Eigenvalues of Hyperfine Hamiltonian

[lelouch_v1]

TL;DR

Couldn't find the 8 eigenvalues of the problem

I was reading a paper on Radical-Pair mechanism (2 atoms with 1 valence electron each) and the author used the hyperfine hamiltonian $$H_{B}=-B(s_{D_z}+s_{A_z})+As_{D_x}I_x+As_{D_y}I_y+as_{D_z}I_z$$ and found the following eigenvalues: a/4 (doubly degenerate) , a/4±B , (-a-2B±2√(A^2+B^2)) , (-a+2B±2√(A^2+B^2))
What i used was the eigenvalue equation $$H{\ket{\psi}}=E{\ket{\psi}}$$ where ψ is a singlet and a triplet state, i.e. $$\ket{S}={\frac{1}{\sqrt{2}}}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow})$$ $$\ket{T_1}=\ket{\uparrow\uparrow}$$ $$\ket{T_{-1}}=\ket{\downarrow\downarrow}$$ $$\ket{T_{0}}=\frac{1}{\sqrt{2}}(\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow})$$ Using these states, i was able to get the first four eigenvalues. Since there are eight eigenvalues, should i right the terms in the hamiltonian as tensor products? Also, since I is the nuclear spin operator, do i take it always as 1/2 when in the z-direction, since i suppose that the nucleus is a hydrogen atom with one proton? Can i write the term $$As_{D_x}I_x+As_{D_y}I_y$$ in terms of the annihilation and creation operators, i.e. $$As_{+}I_{-}+As_{-}I_{+}$$ or is it wrong?
P.S.: In the hamiltonian, the subscript D is used for Donor and the subscript A for Acceptor.

 

[azntoon]

A:Let me try to answer some of your questions.The hyperfine Hamiltonian is usually written in terms of the spin operators, so you don't need to write it in terms of creation and annihilation operators. The $\hat{I}$ operator is usually written in terms of Pauli matrices $\hat{\sigma}$, so if $I=1/2$ then $\hat{I}_z = \sigma_z/2$. The terms $As_{D_x}I_x+As_{D_y}I_y$ are usually called the 'dipolar' term, and are a result of the dipole-dipole interaction between the electron spin and the nuclear spin. This term is often neglected because its contribution is much smaller than that of the Zeeman terms. To get all eight eigenvalues, you need to solve the full matrix equation $\hat{H}\psi = E \psi$. As you mentioned, the states $\psi$ can be chosen as the spin singlet and triplet states.