Positive and negative powers of small parameter in perturbation problem

116 Views Asked by At

I'd like to perturbatively handle an eigenvalue problem similar to this: $$ \lambda f = (\hat{H} + 1/\epsilon^2 \hat{V} + \epsilon {W}) f, $$ where $f$ is a function, $\lambda$ is an eigenvalue, $\epsilon$ is a small parameter, and the rest are linear (differential) operators. The problem is, that if one writes a series for the eigenvalue and the eigenfunction, $$ f = f_0 + \epsilon f_1 + \epsilon^2 f_2 + ...\\ \lambda = \lambda_0 + \epsilon \lambda_1 + \epsilon^2 \lambda_2 + ..., $$ one will get e.g. $$ \lambda_0 f_0 = \hat{H} f_0 + \hat{V} f_2\\ \lambda_1 f_0 + \lambda_0 f_1 = \hat{H} f_1 + \hat{W} f_0 + \hat{V} f_3\\ ... $$ i.e. the different orders of the series start to mix. Is there a way to develop a systematic perturbation theory for this case?

1

There are 1 best solutions below

3
On BEST ANSWER

Since the coefficient of the operator $\hat{V}$ is $\epsilon^{-2}$, this provides the dominant contribution for small $\epsilon$. Hence, you should scale your eigenvalue with $\epsilon^{-2}$, i.e. assume an expansion $$ \lambda = \frac{\lambda_{-2}}{\epsilon^2} + \frac{\lambda_{-1}}{\epsilon} + \lambda_0 + \epsilon \lambda_1 + \ldots. $$ You can leave the expansion of $f$ as it is, since both sides of your equation scale with $f$.