# Selection rules for hydrogen atom transitions

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?

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

$$Y_{l, m}(\theta, \phi) = \Theta(\theta)\textrm{e}^{im\phi}$$
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 $$\phi \rightarrow \phi + \phi_{0}$$?