# Spin Transition Energies

http://arxiv.org/ftp/arxiv/papers/1305/1305.2445.pdf

On p. 22, the following Hamiltonian is given:

$$H = \mu_B g \mathbf{B} \cdot \mathbf{S} + D(S_Z^2+\frac{1}{3}S(S+1)) + E(S_X^2 - S_Y^2)$$

When $B = 0$, the Hamiltonian becomes:

$$H = D(S_Z^2+\frac{1}{3}S(S+1)) + E(S_X^2 - S_Y^2)$$.

We are told that neutral divacancies correspond to $S = 1$, and $D$ and $E$ are the axially symmetric and anisotropic components of the crystal field interaction.

They claim the spin transition energies are $D - E$ and $D + E$ when $B = 0$. Can someone explain how you get that?

I solved for the eigenvalues of the above Hamiltonian after plugging in the spin-1 matrices into the above equation, but I did not obtain this result.

Thanks!

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see this recent question : physics.stackexchange.com/questions/68776 – Trimok Jun 26 '13 at 8:02
Hi Trimok, thank you for this reference. This is the same sort of problem that I am trying to solve. If the $\frac{1}{3} S(S+1)$ term is removed, then I obtain energies $D-E$ and $D+E$. Is it possible that this term vanishes? Otherwise, I think the conclusion in the paper may not be quite right. – Bob Riley Jun 26 '13 at 17:58
The possibles eigenvalues of your hamiltonian, once the $\frac{1}{3} S(S+1)$ term is removed, are $0, D - E,D + E$. Now if we speak about transitions energies between the states corresponding to the eingenvalue 0 and the states corresponding to other eigenvalues, the transition energies are $D - E,D + E$. Adding a constant to the hamiltonian will not change this result. – Trimok Jun 27 '13 at 9:07