Fine-Structure Energy Formula

Derivation

Fine-Structure Energy Formula

D-300 Home PU-306 Threads energy · matter · symmetry · light Depends on Relativistic Kinetic-Energy Correction, Spin-Orbit Coupling from the Rest Frame, Degenerate Perturbation Theory

Statement

For the non-relativistic hydrogen atom treated as unperturbed, the combined first-order shift from the three order-\(\alpha^4\) corrections — the relativistic kinetic term \(-\hat p^4/8m^3c^2\), the spin–orbit coupling \(\xi(r)\,\hat{\mathbf S}\cdot\hat{\mathbf L}\), and the Darwin contact term — collapses onto a single closed form that depends on the principal quantum number \(n\) and the total angular momentum \(j\) only, and not on \(l\): \[ E_{nj}= -\frac{m c^2\alpha^2}{2n^2}\left[1+\frac{\alpha^2}{n^2}\left(\frac{n}{\,j+\tfrac12\,}-\frac34\right)\right]. \]

Why it matters

This is the first place a student sees three physically distinct effects — a kinematic energy correction, a magnetic spin–orbit interaction, and a smearing of the electron's position (Zitterbewegung) — conspire to give something far simpler than any one of them. That the \(l\)-dependence cancels is not a coincidence: it is the shadow of the exact \(SO(4)\)/Dirac symmetry that the split treatment hides.

Numerically it sets the scale of the famous doublets of the alkali and hydrogen spectra (the sodium \(D\) lines, the \(2p\) splitting at \(10.9\) GHz), and it is the launch point for the Lamb shift and hyperfine structure. It is also the cleanest worked demonstration that a degenerate level demands the right basis before perturbation theory means anything.

Assumptions

The electron moves in a pure Coulomb potential \(V=-e^2/4\pi\varepsilon_0 r\).Drop it (finite nuclear size, screening in multi-electron atoms) and \(\langle 1/r^3\rangle\) and \(\psi(0)\) change, so the \(l\)-cancellation is spoiled and quantum defects appear.

The corrections are small compared with the Bohr spacing, \(\alpha^2\ll1\).Drop it (high \(Z\), where \(Z\alpha\to1\)) and first-order perturbation theory diverges from the exact Dirac levels; the expansion parameter \((Z\alpha)^2\) is no longer small.

The unperturbed \(n\)-shell is exactly degenerate and the three perturbations are simultaneously diagonalised by the coupled basis \(|n\,l\,j\,m_j\rangle\).Drop the good-basis requirement and naive non-degenerate PT gives infinities from off-diagonal spin–orbit matrix elements; the answer becomes meaningless.

Spin is a genuine \(s=\tfrac12\) degree of freedom with \(g_s\approx2\), taken as external input.Drop it and the spin–orbit and Darwin terms have no home; the \(j\)-dependence — the whole content of the result — disappears.

Derivation

\[ \hat H'=\hat H'_{\text{rel}}+\hat H'_{\text{so}}+\hat H'_{\text{D}},\qquad \hat H'_{\text{rel}}=-\frac{\hat p^4}{8m^3c^2}. \]

The three terms are the \(O(v^4/c^4)\) pieces obtained by expanding the Dirac Hamiltonian (Foldy–Wouthuysen) to order \(\alpha^4\); they are the complete set at this order. B

\[ \{|n\,l\,m_l\,m_s\rangle\}\ \longrightarrow\ \{|n\,l\,j\,m_j\rangle\},\qquad \hat{\mathbf J}=\hat{\mathbf L}+\hat{\mathbf S}. \]

The \(n\)-shell is degenerate, so PT requires the basis that diagonalises \(\hat H'\). \(\hat H'_{\text{so}}\propto\hat{\mathbf S}\cdot\hat{\mathbf L}\) is off-diagonal in \(m_l,m_s\) but diagonal in \(j,m_j\); \(\hat H'_{\text{rel}}\) and \(\hat H'_{\text{D}}\) are scalars, diagonal in either. The coupled basis is the unique good basis. C

\[ \Delta E^{(1)}_{nlj}=\langle n\,l\,j\,m_j|\hat H'|n\,l\,j\,m_j\rangle =\langle \hat H'_{\text{rel}}\rangle+\langle \hat H'_{\text{so}}\rangle+\langle \hat H'_{\text{D}}\rangle. \]

In the good basis \(\hat H'\) is diagonal, so first-order degenerate PT reduces to a sum of diagonal expectation values on the unperturbed radial–angular states. B

\[ \langle \hat H'_{\text{rel}}\rangle=-\frac{1}{2mc^2}\big\langle (E_n-V)^2\big\rangle =-\frac{1}{2mc^2}\Big[E_n^2-2E_n\langle V\rangle+\langle V^2\rangle\Big]. \]

Write \(\hat p^2/2m=\hat H^{(0)}-V\); acting on an eigenstate \(\hat H^{(0)}\to E_n\), so \(\hat p^2/2m\to E_n-V\) and \(\hat p^4=(2m)^2(E_n-V)^2\). Hermiticity lets \(\hat p^2\) act to each side. B

\[ \Big\langle\frac1r\Big\rangle=\frac{1}{n^2 a},\qquad \Big\langle\frac1{r^2}\Big\rangle=\frac{1}{n^3\left(l+\tfrac12\right)a^2}, \]\[ \Rightarrow\ \langle \hat H'_{\text{rel}}\rangle=-\frac{E_n^2}{2mc^2}\left[\frac{4n}{\,l+\tfrac12\,}-3\right]. \]

Insert the Kramers/Coulomb radial expectation values (\(a\) the Bohr radius) and \(\langle V\rangle=2E_n\), \(\langle V^2\rangle=(e^2/4\pi\varepsilon_0)^2\langle 1/r^2\rangle\); algebra collapses to the bracket. B

\[ \hat H'_{\text{so}}=\frac{1}{2m^2c^2}\frac1r\frac{dV}{dr}\,\hat{\mathbf S}\cdot\hat{\mathbf L},\qquad \hat{\mathbf S}\cdot\hat{\mathbf L}=\tfrac{\hbar^2}{2}\big[j(j{+}1)-l(l{+}1)-\tfrac34\big]. \]

The moving electron sees a magnetic field \(\propto\!\mathbf E\times\mathbf v\); the Thomas-corrected coupling is \(\xi(r)\hat{\mathbf S}\cdot\hat{\mathbf L}\). \(\hat{\mathbf S}\cdot\hat{\mathbf L}=\tfrac12(\hat J^2-\hat L^2-\hat S^2)\) is diagonal in the coupled basis with \(s=\tfrac12\). B

\[ \Big\langle\frac1{r^3}\Big\rangle=\frac{1}{l\left(l+\tfrac12\right)(l+1)\,n^3 a^3}\quad(l\neq0), \]\[ \Rightarrow\ \langle \hat H'_{\text{so}}\rangle=\frac{E_n^2}{mc^2}\,\frac{n\big[j(j{+}1)-l(l{+}1)-\tfrac34\big]}{l\left(l+\tfrac12\right)(l+1)}. \]

\(dV/dr=e^2/4\pi\varepsilon_0 r^2\), so \(\hat H'_{\text{so}}\propto\langle 1/r^3\rangle\). Substituting the Coulomb value gives the \(l\neq0\) shift. B

\[ \hat H'_{\text{D}}=\frac{\pi\hbar^2 e^2}{2m^2c^2\,4\pi\varepsilon_0}\,\delta^3(\mathbf r) \ \Rightarrow\ \langle \hat H'_{\text{D}}\rangle=\frac{\pi\hbar^2 e^2}{2m^2c^2\,4\pi\varepsilon_0}\,|\psi_{n00}(0)|^2\,\delta_{l,0}. \]

The Darwin term smears the electron over a Compton wavelength (Zitterbewegung), sampling \(\nabla^2 V\propto\delta^3(\mathbf r)\); only \(l=0\) states have \(\psi(0)\neq0\), with \(|\psi_{n00}(0)|^2=1/\pi n^3 a^3\). C

\[ j=l+\tfrac12:\quad \langle \hat H'_{\text{rel}}\rangle+\langle \hat H'_{\text{so}}\rangle =\frac{E_n^2}{2mc^2}\left[3-\frac{4n}{\,j+\tfrac12\,}\right]. \]

For \(l\neq0\) the \(\hat{\mathbf S}\cdot\hat{\mathbf L}\) factor is \(l\) (upper branch), and \(-2l-1=-2(l+\tfrac12)\) makes every explicit \(l\) cancel, leaving \(l+1=j+\tfrac12\). The lower branch \(j=l-\tfrac12\) yields the identical bracket. C

\[ E^{(1)}_{\text{fs}}=\frac{E_n^2}{2mc^2}\left[3-\frac{4n}{\,j+\tfrac12\,}\right] \quad(\text{all }l,\ \text{Darwin fills }l=0). \]

For \(l=0\) only \(j=\tfrac12\) exists, \(\langle\hat H'_{\text{so}}\rangle=0\), and the Darwin term exactly reproduces the same bracket with \(j+\tfrac12=1\). One formula now covers every \(l\). B

\[ E_{nj}=E_n+E^{(1)}_{\text{fs}} =-\frac{mc^2\alpha^2}{2n^2}\left[1+\frac{\alpha^2}{n^2}\left(\frac{n}{\,j+\tfrac12\,}-\frac34\right)\right]. \]

Substitute \(E_n=-mc^2\alpha^2/2n^2\) and \(E_n^2/2mc^2=mc^2\alpha^4/8n^4\); factor out \(E_n\). A

Result

\[ E_{nj}= -\frac{m c^2\alpha^2}{2n^2}\left[1+\frac{\alpha^2}{n^2}\left(\frac{n}{\,j+\tfrac12\,}-\frac34\right)\right] \]

Reading. The leading term is the Bohr level; the bracket is a fractional correction of order \(\alpha^2\approx5\times10^{-5}\). Because the correction depends only on \(n\) and \(j\), all states of a given \(n\) and \(j\) but different \(l\) remain degenerate (e.g. \(2s_{1/2}\) and \(2p_{1/2}\) coincide here — the Lamb shift, a QED effect, lifts this). Higher \(j\) sits higher in energy: fine structure is an inverted-then-corrected ladder within each shell.

Units check. \(mc^2\) is an energy; \(\alpha\) is dimensionless; \(n,j\) are dimensionless, so \(n/(j+\tfrac12)\) and \(\tfrac34\) are pure numbers. The whole right side carries the single dimension of energy (J or eV). Both correction terms scale as \(mc^2\alpha^4\), the hallmark of fine structure.

Limiting cases

\(\alpha\to0\): the bracket \(\to1\) and \(E_{nj}\to E_n\), the exact non-relativistic Bohr spectrum.
Maximal \(j=n-\tfrac12\) (circular / stretched state): \(n/(j+\tfrac12)=1\), correction \(=-\tfrac14\alpha^2/n^2\) — the smallest (least negative shift) in the shell.
\(n=1,\ j=\tfrac12\): \(n/(j+\tfrac12)-\tfrac34=\tfrac14\), ground-state shift \(=-\tfrac14 mc^2\alpha^4=-1.8\times10^{-4}\) eV.
Large \(n\): the correction \(\sim mc^2\alpha^4/n^3\) dies faster than the Bohr spacing \(\sim mc^2\alpha^2/n^2\), so fine structure becomes relatively negligible in Rydberg states.
Compared with exact Dirac: expanding the Dirac energy in \((Z\alpha)^2\) reproduces this formula term-by-term at \(O(\alpha^4)\).

Breaks when

High nuclear charge (\(Z\alpha\to1\), e.g. heavy hydrogen-like ions): the expansion parameter is \((Z\alpha)^2\), not \(\alpha^2\); first-order PT fails and one must solve the Dirac equation exactly.
Accidental near-degeneracy with a different \(n\): the derivation assumes the perturbation is small against the \(n\)-shell gap; if an external field or configuration mixing brings another level close, non-degenerate PT within one shell is invalid.
When QED and nuclear structure matter: at the \(2s_{1/2}\)–\(2p_{1/2}\) degeneracy the Lamb shift (\(\sim10^{-6}\) eV) and hyperfine coupling to nuclear spin are larger than neglected terms; the pure fine-structure formula is then incomplete.
Strong external fields (Paschen–Back regime): if the Zeeman energy exceeds the spin–orbit energy, \(j\) is no longer good and \(m_l,m_s\) decouple — the coupled basis of Step 2 is the wrong one.

Failure modes

Using the uncoupled \(|l,m_l,m_s\rangle\) basis: spin–orbit is off-diagonal there, so its "diagonal element" is not the energy shift — a classic degenerate-PT error that quietly gives wrong (or infinite) answers.
Forgetting the Darwin term for \(s\)-states: setting \(\langle\hat H'_{\text{so}}\rangle=0\) for \(l=0\) and stopping — the relativistic term alone gives the wrong \(2s_{1/2}\) energy; the Darwin contact term is exactly what restores the universal \(j\)-formula.
Dropping the Thomas factor of \(\tfrac12\): the naive rest-frame estimate of spin–orbit is twice too big; the relativistic precession of the electron's frame halves it.
Writing \(l+\tfrac12\) instead of \(j+\tfrac12\) in the final formula: confuses fine structure with the intermediate relativistic result and reintroduces spurious \(l\)-dependence.
Treating \(j+\tfrac12\) as an integer offset of \(n\): forgetting the constraint \(j\le n-\tfrac12\), yielding non-existent states.

Discussion

The most striking feature is the disappearance of \(l\). Physically, the relativistic kinetic term is largest for penetrating low-\(l\) orbits (large \(\langle 1/r^2\rangle\)), and spin–orbit is largest there too but with the opposite sense of \(j\)-ordering; the two \(l\)-dependences are engineered by the Coulomb \(1/r\) to cancel, leaving only \(j\). This is a fingerprint of the hidden symmetry of the Coulomb problem, the same symmetry that makes the non-relativistic levels depend only on \(n\).

Each term tells a different story about what "an electron in an atom" means. The kinetic correction is pure special relativity: mass increases with speed, so the deeply-bound fast-moving components are over-bound by the Bohr treatment. Spin–orbit is magnetism: the electron's spin magnetic moment interacts with the field it sees from the orbiting proton, with the Thomas precession halving the naive coupling. The Darwin term is the most quantum of the three — the electron cannot be localised below its Compton wavelength, so it samples an average of the potential around each point, which for the singular Coulomb well shifts only the \(s\)-states that overlap the nucleus.

That all three emerge together, and combine so cleanly, is no accident: they are exactly the \(O(\alpha^4)\) terms of the Foldy–Wouthuysen reduction of the Dirac Hamiltonian. The exact Dirac energy, \(E=mc^2\big[1+(Z\alpha)^2/(n-\delta_j)^2\big]^{-1/2}\) with \(\delta_j=j+\tfrac12-\sqrt{(j+\tfrac12)^2-(Z\alpha)^2}\), expands term-by-term into the Bohr level plus precisely this fine-structure correction. The perturbative split into "kinetic + magnetic + contact" is a bookkeeping of one relativistic object; the \(j\)-only dependence is the split remembering its origin.

Common misconceptions. Fine structure is not the same as the Zeeman effect (that needs an external field) nor as hyperfine structure (that needs nuclear spin, and is \(\sim m_e/m_p\) smaller). And the residual \(l\)-degeneracy predicted here — \(2s_{1/2}=2p_{1/2}\) — is a genuine prediction of this formula that is observed to be violated by the Lamb shift; the violation is a triumph of QED, not a flaw in the algebra.

Worked examples

Example 1 — The hydrogen \(n=2\) fine structure and the \(10.9\) GHz doublet.

\[ E^{(1)}_{\text{fs}}=\frac{E_n^2}{2mc^2}\left[3-\frac{4n}{\,j+\tfrac12\,}\right],\qquad \frac{E_2^2}{2mc^2}=\frac{(3.40\ \text{eV})^2}{2(0.511\times10^6\ \text{eV})}. \]

Insert \(E_2=-13.6/4=-3.40\) eV, \(mc^2=511\) keV. A

\[ \frac{E_2^2}{2mc^2}=\frac{11.56}{1.022\times10^6}\ \text{eV}=1.131\times10^{-5}\ \text{eV}. \]

Arithmetic. A

\[ j=\tfrac12:\ 3-\tfrac{8}{1}=-5\ \Rightarrow\ -5.66\times10^{-5}\ \text{eV};\qquad j=\tfrac32:\ 3-\tfrac{8}{2}=-1\ \Rightarrow\ -1.13\times10^{-5}\ \text{eV}. \]

Evaluate the bracket at \(n=2\) for each \(j\); \(2s_{1/2},2p_{1/2}\) share the first, \(2p_{3/2}\) the second. A

\[ \Delta E_{2p_{3/2}-2p_{1/2}}=(-1.13+5.66)\times10^{-5}=4.53\times10^{-5}\ \text{eV},\quad \nu=\frac{\Delta E}{h}=\frac{4.53\times10^{-5}}{4.136\times10^{-15}\ \text{eV·s}}. \]

Split of the \(2p\) doublet; convert to frequency with \(h\). A

\[ \Delta E_{2p}=4.53\times10^{-5}\ \text{eV}\approx1.10\times10^{10}\ \text{Hz}=10.9\ \text{GHz} \]

Reading. The \(2p_{3/2}\) level sits above \(2p_{1/2}\) by \(45\ \mu\)eV, the measured hydrogen fine-structure interval. Units: eV divided by eV·s gives Hz.

Example 2 — Ground-state shift and the \(\alpha^2\) scale check.

\[ E^{(1)}_{\text{fs}}(1,\tfrac12)=\frac{E_1^2}{2mc^2}\left[3-\frac{4(1)}{1}\right]=-\frac{E_1^2}{2mc^2}. \]

At \(n=1\) the only allowed \(j\) is \(\tfrac12\); the bracket is \(3-4=-1\). A

\[ \frac{E_1^2}{2mc^2}=\frac{(13.6\ \text{eV})^2}{2(0.511\times10^6\ \text{eV})}=\frac{184.96}{1.022\times10^6}\ \text{eV}=1.81\times10^{-4}\ \text{eV}. \]

Insert \(E_1=-13.6\) eV. A

\[ \frac{E^{(1)}_{\text{fs}}}{|E_1|}=\frac{1.81\times10^{-4}}{13.6}=1.33\times10^{-5}\ \stackrel{?}{=}\ \frac{\alpha^2}{4}=\frac{(1/137.036)^2}{4}=1.33\times10^{-5}. \]

Compare the fractional shift with the predicted \(\alpha^2/4\) from the closed form at \(n=1\): \(\tfrac{\alpha^2}{n^2}(\tfrac{n}{j+1/2}-\tfrac34)=\alpha^2(1-\tfrac34)=\alpha^2/4\). B

\[ E^{(1)}_{\text{fs}}(1s_{1/2})=-1.81\times10^{-4}\ \text{eV},\qquad \frac{|E^{(1)}_{\text{fs}}|}{|E_1|}=\frac{\alpha^2}{4} \]

Reading. The ground state is lowered by \(0.18\) meV, and the fractional size is exactly \(\alpha^2/4\), confirming fine structure scales as \(\alpha^2\) relative to Bohr. Units: energy in eV; the ratio is dimensionless.

Problems

List the allowed \(j\) values and the fine-structure ordering of all states in the \(n=3\) shell of hydrogen.
Solution
Sub-shells \(l=0,1,2\). Coupling \(j=l\pm\tfrac12\) (with \(j=\tfrac12\) only for \(l=0\)) gives \(j\in\{\tfrac12,\tfrac32,\tfrac52\}\). States: \(3s_{1/2},3p_{1/2}\) (\(j=\tfrac12\)); \(3p_{3/2},3d_{3/2}\) (\(j=\tfrac32\)); \(3d_{5/2}\) (\(j=\tfrac52\)). Since higher \(j\) lies higher, energy order (lowest first) is \(j=\tfrac12<\tfrac32<\tfrac52\), with \(3s_{1/2}=3p_{1/2}\) and \(3p_{3/2}=3d_{3/2}\) degenerate at this order.
Show that within a shell the maximum-\(j\) state has the smallest fine-structure correction, and find that correction for \(n=4\).
Solution
The bracket \(3-4n/(j+\tfrac12)\) increases (becomes less negative) as \(j\) grows, so max \(j\) gives the least-negative shift. Max \(j=n-\tfrac12\Rightarrow j+\tfrac12=n\), bracket \(=3-4=-1\). Thus \(E^{(1)}_{\text{fs}}=-E_n^2/2mc^2\). For \(n=4\), \(E_4=-13.6/16=-0.850\) eV, \(E_4^2/2mc^2=0.7225/(1.022\times10^6)=7.07\times10^{-7}\) eV, so the shift is \(-7.1\times10^{-7}\) eV.
Using the closed form, compute the energy interval between \(2p_{3/2}\) and \(2s_{1/2}\) in hydrogen, and state what real effect breaks the degeneracy this formula predicts for \(2s_{1/2}\)–\(2p_{1/2}\).
Solution
From Example 1, \(2s_{1/2}\) shift \(=-5.66\times10^{-5}\) eV, \(2p_{3/2}\) shift \(=-1.13\times10^{-5}\) eV, so \(\Delta E_{2p_{3/2}-2s_{1/2}}=4.53\times10^{-5}\) eV \(\approx10.9\) GHz. The formula predicts \(2s_{1/2}=2p_{1/2}\) exactly; the Lamb shift (QED vacuum fluctuations and self-energy) raises \(2s_{1/2}\) above \(2p_{1/2}\) by about \(1.06\) GHz (\(4.4\times10^{-6}\) eV).
For a hydrogen-like ion of charge \(Z\), the corrections scale as \((Z\alpha)^4\) relative to \(mc^2\). Estimate the ratio of the \(2p\) fine-structure splitting in \(\text{C}^{5+}\) (\(Z=6\)) to that in hydrogen.
Solution
The fine-structure interval scales as \(mc^2(Z\alpha)^4/n^3\times\)(\(j\)-factor); at fixed \(n,j\) the ratio is \(Z^4\). Hence \(\Delta E(\text{C}^{5+})/\Delta E(\text{H})=6^4=1296\). Numerically \(1296\times4.53\times10^{-5}\ \text{eV}\approx5.9\times10^{-2}\) eV — nearly a thousandfold larger, which is why fine structure is easily resolved in highly-charged ions.
Verify by direct expansion that the exact Dirac energy reduces to the fine-structure formula. Expand \(E=mc^2\big[1+(\alpha/(n-\delta_j))^2\big]^{-1/2}\) with \(\delta_j=(j+\tfrac12)-\sqrt{(j+\tfrac12)^2-\alpha^2}\) to order \(\alpha^4\).
Solution
Write \(\kappa=j+\tfrac12\). To \(O(\alpha^2)\), \(\delta_j\approx\alpha^2/2\kappa\), so \(n-\delta_j\approx n(1-\alpha^2/2\kappa n)\) and \((\alpha/(n-\delta_j))^2\approx(\alpha^2/n^2)(1+\alpha^2/\kappa n)\). Then \([1+x]^{-1/2}\approx1-\tfrac12 x+\tfrac38 x^2\) with \(x=(\alpha^2/n^2)(1+\alpha^2/\kappa n)\): \(E\approx mc^2\big[1-\tfrac{\alpha^2}{2n^2}-\tfrac{\alpha^4}{2n^3\kappa}+\tfrac38\tfrac{\alpha^4}{n^4}\big]\). Subtract the rest mass \(mc^2\): \(E-mc^2\approx-\tfrac{mc^2\alpha^2}{2n^2}\big[1+\tfrac{\alpha^2}{n^2}(\tfrac{n}{\kappa}-\tfrac34)\big]\), which is exactly \(E_{nj}\) with \(\kappa=j+\tfrac12\). The Bohr term and the full fine-structure bracket are recovered.