Discussion
A hydrogen atom is in its ground state : \(\psi_{100}(\vec{r}) = \frac{e^{-r/a_{o}}}{\sqrt{\pi a_{o}^{3}}}\). The Coulomb potential \(V(\vec{r}) = -e^{2}/r\) of hydrogen atom is perturbed by adding \(H' = \lambda x^{2}\) where \(\lambda\) is a constant parameter. The first order ground state energy shift is :
Answer: B
The shift is \(\lambda a_{o}^{2}\).
First-order perturbation theory gives the shift as \(\lambda\langle x^{2} \rangle\) in the ground state. The ground state is spherically symmetric, so \(\langle x^{2}\rangle = \langle y^{2}\rangle = \langle z^{2}\rangle\) \(= \langle r^{2}\rangle/3\). For the 1s state:
\[ \begin{aligned} \langle r^{2} \rangle &= \frac{4}{a_{o}^{3}}\int_{0}^{\infty} r^{4}e^{-2r/a_{o}}\,dr \\ &= 3a_{o}^{2} \\ \Delta E &= \lambda\cdot\frac{3a_{o}^{2}}{3} = \lambda a_{o}^{2} \end{aligned} \]
(The integral uses \(\int_{0}^{\infty} r^{4}e^{-br}dr = 24/b^{5}\) with \(b = 2/a_{o}\).)
Forgetting to divide by 3 would give \(3\lambda a_{o}^{2}\), and the options \(2\lambda a_{o}^{2}\), \(\lambda a_{o}^{2}/2\) and \(\sqrt{2}\lambda a_{o}^{2}\) do not follow from \(\langle r^{2} \rangle = 3a_{o}^{2}\).
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