- 12,971

- 4,997

Sure, you can take any two ##\ell=0## wave functions. All belong not to the domain of ##\hat{p}^2## as an essentially self-adjoint operator. I think, Griffiths has it right here.We can not just use a specific eigenfunction ##\psi##, for instance the radial part of (##\ell=0##) in combination with the Self_Adjoint test because we need twoindependentwavefunctions. See post #73

The boundery-term in your calculation in #62 for ##\hat{p}^2## does not cancel in the case of two independent wave-functions and one needs to rely on the ##r^2## factor to make it 0.