Active Reading Solution: The φ Probability Distribution

The following discussion should remind you of the effect of e^-iEt/ℏ on a wavefunction, by the way.

Every eigenstate of the hydrogen atom, once you have specified the r and θ dependence, multiplies the resulting function by e^im_lφ.

Note that |e^im_lφ|=1 for any values of m_l or φ. So if the electron is in an energy eigenstate, the position probability |ψ|² will be multiplied by 1 for all values of φ, having no effect at all. Therefore, |ψ|² will be the same for all values of φ; in other words, the position probability will not be a function of φ, implying rotational symmetry about the z-axis..

On the other hand, in a combination of different eigenstates with different values of m_l, you multiply each eigenstate by a different e^im_lφ. They will constructively interfere at some places and destructively interfere at others, giving different position probabilities at different φ-values.