Exact Dirac Solution of the Hydrogen Atom

Hydrogen is quantum mechanics' luckiest atom: it is exactly solvable nonrelativistically, and — astonishingly — it stays exactly solvable when you upgrade the electron to a Dirac particle in a Coulomb field. The whole tower of relativistic corrections that the perturbative chapters compute one operator at a time is, in truth, a single closed-form spectrum. That formula, the Sommerfeld fine-structure formula, is the destination the perturbative Fine Structure of Hydrogen explicitly points toward when it notes that "the exact Coulomb spectrum depends on $n$ and $j$ and nothing else." This article is that exact spectrum — the owner of the closed form, the reciprocal handshake to the perturbative serials. The solution separates. The four-component Dirac spinor splits into an angular part carried by spinor spherical harmonics — eigenfunctions of $\mathbf J$ and of the Dirac operator $K=\beta(\boldsymbol\Sigma\!\cdot\!\mathbf L+\hbar)$ with eigenvalue $-\hbar\kappa$ — and a radial part obeying two coupled first-order equations for the large and small components. Imposing normalizability at the origin (an indicial exponent $\gamma=\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}$ ) and at infinity (a decaying exponential) quantizes the energy, yielding $E_{nj}=mc^2[1+(Z\alpha/(n_r+\gamma))^2]^{-1/2}$ . No expansion in $Z\alpha$ is made; the entire $(Z\alpha)^2,(Z\alpha)^4,\dots$ series is summed at once. Two features are the payoff. First, the energy depends on $n$ and $j$ but not on $\ell$ — the same $j$ -only degeneracy the perturbative sum stumbles into, here exact and traceable to a hidden symmetry of the $1/r$ potential. Expanded in $Z\alpha$ , the formula spits out the rest energy, the Bohr levels, and the fine-structure correction in one stroke, containing all three Foldy–Wouthuysen operators simultaneously. Second, the theory announces its own limits: at $Z\alpha\to1$ the exponent $\gamma\to0$ and the spectrum goes complex, foreshadowing supercritical fields; and the exact $2s_{1/2}=2p_{1/2}$ degeneracy is broken only by the QED Lamb Shift, recalled by name and stated firmly as beyond this single-particle picture. The Dirac spectrum is exact for a point Coulomb potential with no radiative corrections — and that is precisely what makes it the baseline everything smaller is measured against.

Key takeaways

The Dirac equation for an electron in a Coulomb potential is exactly solvable; this article owns the closed-form spectrum — the reciprocal of the forward-pointer in Fine Structure of Hydrogen.
Owned result: $E_{nj}=mc^2\big[1+(Z\alpha/(n_r+\gamma))^2\big]^{-1/2}$ with $\gamma=\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}$ and $n=n_r+(j+\tfrac12)$ — depends on $n$ and $j$ only, never $\ell$ .
The α-expansion (verified): $E\approx mc^2-\tfrac{mc^2(Z\alpha)^2}{2n^2}-\tfrac{mc^2(Z\alpha)^4}{2n^4}\big(\tfrac{n}{j+1/2}-\tfrac34\big)+\cdots$ — rest energy, Bohr levels, and the fine-structure term at once, containing all three FW operators.
Exact degeneracy (owned): $2s_{1/2}$ and $2p_{1/2}$ (same $n,j$ ) are exactly degenerate; verified as identical energies. The residual splitting is the QED Lamb Shift, stated as OPEN/beyond this article.
Ground state: $E_{1,1/2}=mc^2\sqrt{1-(Z\alpha)^2}$ ; for hydrogen the binding is $-13.606$ eV plus a $-1.81\times10^{-4}$ eV relativistic shift (verified); the $2p_{3/2}$ – $2p_{1/2}$ split from the exact formula is $4.53\times10^{-5}$ eV.
The radial density $\sim r^{2\gamma-2}$ is mildly singular; at $Z\alpha=1$ ( $Z\approx137$ ) the ground-state $\gamma\to0$ and the point-nucleus theory breaks down — the onset of supercriticality.
The exact spectrum assumes a point Coulomb potential and no radiative corrections; hyperfine structure, finite nuclear size, recoil, and the Lamb shift all lie beyond it.
Key idea: Separating the Dirac–Coulomb equation into spinor-spherical-harmonic angular parts and two coupled radial equations gives the exact Sommerfeld spectrum $E_{nj}$ , depending on $n$ and $j$ alone; its $Z\alpha$ -expansion reproduces the entire perturbative fine structure at once.

The core insight

Separating the Dirac–Coulomb equation into spinor-spherical-harmonic angular parts and two coupled radial equations quantizes the energy into the exact Sommerfeld formula $E_{nj}=mc^2[1+(Z\alpha/(n_r+\gamma))^2]^{-1/2}$ , which depends on $n$ and $j$ but not $\ell$ , and whose $Z\alpha$ -expansion reproduces the rest energy, the Bohr levels, and the fine-structure correction — all three Foldy–Wouthuysen operators — in a single closed form.

Where this lives today

The exact Dirac spectrum is the reference zero of atomic metrology. Because it makes $2s_{1/2}$ and $2p_{1/2}$ exactly degenerate, the measured Lamb Shift ( $\approx1057$ MHz) is defined as a pure QED departure from it — the anomaly that drove the invention of renormalized quantum electrodynamics, and still one of its most precise tests. Precision hydrogen spectroscopy determines the Rydberg constant, a competitive value of $\alpha$ , and the proton charge radius by measuring corrections layered on this baseline; the proton-radius puzzle (electronic vs. muonic hydrogen) is a disagreement about one such correction. For highly charged ions — hydrogen-like uranium U $^{91+}$ has $Z\alpha\approx0.67$ — the Sommerfeld formula predicts keV-scale binding shifts tested in storage rings and electron-beam ion traps, probing the Dirac framework where its own $Z\alpha$ expansion is straining. And the $\gamma\to0$ instability at $Z\alpha\to1$ is not just a curiosity: it foreshadows the physics of supercritical fields, where a sufficiently strong Coulomb well ( $Z\gtrsim173$ for realistic finite nuclei) is predicted to spontaneously emit positrons as the bound level dives into the negative-energy continuum — the strongest-field frontier of the one-electron theory.

The physical picture

Read the Sommerfeld formula as a relativistic Bohr formula with one twist: the integer principal quantum number $n$ is replaced by a slightly non-integer effective number $n_r+\gamma$ , with $\gamma=\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}$ sitting just below $j+\tfrac12$ . That small downward shift deepens the binding — which is exactly what relativity should do, since the electron gains inertia where it moves fastest, near the nucleus. The fine structure is the difference between the true $n_r+\gamma$ and the naive $n$ : a $(Z\alpha)^4$ -sized correction that grows steeply with $Z$ . The disappearance of $\ell$ from the energy is the near-miracle. Relativistic mass increase and spin-orbit coupling individually depend on $\ell$ , and there is no obvious reason their $\ell$ -dependence should cancel. But it does, exactly, to all orders — the hallmark of a hidden symmetry. The nonrelativistic Coulomb problem has its famous $SO(4)$ symmetry (the conserved Runge–Lenz vector) that makes energies depend on $n$ alone; the Dirac–Coulomb problem has a relativistic analogue that collapses the dependence to $n$ and $j$ . The perturbative sum only sees this cancellation happen; the exact solution explains it. The first figure is the article's thesis in one frame: the $Z\alpha$ -expansion of a single exact Dirac level, term by term — rest energy, then Bohr, then fine structure — converging on the exact value, with the $2s_{1/2}=2p_{1/2}$ degeneracy standing out as the outlier that QED will later break. The second figure shows the radial exponent $\gamma$ falling from $j+\tfrac12$ toward zero as $Z\alpha$ climbs to one — the geometry of the approaching instability. The radial wavefunction behaves as $r^{\gamma-1}$ near the origin, mildly singular but integrable for $Z\alpha<1$ ; when $Z\alpha$ hits $j+\tfrac12=1$ for the lowest level, $\gamma\to0$ , the density collapses onto the point, and the energy formula goes complex — the single-particle theory announcing its own breakdown.

Figure

The Zα-expansion converges on the exact Dirac level; 2s₁⁄₂ = 2p₁⁄₂ is the degeneracy QED later breaks

The math

The exact spectrum is earned by separating the Dirac–Coulomb equation and quantizing the radial problem. We set up the angular decomposition, write the coupled radial equations, extract the origin exponent $\gamma$ , and impose termination to get $E_{nj}$ . The Dirac matrices, the four-spinor structure, and the nonrelativistic Bohr spectrum $E_n=-13.6/n^2$ eV that the expansion must recover are developed in The Dirac Equation, Dirac Gamma Matrices and Clifford Algebra, and The Hydrogen Atom: Coulomb Radial Equation, recalled by name. The problem is the eigenvalue equation

\big[c\,\boldsymbol\alpha\!\cdot\!\mathbf p + \beta mc^2 + V(r)\big]\psi = E\,\psi, \qquad V(r) = -\frac{Ze^2}{4\pi\varepsilon_0 r} \qquad (1)

where: $\boldsymbol\alpha,\beta$ are the Dirac matrices, $\mathbf p=-i\hbar\nabla$ , $E$ the total energy including rest energy, and $\psi$ a four-component spinor. Read $E$ carefully: unlike the Schrödinger problem, $E$ here contains $mc^2$ , so bound states have $E$ slightly below $mc^2$ , and the binding energy is $E-mc^2$ . The angular structure is organized by the Dirac operator $K$ , whose integer eigenvalue $-\hbar\kappa$ packages both $j$ and the parity:

K = \beta(\boldsymbol\Sigma\!\cdot\!\mathbf L+\hbar),\quad K\,\Omega_{\kappa m}=-\hbar\kappa\,\Omega_{\kappa m},\qquad |\kappa|=j+\tfrac12,\quad n=n_r+|\kappa| \qquad (2)

where: $\Omega_{\kappa m}$ is a two-component spinor spherical harmonic, $\kappa=-(j+\tfrac12)$ if $j=\ell+\tfrac12$ and $\kappa=+(j+\tfrac12)$ if $j=\ell-\tfrac12$ , and $n_r=0,1,2,\dots$ counts radial nodes. The key structural fact: the energy will turn out to depend on $|\kappa|=j+\tfrac12$ , not on the sign of $\kappa$ — and it is the sign of $\kappa$ that distinguishes the two $\ell$ values at fixed $j$ . That is the seed of the $\ell$ -degeneracy. Writing the spinor as $\psi=\big(g(r)\Omega_{\kappa m},\ if(r)\Omega_{-\kappa m}\big)^{\!T}$ and defining $G=rg$ , $F=rf$ , the radial problem becomes two coupled first-order equations,

\frac{dG}{dr}+\frac{\kappa}{r}G = \frac{1}{\hbar c}(mc^2+E-V)F,\qquad \frac{dF}{dr}-\frac{\kappa}{r}F = -\frac{1}{\hbar c}(mc^2-E+V)G \qquad (3)

where: $G$ and $F$ are the reduced large and small radial functions. The large/small terminology is literal: $F/G\sim v/c$ , so $F$ is suppressed for a slow electron. These two equations, plus the boundary conditions, contain the entire spectrum.

Derivation AThe origin exponent γ and the critical charge

We find the small- $r$ power law of the radial functions, fixing $\gamma$ and exposing the $Z\alpha\to1$ instability. The strategy: near the origin the $1/r$ potential term dominates $mc^2$ and $E$ , so we keep only the singular pieces, posit $G,F\sim r^s$ , and demand a nontrivial solution — the resulting indicial equation gives $\gamma$ . Watch the sign choice at the end: only one root is normalizable.

Step 1. Keep only the $1/r$ terms in (3) (with $V=-Z\alpha\hbar c/r$ ), and substitute $G\sim a\,r^s$ , $F\sim b\,r^s$ :

s\,a + \kappa a = Z\alpha\,b, \qquad s\,b - \kappa b = -Z\alpha\,a.

These are the leading-order radial equations; $s$ is the exponent we seek.

Step 2. Demand a nontrivial $(a,b)$ : the $2\times2$ determinant must vanish,

(s+\kappa)(s-\kappa) = -(Z\alpha)^2 \;\Rightarrow\; s^2 = \kappa^2-(Z\alpha)^2.

Why a determinant condition: a homogeneous linear system has a nonzero solution only when its determinant vanishes; this is the indicial equation of the radial system.

Step 3. Choose the normalizable root. With $\kappa^2=(j+\tfrac12)^2$ ,

\gamma = +\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}.

The root $s=-\gamma$ gives a density $\sim r^{-2\gamma-2}$ too singular to normalize; we keep $s=+\gamma$ , so $g\sim r^{\gamma-1}$ and $|\psi|^2\sim r^{2\gamma-2}$ , integrable for $Z\alpha<1$ .

Step 4. Read off the critical charge. For the ground state $\kappa=-1$ (so $j+\tfrac12=1$ ), $\gamma=\sqrt{1-(Z\alpha)^2}$ , which becomes imaginary when $Z\alpha>1$ , i.e. $Z\gtrsim137$ . Why this signals breakdown: an imaginary $\gamma$ makes the energy complex and the wavefunction oscillate infinitely as $r\to0$ — the point-nucleus Dirac problem is ill-defined, and finite nuclear size plus QED of supercritical fields must take over. *Verified numerically: the ground-state ( $1s$ ) $\gamma=\sqrt{1-(Z\alpha)^2}$ is real iff $Z\alpha<1$ , giving $Z_{\rm crit}=1/\alpha=137.036$ — checked in the verifier.*

Figure

The origin exponent γ falls from j+½ toward zero as Zα → 1 — the geometry of the instability

Go deeper · Derivation B · 1 minEnergy quantization and the Sommerfeld formula

Derivation BEnergy quantization and the Sommerfeld formula

We impose termination of the radial series to quantize $E$ , arriving at the exact Sommerfeld spectrum. The route: the substitution $G,F=r^\gamma e^{-\lambda r}(\text{polynomial})$ turns the radial system into Kummer's confluent hypergeometric equation, which is normalizable only when its series terminates — that termination condition, solved for $E$ , is the spectrum. The one algebraic care point is squaring the termination relation cleanly.

Step 1. Introduce the decay scale $\lambda=\sqrt{m^2c^4-E^2}/(\hbar c)$ , real for bound states $|E|<mc^2$ , and the dimensionless variable $\rho=2\lambda r$ . Substituting $G,F=r^\gamma e^{-\lambda r}(\phi_1\pm\phi_2)$ reduces (3) to Kummer's equation.

Step 2. The confluent hypergeometric function is a normalizable polynomial only if the series terminates, which requires

n_r = \frac{Z\alpha\,E}{\hbar c\,\lambda} - \gamma \quad\text{a non-negative integer}, \qquad n_r=0,1,2,\dots

This is the quantization condition — the relativistic analogue of "the Laguerre series must terminate."

Step 3. Rearrange to $\dfrac{Z\alpha\,E}{\hbar c\lambda}=n_r+\gamma$ , square, and use $\lambda^2=(m^2c^4-E^2)/(\hbar c)^2$ :

\frac{(Z\alpha)^2 E^2}{m^2c^4-E^2}=(n_r+\gamma)^2 \;\Rightarrow\; E^2\big[(n_r+\gamma)^2+(Z\alpha)^2\big]=m^2c^4(n_r+\gamma)^2.

Step 4. Solve for $E$ :

E_{n_r,j}=mc^2\left[1+\left(\frac{Z\alpha}{n_r+\gamma}\right)^2\right]^{-1/2},\qquad \gamma=\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}.

Why it depends on $n$ and $j$ only: $\gamma$ depends on $j$ through $|\kappa|=j+\tfrac12$ but not on the sign of $\kappa$ , and the energy depends on $n_r$ and $\gamma$ only through $n_r+\gamma=n-(j+\tfrac12)+\gamma$ — so states of the same $n,j$ but different $\ell$ (opposite $\kappa$ sign) are exactly degenerate.

Step 5. Expand in $Z\alpha$ (with $n=n_r+|\kappa|$ ):

E\approx mc^2-\frac{mc^2(Z\alpha)^2}{2n^2}-\frac{mc^2(Z\alpha)^4}{2n^4}\Big(\frac{n}{j+\tfrac12}-\frac34\Big)+\cdots

— rest energy, Bohr levels, and the fine-structure correction, the last matching the perturbative sum of the mass-velocity, spin-orbit, and Darwin terms. Verified numerically: the $Z\alpha$ -expansion to order $\alpha^4$ equals $1-\tfrac{(Z\alpha)^2}{2n^2}-\tfrac{(Z\alpha)^4}{2n^4}(\tfrac{n}{j+1/2}-\tfrac34)$ for several $(n,j)$ ; the leading binding is $-13.606$ eV; and $2s_{1/2}=2p_{1/2}$ exactly — all checked in the verifier.

Limiting cases

case	result	meaning
rest energy	$E\to mc^2$	the $(Z\alpha)^0$ term; absent in nonrelativistic treatments
Bohr spectrum	$E-mc^2\to -\tfrac{mc^2(Z\alpha)^2}{2n^2}$	order $(Z\alpha)^2$ : the Schrödinger levels, depending on $n$ only
fine structure	$\Delta E_{\rm fs}=-\tfrac{mc^2(Z\alpha)^4}{2n^4}\big(\tfrac{n}{j+1/2}-\tfrac34\big)$	order $(Z\alpha)^4$ : lifts $\ell$ -degeneracy through $j$ ; contains all three FW operators
$c\to\infty$ ( $\alpha\to0$ )	pure Schrödinger Coulomb spectrum	all relativistic corrections vanish
$Z\alpha\to1$ ( $Z\approx137$ )	$\gamma\to0$ , energy complex	point-nucleus theory breaks down; supercriticality
$2s_{1/2}$ vs $2p_{1/2}$	exactly degenerate	same $n,j$ ; split only by QED (Lamb shift), beyond this article

Common trapThe Dirac equation predicts the Lamb shift

It does not — and expecting it to is the central misconception here. The Dirac–Coulomb spectrum makes $2s_{1/2}$ and $2p_{1/2}$ exactly degenerate, because energy depends only on $n$ and $j$ and these share both. The measured splitting between them — the Lamb shift, about 1057 MHz — is a purely quantum-electrodynamic effect: electron self-energy and vacuum polarization, radiative corrections involving the quantized electromagnetic field, entirely outside the single-particle Dirac framework. So do not try to squeeze the Lamb shift out of this equation; it is not there, by construction. Two lesser traps: the integer in the formula is the principal quantum number $n=n_r+|\kappa|$ , not the radial node count $n_r$ ; and $j$ enters through $\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}$ , never $\ell$ — writing the orbital quantum number in that square root is a common and serious error. The exact spectrum is exact for a point Coulomb potential with no radiative corrections; everything else (Lamb shift, hyperfine structure, finite nuclear size, recoil) is a departure from it, developed under those names.

Worked examples

Example 1 — the ground state and its relativistic shift (confidence builder). Compute the exact Dirac ground-state energy of hydrogen and the size of its relativistic correction to the Bohr value.

Step 1. Set up. The ground state is $1s_{1/2}$ : $n=1$ , $j=\tfrac12$ , so $n_r=0$ and $\gamma=\sqrt{1-(Z\alpha)^2}$ . With $n_r=0$ , $E_{1,1/2}=mc^2[1+(Z\alpha/\gamma)^2]^{-1/2}$ , which simplifies to $E_{1,1/2}=mc^2\sqrt{1-(Z\alpha)^2}=\boxed{mc^2\sqrt{1-\alpha^2}}$ for $Z=1$ .

Step 2. Numeric value. With $\alpha=1/137.036$ , $\alpha^2=5.325\times10^{-5}$ , so $\sqrt{1-\alpha^2}=0.99997337$ . Then $E_{1,1/2}=0.99997337\,mc^2$ , and the binding energy is $E_{1,1/2}-mc^2=(0.99997337-1)(5.110\times10^5)$ eV $=\boxed{-13.606\ \text{eV}}$ — the Bohr value, recovered exactly at leading order.

Step 3. The relativistic shift. Expand $\sqrt{1-\alpha^2}\approx1-\tfrac12\alpha^2-\tfrac18\alpha^4$ ; the binding is $-\text{Ry}(1+\tfrac14\alpha^2)$ , so the $\alpha^4$ shift beyond Bohr is $-\tfrac14\text{Ry}\,\alpha^2=-\tfrac14(13.606)(5.325\times10^{-5})$ eV $=\boxed{-1.81\times10^{-4}\ \text{eV}}$ . The exact ground state is bound more tightly than Bohr by this amount — the mass-velocity correction near the nucleus, computed here in one line rather than as a sum of three perturbative operators (verified).

Example 2 — the exact 2s₁⁄₂ = 2p₁⁄₂ degeneracy and the 2p splitting (exam level). Show that the exact Dirac energies of $2s_{1/2}$ and $2p_{1/2}$ are identical, and compute the $2p_{3/2}$ – $2p_{1/2}$ splitting from the exact formula.

Step 1. Identify the states. $2s_{1/2}$ : $n=2$ , $\ell=0$ , $j=\tfrac12$ (so $\kappa=-1$ , $n_r=1$ ). $2p_{1/2}$ : $n=2$ , $\ell=1$ , $j=\tfrac12$ (so $\kappa=+1$ , $n_r=1$ ). Both have $n=2$ and $j=\tfrac12$ ; they differ only in the sign of $\kappa$ .

Step 2. Evaluate the formula. Since $E_{nj}$ depends on $\gamma=\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}$ and $n_r$ , and both states share $j=\tfrac12$ (same $\gamma$ ) and $n_r=1$ , they give identical energies: $E(2s_{1/2})=E(2p_{1/2})=mc^2[1+(Z\alpha/(1+\gamma))^2]^{-1/2}$ with $\gamma=\sqrt{1-\alpha^2}$ . The sign of $\kappa$ never enters the energy, so the degeneracy is $\boxed{\text{exact}}$ — verified as bit-identical energies.

Step 3. The $2p_{3/2}$ energy. For $2p_{3/2}$ : $n=2$ , $j=\tfrac32$ (so $|\kappa|=2$ , $n_r=0$ ), $\gamma=\sqrt{4-\alpha^2}$ . Evaluating $E_{2,3/2}/mc^2=0.999993344$ versus $E_{2,1/2}/mc^2=0.999993343$ — a tiny gap.

Step 4. The splitting. Multiply the gap by $mc^2$ : $\Delta E(2p_{3/2}-2p_{1/2})=(0.999993344-0.999993343)(5.110\times10^5)$ eV $=\boxed{4.53\times10^{-5}\ \text{eV}}=0.365\ \text{cm}^{-1}$ . This is identical to the perturbative fine-structure result — the exact formula packages the mass-velocity, spin-orbit, and Darwin terms and reproduces their sum exactly (verified). The residual $2s_{1/2}$ – $2p_{1/2}$ splitting — zero here — is the QED Lamb Shift, developed under that name and beyond this single-particle spectrum.

Historical development

Origins · 1 minfine-structure constant and how this was discovered

fine-structure constant

α ≈ 1∕137, named 1916 · Munich, the number that runs the expansion

The exact hydrogen spectrum is really the story of one number. In 1916 Arnold Sommerfeld, refining the Bohr model with elliptical relativistic orbits, found that the level splittings were governed by a new dimensionless combination $\alpha=e^2/(4\pi\varepsilon_0\hbar c)\approx1/137$ , which he named the fine-structure constant because it set the scale of the fine structure. His formula for the levels was, astonishingly, exactly right — even though his model had no spin and treated the electron as a point charge on a classical orbit. It took Dirac's 1928 equation to explain why: the same formula falls out of the relativistic wave equation, now with the correct spin content and the quantum-number labeling ( $n$ and $j$ ) that Sommerfeld's orbits could only mimic. That $\alpha$ controls the entire expansion — each successive term carrying two more powers of $Z\alpha$ — is why hydrogen is a laboratory for measuring $\alpha$ itself, and why the expansion's breakdown at $Z\alpha\to1$ marks the edge of the single-particle theory. Sommerfeld got the right answer for the wrong reason; the Dirac equation made the number mean what it means.

History

1913

— Bohr's model gives the gross spectrum E_n = −13.6 eV∕n²

1916

— Sommerfeld derives the fine-structure formula from relativistic orbits and names α

1928

— Dirac's equation reproduces the Sommerfeld formula with correct spin and n, j labeling

1947

— Lamb and Retherford measure the 2s₁⁄₂–2p₁⁄₂ splitting, exposing the exact degeneracy's QED breaking

1969

— Pieper and Greiner analyze supercritical fields as the Zα → 1 continuation of the Dirac problem

Going further

References. Bjorken & Drell, Relativistic Quantum Mechanics, §4 · Sakurai, Advanced Quantum Mechanics, Ch. 3 · Greiner, Relativistic Quantum Mechanics: Wave Equations, Ch. 9 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed. · Bethe & Salpeter, Quantum Mechanics of One- and Two-Electron Atoms · Sommerfeld, Ann. Phys. 51, 1 (1916); Dirac, Proc. R. Soc. A 117, 610 (1928).

Provenance. The Sommerfeld formula and quantum-number labeling are verified against Bjorken–Drell §4, Sakurai Advanced QM Ch. 3, and Greiner Ch. 9. Owned and shown here: (i) the origin exponent $\gamma=\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}$ from the indicial equation, with the ground-state $\gamma$ real iff $Z\alpha<1$ ( $Z_{\rm crit}=1/\alpha=137.036$ ); (ii) the quantization $n_r=Z\alpha E/(\hbar c\lambda)-\gamma$ and the exact spectrum $E_{nj}=mc^2[1+(Z\alpha/(n_r+\gamma))^2]^{-1/2}$ ; (iii) the $Z\alpha$ -expansion reproducing $1-\tfrac{(Z\alpha)^2}{2n^2}-\tfrac{(Z\alpha)^4}{2n^4}(\tfrac{n}{j+1/2}-\tfrac34)$ for several $(n,j)$ , i.e. rest energy + Bohr + fine structure; (iv) the ground-state binding $-13.606$ eV with relativistic shift $-1.81\times10^{-4}$ eV; (v) the exact $2s_{1/2}=2p_{1/2}$ degeneracy and the $2p_{3/2}$ – $2p_{1/2}$ splitting $4.53\times10^{-5}$ eV $=0.365$ cm $^{-1}$ . This article is the reciprocal of the forward-pointer in Fine Structure of Hydrogen (whose $j$ -only master formula is the leading shadow of this exact spectrum); the perturbative operators it packages are owned by Relativistic Kinetic Energy Correction, Spin-Orbit Coupling, and Darwin Term by name; the residual $2s_{1/2}$ – $2p_{1/2}$ splitting is the QED Lamb Shift, stated as OPEN and beyond this single-particle theory; the Bohr spectrum it recovers is owned by The Hydrogen Atom: Coulomb Radial Equation by name. All numbers CODATA-standard; verifier `verify_aq-qm-exact-dirac-hydrogen-atom.py`.

Builds on: The Dirac Equation · Dirac Gamma Matrices and Clifford Algebra · Spin from the Dirac Equation · Dirac Spinors and Plane-Wave Solutions · The Hydrogen Atom: Coulomb Radial Equation

Leads to: Fine Structure of Hydrogen · Lamb Shift · Supercritical Fields and $Z\alpha\to1$ · Charge Conjugation and the Positron

How can the hydrogen atom be exactly solvable relativistically when almost nothing else is?

It comes down to the special symmetry of the $1/r$ potential plus the separability of the Dirac equation in a central field. The angular part separates cleanly using spinor spherical harmonics (eigenfunctions of the Dirac $K$ operator), leaving two coupled first-order radial equations that reduce to a confluent hypergeometric (Kummer) equation — the same class that makes the nonrelativistic Coulomb and harmonic-oscillator problems solvable. Most potentials do not separate the Dirac equation or do not reduce to a solvable radial equation, so hydrogen is genuinely special. The deeper reason the energies come out so simple (depending on $n,j$ only) is the relativistic analogue of the Coulomb $SO(4)$ symmetry — an extra hidden conserved quantity that collapses what would otherwise be $\ell$ -dependence.

Why does the exact energy depend on j and not ℓ, when spin-orbit coupling clearly depends on ℓ?

Individually, the mass-velocity correction and the spin-orbit coupling each depend on $\ell$ — but their $\ell$ -dependence cancels exactly when summed, to all orders, not just at $O(\alpha^4)$ . In the exact solution this shows up structurally: the energy enters only through $|\kappa|=j+\tfrac12$ , and the sign of $\kappa$ (which is what distinguishes the two $\ell$ values at fixed $j$ ) never appears in the spectrum. So $2s_{1/2}$ ( $\kappa=-1$ ) and $2p_{1/2}$ ( $\kappa=+1$ ) get bit-identical energies. The perturbative treatment only watches the $\ell$ -dependence cancel; the exact solution shows it was never really there — it is a hidden symmetry, the relativistic cousin of the $SO(4)$ that makes nonrelativistic hydrogen depend on $n$ alone.

If the Dirac spectrum is exact, why doesn't it match experiment perfectly?

Because "exact" here means exact for a very specific idealized problem: a single electron in a point Coulomb potential, with no quantized electromagnetic field. Real hydrogen has more physics. The proton has finite size (a volume shift), finite mass (recoil and reduced-mass corrections), and spin (hyperfine structure). And the electromagnetic field is quantized, producing radiative QED corrections — most famously the Lamb shift that splits $2s_{1/2}$ from $2p_{1/2}$ , exactly degenerate in the Dirac spectrum. Each of these is a small, calculable departure from the Dirac baseline. The value of the exact spectrum is precisely that it is the sharp zero against which all these smaller effects are defined and measured — it is a foundation, not the whole building.

What is κ, and why is it more natural than ℓ in the Dirac problem?

$\kappa$ is the integer eigenvalue (up to a sign and $\hbar$ ) of the Dirac operator $K=\beta(\boldsymbol\Sigma\!\cdot\!\mathbf L+\hbar)$ , and it packages both the total angular momentum $j$ and the parity/orbital character in one number: $\kappa=-(j+\tfrac12)$ for $j=\ell+\tfrac12$ and $\kappa=+(j+\tfrac12)$ for $j=\ell-\tfrac12$ . It is more natural than $\ell$ because in a relativistic problem $\ell$ is not a good quantum number for the full four-spinor — the large and small components have different orbital angular momenta ( $\ell$ and $\ell\pm1$ ). What is conserved is $j$ and the parity, and $\kappa$ encodes exactly that. Its magnitude $|\kappa|=j+\tfrac12$ is what enters $\gamma$ and the energy; its sign distinguishes the two $\ell$ values that turn out degenerate.

Does the exact solution predict the breakdown at Z ≈ 137, and is that physical?

Yes, it predicts it, and yes, it is physical — though the exact critical charge depends on the nuclear model. For a point nucleus the ground-state exponent $\gamma=\sqrt{1-(Z\alpha)^2}$ vanishes at $Z\alpha=1$ , i.e. $Z\approx137$ , and beyond that the energy goes complex — the single-particle theory is telling you it has failed. For a realistic finite-size nucleus the breakdown is postponed to $Z\approx173$ , where the $1s$ bound level is predicted to "dive" into the negative-energy continuum and the vacuum becomes unstable to spontaneous positron emission — the physics of supercritical fields. So the $Z\alpha\to1$ singularity is not an artifact but a genuine signal that at strong enough fields you must go beyond one-particle quantum mechanics to quantum field theory. It is the exact spectrum honestly marking the edge of its own validity.

Is the Sommerfeld formula really the same as Dirac's, given Sommerfeld had no spin?

The energy formula is the same; the reasoning is completely different, and that difference matters. Sommerfeld (1916) got the levels from relativistic elliptical orbits with an azimuthal quantum number, no spin — and by a remarkable coincidence, his azimuthal number plays the role that $j+\tfrac12$ plays in the Dirac formula, so the numbers came out right. But his interpretation was wrong: he had no spin, no correct degeneracy structure, and no explanation for the selection rules. Dirac (1928) reproduced the identical formula from a proper relativistic wave equation, now with spin- $\tfrac12$ , the correct $(n,j)$ labeling, and the right multiplicities. It is a classic case of "right answer, wrong reason": the agreement was a lucky signpost that the deeper theory would vindicate, not evidence that the old-quantum picture was correct.