Perturbative solution to an Eigenvalue Problem with a continuous spectrum

Question

I am trying to find an approximate solution to an eigenvalue equation using techniques from perturbation theory. Roughly speaking, the problem is as follows

$L^s \phi_q^s = \lambda_q^s \phi_q^s$

The operator $L^s$ can be written

$L^s = L^0 + s L^1$

where $s>0$ is small and $L^0$ is self-adjoint acting on some Hilbert space. I am trying to find solutions of the form

$\phi_q^s = \phi_q^0 + s \phi_q^1 + \ldots$

$\lambda_q^s = \lambda_q^0 + s \lambda_q^1 + \ldots$

Inserting the above expansions into the eigenvalue equation, and collecting terms of like orders in $s$ yields

$( L^0 - \lambda_q^0 ) \phi_q^0 = 0$ (1)

$( L^0 - \lambda_q^0 ) \phi_q^1 = (\lambda_q^1 - L^1 ) \phi_q^0 = 0$ (2)

When the Hilbert space is $L^2((a,b))$ ($-\infty < a < b < \infty$) the spectrum of $L^0$ is simple and purely discrete. Solving (1) yields a complete set of orthonormal basis functions with the orthogonality relation $<\phi_n^0,\phi^0_k>=\delta_{n,k}$. By the Fredholm alternative, in order for a solution $\phi_n^1$ of (2) to exist, the RHS of (2) must satisfy

$< \phi_n^0, (\lambda_n^1 - L^1 ) \phi_n^0 > = 0$

Thus $\lambda_n^1$ is given by

$\lambda_n^1 = < \phi_n^0, L^1 \phi_n^0 >$

And $\phi_n^1$ is given by applying the resolvent operator to the RHS of (2), which yields

$\phi_n^1 = \sum_{k \neq n} \frac{< \phi_k^0, L^1 \phi_n^0 >}{\lambda_n - \lambda_k} \phi_k^0$

When the Hilbert space is $L^2(-\infty,\infty)$ the spectrum of $L^0$ is purely absolutely continuous. Solving (1) still yeilds a complete set of orthonormal basis functions with the orthogonality relation $<\phi_p^0,\phi^0_q>=\delta(p-q)$. However, solving (2) is now more complicated. When the operator $L^1$ is such that

$< \phi_p^1 , L^1 \phi_q^0 > = \delta(p-q) f^1(q) + g^1(p,q)$ (3)

Trying a solution analgous to the discrete case works works

$\lambda_q^1 = f^1(q)$

$\psi_q^1 = \int \frac{g^1(p,q)}{\lambda_q - \lambda_p} \phi_p^0 dp$ (the integral converges). (4)

However, I am looking at various $L^1$. And, for certain $L^1$ I do not have (3). Rather, I have

$< \phi_p^1 , L^1 \phi_q^0 > = \delta(p-q) f^1(q) + h(q) \delta'(p-q) + g^1(p,q)$

(yes, that's a derivative of a delta function). Frankly, at this point, I am totally stuck. I've tried a solution of the form (4) with

$g^1(p,q) \to h(q) \delta'(p-q) + g^1(p,q)$.

But that solution blows up. My sense is that I should be looking for some sort of condition on $\lambda_q^1$ which would guarantee that a solution $\phi_q^1$ of (2) exists. But, I know of no such condition.

If it helps, you can think of $L^0$ as $-d^2/dx^2$ so that the eigenfunctions are $\phi_q^0(x)=e^{iqx}/\sqrt{2\pi}$. And $L_1 \phi_q^0 = x \phi_k^0(x)$. If you're wondering, the derivative of the delta function comes about as follows

$< \phi_q^0, x \phi_p^0 > = (1/2 \pi) \int x e^{i(p-q)x} dx = (1/2 \pi i) (d/dp) \int e^{i(p-q)x} dx = (1/2 \pi i)\delta'(p-q)$.

Any guidance on solving this problem would be greatly greatly appreciated.

Answer 1

Operators of the form $$ H = -\frac{d^2}{dx^2} + F x,\quad \text{on } L^2((-\infty,\infty)) $$ are discussed in Cycon--Froese--Kirsch--Simon "Schroedinger Operators". In example 3 at the end of Section 4.1. I know them by the name of Stark Operator. But googling didn't yield any good reference.

You'll have to dig through the references to see how the eigenfunctions of $H$ behave. At least at 0 energy, you can solve the problem explicitly with Airy functions, and you'll see that the behavior is very different than for the free case $H_0 = -\frac{d^2}{dx^2}$. This is to be expected since their spectra satisfy $$ \sigma(H) = (-\infty,\infty) \neq \sigma(H_0) = [0, \infty). $$

Answer 2

I find some statements your post quite curious.

If you look at the Laplace operator on $L^2(-\infty,\infty)$, it has a complete set of generalized eigenvectors, namely $\cos(\sqrt{\lambda}x)$, $\sin(\sqrt{\lambda}x)$, but they are elements of the dual of the Schwartz space and don't fulfill any orthogonality relation, because there is not any inner product (Gelfand triple).

Then, I don't see how a derivative of a delta function could come in, at least I have no clue how your formal calculation below could be made rigorous.