# Error in perturbation series question

1. Jul 6, 2013

### AxiomOfChoice

Suppose we know the perturbation series

$$E = E(\epsilon) = E_0 + \epsilon E_1 + \epsilon^2 E_2 + \ldots$$

converges, where $E_0$ is a discrete eigenvalue of $H_0$ and we are considering a Hamiltonian $H = H_0 + \epsilon H_1$. Does this mean that we know

$$E - E_0 = O(\epsilon)$$

as $\epsilon \to 0$ in the precise sense that we know there exists a $\delta > 0$ and a $C > 0$ such that if $|\epsilon| < \delta$, then $|E - E_0| \leq C|\epsilon|$?

2. Jul 6, 2013

### Staff: Mentor

If the perturbation series converges, this C has to exist - that is just mathematics (taylor series).

3. Jul 6, 2013

### The_Duck

If you really have a convergent series, then I think your statement holds just as a general property of power series, and you can pick any $C > |E_1|$. As I understand it, however, perturbation series typically do *not* converge. They are usually asymptotic series, meaning that if you truncate them after order $\epsilon^N$, the error between the truncated series and the actual eigenvalue vanishes like $\epsilon^{N+1}$ as $\epsilon \to 0$ (which statement you can formalize if you like in the same epsilon-delta style).