# Power series of quantum mechanical eigenfunctions

In Griffith's Quantum Mechanics book, when he tackles time-independent perturbation theory, he states that one can write the eigenfunctions of the perturbed system, $$H = H^0 + \lambda H',$$ as a power series in $$\lambda$$, $$\psi_n = \psi_n^0 + \lambda\psi_n^1 + \lambda^2\psi_n^2 + \cdots$$ I'm trying to understand this intuitively. I am guessing that this is some way of writing the Taylor expansion of $$\psi_n$$. Then I do indeed understand that summing up all the terms gives us the full eigenfunction to arbitrary precision.

I just can't see how the power series in the book is really a Taylor expansion. Given that this is time-independent, the eigenfunctions should be functions of $$x$$, and their general Taylor expansion should look like $$\psi_n(x) = \psi_n(a) + \psi_n'(a)(x - a) + \frac{\psi_n''(a)}{2!}(x - a)^2 + \cdots.$$ He writes that $$\psi_n^1$$ is the first-order correction to the $$nth$$ eigenfunction, $$\psi_n^2$$ the second-order, and so on. Can this statement be reconciled with writing the corrections as $$\psi_n^m(a) = \frac{1}{m!}\frac{d^m\psi_n(x)}{dx^m}\bigg\rvert_{x=a},$$ such that the Taylor expansion becomes $$\psi_n(x) = \psi_n^0(a) + \psi_n^1(a)(x - a) + \psi_n^2(a)(x - a)^2 + \cdots,$$ or what? I would still need to define something like $$x - a \equiv \lambda$$ to get $$\psi_n(x) = \psi_n^0(a) + \lambda\psi_n^1(a) + \lambda^2\psi_n^2(a) + \cdots,$$ but this forces some dependencies on $$\lambda$$ that isn't physically motivated, and screws up the relations between $$x$$, $$a$$ and $$\lambda$$ in the equation, so it seems like a stupid thing to do.

• The eigenfunctions(eigenvectors are functions of $\lambda$. Commented Jul 28, 2023 at 12:08

The perturbed functions are meant to be a Taylor expansions in $$\lambda$$, not in $$x$$. More completely it should be written as: $$\psi_n(x,\lambda) = \psi_n^0(x) + \lambda\psi_n^1(x) + \lambda^2\psi_n^2(x) + \cdots$$
So the $$\psi_n^m(x)$$ are the Taylor coefficients around $$\lambda=0$$. $$\psi_n^m(x) = \frac{1}{m!}\frac{d^m\psi_n(x,\lambda)}{d\lambda^m}\bigg\rvert_{\lambda=0}$$
• Why more completely? In general, the $\psi$s are just elements of some Hilbert space. You don't have to work on $L^2(\mathbb R)$. For example, one might consider a perturbative expansion on $\mathbb C^2$, when e.g. considering some spin-$1/2$ particle (and neglecting the spatial degrees of freedom). Commented Jul 28, 2023 at 12:42
• @TobiasFünke In general you are correct. But as Griffiths talked about eigenfunctions I assume he treated it in the $x$-representation, i.e. in $L^2(\mathbb R)$. Commented Jul 28, 2023 at 12:46
• The whole point in perturbation theory is that the perturbation is small, i.e. $H' \ll H_0$. Therefore an expansion around $\lambda=0$ is the natural choice. Commented Jul 28, 2023 at 12:51
• @Depenau Yes, sure. But with a small $\lambda$ near zero we have a small perturbation. Commented Jul 28, 2023 at 12:55