Weak and strong coupling

In Fig. 6, the energy level schemes are drawn for the weak and the strong coupling case. It is assumed that ν_I < 0 and a_iso > 0. The electron and nuclear Zeeman levels are indexed by their magnetic quantum numbers m_S and m_I and commonly designated by |αα>, |αβ>, |βα> and |ββ>, where |α> and |β> refer to the two spin state in the external magnetic field. In general m_I is not a good quantum number for the characterization of the energy levels since the quantization axis of the nuclear spin deviates from the direction of the external magnetic field due to the HFI. For this reason a new numbering of the levels is introduced, namely |1>, |2>, |3> and |4>. The numbering of the levels in the weak and strong coupling case is shown in the figure.

In the weak coupling case (ν_I > a_iso/2) the direction of quantization of the nuclear spin is given in first order by the external magnetic field B₀. This is illustrated in the vector diagram at the bottom of Fig. 6 for the m_S = +1/2 manifold. The magnetic fields put up by the NZI and the HFI are drawn along the z-axis and have opposite directions. The effective quantization vector, experienced by the nucleus and indicated by the bold arrow, is obtained by the addition of both contributions. The hyperfine field slightly decreases the energy splitting caused by the NZI.

In the strong coupling case (ν_I < a_iso/2) the direction of quantization of the nuclear spin is determined by the hyperfine field a_iso/2. The NZI gives only a small contribution to the effective field. Again the fields act in opposite directions in the m_S = +1/2 manifold.

In the m_S = -1/2 manifold, the nuclear Zeeman and hyperfine fields act in the same direction for both weak and strong coupling. The splitting between the Zeeman states is thus increased as can be inferred from the energy level diagram in Fig. 6.

We now consider the more general case where additional terms in the spin Hamiltonian cause the quantization axis to deviate from the z-axis. This is e.g. the case when the high-field approximation is no more valid and S_x and S_y have to be considered or when the coupling is anisotropic (B <> 0). As a result not only the magnitude of the effective field experienced by the nucleus but also the direction is changed when going from one m_S to another. This has important effects on the transition frequencies observable in the four level system as a change in the m_S spin state can now also affect the spin state of the nucleus. The transition probability of the so-called forbidden transitions increases. In Fig. 6 the six possible transitions between the four levels are indicated both for the weak and strong coupling case. The allowed electron transitions, involving only a spin flip of the electron (Δm_S = ±1, Δm_I = 0) are indicated by full arrows, the forbidden electron transitions where both the electron and the nuclear spins flip (Δm_S = ±1, Δm_I = ±1) by dashed arrows. The hollow arrows represent the nuclear transitions which take place within the m_S manifolds. Note that in our example with isotropic coupling the forbidden transitions have zero transition probability.

The EPR and NMR spectra, obtained for the energy level schemes in Fig. 6, are drawn schematically in Fig. 7 for the weak and strong coupling case. In the EPR spectrum allowed (solid lines) and forbidden (dashed lines) transitions are centered around g_iso. The numbering of the frequencies is taken from the energy level scheme with ν_ij = E_i-E_j. The nuclear transition frequencies ν_α and ν_β are shown as energy differences between the allowed and forbidden transitions. They are given by the absolute values of the transition frequencies ν₁₂ and ν₃₄. In the NMR spectrum in Fig. 7 only positive frequencies are observed with the consequence that ν_α and ν_β are centered around a_iso/2 and separated by 2ν_I in the strong coupling case.

The intensities of the allowed and forbidden electron transitions are given by

$$ I_{a} = \text{cos}^2\eta = \frac {\left|\nu^2_I - \frac {1}{4}\nu_{-}^2\right|} {\nu_{\alpha}\nu_{\beta}}  \ \qquad (21 a)$$

and

$$ I_{f} = \text{sin}^2\eta = \frac {\left|\nu^2_I - \frac {1}{4}\nu_{+}^2\right|} {\nu_{\alpha}\nu_{\beta}}  \ \qquad (21 b)$$