# Wave functions for 2D potential with spin interactions

So consider a 2D system with a circular potential and a spin-orbit interaction:

$V(r) = V_0 \theta(r_0 - r) + c r_0 V_0 L_z S_z \delta(r-r_0)$

where $\theta$ is step function.

So the operators $L_z$ and $S_z$ commute with the Hamiltonian are are therefore conserved quantities. For the same reasons we can write the wave function as a product of radial and orbital parts (and spin parts too):

$R(r) e^{i l \theta}$

where $\theta$ here is the polar angle and $l$ is the orbital quantum number. A spinor can be affixed to the wave function but seems unnecessary as no transitions can occur for the spin.

My question regards adding another spin interaction to $V(r)$ of the type $b_z S_z\theta(r_0-r)$ that only acts within the circular potential. Will the form of the wave functions change as a result of this addition?

My thought is that the wave functions remain the same since once again spin should be preserved so the spinors will not see any further structure. The only purpose of this new interaction will be to alter the effective potential of the circular well - the potential will be shifted either up or down depending on the direction of the spin ($m_s$ = up or down).

So is my reasoning correct? I understand that this problem becomes much more difficult in 3D when the full spin-orbit interaction is used since then you will have a lack of commutation.

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I suppose you are speaking about eigenvectors of the hamiltonian (when you speak about "wavefunctions"). Do you have the usual kinetic term $T= \frac{p_r^2}{2m}+ \frac{p_\theta^2}{2mr^2}$, such as the hamiltonian is $H = T +V$ ? – Trimok Oct 14 '13 at 18:47
yes. If it makes more sense, I am thinking of this as a scattering problem. An electron is elastically scattering from the above potential. – BeauGeste Oct 14 '13 at 21:05
Any insight from perturbations theory? – Bzazz Oct 19 '13 at 13:24