# 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*^{imlφ}.

Note that *|e*^{imlφ}|=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*^{imlφ}. They will constructively interfere at some places and destructively interfere at others, giving different position probabilities at different φ-values.