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I'm reading the book Atomic Physics by Christopher J. Foot and I'm stuck at section 2.2.1 (selection rules). To arrive at the selection rules for $\pi$-transition, the integral of the angular part of the hydrogen wavefunction is evaluated:

$I_{ang}^{\pi}=\int_{0}^{2\pi}\int_{0}^{\pi}Y_{l_2,m_2}^*(\theta,\phi)cos(\theta)Y_{l_1,m_1}(\theta,\phi)sin(\theta)d\theta d\phi$

How do we know that the above system has a cylindrical symmetry so that we could rotate the system about the $z$-axis by an angle $\phi_0$ without changing the integral?

$I_{ang}^{\pi}=e^{i(m_1-m_2)\phi_0}I_{ang}^{\pi}$

Similarly for $\sigma$-transition, the integral will be:

$I_{ang}^{\sigma}=\int_{0}^{2\pi}\int_{0}^{\pi}Y_{l_2,m_2}^*(\theta,\phi)sin(\theta)e^{i\phi} Y_{l_1,m_1}(\theta,\phi)sin(\theta)d\theta d\phi$

My goal here is to obtain the selection rules for both transitions by evaluating the integrals. How do we see the symmetry in these two cases?

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  • $\begingroup$ Think about how $Y_{lm}$ depends on $\phi$. $\endgroup$
    – G. Smith
    Nov 16, 2019 at 17:54
  • $\begingroup$ Thank you I got it now ! $\endgroup$
    – user230272
    Nov 17, 2019 at 23:00

1 Answer 1

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Don't forget the explicit form of the eigenfunctions of angular momentum:

$$Y_{l, m}(\theta, \phi) = \Theta(\theta)\textrm{e}^{im\phi}$$

which ensures that the wavefunctions are single valued under a full rotation $\phi \rightarrow \phi + 2\pi$. Here $\Theta(\theta)$ is the non-trivial $\theta$-dependence. Can you see how your second equation arises if you now make a transformation $\phi \rightarrow \phi + \phi_{0}$?

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  • $\begingroup$ Thank you I got it now ! $\endgroup$
    – user230272
    Nov 17, 2019 at 23:01

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