Zeeman and Hyperfine

Calendar June 6, 2025
Math
NASA

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.

The Spin Hamiltonian H\mathscr{H} governs the spin-physics of recombination. It’s what we need to program before starting to simulate EDMR.

H \mathscr{H} is a combination of the Zeeman Effect, Hyperfine Interactions, the Zero-Field Splitting Effect, and the Exchange Interaction.

H=H^Z+H^HF+H^ZFS+H^EX \mathscr{H} = \hat{H}_{\mathrm{Z}} + \hat{H}_{\mathrm{HF}} + \hat{H}_{\mathrm{ZFS}} + \hat{H}_{\mathrm{EX}}

Normally H\mathscr{H} also has a “Nuclear Quadrupole Interaction Hamiltonian” as well. But since Silicon Carbide (4H-SiC) has no nuclei with nuclear spin I>1/2I > 1/2, it’s set to 0.

The goal of this post is to provide a concise derivation of the first two Hamiltonian terms. The next two are reviewed in the next post.

I use a fancy H\mathscr{H} to represent the full Spin Hamiltonian. Sub-Hamiltonians are normal HH’s but with hats (e.g. H^Z\hat{H}_Z).

The Zeeman Effect#

H^Z\hat{H}_Z determines how spin-state energies split in the presence of an external B⃗\vec{B} field. It’s the dominant term in the Spin Hamiltonian that drives recombination.

A spinning charged particle is a magnetic dipole. Its magnetic dipole moment, μ⃗\vec{\mu}, is proportional to its spin angular momentum, S⃗\vec{S}

μ⃗=γS⃗. \vec{\mu} = \gamma \vec{S}.

The proportionality constant, γ\gamma is the gyromagnetic ratio. From the Dirac equation it can be shown

μ⃗=γS⃗=−gq2meS⃗=−gμBℏS⃗, \vec{\mu} = \gamma \vec{S} = -g \frac{q}{2m_e} \vec{S} = -\frac{g\mu_B}{\hbar} \vec{S},

where g≈2.0023g \approx 2.0023 is the Landé gg-factor for the free electron and μB≈9.2⋅10−24J/T\mu_B \approx 9.2 \cdot 10^{-24} J/T is the Bohr Magneton.

When a magnetic dipole is placed in a magnetic field B⃗\vec{B}, it experiences a torque, μ⃗×B⃗\vec{\mu} \times \vec{B}, which tends to line it up parallel to the field like a compass. The energy associated with this torque is

H=−μ⃗⋅B⃗=−γB⃗⋅S⃗. \begin{align} H = -\vec{\mu} \cdot \vec{B} = -\gamma \vec{B} \cdot \vec{S}. \end{align}

Therefore, the Zeeman Hamiltonian is

H^Z=−μ⃗⋅B⃗0=gμBℏS⃗⋅B⃗0=gμBℏB0Sz \hat{H}_Z = -\vec{\mu} \cdot \vec{B}_0 = \frac{g\mu_B}{\hbar} \vec{S} \cdot \vec{B}_0 = \frac{g\mu_B}{\hbar}B_0 S_z

where Sz=ℏ2σ^zS_z = \frac{\hbar}{2} \hat{\sigma}_z and σz\sigma_z is the Pauli-zz spin matrix. Applying H^Z\hat{H}_Z on an arbitrary spin state ∣s,ms⟩|s, m_s\rangle gives

H^Z∣s,ms⟩=gμBℏB0⋅Sz∣s,ms⟩=gμBℏB0⋅msℏ∣s,ms⟩=msgμBB0∣s,ms⟩. \begin{align} \hat{H}_Z |s, m_s\rangle &= \frac{g\mu_B}{\hbar}B_0 \cdot S_z |s, m_s\rangle \\ &= \frac{g\mu_B}{\hbar}B_0 \cdot m_s\hbar|s, m_s\rangle \\ &= m_s g \mu_B B_0 |s, m_s\rangle. \end{align}

From Equation (4) above, we can calculate H^Z\hat{H}_Z’s effect on any spin state ∣s,ms⟩|s, m_s\rangle.



Ordinarily we use an anisotropic gg-tensor. Magnetic fields in different directions act differently on ∣s,ms⟩|s, m_s\rangle. So we would replace with a 3×33 \times 3 gg tensor,

H^Z=μBS⃗⋅g⋅B⃗0.\hat{H}_Z = \mu_B \vec{S} \cdot g \cdot \vec{B}_0.

This can be diagonalized along ∣s,ms⟩|s, m_s\rangle’s principle axis to make things easier.

gdiag=(gx000gy000gz). g_{\mathrm{diag}} = \begin{pmatrix} g_x & 0 & 0 \\ 0 & g_y & 0 \\ 0 & 0 & g_z \end{pmatrix}.

For our simulation though, we’ll start with g≈2.0023g \approx 2.0023. Our Zeeman Hamiltonian is given by

H^Z∣s,ms⟩=msgμBB0∣s,ms⟩. \boxed { \hat{H}_Z |s, m_s\rangle = m_s g \mu_B B_0 |s, m_s\rangle. }

Hyperfine Interaction#

The hyperfine interaction comes from the magnetic field generated by the nucleus B⃗(r⃗)\vec{B}(\vec{r}) acting on the magnetic moment of an orbiting electron. We will calculate B⃗(r⃗)\vec{B}(\vec{r}) and use Equation (1) to calculate H^HF\hat{H}_{HF}.

Maxwell’s Equations say

∇⋅B⃗=0,∇×B⃗=μ0J⃗. \nabla \cdot \vec{B} = 0, \quad \nabla \times \vec{B} = \mu_0 \vec{J}.

Vector Potential#

The first equation tells us there’s no magnetic monopoles; the magnetic field B⃗\vec{B} has zero divergence. From vector calculus this means there exists a vector field A⃗\vec{A} such that

B⃗=∇×A⃗. \vec{B} = \nabla \times \vec{A}.

AA isn’t unique, but to make math simple we can impose the Coulomb gauge condition ∇⋅A⃗=0. \nabla \cdot \vec{A} = 0. Substituting yields

∇2A⃗=−μ0J⃗. \begin{align} \nabla^2 \vec{A} = -\mu_0 \vec{J}. \end{align}

Current Density#

J⃗\vec{J} is called the “current density.” We know current is the amount of charge per unit time that travels through a wire (i.e. electron flux). If electrons aren’t trapped in a wire, they flow throughout space. At every point in this space we can assign a small vector that says

“Here’s how much charge is flowing through this point, and in what direction.”

This vector field is the current density J⃗\vec{J}. If we model the localized nuclear magnetic moment μ⃗n\vec{\mu}_n at the origin, it generates an effective current density

J⃗n(r⃗’)=∇’×[μ⃗nδ3(r⃗’)]. \vec{J}_n(\vec{r}’) = \nabla’ \times [\vec{\mu}_n \delta^3 (\vec{r}’)].

The general solution for A⃗\vec{A} in Equation (5) is

A⃗(r⃗)=μ04π∫d3r’ J⃗n(r⃗’)∣r⃗−r⃗’∣=μ04πμ⃗n×∇(1r)=μ04πμ⃗n×r⃗r3. \vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int d^3 r’ \, \frac{\vec{J}_n (\vec{r}’)}{|\vec{r} - \vec{r}’|} = \frac{\mu_0}{4\pi} \vec{\mu}_n \times \nabla \left( \frac{1}{r} \right) = \frac{\mu_0}{4\pi}\frac{\vec{\mu}_n \times \vec{r}}{r^3}.

Then, using the identity

∇×(a⃗×r⃗f(r))=a⃗∇⋅(r⃗f)−(a⃗⋅∇)(r⃗f), \nabla \times (\vec{a} \times \vec{r}f(r)) = \vec{a} \nabla \cdot (\vec{r}f) - (\vec{a} \cdot \nabla)(\vec{r} f),

with a⃗=μ⃗n\vec{a} = \vec{\mu}_n and f(r)=1/r3f(r) = 1/r^3, we get the canonical expression for a magnetic field from a dipole in classical electrodynamics:

B⃗(r⃗)=μ04πr3[3(μ⃗n⋅r^)r^−μ⃗n]+2μ03μ⃗nδ3(r⃗). \begin{align} \vec{B}(\vec{r}) = \frac{\mu_0}{4\pi r^3} \left[ 3(\vec{\mu}_n \cdot \hat{r}) \hat{r} - \vec{\mu}_n \right] + \frac{2\mu_0}{3}\vec{\mu}_n \delta^3 (\vec{r}). \end{align}

Hamiltonian Density#

A magnetic moment μ⃗e\vec{\mu}_e at r⃗\vec{r} has energy (Equation (1))

U(r⃗)=−μ⃗e⋅B⃗(r⃗). U(\vec{r}) = -\vec{\mu}_e \cdot \vec{B}(\vec{r}).

Substituting B⃗(r⃗)\vec{B}(\vec{r}) with the expression in Equation (6), we can separate the energy into two terms:

Udipolar(r⃗)=−μ⃗e⋅[μ04πr3(3(μ⃗n⋅r^)r^−μ⃗n)],Ucontact(r⃗)=−μ⃗e⋅[2μ03μ⃗nδ3(r⃗)]. \begin{align} U_{\text{dipolar}}(\vec{r}) &= -\vec{\mu}_e \cdot \left[ \frac{\mu_0}{4\pi r^3} (3(\vec{\mu}_n \cdot \hat{r}) \hat{r} - \vec{\mu}_n) \right],\\ U_{\text{contact}}(\vec{r}) &= -\vec{\mu}_e \cdot \left[ \frac{2\mu_0}{3} \vec{\mu}_n \delta^3 (\vec{r})\right]. \end{align}

Promoting magnetic moments to their “quantum operators”,

μ⃗e=−geμBS^ℏ,μ⃗n=+gnμNI^ℏ, \begin{align} \vec{\mu}_e &= -g_e \mu_B \frac{\hat{S}}{\hbar}, \\ \vec{\mu}_n &= +g_n \mu_N \frac{\hat{I}}{\hbar}, \end{align}

where the Bohr Magneton μB=eℏ/(2me)\mu_B = e\hbar / (2m_e) and the Nuclear Magneton μN=eℏ/(2mp)\mu_N = e\hbar / (2m_p). Now we can substitute everything into Equation (1) to calculate the Hamiltonian of the electron, in the magnetic field generated from the nuclei’s magnetic dipole moment at r⃗\vec{r}.

H(r⃗)=μ04πgeμBgnμNℏ2(S⃗⋅I⃗−3(S⃗⋅r^)(I⃗⋅r^)r3+8π3δ3(r⃗)S⃗⋅I⃗). H(\vec{r}) = \frac{\mu_0}{4\pi} \frac{g_e \mu_B g_n \mu_N}{\hbar^2} \left( \frac{\vec{S} \cdot \vec{I} - 3(\vec{S} \cdot \hat{r})(\vec{I} \cdot \hat{r})}{r^3} + \frac{8\pi}{3} \delta^3(\vec{r}) \vec{S} \cdot \vec{I} \right).

The first term before the parenthesis is just a constant CC. If an electron occupies a spatial wave function, ψ(r⃗)\psi(\vec{r}), the full Hyperfine Hamiltonian is

H^HF=∫d3r ∣ψ(r⃗)∣2H(r⃗). \hat{H}_{HF} = \int d^3 r \, |\psi(\vec{r})|^2 H(\vec{r}).

Now we can solve. We’ll write the Hamiltonian in index form,

H(r⃗)=C[1r3(δkl−3r^kr^l)SkIl+8π3δ3(r⃗)SkIk], H(\vec{r}) = C \left[ \frac{1}{r^3}(\delta_{kl} - 3\hat{r}_k \hat{r}_l) S_k I_l + \frac{8\pi}{3}\delta^3 (\vec{r}) S_k I_k \right],

where we sum over repeated indices. Now the final expression to solve is

H^HF=C[SkIl∫d3∣ψ(r⃗)∣2δkl−3r^kr^lr3⏟dipolar term+8π3SkIk∫d3r∣ψ(r⃗)∣2δ3(r⃗)⏟contact (isotropic) term]. \hat{H}_{HF} = C \left[ \underbrace{S_k I_l \int d^3 |\psi(\vec{r})|^2 \frac{\delta_{kl} - 3\hat{r}_k\hat{r}_l}{r^3}}_{\text{dipolar term}} + \underbrace{\frac{8\pi}{3}S_k I_k \int d^3 r |\psi(\vec{r})|^2 \delta^3 (\vec{r})}_{\text{contact (isotropic) term}} \right].

The contact term reduces to

8π3SkIk∫d3∣ψ(r⃗)∣2δ3(r⃗)=8π3∣ψ(0)∣2SkIk=AisoδklSkIl, \begin{align} \frac{8\pi}{3} S_k I_k \int d^3 |\psi(\vec{r})|^2 \delta^3(\vec{r}) &= \frac{8\pi}{3} |\psi(0)|^2 S_k I_k \\ &= A_{\text{iso}} \delta_kl S_k I_l, \end{align}

where Aiso=C8π3∣ψ(0)∣2.A_{\text{iso}} = C \frac{8\pi}{3}|\psi(0)|^2. To solve the dipolar term, we define

Akldip=C∫d3r∣ψ(r⃗)∣2δkl−3r^kr^lr3. A_{kl}^{\text{dip}} = C \int d^3 r |\psi(\vec{r})|^2 \frac{\delta_{kl} - 3\hat{r}_k \hat{r}_l}{r^3}.

Then the dipolar term reduces to SkIlAkldip.S_k I_l A_{kl}^{\text{dip}}. We define the AijA_{ij} tensor as

Aij=Aisoδij+Aijdip. A_{ij} = A_{\text{iso}} \delta_{ij} + A_{ij}^{\text{dip}}.

Finally,

H^HF=S^⋅A⋅I^=S^iAijI^j. \boxed{ \hat{H}_{HF} = \hat{S} \cdot A \cdot \hat{I} = \hat{S}_i A_{ij} \hat{I}_j. }

For multiple nuclei, like in Silicon Carbide, we sum over every jj inequivalent nucleus.

H^HF=∑jS⃗^⋅Aj⋅I⃗^j. \hat{H}_{HF} = \sum_j \hat{\vec{S}} \cdot A_j \cdot \hat{\vec{I}}_j.

For a 2-nuclei (Silicon + Carbon) ×\times 2-electron (Defect + Carrier) system, the Hilbert space is 24=162^4 = 16 dimensional. The Hyperfine Hamiltonian is a 16×1616\times 16 matrix.

Making a B0B_0 vs. Energy plot for Hyperfine like with Zeeman above is difficult. However, in a future post, we’ll generate it. I will be writing some “eigen-energy” simulations. These will let us see what the Hyperfine Interaction does to the 16 basis states defining our spin-system at play.