1$ permits $\kappa=-1$ and the $1S_{1/2}$ state, but no $1P_{1/2}$ state. Gallone and Michelangeli (2018) give an independent spectral statement of this sign-dependent restriction.

Each allowed $\kappa$ has $2j+1=2|\kappa|$ magnetic substates. Summing them gives $2n^2$ states in the principal shell. The degree $N$ is often called a radial quantum number; it should not be identified without qualification with the number of nodes of the upper radial component.

Ground-state spinor as a check

For $\kappa=-1$, $N=0$,

$$\begin{aligned} E_{1S}&=M\sqrt{1-a^2},\\ \gamma&=\sqrt{1-a^2},\qquad \lambda=\frac{mc a}{\hbar}. \end{aligned}$$

The reduced radial functions are

$$\begin{aligned} G(r)&=\mathcal N(2\lambda r)^\gamma e^{-\lambda r},\\ F(r)&=-\frac{a}{1+\gamma}G(r). \end{aligned}$$

Writing $\tau=a/(1+\gamma)$, normalization gives

$$\mathcal N =\left[ \frac{2\lambda}{(1+\tau^2)\Gamma(2\gamma+1)} \right]^{1/2}.$$

This pair satisfies both first-order radial equations. The lower-component probability is

$$\begin{aligned} P_{\rm lower} &=\frac{\tau^2}{1+\tau^2} =\frac{1-\sqrt{1-a^2}}{2}\\ &=\frac{a^2}{4}+O(a^4). \end{aligned}$$

It is small at weak coupling, as required by the Pauli limit. It is not the probability of occupying a negative-energy eigenstate: this entire stationary solution has the positive bound energy $E_{1S}$.

Weak-coupling spectrum and fine structure

Set $k=j+1/2=|\kappa|$. The exact energy depends on $n-k+\sqrt{k^2-a^2}$, and its expansion is

$$\begin{aligned} E_{nj} ={}&M-\frac{Ma^2}{2n^2}\\ &+Ma^4\left[ \frac{3}{8n^4}-\frac{1}{2n^3(j+\tfrac12)} \right]\\ &+O(Ma^6). \end{aligned}$$

This agrees with the independent kinetic, spin–orbit, and Darwin matrix elements on Relativistic Corrections to Hydrogen. That page owns their detailed cancellation and the worked $n=2$ decomposition. For fixed $n,j$, both allowed signs of $\kappa$ have equal exact point-Dirac energies. For example, $2S_{1/2}$ and $2P_{1/2}$ are degenerate in this model.

The atomic fine-structure hierarchy separates this result from recoil, nuclear-size, hyperfine, and radiative effects. In particular, the Lamb shift is absent from the minimal fixed-source Dirac equation. An exact solution of that equation is not an exact prediction of the physical hydrogen spectrum.

The point-origin boundary beyond weak coupling

The reduced radial norm uses $dr$. The $r^{-\gamma}$ branch becomes square integrable when $\gamma<1/2$. For $\sqrt3/2<a<1$, this occurs in the $\kappa=\pm1$ channels, and square integrability alone no longer specifies the operator. The usual formula above continues to describe the distinguished self-adjoint extension, selecting the $r^{+\gamma}$ boundary behavior. The alternative branch has a divergent radial Coulomb-form integral. Gallone and Michelangeli (2018) explain the domain dependence of the spectrum.

This extension must not be called a Friedrichs extension of the full Dirac operator, which is not bounded below. The conditions for a real exponent, a unique self-adjoint closure, and physical pair production are different. For instance, the formal $1S$ energy tends to zero as $a\to1^-$, not to the negative continuum edge $-M$. Finite-nucleus supercritical resonances and pair creation require a different physical and spectral analysis.

Exercises

Allowed n = 2 states. Enumerate $\kappa$, $N$, and the upper-component spectroscopic labels, then count the states including $m_j$.

Solution

The allowed values are $\kappa=-1,+1,-2$. They give respectively $(N,\text{label})=(1,2S_{1/2})$, $(1,2P_{1/2})$, and $(0,2P_{3/2})$. The multiplicities are $2,2,4$, summing to $8=2n^2$. The excluded $\kappa=+2,N=0$ would incorrectly add a $2D_{3/2}$ state.

The ground-state lower component. Show that $\langle\beta\rangle=\sqrt{1-a^2}$ for the ground state.

Solution

The upper and lower probabilities sum to one, and $\beta$ has eigenvalues $+1$ and $-1$ on their respective blocks. Hence

$$\langle\beta\rangle =1-2P_{\rm lower} =\sqrt{1-a^2} =\frac{E_{1S}}{mc^2}.$$

This is consistent with differentiating the exact energy with respect to $mc^2$ at fixed Coulomb coupling.

Why the origin criterion changes. Use the reduced radial norm to determine when $r^{-\gamma}$ is locally integrable. Explain why this is not the condition $\gamma>0$.

Solution

The integral is $\int_0^\varepsilon r^{-2\gamma}\,dr$. It is finite only for $\gamma<1/2$; at $\gamma=1/2$ it diverges logarithmically. A real positive exponent merely requires $a<|\kappa|$. For $|\kappa|=1$, the additional integrable branch appears at $a>\sqrt3/2$, before the exponent ceases to be real at $a=1$.

References

• Gallone, Matteo, and Alessandro Michelangeli. “Discrete spectra for critical Dirac–Coulomb Hamiltonians.” Journal of Mathematical Physics 59, 062108 (2018). doi:10.1063/1.5011305. arXiv:1710.11389.
• 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.
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0.