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 $$ H = -\frac{1}{2}\nabla^2-\frac{1}{r} $$


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

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.


1 Answer 1


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 $-1/2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.