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We want to find the energy of a hydrogen atom ($Z=1$) in the ground state $$ \psi_{100} = \frac{1}{\sqrt{\pi}}e^{-r}\ \ \ \ \ \ (\mbox{atomic units}) $$ with Hamiltonian $$ H = -\frac{1}{2}\nabla^2-\frac{1}{r} $$

Then

$$ \begin{align*} \langle \psi_{100}|H|\psi_{100}\rangle &=\int_0^\infty \int_0^{2\pi} \int_0^\pi r^2\sin\theta\frac{1}{\sqrt{\pi}}e^{-r}\left(-\frac{1}{2}\frac{d^2}{dr^2}-\frac{1}{r}\right)\frac{1}{\sqrt{\pi}}e^{-r} d\theta d\phi dr \\ &=\int_0^\pi \sin\theta d\theta \int_0^{2\pi}d\phi \int_0^\infty \frac{r^2e^{-r}}{\sqrt{\pi}}\left(-\frac{1}{2\sqrt{\pi}}e^{-r}-\frac{1}{r\sqrt{\pi}}e^{-r}\right)dr \\ &= 4\pi \cdot \frac{1}{\pi}\int_0^\infty \left(-\frac{r^2}{2}e^{-2r}-re^{-2r}\right)dr \\ &= 4\left(-\frac{3}{8}\right) \\ &= -\frac{3}{2} \end{align*} $$

However, I've read everywhere that $E = -\frac{Z^2}{2n^2}$, and so for a hydrogen atom in the ground state we should have $E=-\frac{1}{2}$. So why am I getting $-\frac{3}{2}$? I've double-checked with Mathematica.

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Your problem lies in assuming that $$ \nabla^2 = \frac{\partial^2}{\partial r^2} + \cdots $$

This is not the case, you need to use $$ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\cdots $$ Then will you obtain the correct answer of $-1/2$.

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