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.
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?
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.)
(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.
(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! )
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?
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)