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. In Part 2 we derived the Zero-Field Splitting Hamiltonian analogously. The goal of this post is to derive the last sub-Hamiltonian for the Exchange Interaction and consequently derive the complete Spin Hamiltonian $\mathscr{H}$.
I use a fancy $\mathscr{H}$ to represent the Spin Hamiltonian. Sub-Hamiltonians are normal $H$’s but with hats (e.g. $\hat{H}_Z$).
The Exchange Interaction#
Quantum mechanics says electrons are indistinguishable fermions. Their combined spatial + spin wave function must be antisymmetric. This symmetry links neighboring electrons spins and positions. $\hat{H}_{EX}$ is another term that shifts energy levels.
Consider two electrons in orbitals $\phi_a (\vec{r}), \phi_b(\vec{r})$. The two-electron Hamiltonian is
$$ H = \sum_{i=1}^2 \left[\underbrace{-\frac{\hbar^2}{2m} \nabla_i^2}_{\text{kinetic energy}} + \underbrace{V(\vec{r}_i)}_{\text{potential energy}}\right] + \underbrace{\frac{e^2}{4\pi\epsilon_0} |\vec{r}_1 - \vec{r}_2|}_{\text{Coulomb repulsion}}. $$
Here, $V(\vec{r})$ is the single-particle potential. Because electrons are fermions, their state must be antisymmetric under spatial and spin exchange. We factorize into spatial $\Psi(\vec{r}_1, \vec{r}_2)$ and spin $\chi(s_1, s_2)$ parts:
$$ \Psi_{\text{tot}} (1, 2) = \Psi(\vec{r}_1, \vec{r}_2)\chi(s_1, s_2), $$
where antisymmetry requires
$$ \Psi_{\text{tot}}(2, 1) = -\Psi_{\text{tot}}(1, 2). $$
If spatial $\Psi$ is symmetric, then the spin state $\chi$ must be antisymmetric (a singlet). If $\Psi$ is antisymmetric, $\chi$ must be symmetric (a triplet).
We build orthonormal spatial orbitals $\phi_a, \phi_b$. Let’s choose
$$ \begin{align*} \Psi_S(\vec{r}_1, \vec{r}_2) &= \frac{1}{\sqrt{2}}[\phi_a(1) \phi_b(2) + \phi_b(1)\phi_a(2)] \\ \Psi_A(\vec{r}_1, \vec{r}_2) &= \frac{1}{\sqrt{2}}[\phi_a(1)\phi_b(2) - \phi_b(1)\phi_a(2)]. \end{align*} $$
The corresponding spin states are
$$ \begin{align*} \chi_S &= \frac{1}{\sqrt{2}}(|\uparrow \downarrow\rangle - |\downarrow \uparrow\rangle) \quad \text{(singlet)}, \\ \chi_A &= \frac{1}{\sqrt{2}}(|\uparrow \downarrow \rangle + |\downarrow \uparrow\rangle) \quad \text{(triplet)}. \end{align*} $$
We can diagonalize $H$ into the two-dimensional subspace spanned by $\phi_a\phi_b$. Both electrons share the same one-particle energy,
$$ E_0 = \langle \phi_a | \hat{h} | \phi_a\rangle + \langle \phi_b|\hat{h}|\phi_b\rangle, \quad \hat{h} = -\frac{\hbar^2}{2m}\nabla^2 + V(\vec{r}). $$
Therefore, any energy splitting arises entirely from the Coulomb term.
Energy Splitting#
Let’s define the “direct” and “exchange” integrals:
$$ \begin{align*} K &\equiv \int \int d^3r_1 d^3r_2 \, |\phi_a (1)|^2 \frac{e^2}{4\pi\epsilon_0 r_{12}} |\phi_b(2)|^2, \\ J &\equiv \int \int d^3r_1 d^3r_2 \phi_a^* (1) \phi_b (1) \frac{e^2}{4\pi\epsilon_0r_{12}} \phi_b^* (2) \phi_a (2). \end{align*} $$
By magic,
$$ \begin{align} \langle \Psi_S | H | \Psi_S \rangle = E_0 + K + J, \\ \langle \Psi_A | H | \Psi_A \rangle = E_0 + K - J. \end{align} $$
Thus the singlet sits at energy $E_S = E_0 + K + J$, while each triplet has $E_T = E_0 + K - J$. Their energy difference/splitting is
$$ \begin{align} \Delta E \equiv E_S - E_T = 2J. \end{align} $$
The energy splitting due to the Zero Field $\hat{H}_{ZFS}$ is analogously also $2D$. We now wish to replace this two-level spatial + spin system with a pure spin Hamiltonian that reproduces the same energy shift between singlet and triplet.
Exchange Operator#
Recall that the operator exchanging two spins is
$$ \hat{P}_{12}\chi(1, 2) = \chi(2, 1). $$
Its eigenvalues are +1 on the symmetric (triplet) subspace and -1 on the antisymmetric (singlet) subspace. Define the total spin operator and its square
$$ \begin{align*} \hat{\vec{S}}_{\text{tot}} &= \hat{\vec{S}}_1 + \hat{\vec{S}}_2, \\ \hat{S}_{\text{tot}}^2 &= \hat{S}_1^2 + \hat{S}_2^2 + 2\hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2. \end{align*} $$
for spin-1/2 particles, $\hat{S}_1^2 = \hat{S}_2^2 = \hat{S}_x^2 + \hat{S}_y^2 + \hat{S}_z^2 = \frac{3\hbar^2}{4}\mathbb{I}$. Hence,
$$ \hat{S}_{\text{tot}}^2 = \frac{3\hbar^2}{2}\mathbb{I} + 2\hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2. $$
Now for both triplet ($S_{\text{tot}} = 1$) and singlet ($S_{\text{tot}} = 0$) states,
$$ \begin{align} \hat{S}_{\text{tot}}^2 |10\rangle &= S(S+1)\hbar^2 |10\rangle = 2\hbar^2 |10\rangle \\ \hat{S}_{\text{tot}}^2 |00\rangle &= S(S+1)\hbar^2 |00\rangle = 0 \end{align} $$
But we need an exchange operator that satisfies
$$ \begin{align} \hat{P}_{12}|10\rangle &= +|10\rangle \\ \hat{P}_{12}|00\rangle &= -|00\rangle. \end{align} $$
We define an ansatz $\hat{P}_{12} = a \hat{S}_{\text{tot}}^2 + b \mathbb{I}$. Plugging Equations (4) and (5) into (6) and (7) gives
$$ \begin{align*} \hat{P}_{12} |10\rangle &= a(2) + b |10\rangle = +1 |00\rangle \\ \hat{P}_{12} |00\rangle &= a(0) + b |00\rangle = -1 |00\rangle. \end{align*} $$
Solving for a and b gives $a = 1$ and $b = -1$. So we get the exchange operator:
$$ \hat{P}_{12} = 1 \cdot \hat{S}_{\text{tot}}^2 - 1 \cdot \mathbb{I} = \hat{S}_{\text{tot}}^2 - \mathbb{I}. $$
The final expression for the exchange operator is
$$ \hat{P}_\text{12} = \frac{1}{2} \left(1 + 4 \hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2 \right). $$
Its eigenvalues are correctly -1 on the singlet and +1 on the triplet manifold. Hence, an operator proportional to $\hat{P}_{12}$ will split singlet/triplet.
Proportionality Coefficient#
Define the proportionality coefficient $J_{ex}$ such that
$$ \hat{H}_{EX} = J_{ex} \hat{P}_{12} = \frac{J_{ex}}{2} \left(1 + 4\hat{\vec{S}}_1 \cdot \hat{\vec{S}}_2\right). $$
Acting on the singlet and triplet manifolds:
$$ \begin{align} \hat{H}_{EX} \chi_S &= J_{ex}(-1)\chi_S \\ \hat{H}_{EX} \chi_T &= J_{ex}(+1)\chi_T. \end{align} $$
To reproduce the $\Delta E = 2J$ energy splitting from Equation (2) demands $J_{ex} = -J$. Then
$$ E_{S} = -J_{ex} = +J \qquad E_{T} = +J_{ex} = -J; $$ $$ \Delta E = 2J, $$
as desired. We set $J$ to be the exchange coupling matrix proportional to the degree of overlap between wave functions. The Exchange Hamiltonian can finally be expressed as
$$ \hat{H}_{EX} = \hat{\vec{S}}_a \cdot J \cdot \hat{\vec{S}}_b. $$
Assuming the exchange coupling is isotropic, we get
$$ \boxed{ \hat{H}_{EX} = -J \hat{\vec{S}}_a \cdot \hat{\vec{S}}_b. } $$
This is the final component of the 16 $\times$ 16 Spin Hamiltonian. Together, the Zeeman, Hyperfine, ZFS, and Exchange Hamiltonians can describe the mechanics of spin-dependent recombination.
Basis States#
For a two-electron (carrier + defect) and two-nuclei (silicon and carbon) spin system, there are 16 orthonormal basis states. Our Spin Hamiltonian is a matrix of shape $16 \times 16$.
To make future calculations easier, we’ll define our basis states with the electrons coupled ($|s, m\rangle$) and the nuclei in their Zeeman $\{\uparrow, \downarrow \} \equiv \{ +\frac{1}{2}, -\frac{1}{2} \}$ basis.
Triplet $|1, +1\rangle$ manifold#
$$ \begin{cases} |1,1\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |1,1\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \\ |1,1\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |1,1\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \end{cases} $$
Triplet $|1, 0\rangle$ manifold#
$$ \begin{cases} |1,0\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |1,0\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \\ |1,0\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |1,0\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \end{cases} $$
Singlet $|0, 0\rangle$ manifold#
$$ \begin{cases} |0,0\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |0,0\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \\ |0,0\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |0,0\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \end{cases} $$
Triplet $|1, -1\rangle$ manifold#
$$ \begin{cases} |1,-1\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |1,-1\rangle \otimes |+\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \\ |1,-1\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |+\tfrac{1}{2}\rangle_2 \\ |1,-1\rangle \otimes |-\tfrac{1}{2}\rangle_1 \otimes |-\tfrac{1}{2}\rangle_2 \end{cases} $$
The first $|s, m\rangle$ states represent the two electrons in their coupled basis. The second $|\pm\frac{1}{2}\rangle_1 \otimes |\pm \frac{1}{2}\rangle_2$ states represent the two nuclei in their Zeeman basis.
Now we are ready to programmatically calculate the Spin Hamiltonian $\mathscr{H}$. We’ll evaluate its effect on each basis state above. To do this, we program the Zeeman, Hyperfine, ZFS, and Exchange Hamiltonian’s effect on each basis state, and then sum them together.