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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.

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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

1 Answer 1

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.

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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

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