Relativistic Correction to the Hydrogen Atom and Spherically Symmetric Operators

Question

We know, the lowest order relativistic correction to the Hamiltonian for the Hydrogen atom is $$H^{'}=-\frac{p^{4}}{8m^{3}c^{2}}$$ Where, $$p=-\frac{\hbar}{i}\nabla$$ So, is $p^{2}=-\hbar^{2}\nabla^{2}$ and $p^{4}=\hbar^{4}\nabla^{2}(\nabla^{2})$?

In various sources, for calculating the first order correction to the energy, a fact has been utilized that the perturbation is spherically symmetric.

a) First of all, what should a spherically symmetric operator look like?

b) Secondly, does this mean that the operator $\nabla^{2}(\nabla^{2})$ is spherically symmetric? If so, how to prove that? Or is there any easy way to intuitively understand that? Any help is appreciated.

Answer 1

An operator is called spherically symmetric when it is invariant under rotations. The Laplace operator is invariant under rotations and translation (the Euclidean transformations).

In fact, the algebra of all scalar linear differential operators, with constant coefficients, that commute with all Euclidean transformations, is the polynomial algebra generated by the Laplace operator. 1

For a hint to how to show that indeed the Laplace operator is invariant under rotations see e.g. this math.SE post. For some intuitive hints on why the Laplace operator is spherically symmetric see this & this posts on math.SE and physics.SE.