# Dirac Delta function as a Fourier transform

1. Mar 9, 2012

### andrewkirk

It is fairly easy to demonstrate that the Dirac delta function is the Fourier transform of the plane wave function, and hence that:

$\delta(x)=∫_{-∞}^{∞}e^{ikx}dk$ (eg Tannoudji et al 'Quantum Physics' Vol 1 p101 A-39)

Hence it should be the case that $∫_{-∞}^{∞}e^{ik}dk = \delta(1) = 0$

However the integral on the LHS does not even seem to exist, let alone equal zero. We can write:

$∫_{-∞}^{∞}e^{ik}dk = ∫_{-∞}^{∞}cos(k)+i sin(k) dk$

$= ∫_{-∞}^{∞}cos(k)dk+i ∫_{-∞}^{∞}sin(k)dk$

and neither of these integrals exist.

Yet, based on the Fourier argument, they must both exist and equal zero.

I have probably done something silly to get myself into this contradiction. Can somebody please help me by showing what it is?

2. Mar 9, 2012

### maverick_starstrider

There are a number of things going on here I don't really understand but chronologically why does delta(1)=0? Also there is nothing trivial about a Dirac delta function, it's always been a trouble maker in terms of rigorous mathematics.

3. Mar 9, 2012

### strangerep

It seems you're missing the crucial piece of information that the Dirac delta is not an ordinary function in the usual sense. Rather it's a generalized function known as a "linear functional" or "distribution".

http://en.wikipedia.org/wiki/Distribution_(mathematics)

It only makes mathematical sense when integrated with another function, e.g.,
$$\int_{-\infty}^{\infty}\!\! dx f(x) \delta(x-a) ~=~ f(a)$$

4. Mar 9, 2012

### ParticleGrl

Rigorously, the dirac delta function is actually a distribution, not a function. Even still, you can do a bit of 'physics math' by adding a term like -epsilon*k^2 to the exponential in the integral to help it converge, and then limiting it away after the calculation.

5. Mar 13, 2012

### lugita15

That is indeed the Cauchy principle value of the integral, so you can at least make some sense of what you got.

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