Spectral theorem for discontinuous operators

[jdstokes]

Hi all,

I'm trying to extract a complete set of states, by applying the spectral theorem to the following differential operator:

$L = -\frac{d^2}{dx^2} + \mathrm{rect}(x)$

where rect(x) is the (discontinuous) rectangular function:

http://en.wikipedia.org/wiki/Rectangular_function

I have a feeling that this may not be possible because of the discontinuity in rect(x).

On the one hand, tt should be possible to approximate rect(x) by a sequence of functions for which the spectral theorem applies. But on the other hand, I don't think eigenspectra of this sequence is guaranteed to converge to that of L.

Can anyone more familiar with functional analysis confirm my suspicion?

 

[ice109]

why are you using the "rectangular function" which isn't a real function anyway? just break up the domain of the operator in two 3 parts - where rect(x) is zero, where rect(x) = k, and finally where rect(x) is zero again.

outside the square the two eigenfunctions will be exp(ikx) and exp(-ikx) and inside they will be exp(mx) and exp(-mx). they're not orthogonal though, they're dirac orthogonal.

 

[HallsofIvy]

In what sense is a "rectangular function" not a function? What definition of function are you using? By the most common definition of "function", that certainly is a function.

 

[ice109]

In what sense is a "rectangular function" not a function? What definition of function are you using? By the most common definition of "function", that certainly is a function.

yea i guess you're right, i don't know what i was thinking saying that. something along the lines that the lines up sides made the function non-single valued.

 

[HallsofIvy]

I thought that for a moment but in the picture given the value at the "break point" is specifically the average of the two constant values.

 

[jdstokes]

why are you using the "rectangular function" which isn't a real function anyway? just break up the domain of the operator in two 3 parts - where rect(x) is zero, where rect(x) = k, and finally where rect(x) is zero again.

outside the square the two eigenfunctions will be exp(ikx) and exp(-ikx) and inside they will be exp(mx) and exp(-mx). they're not orthogonal though, they're dirac orthogonal.

The eigenfunctions are defined piecewise consisting of $\exp(\pm ikx)$ and $\exp(\pm \kappa x)$, with the functions and their first derivatives matched at $\pm 1/2$.

Are you absolutely sure that these piecewise-defined eigenfunctions form a complete set (ie Dirac orthonormal)? In order for this to be the case we must have (with appropriate normalisation)

$\left(\int_{-\infty}^{-1/2} + \int_{-1/2}^{1/2} +\int_{1/2}^{\infty}\right)\psi_k \psi_l dx= \delta (k - l)$.

It is not immediately obvious to me that this will work due to the way $\psi_k$ is defined and those funny integration limits. In fact I'm starting to think that the spectral theorem is not applicable here.