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If you are given a potential, which doesn't change if you change the azimuthal angle, we might call that has spherical symmetry. Which means we must find symmetry of angular momentum $L_Z, L^2$.

My wondering is what about we have different shaped potential for example $-e^2/r+ \lambda (y^2 + x)$, and we want to tell if there are continuous spherical symmetry or not? Or more specially if we had to identify the kinds of symmetry can be found in that potential. e and $\lambda$ are some parameters.

Could you explain what would be the correct strategy to understand existing symmetry and how to deal with mathematical? Any suggested example online or from books are highly appreciated.

More explicitly, the potential can be written in terms of spherical coordinate as:

$$V\left(\mathbf{r}\right)=-\frac{e^2}{r} + \lambda ((r \sin \theta \sin\phi)^2 + r \sin\theta \cos \phi)$$

From the potential, it seems like it is not symmetric with respect to $\theta$ and $\phi$. Does it mean that it's not spatial symmetric? What other continuous can the potential have?

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    $\begingroup$ If the potential doesn’t change with azimuthal angle, it is called cylindrically symmetric. If it doesn’t change with either azimuthal or polar angle, then it is spherically symmetric. $\endgroup$ – G. Smith Jul 22 '19 at 2:24
  • $\begingroup$ Thanks for the comment. I understand it better now. I also have updated the question. Could you check again please? $\endgroup$ – user193422 Jul 22 '19 at 2:34
  • $\begingroup$ @G.Smith Is there a distinction between axial and cylindrical symmetry? I tend to use 'axial symmetry' or 'axisymmetric' for rotational symmetry about an axis, and 'cylindrical symmetry' for 'axial symmetry plus translational symmetry along this axis' but I don't know if this is widespread usage. $\endgroup$ – Andrew Steane Aug 7 '19 at 11:52
  • $\begingroup$ @AndrewSteane I think your usage is better. I should have called it axially symmetric. $\endgroup$ – G. Smith Aug 7 '19 at 15:38
  • $\begingroup$ Also spatial symmetry and spherical symmetry are different. $\endgroup$ – Qmechanic Aug 13 '19 at 19:16
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A necessary and sufficient condition is that $$ [H,L_i] = 0 $$ for all $i$, i.e. that the hamiltonian (and, specifically, the potential) commute with each of the components of the angular momentum.

Everything else is just frills.

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  • $\begingroup$ There are definitely refinements. There could be an $SO(2)$ symmetry around some non-trivial axis, in which case none of the $L_i$ commute with $H$ but a linear combination of the $L_i$ does commute with $H$. $\endgroup$ – Hans Moleman Jul 29 '19 at 18:40

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