Discussion
A two-level quantum system has energy eigenvalues \(E_{1}\) and \(E_{2}\). A perturbing potential \(H' = \lambda\Delta\sigma_{x}\) is introduced, where \(\Delta\) is a constant having dimensions of energy, \(\lambda\) is a small dimensionless parameter, and \(\sigma_{x} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). The magnitudes of the first and the second order corrections to \(E_{1}\) due to \(H'\), respectively, are
Answer: A
The corrections are 0 and \(\frac{\lambda^{2}\Delta^{2}}{|E_{1} - E_{2}|}\).
In the unperturbed basis \(\sigma_{x}\) has zeros on the diagonal, so the first-order correction \(\langle 1|H'|1\rangle\) vanishes. The second-order correction comes from the off-diagonal element \(\langle 2|H'|1\rangle = \lambda\Delta\):
\[ \begin{aligned} E_{1}^{(2)} &= \frac{|\langle 2|H'|1\rangle|^{2}}{E_{1} - E_{2}} \\ &= \frac{\lambda^{2}\Delta^{2}}{E_{1} - E_{2}} \end{aligned} \]
Its magnitude is \(\frac{\lambda^{2}\Delta^{2}}{|E_{1} - E_{2}|}\). The exact eigenvalues expand to the same result:
\[ \begin{aligned} E_{\pm} &= \frac{E_{1} + E_{2}}{2} \\ &\pm \sqrt{\left(\frac{E_{1} - E_{2}}{2}\right)^{2} + \lambda^{2}\Delta^{2}} \end{aligned} \]
A non-zero first-order term would need diagonal elements, which \(\sigma_{x}\) does not have, and the extra \(\frac{1}{2}\) in the last option is not in the formula.
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