Mathematics Asked on December 6, 2021
In a physical set-up, one can consider an electrostatic problem where the charge density at each point in space is a random variable, and try to find the electric potential or electric field. To be
more concrete, Consider a Poisson equation
$$nabla^2phi (mathbf{r}) = – rho (mathbf{r})$$
with free boundaries where $rho$ is given by an uncorrelated white Gaussian noise, i.e.
$$langle rho(mathbf{r}) rangle = 0, qquadqquadleftlangle rho(mathbf{r}) rho(mathbf{r}’) rightrangle = A delta(mathbf{r}-mathbf{r}’).$$
If I write down solutions in the Fourier space, they look like $mathbf{q}^{-2} rho(mathbf{q})$ and on average the potential is zero. However, The correlations of $phi$ then read
$$langle phi(mathbf{q}) phi(mathbf{q}’) rangle = frac{langle rho(mathbf{q}) rho(mathbf{q}’) rangle}{mathbf{q}^2 {mathbf{q}’}^2} = frac{A , (2pi)^d delta(mathbf{q}+mathbf{q}’)}{mathbf{q}^4}.$$
I think this is not a well-defined expression because of $q^{-4}$ term, and I cannot make sense of it. My guess is that this is because the charge density $rho$ can assume configurations in which it is not localized in space, and attempting to solve the Poisson equation in that scenario leads to this failure (similar to $nabla^2 phi = cnst.$).
On the other hand, such configurations must occur with a vanishing probability (i.e. all space being filled with charges is improbable) and so I would naively expect them not to contribute to the correlations. How can I resolve this apparent contradiction?
If I understand correctly, the steps in the calculations are as follows:
You start with the Poisson equation $$ nabla^2phi (mathbf{r}) = - rho (mathbf{r}) \ $$ Then take the Fourier transform of it using $f(mathbf{q}) = int f(mathbf{r}) e^{-imathbf{q}cdotmathbf{r}} dmathbf{r}$ (just using the same notation for the function and its transform; the argument separates them), getting $$ -mathbf{q}^2phi(mathbf{q}) = -rho(mathbf{q}) \ $$ You solve this equation getting $$ phi(mathbf{q}) = frac{rho(mathbf{q})}{mathbf{q}^2} \ $$
Then you form the correlator $$ left< phi(mathbf{q}) phi(mathbf{q}') right> = left< frac{rho(mathbf{q})}{mathbf{q}^2} frac{rho(mathbf{q}')}{mathbf{q}'^2} right> = frac{left< rho(mathbf{q}) rho(mathbf{q}') right>}{mathbf{q}^2 mathbf{q}'^2} $$ where $$ left< rho(mathbf{q}) rho(mathbf{q}') right> = left< int rho(mathbf{r}) e^{-imathbf{q}cdotmathbf{r}} dmathbf{r} int rho(mathbf{r'}) e^{-imathbf{q}'cdotmathbf{r}'} dmathbf{r}' right> = iint left< rho(mathbf{r}) rho(mathbf{r'}) right> e^{-imathbf{q}cdotmathbf{r}} e^{-imathbf{q}'cdotmathbf{r}'} dmathbf{r} , dmathbf{r}' \ = iint A delta(mathbf{r} - mathbf{r'}) e^{-imathbf{q}cdotmathbf{r}} e^{-imathbf{q}'cdotmathbf{r}'} dmathbf{r} , dmathbf{r}' = int A e^{-i(mathbf{q}+mathbf{q}')cdot mathbf{r}} dmathbf{r} = A (2pi)^d delta(mathbf{q}+mathbf{q}') $$
There is one error above. The solution to the equation $-mathbf{q}^2phi(mathbf{q}) = -rho(mathbf{q})$ actually has a couple of extra terms: $$ phi(mathbf{q}) = frac{rho(mathbf{q})}{mathbf{q}^2} + adelta(mathbf{q}) + mathbf{b}cdot nabladelta(mathbf{q}), \ $$ where $a$ and $mathbf{b}$ are constants (scalar and vector respectively).
I'm not sure that helps, however.
Answered by md2perpe on December 6, 2021
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