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Notice that your first fomula, $\delta(g(x)):=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}$ can be written as $\delta(x-x_0)dx=\delta(y-y_0)dy$ for a diffeomorphism (one-one onto, thus no summation) , and this already shows that the integral you give is independent of parameterization.


How is this last statement true? On the position basis, the (1-D) position operator is multiplication by $x$. A well known property of the Dirac delta distribution is $$f(x)\delta(x) = f(0)\delta(x)$$ since $\delta(x)$ is zero everywhere except $x=0$ were it has unit area. Thus $$x\delta(x) = 0\delta(x)$$ So, operating on $\delta(x)$ with the ...


For that to be true you will have to go beyond the Hilbert space and consider the rigged Hilbert space, that is the Hilbert space plus distributions. Hence this can be made formal if you identify Dirac's bras and kets with the (anti)linear rigged Hilbert space. In more intuitive terms, the delta function $\delta(\mathbf r)$ can be realised as limits of a ...


In your decompositon you have $$ \psi = 1/r^3$$ and $$ \vec{E} = \vec{r} \, .$$ Then, as you wrote, $$ \vec{\nabla} \cdot \left(\vec{r} \frac{1}{r^3}\right) = \vec{\nabla}\psi \cdot \vec{r} + \psi \vec{\nabla}\cdot \vec{r} \, .$$ Insert $$ \vec{\nabla} \psi = -\frac{3}{r^5} \vec{r}$$ and $$\vec{\nabla} \cdot \vec{r} = 3$$ and you are done. Note ...

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