Simple Integral: Complex exp -> delta function

[Livethefire]

Simple Integral: Complex exp --> delta function

My Professor has written this down but I'm having some trouble of precisely where this is coming from:

[tex] \int\psi^*_{f}(\boldsymbol k')\psi_{f}(\boldsymbol k) d^3\boldsymbol r = (2\pi)^3\delta(\boldsymbol k- \boldsymbol k') [/tex]

where

[tex] \psi_{f} = e^{i\boldsymbol k\centerdot \boldsymbol r} [/tex] (Plane wave)

I understand that in general (due to orthogonality) we have:

[tex] < \psi_{m} \vert \psi_{n} > =\delta_{mn} [/tex]

and also that a symmetrical integral about a complex exponential yields a sinc function. Qualitatively in the limit that the boundaries become infinite the sinc function becomes a delta function.

What is confusing me here is where are the pi's coming from?
Is there any explicit way to calculate this? Maybe using a spherical coordinate volume element or is there an easier way?

N.B. This calculation pertains to the calculation of density of states (note the f's on the wavefunctions). I know pi's crop up here and there in those types of equations but I can't seem to piece this together.

Thanks.

 

[dingo_d]

it comes from the definition of the delta function ;)

There is a lot abut that in Arfken-Weber

 

[Livethefire]

To be clear, is this what you mean?

https://www.amazon.com/dp/0120598256/?tag=pfamazon01-20

I must go check my mathematics text...

Thanks alot! Ill check that out in the library tomorrow

 

[Ben Niehoff]

I would just look at the one-dimensional integral,

[tex]\int e^{i(k-k')r} \; dr[/tex]

The way to attack this is to put in a small piece to make it converge:

[tex]\int e^{i(k-k')r - \varepsilon r^2} \; dr[/tex]

Do that integral, and then take the limit as [itex]\varepsilon \rightarrow 0[/itex].

 

[Livethefire]

The limits on the integral are surely -infinity to infinity for all space, but that poses problems also.

Regardless, indefinite :

[tex] \int e^{i(k-k')r-\epsilon r^2} dr [/tex]
[tex] \lim_{\substack{\epsilon \rightarrow 0}} \frac{e^{i(k-k')r-\epsilon r^2}}{i(k-k')-2\epsilon r} + C[/tex]
[tex]\frac{e^{i(k-k')r}}{i(k-k')} + C[/tex]

Any pushes in the right direction?

 

[Avodyne]

It's a definite integral. Let's set k' to zero for simplicity. You want to compute
[tex]\int_{-\infty}^{+\infty}dx\;e^{ikx-\epsilon x^2}[/tex]
This is a so-called "gaussian integral". Can you evaluate it?

 

[Livethefire]

Wow that was bad maths in #5. I just woke up and saw it was clearly wrong - must have been tired. Is there any "clear" way to explain why that methodology doesn't work other than: differentiating the result is a quotient (ultimately due to the fact the function isn't linear) and thus more terms will appear- therefore it is wrong?

As for the gaussian integral, I know the standard gaussian integral is root pi. That integration is giving me trouble, I know some of that "standard results".

Incidently I came across this concept in Landau and Lifshiz Vol. 3. I understand conceptually where the (2pi)^3 and delta are coming from but I still can't evaluate it for myself.

 

[Livethefire]

One standard result is from:
http://en.wikipedia.org/wiki/Integral_of_a_Gaussian_function"

yielding

[tex] \frac{\sqrt \pi }{\sqrt \epsilon } \exp \left ( \frac{-k^2}{4\epsilon} \right ) [/tex]

 

[henry_m]

These Gaussian integrals are quite a good way to motivate the result, so make sure you can do it to see 'why' it should be true. (There are, incidentally, numerous other ways you could modify the integral to make it converge, all of which should lead to the right sort of thing).

But the only 'proper' way of doing it is to consider what happens when you multiply by an arbitrary (sufficiently 'nice') function f(k) and integrate over all k. Then you exchange the order of integration and should recognise the Fourier transform and its inverse.

 

[vanhees71]

A very nice book on the subject is

M. J. Lighthill, Introduction to Fourier Analysis and Generalized Functions, Cam. Uni. Press 1958