Fine Structure and the Real Hydrogen Atom

The hydrogen spectrum$E_n=-13.6\text{eV}/n^2$ comes from a nonrelativistic electron in a pure Coulomb potential with no spin. High-resolution spectroscopy shows each of those levels is split into closely spaced components, the fine structure, smaller than the gross structure by a factor of order ${\alpha }^2\approx 5\times 10^{-5}$, where $\alpha =e^2/4\pi {ϵ}_0\hslash c\approx 1/137.036$ is the fine-structure constant.¹ The splittings are the first corrections to a model that treated the electron as slow, spinless, and pointlike. Two of them, the relativistic kinetic correction and spin–orbit coupling, are both of order ${\alpha }^2{E}_n$; a third, the Darwin term, matters only for $s$ states. Treated together as a perturbation on the Bohr Hamiltonian they combine into a single formula depending on $n$ and the total angular momentum $j$.

The hierarchy of corrections

The corrections form a graded sequence, each smaller than the last by a power of $\alpha$ or the electron-to-proton mass ratio. Reading the Bohr energy as $E_n\sim {\alpha }^{2}m{c}^2/n^2$ fixes the scale of everything below it.

Because the perturbations act on the degenerate hydrogen levels (each $n$ level is $n^2$-fold degenerate in $\ell$ and $m_{\ell }$, and doubly so in spin), the degenerate theory governs. The saving grace is that both fine-structure operators commute with $L^2$, $J^2$, and $J_z$, so the coupled basis $|n,\ell ,j,m_j⟩$ is the good basis and each shift is a diagonal expectation value.

The relativistic kinetic correction

The kinetic energy $T=p^2/2m$ is the nonrelativistic limit of the relativistic expression $T=\sqrt{p^2{c}^2+m^2{c}^4}-m{c}^2$. Expanding in $p/mc$,

identifies the leading correction as the perturbation ${\hat{H}}_r^{\prime }=-\frac{{\hat{p}}^4}{8m^3{c}^2}$. Its first-order shift needs $⟨{\hat{p}}^4⟩$ in the unperturbed state. Rather than compute a quartic momentum integral, use the unperturbed equation ${\hat{p}}^2/2m=E_n-V$ with $V=-e^2/4\pi {ϵ}_{0}r$ to write ${\hat{p}}^2\psi =2m(E_n-V)\psi$, so

The needed radial averages are standard results for hydrogen, $⟨1/r⟩=1/(n^2{a}_0)$ and $⟨1/r^2⟩=1/[(\ell +\frac{1}{2})n^3{a}_0^2]$. Substituting and using $E_n=-e^2/8\pi {ϵ}_0{a}_0{n}^2$ to eliminate $a_0$ gives a compact result.

It is negative (the true kinetic energy is less than the nonrelativistic estimate at fixed momentum spread) and grows with $n$ and with decreasing $\ell$, because low-$\ell$ orbits penetrate closer to the nucleus where the electron moves fastest.

Spin–orbit coupling

In the electron's instantaneous rest frame the proton circulates and produces a magnetic field $\vec{B}$; the electron's intrinsic magnetic moment $\vec{\mu }=-(e/m)\vec{S}$ has energy $-\vec{\mu }\cdot \vec{B}$ in that field. Transforming the proton's Coulomb field to the electron frame and including the Thomas precession factor of $\frac{1}{2}$, which corrects for the non-inertial rotation of the electron frame, gives the spin–orbit Hamiltonian

The operator $\vec{S}\cdot \vec{L}$ is not diagonal in the uncoupled basis $|\ell ,m_{\ell }⟩|s,m_s⟩$, since it mixes $m_{\ell }$ and $m_s$ while conserving their sum. It is diagonal in the coupled basis, where the total angular momentum $\vec{J}=\vec{L}+\vec{S}$ has definite $j$. Squaring $\vec{J}=\vec{L}+\vec{S}$,

with $s=\frac{1}{2}$. This is the algebraic reason the coupled basis is good: the addition of angular momenta diagonalizes the perturbation.

The first-order shift uses $⟨1/r^3⟩=1/[\ell (\ell +\frac{1}{2})(\ell +1)n^3{a}_0^3]$, valid for $\ell \ge 1$.

A given $\ell \ge 1$ couples with spin to $j=\ell \pm \frac{1}{2}$; the $j=\ell +\frac{1}{2}$ level (spin aligned with orbit) is pushed up and $j=\ell -\frac{1}{2}$ down, splitting each $\ell$ shell into a doublet.

The combined fine-structure formula

The relativistic and spin–orbit shifts have different-looking $\ell$ and $j$ dependence, yet their sum collapses. Adding the two theorems and using $j=\ell \pm \frac{1}{2}$ to trade $\ell$ for $j$ (each fixed $j$ receives the $\ell =j-\frac{1}{2}$ or $\ell =j+\frac{1}{2}$ contribution), the $\ell$ dependence cancels exactly.

The cancellation of $\ell$ is not an accident: it reflects the exact $SO(4)$ symmetry of the Coulomb problem carried into the relativistic correction, and the same $n,j$-only formula follows from the exact Dirac equation expanded to order ${\alpha }^2$. Levels of equal $n$ and $j$ but different $\ell$, such as $2S_{\frac{1}{2}}$ and $2P_{\frac{1}{2}}$, remain degenerate at this order.

The Darwin term

The spin–orbit formula carries $\ell$ in its denominator and fails for $s$ states ($\ell =0$), which nonetheless must obey the $n,j$ formula since a $j=\frac{1}{2}$ $s$ state is degenerate with the corresponding $p$ state. The gap is filled by the Darwin term, a correction with no classical analogue that arises from the electron's inability to be localized more finely than its Compton wavelength (the Zitterbewegung smearing of the point charge over a region of size $\hslash /mc$). It smears the potential, and the leading effect is proportional to its Laplacian,

using ${\nabla }^2(1/r)=-4\pi {\delta }^3(\vec{r})$. Because it is a contact term it acts only where the wavefunction is nonzero at the origin, which is only for $\ell =0$. With $|{\psi }_{n00}(0){|}^2=1/\pi n^3{a}_0^3$,

and this exactly equals the value the fine-structure formula assigns to an $\ell =0$, $j=\frac{1}{2}$ state. The three corrections therefore combine into one formula valid for all $\ell$: relativistic and Darwin for $s$ states, relativistic and spin–orbit for $\ell \ge 1$.

The Lamb shift

The degeneracy of $2S_{\frac{1}{2}}$ and $2P_{\frac{1}{2}}$ survives the entire fine-structure calculation and the exact Dirac equation, but not experiment. Lamb and Retherford (1947) measured a splitting of about $1057\text{MHz}$ ($4.4\times 10^{-6}\text{eV}$), with $2S_{\frac{1}{2}}$ lying above $2P_{\frac{1}{2}}$. The Lamb shift is a quantum-electrodynamic effect: the electron interacts with the fluctuating vacuum electromagnetic field, which slightly smears its position and shifts $s$ states (nonzero at the origin) more than $p$ states. It sits at order ${\alpha }^{5}m{c}^2$, one power of $\alpha$ below fine structure, and it is beyond the scope of a Schrödinger-equation perturbation, requiring the quantized radiation field. It stands here as the boundary where the single-particle theory ends.

Hyperfine structure and the 21 cm line

The proton is not a static point charge but a spin-$\frac{1}{2}$ particle with its own magnetic moment ${\vec{\mu }}_p=g_p(e/2m_p){\vec{S}}_p$, with $g_p\approx 5.59$. The electron's magnetic moment couples to the field of the proton's dipole. For the $\ell =0$ ground state the dominant piece is the Fermi contact interaction, proportional to ${\vec{S}}_p\cdot {\vec{S}}_e$ evaluated at the electron density at the origin,

The two spins add to a total $\vec{F}={\vec{S}}_p+{\vec{S}}_e$ with ${\vec{S}}_p\cdot {\vec{S}}_e=\frac{1}{2}(F^2-S_p^2-S_e^2)$, giving $+\frac{1}{4}{\hslash }^2$ for the triplet ($F=1$) and $-\frac{3}{4}{\hslash }^2$ for the singlet ($F=0$). The ground state splits in two, with the singlet lower.

The corrections continue below the hyperfine scale (the proton's finite size, higher QED orders), but each is another power of $\alpha$ or the mass ratio smaller, and hydrogen spectroscopy is now the most precisely tested prediction in physics. The next lesson places the atom in external fields, where the same coupled basis competes with laboratory-scale perturbations and the good states depend on which dominates.