Hyperfine Splitting and the 21 cm Line

Derivation

Hyperfine Splitting and the 21 cm Line

D-301 Home PU-306 Threads energy · fields · matter · light Depends on Fine-Structure Energy Formula, spin-half-algebra

Statement

Treating the magnetic-dipole coupling between the electron spin \(\vec{S}\) and the proton spin \(\vec{I}\) as a perturbation on the hydrogen \(1s\) ground state, we derive the Fermi contact interaction \(H_{\mathrm{hf}} = \frac{2\mu_0}{3}\,g_e g_p\,\frac{\mu_B\mu_N}{\hbar^2}\,\vec{S}\cdot\vec{I}\,\delta^{(3)}(\vec{r})\), show that it splits the ground state into a spin singlet (\(F=0\)) and triplet (\(F=1\)), and obtain the splitting \(\Delta E_{\mathrm{hf}} = \frac{4}{3}g_p\frac{m_e}{m_p}\alpha^4 m_e c^2 \approx 5.88\ \mu\mathrm{eV}\), corresponding to the \(\nu = 1420\ \mathrm{MHz}\), \(\lambda = 21\ \mathrm{cm}\) radio line.

Why it matters

The 21 cm line is the single most important spectral feature in radio astronomy. Because it arises from a spin flip in neutral atomic hydrogen — the most abundant baryonic species in the universe — it maps cold gas that emits no optical light: galactic rotation curves (and hence dark matter), the structure of the interstellar medium, and, redshifted, the neutral hydrogen of the cosmic dawn and reionization epoch.

Physically it is the cleanest laboratory for a magnetic dipole–dipole coupling in a bound system. The same machinery — a spin-spin operator projected onto a coupled basis \(\vec{F}=\vec{S}+\vec{I}\) — recurs in NMR, ESR, atomic clocks (the hyperfine transition of \(^{133}\mathrm{Cs}\) defines the second), and precision QED tests through the measured value of \(\Delta E_{\mathrm{hf}}\).

Assumptions

The nucleus is a point magnetic dipole \(\vec{\mu}_p = g_p\mu_N\vec{I}/\hbar\).Finite proton size (the Zemach radius) shifts \(\Delta E_{\mathrm{hf}}\) at the \(\sim40\) ppm level; dropping the point approximation is required to match the measured 1420.405751 MHz to that precision.

The unperturbed state is the exact non-relativistic Coulomb \(1s\) orbital, with \(g_e=2\).Without it \(|\psi(0)|^2 = 1/(\pi a_0^3)\) is not exact; relativistic, reduced-mass and QED (anomalous moment, vacuum polarization) corrections enter at order \(\alpha\) and account for the residual \(\sim0.1\%\) discrepancy of the leading formula.

First-order (non-degenerate within each \(F\)) perturbation theory suffices.If dropped, one must diagonalise \(H_{\mathrm{hf}}\) together with the fine structure and any external field; the clean singlet/triplet pattern only holds when hyperfine \(\ll\) fine structure \(\ll\) gross structure, which fails at high field (Paschen–Back regime).

The dipole–dipole tensor term averages to zero for an \(\ell=0\) state.For \(\ell\neq0\) the \(\frac{3(\vec{\mu}\cdot\hat r)\hat r-\vec{\mu}}{r^3}\) term survives and the contact term vanishes (the wavefunction vanishes at the origin); the whole derivation below is specific to s-states.

Derivation

\[ \vec{B}_p(\vec{r}) = \frac{\mu_0}{4\pi}\left[\frac{3(\vec{\mu}_p\cdot\hat r)\hat r-\vec{\mu}_p}{r^3}\right] + \frac{2\mu_0}{3}\vec{\mu}_p\,\delta^{(3)}(\vec{r}) \]

Exact field of a point magnetic dipole; the delta term is fixed by demanding the correct volume integral \(\int_{\text{sphere}}\vec{B}\,d^3r = \tfrac{2}{3}\mu_0\vec{\mu}_p\), which the naive \(1/r^3\) tail alone gets wrong. B

\[ H_{\mathrm{hf}} = -\vec{\mu}_e\cdot\vec{B}_p, \qquad \vec{\mu}_e = -g_e\frac{\mu_B}{\hbar}\vec{S},\quad \vec{\mu}_p = +g_p\frac{\mu_N}{\hbar}\vec{I} \]

The perturbation is the Zeeman energy of the electron moment in the proton's field; the moments are written via the \(g\)-factors, with the electron moment antiparallel to its spin (negative charge). A

\[ H_{\mathrm{hf}}^{\text{contact}} = -\frac{2\mu_0}{3}\,\vec{\mu}_e\cdot\vec{\mu}_p\,\delta^{(3)}(\vec{r}) = \frac{2\mu_0}{3}\,g_e g_p\,\frac{\mu_B\mu_N}{\hbar^2}\,\vec{S}\cdot\vec{I}\;\delta^{(3)}(\vec{r}) \]

Keep only the contact term (the tensor term integrates to zero against a spherically symmetric \(1s\) state, Assumption 4); substituting the moments turns the sign, since \((-)(+)(-)=(+)\). B

\[ \Delta E = \langle\psi_{100}|H_{\mathrm{hf}}^{\text{contact}}|\psi_{100}\rangle = \frac{2\mu_0}{3}\,g_e g_p\,\frac{\mu_B\mu_N}{\hbar^2}\,|\psi_{100}(0)|^2\,\langle\vec{S}\cdot\vec{I}\rangle \]

First-order shift; the delta function collapses the spatial integral to the probability density at the origin, factorising the spin part out. B

\[ \psi_{100}(\vec r) = \frac{1}{\sqrt{\pi a_0^3}}\,e^{-r/a_0} \;\Longrightarrow\; |\psi_{100}(0)|^2 = \frac{1}{\pi a_0^3} \]

The hydrogen ground state is finite and non-zero at the nucleus — the defining feature of s-states that makes the contact term act. A

\[ \vec{F}=\vec{S}+\vec{I}\;\Rightarrow\; \vec{S}\cdot\vec{I} = \tfrac{1}{2}\left(F^2-S^2-I^2\right) = \frac{\hbar^2}{2}\big[F(F+1)-S(S+1)-I(I+1)\big] \]

\(\vec{S}\cdot\vec{I}\) is not diagonal in the uncoupled \(|m_S,m_I\rangle\) basis but is diagonal in the coupled \(|F,m_F\rangle\) basis; square the total-spin definition to express it through Casimir eigenvalues. A

\[ S=I=\tfrac12:\quad \langle\vec{S}\cdot\vec{I}\rangle_{F=1} = +\frac{\hbar^2}{4},\qquad \langle\vec{S}\cdot\vec{I}\rangle_{F=0} = -\frac{3\hbar^2}{4} \]

Insert \(S(S+1)=I(I+1)=\tfrac34\) and \(F(F+1)=2\) (triplet) or \(0\) (singlet); the two spin states now carry different energy. A

\[ \Delta E_{\mathrm{hf}} = E_{F=1}-E_{F=0} = \frac{2\mu_0}{3}\,g_e g_p\,\frac{\mu_B\mu_N}{\hbar^2}\,\frac{1}{\pi a_0^3}\left[\frac{\hbar^2}{4}-\left(-\frac{3\hbar^2}{4}\right)\right] = \frac{2\mu_0}{3}\,g_e g_p\,\frac{\mu_B\mu_N}{\pi a_0^3} \]

Subtract the two level shifts; the bracket is exactly \(\hbar^2\), cancelling the \(\hbar^{-2}\) and leaving a spin-independent gap. B

\[ \text{With }g_e=2,\ \mu_B=\frac{e\hbar}{2m_e},\ \mu_N=\frac{e\hbar}{2m_p},\ \mu_0=\frac{1}{\epsilon_0 c^2},\ a_0=\frac{4\pi\epsilon_0\hbar^2}{m_e e^2},\ \alpha=\frac{e^2}{4\pi\epsilon_0\hbar c}: \]

\[ \Delta E_{\mathrm{hf}} = \frac{\mu_0 g_p e^2\hbar^2}{3\pi m_e m_p a_0^3} = \frac{4}{3}\,g_p\,\frac{m_e}{m_p}\,\alpha^4\, m_e c^2 \]

Substitute the fundamental constants and collapse; every factor combines into powers of \(\alpha\) and the electron rest energy, exposing the scaling \(\Delta E_{\mathrm{hf}}\propto (m_e/m_p)\alpha^4 m_e c^2\). C

Result

\[ \Delta E_{\mathrm{hf}} = \frac{4}{3}\,g_p\,\frac{m_e}{m_p}\,\alpha^4\, m_e c^2 \approx 5.88\ \mu\mathrm{eV},\qquad \nu = \frac{\Delta E_{\mathrm{hf}}}{h}\approx 1420\ \mathrm{MHz},\qquad \lambda=\frac{c}{\nu}\approx 21\ \mathrm{cm} \]

Reading. The magnetic energy of the electron sitting in the proton's contact field depends on whether the two spins are parallel (triplet, higher) or antiparallel (singlet, lower). The gap is suppressed relative to the gross structure \(\alpha^2 m_e c^2\) by a further \(\alpha^2 (m_e/m_p)\) — one factor \(\alpha^2\) from the magnetic (rather than Coulomb) coupling and \(m_e/m_p\) from the small nuclear moment. A spin flip from triplet to singlet emits one 21 cm photon.

Units check. \(g_p\) and \(\alpha\) are dimensionless, \(m_e/m_p\) is dimensionless, and \(m_e c^2\) is an energy — so the right side is an energy. Independently, \(\mu_0[\mathrm{T\,m/A}]\cdot\mu_B\mu_N[\mathrm{J^2/T^2}]\,/\,a_0^3[\mathrm{m^3}] = \mathrm{T\,m\,A^{-1}\,J^2\,T^{-2}\,m^{-3}}\); using \(\mathrm{T=kg\,A^{-1}s^{-2}}\) and \(\mathrm{J=kg\,m^2 s^{-2}}\) this reduces to \(\mathrm{J}\). Numerically \(\nu = \Delta E/h = 5.88\times10^{-6}\times1.602\times10^{-19}/6.626\times10^{-34}\ \mathrm{Hz}=1.42\times10^{9}\ \mathrm{Hz}\). ✓

Limiting cases

Center of gravity: weighting each level by its multiplicity \((2F+1)\), \(3\cdot(+\tfrac14)+1\cdot(-\tfrac34)=0\): the hyperfine interaction does not shift the mean energy, only splits it. The unperturbed \(1s\) energy is the multiplicity-weighted centroid.
Heavy nucleus \(m_p\to\infty\): \(\mu_N\propto 1/m_p\to0\), so \(\Delta E_{\mathrm{hf}}\to0\) — an infinitely heavy nucleus has no measurable moment and the splitting closes.
Hydrogenic ion, charge \(Z\): \(|\psi(0)|^2\propto Z^3/a_0^3\), so \(\Delta E_{\mathrm{hf}}\propto Z^3\) (times any change of nuclear \(g\)); the splitting grows steeply with nuclear charge.
Muonium (\(\mu^+e^-\)): replace \(g_p\mu_N\) by the muon moment \(\propto 1/m_\mu\); the splitting rescales cleanly and is a pure-lepton QED test with no proton-structure uncertainty.

Breaks when

Non-zero orbital angular momentum (\(\ell\neq0\)). The \(1s\)-specific contact term vanishes because \(\psi(0)=0\); the physics shifts entirely to the dipole–dipole tensor term and orbital hyperfine coupling, and the formula \(\propto|\psi(0)|^2\) gives zero.
Strong external magnetic field (Paschen–Back / Back–Goudsmit regime). When the electron Zeeman energy \(g_e\mu_B B\) exceeds \(\Delta E_{\mathrm{hf}}\), \(\vec{S}\) and \(\vec{I}\) decouple, \(F\) ceases to be a good quantum number, and one must diagonalise \(A\,\vec{S}\cdot\vec{I} + g_e\mu_B B S_z\) directly.
Precision beyond \(\sim0.1\%\). The leading formula misses the electron anomalous moment (\(g_e\ne2\)), reduced-mass, relativistic, recoil, and finite-size (Zemach radius) corrections; each must be added to reach the measured 1420.405751 MHz.

Failure modes

Dropping the contact delta term and using only the \(1/r^3\) dipole tail — this integrates to zero for the \(1s\) state and predicts no splitting at all.
Sign confusion in \(\vec{\mu}_e\): forgetting that the electron moment is antiparallel to its spin (\(g_e>0\) with a minus sign), which flips triplet and singlet and gives an inverted spectrum.
Using \(\vec{S}\cdot\vec{I}\) in the uncoupled basis: treating \(|m_S,m_I\rangle\) as eigenstates. \(S_xI_x+S_yI_y\) mixes them; only \(|F,m_F\rangle\) diagonalises the operator.
Miscounting eigenvalues: writing \(\langle\vec{S}\cdot\vec{I}\rangle=\pm\tfrac14\hbar^2\) for both levels, missing that the singlet is \(-\tfrac34\hbar^2\); this makes the gap \(\tfrac12\hbar^2\) instead of \(\hbar^2\) and halves \(\Delta E\).
Confusing \(\mu_N\) with \(\mu_B\) (or omitting \(g_p\)): using the electron magneton for the proton inflates the splitting by \(m_p/m_e\approx1836\).
Calling it a fine-structure effect: hyperfine is spin–spin (\(\propto\alpha^4 m_e/m_p\)), not the spin–orbit fine structure (\(\propto\alpha^4\)); they differ by the nuclear mass ratio.

Discussion

The hierarchy of hydrogen energy scales reads directly off the powers of \(\alpha\): gross structure \(\sim\alpha^2 m_e c^2\) (Coulomb binding), fine structure \(\sim\alpha^4 m_e c^2\) (spin–orbit and relativistic corrections), and hyperfine \(\sim\alpha^4 (m_e/m_p) m_e c^2\). Hyperfine is smaller than fine structure by exactly the electron-to-proton mass ratio — a factor \(\approx1/1836\) — which is why "hyperfine" splitting is roughly a thousand times finer than "fine" splitting. This single ratio explains why the 21 cm line sits in the radio rather than the microwave or optical.

The contact interaction is a genuinely relativistic and quantum object: the \(\frac{2}{3}\mu_0\vec{\mu}\,\delta^{(3)}(\vec{r})\) term is exactly what falls out of the non-relativistic reduction of the Dirac equation coupled to the nuclear moment. It encodes the fact that an s-electron spends part of its life literally on top of the nucleus, sampling the strong near-field where the pointlike moment's field is a delta function. Everything hinges on \(|\psi(0)|^2\neq0\), the same quantity that governs the Lamb shift and isotope shifts.

Astrophysically, the transition is famously forbidden as an electric dipole and proceeds only by magnetic dipole radiation, with an Einstein coefficient \(A_{10}\approx2.9\times10^{-15}\,\mathrm{s^{-1}}\) — a mean radiative lifetime of \(\sim11\) million years. No terrestrial gas is dilute enough to emit it before collisions redistribute the spins, but the interstellar medium is, so the line is ubiquitous in the galaxy. The observed brightness is governed by the "spin temperature" \(T_s\), defined through the level populations \(n_1/n_0 = (g_1/g_0)e^{-\Delta E/k_B T_s} = 3\,e^{-\Delta E/k_B T_s}\); because \(\Delta E/k_B = 0.068\) K is tiny, the exponential is essentially unity and the ratio is pinned near the degeneracy value 3, making the line always available for absorption and emission.

Beyond the leading formula, \(\Delta E_{\mathrm{hf}}\) is one of the most precisely measured quantities in physics (13 significant figures). Comparing theory to experiment isolates proton structure through the Zemach radius, and the analogous positronium and muonium splittings — free of hadronic uncertainty — provide clean tests of bound-state QED, including the electron anomalous moment and radiative recoil. The persistent \(\sim3\) ppm "hyperfine puzzle" in hydrogen, where theory and experiment disagree at the level of the proton polarizability contribution, remains an active frontier. Common misconceptions: the splitting is often attributed to spin–orbit coupling or to the proton "orbiting" the electron; it is neither — it is the magnetostatic energy of two fixed magnetic dipoles whose relative orientation is set by their coupled spin state, sampled at the point \(r=0\).

Worked examples

Example 1 — Frequency and wavelength of the line.

\[ \Delta E_{\mathrm{hf}} = \frac{4}{3}\,g_p\,\frac{m_e}{m_p}\,\alpha^4\, m_e c^2 \]

Start from the closed form; solve symbolically before inserting numbers. A

\[ g_p=5.586,\quad \frac{m_e}{m_p}=\frac{1}{1836.15},\quad \alpha=\frac{1}{137.036},\quad m_ec^2=0.5110\ \mathrm{MeV} \]

List the inputs with units before substituting. A

\[ \Delta E_{\mathrm{hf}} = \tfrac{4}{3}(5.586)(5.446\times10^{-4})(2.836\times10^{-9})(5.110\times10^{5}\ \mathrm{eV}) = 5.88\times10^{-6}\ \mathrm{eV} \]

Multiply; \(\alpha^4=2.836\times10^{-9}\). A

\[ \nu=\frac{\Delta E_{\mathrm{hf}}}{h}=\frac{5.88\times10^{-6}\times1.602\times10^{-19}\ \mathrm{J}}{6.626\times10^{-34}\ \mathrm{J\,s}} = 1.42\times10^{9}\ \mathrm{Hz},\qquad \lambda=\frac{c}{\nu}=\frac{3.00\times10^{8}}{1.42\times10^{9}}\ \mathrm{m} \]

Convert energy to frequency via Planck, then to wavelength via \(c=\nu\lambda\). A

\[ \boxed{\ \Delta E_{\mathrm{hf}}\approx5.88\ \mu\mathrm{eV},\quad \nu\approx1421\ \mathrm{MHz},\quad \lambda\approx21.1\ \mathrm{cm}\ } \]

Reading. The leading formula lands within \(0.06\%\) of the measured 1420.405751 MHz; the small residual is the QED and proton-structure corrections deliberately omitted here.

Example 2 — Spin-state populations at cloud temperature.

\[ \frac{n_{F=1}}{n_{F=0}} = \frac{g_1}{g_0}\,e^{-\Delta E_{\mathrm{hf}}/k_B T} = 3\,e^{-\Delta E_{\mathrm{hf}}/k_B T} \]

Boltzmann ratio with degeneracies \(g_1=2F+1=3\) (triplet) and \(g_0=1\) (singlet). A

\[ \frac{\Delta E_{\mathrm{hf}}}{k_B} = \frac{9.42\times10^{-25}\ \mathrm{J}}{1.381\times10^{-23}\ \mathrm{J/K}} = 0.0682\ \mathrm{K} \]

Convert the gap to a temperature; this "hyperfine temperature" is the key small number. A

\[ T=100\ \mathrm{K}:\quad \frac{\Delta E_{\mathrm{hf}}}{k_B T}=\frac{0.0682}{100}=6.82\times10^{-4}\ \Rightarrow\ \frac{n_1}{n_0}=3\,e^{-6.82\times10^{-4}}=3(0.99932) \]

Evaluate the exponential; the argument is tiny so the ratio barely departs from the degeneracy limit. A

\[ \boxed{\ \frac{n_{F=1}}{n_{F=0}}\approx2.998\ \approx\ 3\ } \]

Reading. For any astrophysical temperature above a fraction of a Kelvin the two states are populated essentially in their statistical ratio 3:1 — the transition is "always on." This is why 21 cm surveys trace hydrogen column density almost independently of the gas temperature, and why the spin temperature is set by collisions and radiation rather than by the level energetics.

Problems

Verify the coupled-basis eigenvalues. Using \(\vec{S}\cdot\vec{I}=\tfrac12(F^2-S^2-I^2)\) with \(S=I=\tfrac12\), compute \(\langle\vec{S}\cdot\vec{I}\rangle\) for \(F=1\) and \(F=0\), and give the energy shift of each level relative to the unperturbed \(1s\) energy in units of the coupling constant \(A\equiv\Delta E_{\mathrm{hf}}/\hbar^2\) (so \(E_F = A\langle\vec{S}\cdot\vec{I}\rangle\)).
Solution
With \(S(S+1)=I(I+1)=\tfrac34\): for \(F=1\), \(F(F+1)=2\), so \(\langle\vec{S}\cdot\vec{I}\rangle=\tfrac{\hbar^2}{2}(2-\tfrac34-\tfrac34)=+\tfrac{\hbar^2}{4}\); for \(F=0\), \(F(F+1)=0\), so \(\langle\vec{S}\cdot\vec{I}\rangle=\tfrac{\hbar^2}{2}(0-\tfrac32)=-\tfrac{3\hbar^2}{4}\). The shifts are \(E_{F=1}=+\tfrac14 A\hbar^2\) and \(E_{F=0}=-\tfrac34 A\hbar^2\); their difference is \(A\hbar^2=\Delta E_{\mathrm{hf}}\). ✓
Center-of-gravity theorem. Show that the multiplicity-weighted mean shift vanishes, and state why the unperturbed \(1s\) energy equals the centroid rather than either level.
Solution
Weight each level by \(g_F=2F+1\): \(\dfrac{g_1 E_1 + g_0 E_0}{g_1+g_0} = \dfrac{3(+\tfrac14 A\hbar^2)+1(-\tfrac34 A\hbar^2)}{3+1} = \dfrac{\tfrac34-\tfrac34}{4}A\hbar^2 = 0.\) The trace of \(\vec{S}\cdot\vec{I}\) over all four spin states is zero, so the perturbation only redistributes energy. The measured \(1s\) energy (e.g. from the Lyman series) is the centroid; the triplet lies \(+\tfrac14\Delta E\) above it and the singlet \(-\tfrac34\Delta E\) below.
Deuterium. The deuteron has spin \(I=1\). List the allowed total-spin quantum numbers \(F\) for the \(1s\) electron coupled to the deuteron, and compute \(\langle\vec{S}\cdot\vec{I}\rangle\) for each.
Solution
\(\vec{F}=\vec{S}+\vec{I}\) with \(S=\tfrac12,\ I=1\) gives \(F=\tfrac32\) and \(F=\tfrac12\). Using \(\vec{S}\cdot\vec{I}=\tfrac{\hbar^2}{2}[F(F+1)-S(S+1)-I(I+1)]\) with \(S(S+1)=\tfrac34,\ I(I+1)=2\): for \(F=\tfrac32\), \(F(F+1)=\tfrac{15}{4}\), \(\langle\vec{S}\cdot\vec{I}\rangle=\tfrac{\hbar^2}{2}(\tfrac{15}{4}-\tfrac34-2)=+\tfrac{\hbar^2}{2}\); for \(F=\tfrac12\), \(F(F+1)=\tfrac34\), \(\langle\vec{S}\cdot\vec{I}\rangle=\tfrac{\hbar^2}{2}(\tfrac34-\tfrac34-2)=-\hbar^2\). The gap is \(\tfrac32\hbar^2\) times the coupling; the deuterium line sits near 327 MHz.
Cosmological redshift. The rest-frame 21 cm line (\(\nu_0=1420.4\) MHz) is observed at \(\nu_{\mathrm{obs}}=355\) MHz. Find the redshift \(z\) of the emitting hydrogen.
Solution
\(1+z=\dfrac{\nu_0}{\nu_{\mathrm{obs}}}=\dfrac{1420.4}{355}=4.001\), so \(z=3.00\). This is exactly the kind of high-redshift neutral hydrogen (the epoch of reionization is probed at \(z\sim6\text{–}20\), pushing the line to \(\sim70\text{–}200\) MHz) that experiments like HERA and the SKA target.
Effective internal field. Model the singlet–triplet gap as the electron moment flipping in an effective field \(B_{\mathrm{eff}}\), i.e. \(\Delta E_{\mathrm{hf}}=g_e\mu_B B_{\mathrm{eff}}\). Estimate \(B_{\mathrm{eff}}\).
Solution
\(B_{\mathrm{eff}}=\dfrac{\Delta E_{\mathrm{hf}}}{g_e\mu_B}=\dfrac{9.42\times10^{-25}\ \mathrm{J}}{2\times9.274\times10^{-24}\ \mathrm{J/T}}=0.0508\ \mathrm{T}\approx 51\ \mathrm{mT}.\) So the proton's contact field, sampled by the s-electron over its orbit, is equivalent to an internal magnetic field of order \(0.05\) T — modest on the atomic scale precisely because the nuclear magneton is \(1836\) times smaller than the Bohr magneton.