This post is math heavy. I’ll try to walk through it all elegantly though. The important equations are boxed if you’d like to skip through. On mobile some equations may go overfull.
In Part 1 we derived the Zeeman Hamiltonian and the Hyperfine Hamiltonian. The goal of this post is to provide a concise derivation of the next Hamiltonian term for Zero-Field Splitting. The next post will do the last Exchange Interaction term and finally give a complete Spin Hamiltonian description.
I use a fancy $\mathscr{H}$ to represent the full Spin Hamiltonian. Sub-Hamiltonians are normal $H$’s but with hats (e.g. $\hat{H}_Z$).
Zero-Field Splitting#
The Zero Field Splitting (ZFS) comes from the magnetic field generated by one electron’s spin acting on another electron’s magnetic moment. Even when no external $B$ field is applied, this dipole-dipole interaction still shifts the spin energy levels. We can treat ZFS directly as an electron-electron analog of the hyperfine interaction.
We start immediately from Equation (6) in Part 1:
$$ \begin{align} \vec{B}(\vec{r}) = \underbrace{\frac{\mu_0}{4\pi r^3} \left[ 3(\vec{\mu}_n \cdot \hat{r}) \hat{r} - \vec{\mu}_n \right]}_{\text{dipolar term}} + \underbrace{\frac{2\mu_0}{3}\vec{\mu}_n \delta^3 (\vec{r})}_{\text{contact (isotropic) term}}. \end{align} $$
The contact term, with the delta function $\delta^3(\vec{r})$ exists for a point-like nucleus. For Hyperfine, the nucleus sits at a point. When we take the curl of $\vec{A}$, we pick up the singular term $\frac{2\mu_0}{3}\vec{\mu}_n \delta^3(\vec{r})$.
However, for electron-electron interactions, electrons don’t coincide exactly. There is no $\delta$-contact term. Only the $1/r^3$ dipolar piece survives from taking the curl. Hence, the magnetic field generated by an electron is
$$ \vec{B}_1(\vec{r}) = \frac{\mu_0}{4\pi r^3} [3(\vec{\mu}_1 \cdot \hat{r}) \hat{r} - \vec{\mu}_1], \quad \hat{r} = \frac{\vec{r}}{r}. $$
$\vec{\mu}_1$ is the magnetic moment of electron 1 and $\vec{B}_1(\vec{r})$ is the magnetic field generated by electron 1 at point $\vec{r}$. From Equation (1) in Part 1, we can calculate the energy experienced by a second electron at $\vec{r}$ from this field.
A second electron has magnetic moment $\vec{\mu}_2$. So we get
$$ \begin{align} U = -\vec{\mu}_2 \cdot \vec{B}_1(\vec{r}) = \frac{\mu_0}{4\pi r^3} [ -\vec{\mu}_1 \cdot \vec{\mu}_2 + 3 (\vec{\mu}_1 \cdot \hat{r})(\vec{\mu}_2 \cdot \hat{r}) ]. \end{align} $$
Now we can promote $U \rightarrow \hat{H}_{ZFS}$. Given
$$ \hat{\vec{\mu}}_i = -g_e \mu_B \frac{\hat{\vec{S}}_i}{\hbar}, \quad \mu_B = \frac{e\hbar}{2m_e}. $$
Therefore, replacing $\vec{\mu}_{1,2}$ with the quantum operators $\hat{\vec{\mu}}_{1,2}$, the pre-factor in Equation (2) becomes
$$ -\frac{\mu_0}{4\pi} \frac{(g_e\mu_B)^2}{\hbar^2}. $$
And substituting yields
$$ \begin{align} \hat{H}_{ZFS} &= -\frac{\mu_0}{4\pi} \frac{(g_e\mu_B)^2}{\hbar^2} \frac{1}{r^3} \left[-\hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2 + 3(\hat{\vec{S}}_1 \cdot \hat{r})(\hat{\vec{S}}_2 \cdot \hat{r})\right] \\ &= \frac{\mu_0}{4\pi} \frac{(g_e \mu_B)^2}{\hbar^2} \frac{1}{r^3} \left[ \hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2 - 3(\hat{\vec{S}}_1 \cdot \hat{r})(\hat{\vec{S}}_2 \cdot \hat{r})\right]. \end{align} $$
Now we simplify into an analogous form of the Hyperfine Hamiltonian. We have the equalities
$$ \begin{align} (\vec{S}_1 \cdot \vec{S}_2) &= \sum_{i \in {x, y, z}} S_{1i} S_{2i} \\ (\vec{S}_1 \cdot \hat{r}) (\vec{S}_2 \cdot \hat{r}) &= \left(\sum_i S_{1i}\hat{r}_i \right) \left( \sum_j S_{2j}\hat{r}_j\right) \\ &= \sum_{i, j} S_{1i} (\hat{r}_i \hat{r}_j) S_{2j} \\ &= \sum_{i, j} S_{1i} \frac{r_i r_j}{r^2} S_{2j}. \end{align} $$
Plugging this into Equation (4) gives
$$ \begin{align} \hat{H}_{ZFS} &= \frac{\mu_0}{4\pi} \frac{(g_e \mu_B)^2}{\hbar^2} \frac{1}{r^3} \left( \sum_i S_{1i}S_{2i} - 3 \sum_{i, j} S_{1i} \frac{r_i r_j}{r^2} S_{2j} \right) \\ &= \frac{\mu_0}{4\pi} \frac{(g_e \mu_B)^2}{\hbar^2} \left( \sum_i \frac{\delta_{i,j}}{r^3} S_{1i}S_{2j} - 3\sum_{i, j} \frac{r_i r_j}{r^5} S_{1i}S_{2j} \right) \\ &= \sum_{i, j} S_{1i} \left[ \frac{\mu_0}{4\pi} \frac{(g_e \mu_B)^2}{\hbar^2} \left( \frac{\delta_{i, j}}{r^3}- \frac{3r_ir_j}{r^5} \right) \right] S_{2j}. \end{align} $$
To simplify, we define the tensor $D_{ij}$
$$ \begin{align} D_{ij}(\vec{r}) = \frac{\mu_0}{4\pi} \frac{(g_e \mu_B)^2}{\hbar^2} \left( \frac{\delta_{i,j}}{r^3} - \frac{3r_i r_j}{r^5} \right). \end{align} $$
Then we get a similar equation to the Hyperfine Hamiltonian:
$$ \begin{align} \hat{H}_{ZFS} = \sum_{i, j \in \{x, y, z\}} S_{1i} \, D_{ij}(\vec{r}) \, S_{2j} = \hat{\vec{S}} \cdot D \cdot \hat{\vec{S}}. \end{align} $$
$D$ and $E$#
However, our work with ZFS isn’t done. Let’s diagonalize $D$ via the defect’s principal axis. We start by simplifying Equation (12). If we take the trace of the expression in the parenthesis, we find
$$ \begin{align*} \sum_i \left( \frac{\delta_{i,i}}{r^3} - \frac{3r_i r_i}{r^5} \right) &= \sum_i \frac{1}{r^3} - 3\sum_i \frac{r_i^2}{r^5} \\ &= \frac{3}{r^3} - \frac{3}{r^2}(r_x^2 + r_y^2 + r_z^2) = 0. \end{align*} $$
So $D_{ij}$ is traceless. If we diagonalize, we’ll get
$$ \begin{align} D_{ij} = \begin{pmatrix} D_x & 0 & 0 \\ 0 & D_y & 0 \\ 0 & 0 & D_z \end{pmatrix}, \end{align} $$
where $D_x + D_y + D_z = 0.$
Now things will get tricky for a bit. Because the Silicon Vacancy $V_{\text{Si}}^- \in C_{3V}$ point group, we have $D_x = D_y$ symmetry. Something about the $C_{3V}$ crystal structure makes the $D$ tensor symmetric along $x \leftrightarrow y$ directions; not sure. Regardless, we can parameterize now to fewer variables:
$$ \begin{cases} 2D_x + D_z = 0 \\ D_x = D_y \end{cases} \quad \Rightarrow \quad D_x = D_y = -\frac{1}{2}D_z. $$
Define scalars $D, E$ such that
$$ \begin{align} D &= \frac{3}{2}D_z \\ E &= \frac{1}{2}(D_x - D_y) \end{align} $$
$D$ measures how much the crystal field stretches or compresses along the $z$ axis. $E$ measures how much the environment deviates from perfect axial symmetry. In practice, $E \simeq 0$.
Now the matrix in Equation (14) only has two unknowns:
$$ D_{\text{diag}} = \begin{pmatrix} D_x & 0 & 0 \\ 0 & D_y & 0 \\ 0 & 0 & D_z \end{pmatrix} = \begin{pmatrix} -\frac{D}{3} + E & 0 & 0 \\ 0 & -\frac{D}{3} - E & 0 \\ 0 & 0 & \frac{2}{3}D \end{pmatrix}. $$
Raising and Lowering#
From Equation (13), $\hat{H}_{ZFS} = \hat{\vec{S}} \cdot D \cdot \hat{\vec{S}}$. We can expand,
$$ \begin{align*} \hat{H}_{ZFS} &= \hat{\vec{S}} \cdot D \cdot \hat{\vec{S}} \\ &= \left(-\frac{D}{3} + E\right) \hat{S}_x^2 - \left( \frac{D}{3} + E \right) \hat{S}_y^2 + \frac{2}{3}D \hat{S}_z^2 \\ &= D \left( \hat{S}_z^2 - \frac{1}{3} \left( \hat{S}_x^2 + \hat{S}_y^2 + \hat{S}_z^2 \right) \right) + E(\hat{S}_x^2 - \hat{S}_y^2), \end{align*} $$
with the $\hat{S}_x, \hat{S}_y$ Cartesian spin operators being related to the spin raising and lowering operators via
$$ \begin{align} \hat{S}_{+} |s, m\rangle &= (\hat{S}_x + i\hat{S}_y ) |s, m\rangle \\ &= \sqrt{(s(s+1) - m(m+1))}|s, m+1\rangle \\ \hat{S}_{-} |s, m\rangle &= (\hat{S}_x - i\hat{S}_y ) |s, m\rangle \\ &= \sqrt{(s(s+1) - m(m-1))}|s, m-1\rangle. \end{align} $$
Finally, putting everything together gives a general expression for $\hat{H}_{ZFS}$ acting on a general $|s, m\rangle$:
$$\boxed{ \begin{align*} \hat{H}_{ZFS}|s, m\rangle &= Dm^2 |s, m\rangle - \frac{D}{3}(s(s+1))|s, m\rangle \\ &+ \frac{E}{2} (s(s+1) - m(m+1))|s, m+2\rangle \\ &+ \frac{E}{2} (s(s+1) - m(m-1))|s, m-2\rangle. \end{align*}} $$
$\hat{H}_{ZFS}$ splits energy levels slightly even without an external magnetic field – there’s no $B$ term anywhere. However, it introduces forbidden $|s, m\pm 2\rangle$ transitions that cause a very faint “half-field” response in EDMR spectra.
To build a perfect simulation with the half-field response, we’ll need to treat ZFS with care.