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 1


$SU(2)$. 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).


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