The topic was discussed in a paper “Dynamic nuclear polarization by photoexcited-triplet electron spins in polycrystalline samples” by K. Takeda, K. Takegoshi, and T. Terao. I am going to give some supplementary on how the calculation can be done.

In a spin-1 system, the spin operator $S_x, S_y, S_z$ and the rotation operator are easy to calculate.

The zero-field splitting Hamiltonian in the ordinary basis of $S_z$ is

$\displaystyle H_{zfs} = \begin{pmatrix} D/3 & 0 & E \\ 0 & -2D/3 & 0 \\ E & 0 & D/3 \end{pmatrix}$

The eigen vectors in the paper are

$\displaystyle \left|X\right> = \frac{1}{\sqrt{2}}( \left|-\right> - \left|+\right> )$

$\displaystyle \left|Y\right> = \frac{i}{\sqrt{2}}( \left|-\right> + \left|+\right> )$

$\displaystyle \left|Z\right> = \left|0\right>$

with eigen values $X = D/3 -E$, $Y = D/3 +E$, and $Z = -2D/3$. The convention that the $\left| Y \right>$ has an imaginary number $i$ is not necessary. In the paper, there are typos on the eigen values.

In other way, the zero-field splitting Hamiltonian can be diagonalized into

$\displaystyle H_{zfs} = P\cdot H_{XYZ} \cdot P^{-1}$

$\displaystyle H_{XYZ} = \begin{pmatrix} X & 0 & 0 \\ 0 & Y & 0 \\ 0 & 0 & Z \end{pmatrix}$

$\displaystyle P = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & i & 0 \\ 0 & 0 & \sqrt{2} \\ 1 & i & 0 \end{pmatrix}$

where $P$ is the transformation changing from ordinary basis to eigen basis.

Given that the population of the eigen basis is $w = ( w_X, w_Y, w_Z)$, the density matrix is

$\displaystyle \rho_{XYZ} = \sum_{i = X, Y, Z} \left| i \right>\left< i \right| w_i$

The corresponding population in ordinary basis is the diagonal elements of

$\rho = P \cdot \rho_{XYZ} \cdot P^{-1}$

In the paper, it gives $w = (0.76, 0.16, 0.08)$. In order to get the population when the molecule X-axis is aligned with the Z-axis Lab frame, the $\rho$ has to be rotated by 90 degree, and we have the population $w' = ( 0.12, 0.76, 0.12)$.

The Zeeman Hamiltonian is $H_B = \omega S_Z$, where $\omega = \gamma_e B \hbar$. In the paper, the author fix the crystal and rotating the external magnetic field. To express the $H_B$ into eigen basis. We simply apply a transformation as the density matrix. But first we need to rotate the magnetic field. The rotation is simple, as the direction of the magnetic field is $(\theta, \phi)$. Thus, the Zeeman Hamiltonian is $H_B = \omega \hat{n} \cdot \vec{S}$.

The transformation is

$H_{B_XYZ} = P^{-1} \cdot H_B \cdot P$

again, the total Hamiltonian is difference from the author.

$\displaystyle H = \begin{pmatrix} X & -i \omega cos(\theta) & i \omega cos(\phi) sin(\theta) \\ i \omega cos(\theta) & Y & -i\omega sin(\phi) sin(\theta) \\ -i\omega cos(\phi) sin(\theta) & i \omega sin(\phi) sin(\theta) & Z \end{pmatrix}$

The difference is because the normal vector $\hat{n} = ( sin(\phi) sin(\theta), cos(\phi) sin(\theta), cos(\theta) )$ is used in the paper.

After that, the calculation for the power sample is the same.

Instead of using integration, we can generated the power using isotropic distribution. I generated 10k molecule, the result of 9.65 GHz microwave is

where the y axis is the magnetic field in Tesla. The result is similar to the paper. I think that their calculation somehow changed the eigen value, so that the eigen vectors were matching with the eigen energy. And the notation of the normal vector and the constant factor of the eigen vectors does not matter.