In the absence of hyperfine interactions, the energy levels of a lanthanum atom under an external magnetic field can be calculated straightforwardly. Each energy level E(mJ) depends on the magnetic quantum number mJ according to the formula E=−gμNB0mJE=−gμNB0mJ, where g is the Landé g-factor, μN is the nuclear magneton, and B0 is the external magnetic field strength.

E(−32)=−1×3.15×10−8 eV/T×2 T×−32=9.45×10−8 eV E(−23)=−1×3.15×10−8eV/T×2T×−23=9.45×10−8eV

E(−12)=−1×3.15×10−8 eV/T×2 T×−12=3.15×10−8 eV E(−21)=−1×3.15×10−8eV/T×2T×−21=3.15×10−8eV

E(12)=−1×3.15×10−8 eV/T×2 T×12=−3.15×10−8 eV E(21)=−1×3.15×10−8eV/T×2T×21=−3.15×10−8eV

With hyperfine interactions considered, additional terms involving the hyperfine coupling constant AA and the nuclear magnetic quantum number mImI are added to the energy expressions.

E(−32)=−1×3.15×10−8 eV/T×2 T×−32+42×10−8 eV×12×(−32)=−22.05×10−8 eVE(−23)=−1×3.15×10−8eV/T×2T×−23+42×10−8eV×21×(−23)=−22.05×10−8eV

E(−12)=−1×3.15×10−8 eV/T×2 T×−12+42×10−8 eV×12×(−12)=−7.35×10−8 eVE(−21)=−1×3.15×10−8eV/T×2T×−21+42×10−8eV×21×(−21)=−7.35×10−8eV

E(12)=−1×3.15×10−8 eV/T×2 T×12+42×10−8 eV×12×12=7.35×10−8 eVE(21)=−1×3.15×10−8eV/T×2T×21+42×10−8eV×21×21=7.35×10−8eV