Perturbation theory in quantum mechanics

Remarks

The first two corrections are $E_1 = \ip{\psi_0}{V\psi_0}$ and, when $H_0$ has an orthonormal eigenbasis $(\psi_j)$ with eigenvalues $(E^{(j)})$, $E_2 = \sum_{j \ne 0} \frac{\abs{\ip{\psi_j}{V\psi_0}}^2}{E_0 - E^{(j)}}$.

Example: let $H_0 = \frac12\left(-\frac{d^2}{dx^2} + x^2\right)$ on $L^2(\R)$, with ground energy $E_0 = \frac12$, and $V = x$. Then $E_1 = 0$ by symmetry, and only the first excited state contributes to $E_2$, with $\abs{\ip{\psi_1}{x\psi_0}}^2 = \frac12$, so $E_2 = -\frac12$. This agrees with the exact answer: $\frac{x^2}{2} + \lambda x = \frac{(x + \lambda)^2}{2} - \frac{\lambda^2}{2}$, so the ground energy is $\frac12 - \frac{\lambda^2}{2}$.

Non-example of the hypothesis: let $E_0$ be a degenerate eigenvalue. Then $E_0 - E^{(j)}$ vanishes for some $j \ne 0$, and the formulas break down. One must first diagonalise $V$ on the eigenspace, which is degenerate perturbation theory.

When $V$ is bounded, the series converge for small $\abs{\lambda}$ to an eigenvalue and an eigenvector of $H(\lambda)$, by the analytic theory of eigenvalue perturbation. For unbounded perturbations they may diverge: for $V = x^4$ in the example above, the energy series has radius of convergence $0$, although its first terms are still good approximations.