# Symmetry of the hamiltonian $H = \frac{1}{2m}p^2 + V(r) + a \, \vec{s} \cdot \vec{l}$

Consider the hamiltonian \begin{align} H& = H_0 + a\, \vec{s} \cdot \vec{l} \\& = \frac{1}{2m}p^2+ V(r) + a\, \vec{s} \cdot \vec{l}, \end{align} where $$V(r)$$ denotes an arbitrary central potential.

What is the symmetry of this hamiltonian? $$\rm SU(2)$$ or $$\rm SO(3)$$? Without the spin-orbit coupling part (or even the spin-degree of freedom), the symmetry of $$H_0$$ should be $$\rm SO(3)$$, right?

I suspect that it is $$\rm SU(2)$$, but cannot prove it.

## 1 Answer

$$SU(2)$$. Indeed, the set of the 3 spin-operator components is affected by the same spatial rotation as the ones of the angular momentum operator when $$SU(2)$$ acts and the scalar product between the two (triples of) operators is rotationally invariant. The rest of the total Hamiltonian is rotationally invariant so that it is both $$SU(2)$$ and $$SO(3)$$ invariant.

It is actually disputable if a $$SU(2)$$ symmetry exists at all (referring to physical space transformations). Since symmetries are always projective from the Wigner theorem and one cannot distinguish between the action of $$SO(3)$$ and that of $$SU(2)$$ on pure states (unit vectors up to phases).