In my teacher's notes there is a discussion of the Hamiltonian for a central force field with potential $V(r)$.

The Hamiltonian is formulated in spherical polar coordinates: $$H=\frac{p_r^2}{2m}+\frac{p_\phi^2}{2mr^2\sin^2\theta}+\frac{p_\theta^2}{2mr^2}+V(r)$$

Then the conservation of $L_z$ is trivial, because $\phi$ is cyclic. However it is asserted, as if it were evident, that $L^2$ is also conserved. One consequence is that the Hamiltonian can be written in the form $H=\frac{p_r^2}{2m}+\frac{L^2}{2mr^2}+V(r)$.

Now, I know that for any radial force it is trivial to prove that $\vec L$ is conserved, but how can you prove that $L^2$ is conserved just from looking at the previous Hamiltonian? Moreover, does it suffices to prove that $L^2$ and $L_z$ are conserved in order to find that all the three components of $\vec L$ are conserved?

I'm a bit confused, so any help is appreciated


1 Answer 1


By comparing the two expressions for $H$ you can infer an expression for $L^2$, which you need to prove, by transforming the standard expression for $L^2$ to spherical polar coordinates.

Conservation of $L^2$ and $L_z$ is enough to infer the spectrum, but one cannot deduce from it conservation of $L$.

  • $\begingroup$ OK, I have found the expression of $L^2$, even if the derivation is a bit messy $L^2 = m^2 r^4 (\dot \theta^2 + \dot \phi^2 \sin^2\theta ) = p_\theta^2 +p_\phi^2 / \sin^2\theta$. But why is $L^2$ conserved? And what's "the spectrum"? $\endgroup$
    – Ralph
    Jul 11, 2012 at 18:07
  • 1
    $\begingroup$ To see conservation, work out $\frac{d}{dt} L^2$. This is easiest if you know Poisson brackets. -- The spectrum - thats the set of labels you put on the spherical function in the separation of variables. As I don't know the context of your question, it is not easy to know what needs explanation. $\endgroup$ Jul 11, 2012 at 19:13
  • $\begingroup$ Thanks, so the idea was to find that $[L^2,H]=0$. I am just an undergraduate studying classical mechanics for the first time, so there are many techniques I am unfamiliar with $\endgroup$
    – Ralph
    Jul 11, 2012 at 21:08

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