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.

  • $\begingroup$ It is a good exercise to do it for arbitrary perturbation potential, starting from the Dirac equation. The resulting form is actually used to model spin-orbit coupling in semiconductors, although replacing $mc^2$ by $E_g$. $\endgroup$
    – Roger V.
    Commented Feb 16, 2023 at 15:07

1 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.

  • $\begingroup$ Thank you for the resources. I will look into those. $\endgroup$ Commented Feb 16, 2023 at 16:13

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