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The $a,a^\dagger$ act on the Fock space. If you write a random $\delta(\vec x - \vec y)$ is it neither an element of a Fock space nor an operator on it - the equation doesn't really make sense without further context. However, $\delta$ is just a distribution on functions of spacetime and not operator-valued itself, so the meaning of $a(x)\delta(x-y)$ is ...

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I think I have found the answer. Let us consider the case of a uniformly charged sphere of radius $R$ and charge density $\rho$. The field inside this sphere is $E_{in}=\frac{\rho\times r}{3\epsilon_0}$. Here $r$ is the distance from the centre and $r < R$. If we calculate the divergence of $E_{in}$ then $$\nabla.E(r)=\frac{\rho}{\epsilon_0}$$ ...

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$\rho(\vec{s})$ is function of $\vec{s}$ where $\vec{s}$ is position vector of the location of charge density.Whereas $\vec{r}$ is position at which you want to calculate $E(\vec{r})$. So what you are essentially doing is calculating electric field due to some elementary volume $d^3s$ located at the position $\vec{s}$ and finally integrating over all such ...

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Dirac's delta is a function that describes a distribution (of charge, in this instance) which is concentrated at one point: precisely what you need. So essentially, the equations on the given proof outline read in plain english as follows: (1) Coulomb's law of a point charge (2) Coulomb's law integrated for a smoothly distributed charge with density $\rho$ ...

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