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

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?