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When deriving the solutions for the "simple" quantum mechanical hydrogen problem, one normally uses spherical coordinates $(r,\theta,\phi)$, since the problem has rotational symmetry. The solution has the form $$\psi_{lmn}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)$$ where $R(r)$ describes the radial part and $Y_{lm}(\theta,\phi)$ are the spherical harmonics.

When deriving this solution, one normally aligns the problem to the $z$ coordinate and gets a representation for $Y_{lm}(\theta,\phi)$ with the angle $\theta$ measured from the $z$-axis.

One "surprising" observation is, that the spherical harmonics and therefore the solutions of the wavefunction and its absolute squared are not rotational symmetric. I. e. it has these "lumps" of high probability density when forming the absolute squared wave function.

My question is, how does this asymmetry and distinguishing the $z$-axis relate to the "image" of hydrogen? If I actually had a single hydrogen atom and would want o measure the probability density of the electron: To which axis would the wave function be aligned?


marked as duplicate by Qmechanic quantum-mechanics Feb 9 at 20:31

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