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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.

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It's not possible to do this. The perturbation you consider is too singular. If you had that $(L^0 + i)^{-1} L^1$ was trace class, you could use wave operators .... but that's about to where what you want is possible. –  Helge Jul 29 '11 at 19:42
    
Until I read you post, I had not heard of the "trace class" before. But, after a google search, if my understanding is correct, unless I can show $\int < |(L^0 + i)^{-1} L^1 | \psi_p^0, \psi^p_0 > dp < \infty$ Then my search for a solution is hopeless. –  psyduck Jul 29 '11 at 20:19
    
sorry. $\psi_p^0$ and $\psi_0^p$ were supposed to read $\phi_p^0$ above. –  psyduck Jul 29 '11 at 20:21
    
When you say that the one attempt "blows up"... in what sense, exactly? One reason for asking is that even when a series expression doesn't literally "converge", it still may have some validity as an asymptotic expansion, in various regimes. Poincare and others already found examples in various scenarios in celestial mechanics, where "approximate" solutions were eventually shown to not be parts of convergent infinite series... but, nevertheless, were perfectly fine asymptotic expansions (which is how they'd been used all along, in reality). Just a thought... –  paul garrett Jul 29 '11 at 20:23
    
By "blow up" I mean that if I try to evaluate an integral of the form $\int \frac{e^{ipx} h(p) }{p^2 - q^2} \delta'(p-q) dp = - \int \delta(p-q) \frac{d}{dp} ( \frac{e^{ipx} h(p) }{p^2 - q^2} ) dp$ the result is infinity. –  psyduck Jul 29 '11 at 20:45

2 Answers 2

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 $$ \sigma(H) = (-\infty,\infty) \neq \sigma(H_0) = [0, \infty). $$

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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.

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