Fourier Integral and the Dirac Delta

**topsquark** · November 30th 2006, 08:45 AM

This is going to end up being a simple question with a simple answer, but I just can't tweak the answer out of my Math Methods text.

I have the following statement:
$\dot{\Delta}(\vec{x},0) = i \int \frac{d^3k}{2(2 \pi )^3} \left [ e^{i \vec{k} \cdot \vec{x}} + e^{-i \vec{k} \cdot \vec{x}} \right ]$ ( $\vec{k}$ and $\vec{x}$ are real 3-vectors and $\vec{k} \cdot \vec{x}$ is the usual dot product.)

I'm supposed to get that
$\dot{\Delta}(\vec{x},0) = -i \delta ^3 (\vec{x})$

But looking at the integral I'm thinking it ought to be:
$\dot{\Delta}(\vec{x},0) = \frac{i}{2} \left ( \delta ^3 (\vec{x}) + \delta ^3 (-\vec{x}) \right )$
which would give me 0, which obviously isn't true. So (sigh) what am I doing wrong?

Thanks!
-Dan

**CaptainBlack** · November 30th 2006, 12:12 PM

Originally Posted by topsquark

This is going to end up being a simple question with a simple answer, but I just can't tweak the answer out of my Math Methods text.

I have the following statement:
$\dot{\Delta}(\vec{x},0) = i \int \frac{d^3k}{2(2 \pi )^3} \left [ e^{i \vec{k} \cdot \vec{x}} + e^{-i \vec{k} \cdot \vec{x}} \right ]$ ( $\vec{k}$ and $\vec{x}$ are real 3-vectors and $\vec{k} \cdot \vec{x}$ is the usual dot product.)

I'm supposed to get that
$\dot{\Delta}(\vec{x},0) = -i \delta ^3 (\vec{x})$

But looking at the integral I'm thinking it ought to be:
$\dot{\Delta}(\vec{x},0) = \frac{i}{2} \left ( \delta ^3 (\vec{x}) + \delta ^3 (-\vec{x}) \right )$
which would give me 0, which obviously isn't true. So (sigh) what am I doing wrong?

Thanks!
-Dan

Do you think this might be related to:

$<br /> \delta(x)=\delta(-x)\ ?<br />$

and that you are using the other sense convention for what is the forward FT

RonL

**topsquark** · November 30th 2006, 12:50 PM

Originally Posted by CaptainBlack

Do you think this might be related to:

$<br /> \delta(x)=\delta(-x)\ ?<br />$

and that you are using the other sense convention for what is the forward FT

RonL

Okay, I see what you are saying about the convention. And (duh!) I forgot that $\delta$ is even. I knew it was something simple! Thanks.

-Dan

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