# Hamiltonian with position spin coupling

I am solving a Hamiltonian including a term $(x\cdot S)^2$. The Hamiltonian is like this form: $$H=L\cdot S+(x\cdot S)^2$$ where $L$ is angular momentum operator and $S$ is spin operator. The eigenvalue for $L^2$ and $S^2$ are $l(l+1)$ and $s(s+1)$.

If the Hamiltonian only has the first term, it is just spin orbital coupling and it is easy to solve. The total $J=L+S$, $L^2$ and $S^2$ are quantum number. However, when we consider the second term position and spin coupling $(x\cdot S)^2$, it becomes much harder. The total $J$ is still a quantum number. We have $[(x\cdot S)^2, J]=0$. However, $[(x\cdot S)^2,L^2]≠0$, $L$ is not a quantum number anymore.

Anybody have ideas on how to solve this Hamiltonian?

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Is $x$ position? If so something seems fishy about the units, unless there are coefficients on each of those terms that you haven't included. –  David Z Nov 23 '11 at 22:06
@David: This is obviously in natural units. The more worrying omission is the kinetic term: there is an x operator without a p-operator in H. I assume that $H= p^2 +V(r)+ L\cdot S + (x\cdot S)^2$, and that the unperturbed problem is just taken for granted. –  Ron Maimon Nov 24 '11 at 6:37
@Ron: It's just as obvious that this is an interaction Hamiltonian as it is that it's in natural (Planck) units ;-) Both of those things should be specified in the question. (I assumed that the kinetic term was implicit without even realizing it, until you posted your comment) –  David Z Nov 24 '11 at 7:11
Thanks for the reply. The only way I thought was to expand the Hamiltonian in $J_z$ component. In this way, the Hamiltonian can be written in a matrix form, and then diagonalized numerically. The calculation will be complicated. I am wondering is there some algebraic method? –  user6333 Nov 24 '11 at 18:02
@David: I didn't mean Planck units, I meant natural units for the atomic problem, where after setting hbar and m_e to 1, you still have one more unit of energy. I assumed that there is a kinetic term too, but is there an external potential? Is this in an atom, or in free space? –  Ron Maimon Nov 24 '11 at 20:03

I would solve it in the matrix way. Using the Pauli matrices multiplied by $\frac{\hbar}{2}$, this is the base:

$$|n;S_z = +\rangle, |n ; S_z = -\rangle$$

note that

$$S\cdot L = S_xL_x +S_yL_y +S_zL_z$$

$$X\cdot S = XS_x$$

$$[L,S] =0$$

you get some matrix in the $S_z$ base ($2\times 2$) and diagonalize it (finding eigenstates and eigenvalue).

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This problem appears interesting for the following reason. Let us write it down in Cartesian coordnates:

$$-\frac{1}{2}\left(\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right)+\frac{1}{2}(x\cdot S)^2\psi+L\cdot S\psi=E\psi$$

where I have introduced a 1/2 factor for later convenience. Now, I concentrate on x and I consider the operator

$$-\frac{1}{2}\frac{\partial^2}{\partial x^2}+\frac{1}{2}(x\cdot S)^2$$

One can introduce creation and annihilation operators in a similar way as for the harmonic oscillator

$$A_S=\frac{1}{\sqrt{2}}\left(\frac{\partial}{\partial x}+xS\right)$$

and the corresponding eigenvectors will be labeled as $|n,S\rangle$. The next step is to write down $L\cdot S=\frac{1}{2}(J^2-L^2-S^2)$ and we can restate this problem in the form

$$\left(A_S^\dagger A_S+\frac{1}{2}\right)\psi-\frac{1}{2}\left(\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right)+\frac{1}{2}(J^2-L^2-S^2)\psi=E\psi$$

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I would recommend to start from inspecting your Hamiltonian symmetry. It is easy to note that it has rotational symmetry around $x$ axis. Thus, the states may be classified in accordance with $J_x$.

Probably, you could start from Hamiltonian without $xS$ term, write the solution in terms of $J$ and (!) $J_x$ and then examine $(xS)^2$ operator in this basis.

Maybe, it is more convenient to write your additional term as $(zS)$ which makes it easy to use standard basis from the textbooks.

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I think that OP by the notation $x\cdot S$ means $\vec{r}\cdot\vec{S}$. –  Qmechanic Nov 25 '11 at 11:31
Probably. At the first glance it seems that standard approach for spherically symmetric potential would work in that case. Dot product of two [pseudo]vectors usually gives operator invariant under rotations. –  Misha Nov 25 '11 at 11:44
Surely the problem is very badly stated and also OP does not appear to have very clear ideas about. –  Jon Nov 25 '11 at 20:44