Perturbative solution to an Eigenvalue Problem with a continuous spectrum 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.
 A: 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
$$
 \sigma(H) = (-\infty,\infty) \neq \sigma(H_0) = [0, \infty).
$$
A: 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.
