1
$\begingroup$

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

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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 $L$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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
    Commented 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. $\endgroup$
    – Arnold Neumaier
    Commented 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
    Commented Jul 11, 2012 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.