in dynamic nuclear polarization by microwave induced cross-polarization ( polarization transfer from one spin to another, for example, from $^1H$ to $^{13}C$ ), the condition is called Hartmann-Hahn matching.

The Hamiltonian of the 2 kind of spin in lab frame is:

$H_0 = \omega_s S_z + \omega_I I_z + (A+B) S_z I_z$

where A and B are scalar coupling when the 2 spin in contact and the dipolar interaction.

$A = \frac{4\pi}{3} \gamma_s \gamma_I \rho(x)$

$B = \frac{1}{r_{sI}^2} \gamma_s \gamma_I ( 1- 3 cos^2(\theta_{sI}))$

the angle in dipolar interaction is relative to the direction of the external B-field. when the angle is at:

$cos^2(\theta_{sI}) = 1/3$

the dipolar interaction disappear and the spectrum line get thinner, and higher resolution. this is called the Magic Angle.

when a transverse oscillating field applied, and only affect the S spin, then the total Hamiltonian is :

$H= H_0 + \omega_R U^{-1}S_x U$

$U= e^{-\frac{i}{\hbar} \omega t S_z}$

where U is the rotation operator. a standard method is switch to the rotating frame along with the transverse field. the Hamiltonian in the rotating frame is:

$H_R = \omega S_z+U H U^{-1}$

$H_R =(\omega + \omega_s) S_z + \omega_R S_x + \omega_I I_z + (A+B) S_z I_z$

we can see that, during the Spin-Lock, i.e. $\omega + \omega_s = 0$, the longitudinal component of  S spin gone. But in general, it is not the case, thus, we have the rotation axis of S spin is titled. we can simplify the Hamiltonian by transform it in a titled coordinate, by another unitary transform which rotate on Sy axis.

$\tilde{U} = e^{\frac{i}{\hbar} \phi S_y }$

$tan(\phi) = \frac{\omega_R}{\omega+\omega_s}$

in this tilted axis, the rotating axis is on the $\tilde{S_z}$ axis with magnitude:

$\omega_{eff} = \sqrt{ (\omega + \omega_s)^2 + \omega_R^2 }$

the tilted Hamiltonian is :

$\tilde{H} = \omega_{eff} \tilde{S_z} + \omega_I I_z + (A+\tilde{B}) I_z ( \tilde{S_z} cos(\phi) - \tilde{S_x} sin(\phi) )$

For the interaction terms are small, the energy level is just like ordinary 2 spin system. but when

$\omega_{eff} = \omega_I$

which is the Hartmann-Hahn matching, the flip-flop exchange of the spin no need any energy and then the spin transfer. on the other hand, if:

$\omega_{eff} = - \omega_I$

the flip-flip forbidden transition happen.

In the case of electron spin to proton spin, if we apply a ESR freqeuncy, which is GHz order, so it is microwave, the power of the microwave have to be matched to the proton Larmor frequency.

$(\omega_{\mu w} - \gamma_s H )^2 + k P_{\mu w} = \gamma_I^2 H^2$

here i used the microwave wave B-field strength is proportional to the  voltage applied, and power is proportional to the square of voltage.