Discussion
A hydrogen atom is perturbed by a potential \(H_{1} = bx^{2}\). The first order correction to ground state energy is (\(a_{0}\) is the Bohr radius) :
Answer: A
The first-order correction is \(ba_{0}^{2}\).
First-order perturbation theory gives \(E^{(1)} = \langle 100|bx^{2}|100\rangle = b\langle x^{2}\rangle\). The hydrogen ground state is spherically symmetric, so \(\langle x^{2}\rangle\), \(\langle y^{2}\rangle\) and \(\langle z^{2}\rangle\) are equal, each \(\frac{1}{3}\langle r^{2}\rangle\), and
\[ \begin{aligned} \langle r^{2}\rangle &= \frac{4}{a_{0}^{3}}\int_{0}^{\infty} r^{4}e^{-2r/a_{0}}dr = 3a_{0}^{2} \\ \langle x^{2}\rangle &= a_{0}^{2} \end{aligned} \]
so \(E^{(1)} = ba_{0}^{2}\). Using \(\langle r^{2}\rangle\) directly would give \(3ba_{0}^{2}\), not offered, and the halved or quartered values come from mistakes in the radial integral.
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