Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am going through a book and having trouble with reproducing some results mentioned. The aim is to solve for $D_{s}$ from equation (1) below

$\int D_{s}(\vec{x}-\vec{a})D_{s}(\vec{y}-\vec{b})Q_{ss}(\vec{x}-\vec{y})d\vec{x}d\vec{y}=\sigma_{\mathrm{L}}^{2}\delta(\vec{a}-\vec{b})\ \ \ \ \ \ \ (1)$

The book says it can be done by writing $D_{s}$ and $Q_{ss}$ in terms of their fourier transforms as shown in equations (2) and (3) below.

$D_{s}(\vec{x}-\vec{a})=\frac{1}{4\pi^{2}}\int\exp(-i\vec{k}\cdot(\vec{x}-\vec{a}))\tilde{D}_{s}(\vec{k})d\vec{k}\ \ \ \ \ \ \ (2)$

$Q_{ss}(\vec{x}-\vec{y})=\frac{1}{4\pi^{2}}\int\exp(-i\vec{k}\cdot(\vec{x}-\vec{y}))\tilde{Q}_{ss}(\vec{k})d\vec{k}\ \ \ \ \ \ \ (3)$

Where $\tilde{D}_{s}$ and $\tilde{Q}_{ss}$ are the fourier transforms respectively.

The above then results in equation (4)

$|\tilde{D}_{s}(\vec{k})|^{2}\tilde{Q}_{ss}=\sigma_{\mathrm{L}}^{2} \ \ \ \ \ \ \ (4)$

I do not understand how to go from (1) from (4). I tried some simple substitutions but am unable to prove it.

share|cite|improve this question
a) You've used the label (4) twice. b) Is there a reason for mentioning $Q_{LL}$? It never occurs anywhere except in (1) and (2), so can't we just replace (1) and (2) by eliminating $Q_{LL}$? – joriki Apr 12 '11 at 22:24
Thank you I have corrected it now. – Infinity Apr 12 '11 at 22:38
up vote 2 down vote accepted

You didn't mention that $D_s$ is real, but I believe the result only follows under that assumption. It's a consequence of the fact that convolutions in direct space correspond to multiplications in Fourier space. The integral over $\vec{y}$ in (1) is a convolution integral and yields $(D_s*Q_{ss})(\vec{x}-\vec{b})$ (the argument of the result is the sum of the arguments in the convolution). In order for the remaining integral over $\vec{x}$ to also be a convolution integral, we now have to replace $D_s(\vec{x}-\vec{a})$ by $D_s(\vec{a}-\vec{x})$, which corresponds to replacing $\tilde{D}_s$ by its complex conjugate iff $D_s$ is real. If it is, the result follows (up to constants, which I didn't check), since the Fourier transform of the delta function is a constant.

By the way, it's not true that $D_s$ can then be calculated by taking the inverse Fourier transform, since you only have its magnitude, not its phase. That is necessarily so, since shifting $D_s$ in direct space leaves the integral in (1) invariant.

P.S.: The above talk about "replacing" $D_s$ and its Fourier transform is a bit informal -- more precisely, let $E_s(\vec{x})=D_s(-\vec{x})$; then the remaining integral over $\vec{x}$ is a convolution of $E_s$ with $D_s*Q_{ss}$, resulting in $E_s*D_s*Q_{ss}$, and the Fourier transform of $E_s$ is the complex conjugate of the Fourier transform of $D_s$ iff $D_s$ is real.

share|cite|improve this answer
Thank you very much, that makes sense I did not pick up that it was a convolution ($D_s$ is indeed real). And yes, I was wrong, inverse fourier transform would not be possible. – Infinity Apr 12 '11 at 22:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.