Why should the perturbation be small and in what sense?

Question

In time-independent perturbation theory, one writes $$\hat{H}=\hat{H}_0+\lambda \hat{H}^\prime$$ where $\lambda H^\prime$ is a "small" perturbation.

1. Why should the perturbation be small for perturbation theory to work?

2. Both $\hat{H}_0$ and $\hat{H}^\prime$ are operators. Therefore, what does it mean to say the perturbation is "small"? I think, saying $\lambda \hat{H}^\prime\ll \hat{H}_0$ is meaningless.

3. Is it that the matrix elements of $\lambda\hat{H}^\prime$ much smaller than that of $\hat{H}_0$ in the eigenbasis of $\hat{H}_0$? If yes, why is such a mathematical requirement necessary? In other words, what if the matrix elements of $\lambda\hat{H}^\prime$ are comparable to that of $\hat{H}_0$?

Answer 1

Answer to (2); saying that the perturbation is "small"...all that means is that $\lambda << 1$.

As a silly example: if $\hat{H}_{0} = \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$ and $\hat{H}^{\prime} = \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$, and if you've got a tiny $\lambda = 0.000001$, then the system you're looking at: $$ \hat{H} = \left[ \begin{matrix} 1 & 0.000001 \\ 0.000001 & 1 \end{matrix} \right] $$

Can you see how the overall system is "almost" the same as $\hat{H}_{0}$? This is the sense in which the perturbation is small. You can then safely do expansions in terms of the parameter $\lambda$.

(As somebody else mentioned, you can often set $\lambda = 1$ after you've down all the calculations.)

Answer 2

When one assumes the solution to the perturbed system is of the form

$$|\psi\rangle=\sum_{n=0}^\infty \lambda^n|\psi_n\rangle$$

where $|\psi_0\rangle$ is an eigenstate of $H_0$, one hopes that the expression is meaningful and that only the first few terms are significant which is to say that $|\psi\rangle$ is in some sense close to $|\psi_0\rangle$, i.e., that the unperturbed Hamiltonian is in some sense dominant.

Answer 3

Actually, perturbation theory can work very well even if lambda is not small. For example, the harmonic oscillator with linear potential term added yield exact results to first order in perturbation theory.

The proper way to look at it is that you assume the hamiltonian, the eigenfunctions and the energies are analytic functions of $\lambda$ , and can be Taylor expanded. Then you solve for the expansion coefficients. Such expansions can work well even at large $\lambda$ provided that the higher order terms cancel out.

Whether performing such a perturbation expansion is legitimate is mostly a matter of trial and error, since rigorous mathematical estimates are exceedingly hard to obtain.