Suppose we have an electron in a hydrogen atom that satisfies the time-independent Schrodinger equation:(adsbygoogle = window.adsbygoogle || []).push({});

$$-\frac{\hbar ^{2}}{2m}\nabla ^{2}\psi - \frac{e^{2}}{4\pi \epsilon_{0}r}\psi = E\psi$$

How can it be that the Hamiltonian is spherically-symmetric when the energy eigenstate isn't? I was thinking along the lines of rotations with angular momentum operators but I'm not sure I can come up with a nice explanation. Can someone help me see why this is the case?

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# I Symmetry of Hamiltonian and eigenstates

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