I've got another one for you.
I ran across this theorem in my Field Theory text. The theorem is this:
Given a sufficiently smooth function f(x) we have that:
The term "sufficiently smooth" does not require that f(x) is a function, but as most wavefunctions in Physics are I would be satisfied with a proof of this case. (Note: Obviously we are concerned only with f(x) that have limits that existat .)
The only f(x) I can figure out how to integrate (as a test of the theorem) is a Gaussian. All the other functions I know of with the required properties at infinity is discontinous at at least one point on the real line, so it doesn't qualify.
-Dan
It comes in handy when deriving the interaction matrix (S matrix) using the path integral method. In the long run we are going to take the natural log of a quantity that is the integral of this, so it is convenient to change all factors into exponential terms. That's as much as I can give you without quoting some seriously nasty integrals. (Which I'm not about to do without LaTeX!)
-Dan
Pffl. I'm running into a problem almost immediately.
Although many wavefunctions are parity eigenstates I cannot assume this, so I have no way to simplify the first integral in general. My Laplace transforms abilities are introductory. Is there a formula in terms of for the first integral?
If I DO assume that F(x) is even/odd then I get:
So when I take I get nothing that resembles , it's in terms of the transform not the original function?
-Dan
I was thinking about this approach also. I tend to make things more complicated than they need to be so I thought I'd simply try an integration by parts:
To finish this I need to operate on this, so I get:
But how does one take this limit?
(I'm also having a problem arguing the limit of the integral, but I'll worry about that later.)
-Dan
Actually, no. I subbed into the integral to get it into the form I posted.
So
(I did this to put it into the "correct" form for a Laplace transform.)
As to the "largeness" argument, I am assuming a function that approaches a constant value at either infinity (ie I am assuming exists) and since I'm assuming f(x) is a function I am also assuming no discontinuities. (Perhaps I wasn't clear enough about that in my first post.) I believe (please correct me if I'm wrong) that guarentees the existence of the integral. If not then I am simply forced to add the assumption that the integral does exist and I'll work out later what restrictions that puts on the Physics.
-Dan
Understood. I am going on a two week stint in Myrtle Beach with my parents (they're staying there until April) so I might not be around much myself.
(Chuckles) Don't worry. I suspect the problem will still be around when you get back. (Unless someone has a flash of inspiration anyway. But as I'll be away from my personal "library" it won't likely be me! )
-Dan
I got a tip from another forum on this. I want to bounce it off you guys.
For starters this is NOT rigorous, however it does explain why the assumption of a "sufficiently smooth" function was made as opposed to a more precise statement.
Split the integral into two parts, x < 0 and x > 0. I am going to look at the latter integral.
(*) Now make the substitution :
If the function is "sufficiently smooth" we may interchange the limit and integration operations:
Now, if it exists is just the number and is a constant, so we may take it outside of the integration.
where I have used an obvious notation.
The x < 0 integration may be done in the same way. Thus we have:
One comment. Go back to the (*) line. Being more careful, this integral should be written as:
So when I make the y substitution:
In order to get the expression I wrote I implicitly took . The argument is that , however small, is always a positive number so upon taking the N limit should still go to infinity. This, aside from switching the limit and the integration, is the weakest point in the argument. Any comments?
Thanks!
-Dan
Rough outline:
1. Every function can be written as the sum of an odd function and an even function zero at the origin and a constant.
2. The result is trivially true for a sufficiently well behaved odd function.
3. The result is a result of the fundamental theorem of calculus, and the
Laplace transform of a derivative for an even function (sufficiently well behaved) zero at the origin.
3a. The result is true for a constant (we need to use the Laplace transform of a constant result here)
4. Hence the result is true for a sufficiently well behaved function.
I can amplify the details if you need (though I doubt that I will feel like
going for full rigour)
RonL
No, I don't need "full rigor." (Sounds dangerous anyway! ) I just wanted to check that there were no glaring logical arguments in the proof. For some reason when I integrate over arbitrary functions my intuition and confidence goes right out the window. Thanks for the review!
-Dan