advanced28 min studyLesson 453 in Physics

Hyperfine splitting and nuclear spin

Couple nuclear and electronic angular momenta and distinguish magnetic-dipole from electric-quadrupole hyperfine terms.

On this page

What you’ll understand

  • Construct the allowed F levels and their degeneracies.
  • Evaluate magnetic-dipole hyperfine shifts with a declared unit convention.
  • Recognize when quadrupole terms or strong-field mixing invalidate a simple formula.

Fine structure resolves electronic angular momentum, but the nucleus can carry angular momentum as well. Its electromagnetic moments couple to the electronic environment and create hyperfine structure. The smaller nuclear magneton often makes these energies much less than electronic fine-structure separations, although the hierarchy must be checked for the atom and level in question. Hyperfine frequencies are useful probes of both nuclear properties and electronic wavefunctions; they are not determined by nuclear spin alone.

Build the total atomic angular momentum

Denote nuclear spin by I and total electronic angular momentum by J. In zero field, their coupled total is F=I+J\mathbf F=\mathbf I+\mathbf J. If a single electronic J manifold is isolated, F ranges from ∣I−J∣|I-J| to I+JI+J in unit steps. Each F level contains 2F+12F+1 substates. Summing those counts gives (2I+1)(2J+1)(2I+1)(2J+1), ensuring that coupling reorganizes rather than creates states. This counting includes nuclear spin degeneracy absent from an electronic term symbol.

An effective magnetic-dipole interaction is proportional to I⋅J\mathbf I\cdot\mathbf J. The angular factor follows from the same squared-vector identity used for spin-orbit coupling. Its coefficient A contains nuclear magnetic moment and electronic matrix elements. For an s electron, a contact contribution samples electron density at the nucleus; orbital and spin-dipole contributions also occur in general. A nonzero nuclear spin need not imply a large measured splitting, because moment size and electronic structure both matter.

Hhfs=AI⋅Jℏ2,EF=A2K,K=F(F+1)−I(I+1)−J(J+1)H_{\mathrm{hfs}}=A\frac{\mathbf I\cdot\mathbf J}{\hbar^2},\qquad E_F=\frac A2K,\qquad K=F(F+1)-I(I+1)-J(J+1)
Here A is an energy, not a frequency, and I, J, F are dimensionless quantum numbers. This model retains only magnetic-dipole hyperfine coupling in one electronic J manifold.↔ Scroll wide equations horizontally.

Declare frequency conventions before calculating

Spectroscopic tables often quote A/hA/h in megahertz rather than A in joules. Then the level formula predicts EF/hE_F/h directly in the same frequency unit. A table quoting A/ℏA/\hbar instead would use angular frequency and differ by a factor of 2π2\pi. Keeping an explicit h in a calculation prevents a familiar numerical error. The weighted sum of dipole shifts over all F substates vanishes relative to the electronic center of gravity.

For J=1/2J=1/2, only two F values occur when I≥1/2I\geq1/2, and their interval is A(I+1/2)A(I+1/2). Positive A puts the larger F higher; negative A reverses the ordering. Nuclear spin zero removes magnetic-dipole and electric-quadrupole hyperfine splitting, although isotope mass and charge-radius changes still move electronic energies. A spectral isotope shift and a hyperfine interval therefore encode different physical information and should be reported separately.

Δν=∣A∣h(I+12)(J=1/2),∑F(2F+1)EF=0\Delta\nu=\frac{|A|}{h}\left(I+\frac12\right)\quad(J=1/2),\qquad \sum_F(2F+1)E_F=0
The first expression is the positive ordinary-frequency separation for the dipole-only doublet. The second is the dipole center-of-gravity identity, with energies measured relative to the unsplit electronic level.↔ Scroll wide equations horizontally.

Know what the dipole model omits

A nucleus with spin at least one can have an electric quadrupole moment, which couples to the electronic electric-field gradient. A first-order quadrupole splitting within a J manifold additionally requires J≥1J\geq1. Thus a spin-one nucleus does not by itself produce a first-order quadrupole pattern in a J=1/2J=1/2 state. Higher-order mixing with other electronic levels can create additional small corrections. Fitting every observed interval with a single dipole constant can conceal these effects.

An external magnetic field competes with hyperfine coupling. When the electronic Zeeman energy is much smaller than the hyperfine interval, F remains an approximate good quantum number. At larger fields the nucleus and electronic angular momentum progressively decouple, and a matrix treatment or an appropriate strong-field expansion is required. Optical electric-dipole transitions usually conserve nuclear spin and obey total-F triangle constraints; the ground-state hydrogen hyperfine radio transition instead proceeds magnetically. A Bloch sphere cannot reproduce these multilevel couplings or their transition rates.

μBB≪ΔEhfsfor a weak-field hyperfine basis\mu_B B\ll\Delta E_{\mathrm{hfs}}\quad\text{for a weak-field hyperfine basis}
The Bohr magneton mu_B has units joule per tesla and B is in tesla. This comparison is a regime criterion, with actual matrix elements and level separations needed near crossings.↔ Scroll wide equations horizontally.

A nuclear spin three-halves doublet

A J=1/2J=1/2 electronic level has I=3/2I=3/2 and A/h=200 MHzA/h=200\,\mathrm{MHz}. Find the hyperfine frequencies relative to the level center and their separation.

  1. Coupling gives F=1,2F=1,2. Compute I(I+1)=15/4I(I+1)=15/4 and J(J+1)=3/4J(J+1)=3/4, so their sum is 9/29/2.

  2. For F equal to one, K=2−9/2=−5/2K=2-9/2=-5/2, giving E1/h=−250 MHzE_1/h=-250\,\mathrm{MHz}. For F equal to two, K=6−9/2=3/2K=6-9/2=3/2, giving E2/h=150 MHzE_2/h=150\,\mathrm{MHz}.

  3. The separation is 400 MHz400\,\mathrm{MHz}. Check the degeneracy-weighted center: 3(−250)+5(150)=03(-250)+5(150)=0, and compare with 200(3/2+1/2)=400 MHz200(3/2+1/2)=400\,\mathrm{MHz}.

The result

The two hyperfine levels lie at −250 and +150 MHz relative to their center; the interval is 400 MHz.

Explore the model

Interactive lab

Interactive lab · quantum

A restricted pair of magnetic substates

Treat the basis labels only as two selected substates. Reset and set mixing angle 60∘60^\circ, phase 0∘0^\circ: predict population one of one quarter. Repeat from Reset with both controls 120∘120^\circ, 0∘0^\circ: predict three quarters. Playback changes phase coherently, not hyperfine populations or nuclear spin; this is not a full hyperfine atom.

A restricted pair of magnetic substatesA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTwo-state population geometry omits the multiplicity and coupling constants needed for an actual hyperfine spectrum.

Check your understanding

Which condition is required for a first-order electric-quadrupole hyperfine splitting within an isolated level?

Try it yourself

For I=J=1/2I=J=1/2 with A/h=−120 MHzA/h=-120\,\mathrm{MHz}, find the F levels, their signed shifts, and which level is lower.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute K separately for F = 0 and F = 1; retain the negative sign of A.

Compare with the solution

The allowed values are F=0,1F=0,1. Their K values are −3/2-3/2 and 1/21/2, giving shifts +90 MHz+90\,\mathrm{MHz} and −30 MHz-30\,\mathrm{MHz}. Thus F equal to one is lower, and the positive separation is 120 MHz120\,\mathrm{MHz}. The weighted center check is 1(90)+3(−30)=01(90)+3(-30)=0. First-order quadrupole coupling is absent because both spins are one-half.

Take it with you

  • Hyperfine coupling reorganizes the product of nuclear and electronic spin spaces.
  • Constants quoted in MHz usually mean A divided by h; frequency and angular frequency differ.
  • Quadrupole terms and external-field mixing require assumptions beyond a dipole-only F formula.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.