Chem 3240 · Lecture 6.2
\[\hat{H}^0 \mid n^0\rangle = E^0_n \mid n^0\rangle\]
\[E_n = E^0_n + \lambda E^1_n + \lambda^2 E^2_n + \dots\]
\[\mid n\rangle = \mid n^0\rangle + \lambda \mid n^1\rangle + \lambda^2 \mid n^2\rangle + \dots\]
\[\hat{H}^0\mid n^0\rangle = E^0_n\mid n^0 \rangle\]
\[\hat{H}^0\mid n^1\rangle + \hat{H}^1\mid n^0\rangle = E^0_n\mid n^1 \rangle + E^1_n\mid n^0 \rangle\]
\[E_n = E^0_n + H_{nn} + \sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n - E^0_k}\]
\[E_n^1 = \langle n^0 \mid \hat{H}^1\mid n^0 \rangle\]
\[E_n^2 = \sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n - E^0_k}\]
\[E_n^1 = \langle n \mid V_0 \mid n \rangle = V_0 \cdot \frac{2}{L}\int_0^L \sin^2\frac{n\pi x}{L}\,dx = V_0\]
\[E_n^1 = \langle n\mid \gamma x^3 \mid n\rangle = 0\]
Perturbation theory turns a hard problem into a solvable one plus a small correction, expanding the energy as \(E_n = E^0_n + H_{nn} + \sum_{k\neq n}\frac{\mid H_{nk}\mid^2}{E^0_n - E^0_k}\). The first-order shift is a diagonal average; the second-order shift mixes in other states, with nearby levels dominating.