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  Density of states for the diffusion

+ 4 like - 0 dislike

For the wave equation, the propagator in the Fourier domain is written as $$G(\mathbf{k},\omega)=-\frac{1}{\frac{\omega^2}{c^2}-\mathbf{k}^2+\mathrm i\epsilon}.$$ When $\omega/c$ is close to $\|\mathbf{k}\|=k$, we have the following approximation (Stokholski-Plemelj theorem) $$\frac{1}{\pi}\operatorname{Im} G(\mathbf{k},\omega)=-\delta\left(\frac{\omega^2}{c^2}-\mathbf{k}^2\right)=-\frac{c^2}{2\omega}\delta\left(\omega-ck\right),$$ such that, if we integrate over $\mathbf{k}$ we get the famous expression of the density of states $$\rho(\omega)=-\frac{2\omega}{\pi c^2}\operatorname{Im}G(\mathbf{r}=0,\omega).\tag{1}$$

Can one do the same computation with the diffusive propagator (diffuson) $$G(\mathbf{k},\omega)=\frac{1}{\frac{\mathrm i\omega}{D}-\mathbf{k}^2} ?$$

If $\omega$ is close to zero we have $$\frac{1}{\pi}\operatorname{Im}G(\mathbf{k},\omega=0)=\delta(\mathbf k^2)$$ or taking the real part $$\frac{1}{\pi}\operatorname{Re}G(\mathbf{k}=0,\omega)=D\delta(\omega),$$ but I can't really see how one could obtain a relation such as (1).


Considering that $$\operatorname{Im}G(\mathbf{r}=0,\omega)=-\frac{1}{4\pi}\sqrt{\frac{\omega}{2D}}$$ in three dimensions, it is tempting to use a similar relation to get


But is this really justified ?

This post imported from StackExchange Physics at 2015-06-17 16:39 (UTC), posted by SE-user Tom-Tom

asked Jun 17, 2015 in Theoretical Physics by Tom-Tom (50 points) [ revision history ]
edited Jun 18, 2015 by Tom-Tom

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