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Within the Hartree-Procedure (not Hartree-Fock) one is supposed to calculate integrals, namely

$$V_i(\vec{x}) = \sum_{j \neq i} \int \mathrm{d}^3 x_j \frac{e^2}{4\pi\epsilon_0 | \vec{x} - \vec{x}_j|}|\varphi_j(\vec{x}_j)|^2 \, ,$$

with $\varphi_j$ denoting the guessed wave-functions, and $i = 1, \ldots N$.

I guess I am just not seeing my error in thinking, but aren't integrals like that divergent? (The singularity at $\vec{x} = \vec{x}_j$ should only get fixed by the functional determinant in spherical coordinates, when $\vec{x} = 0$?)

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  • $\begingroup$ What does $i$ refer to? $\endgroup$ – my2cts Apr 25 at 17:07
  • $\begingroup$ The index $i$ refers to an electron in an atom. $V_i$ is the electrostatic potential that the electron $i$ "sees", coming from the probability distribution of all the other electrons. (I think, since this equation refers to the Hartree-equations (not Hartree-Fock) and we don't care about indistinguishability yet, it makes sense to speak about the $i$-th electron. ... This is all confusing. $\endgroup$ – Antihero Apr 25 at 17:34
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It does not matter for the indefinite integral where you put the origin of the coordinates. Therefore you can put it at $\vec x$ every time. In this way you dan see that the integral converges for finite x.

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  • $\begingroup$ But isn't this contradicting the fact, that the integral is divergent, if we choose an other origin? Shouldn't the value be independent of the choice of origin? $\endgroup$ – Antihero Apr 25 at 17:37
  • $\begingroup$ It's not. I am Sorry. Now I see my error in thinking: The integral $\int_{[-1,1]} 1/x \mathrm{d}x$ is divergent due to the singularity at $x = 0$. However, the integral $\int_{[-1,1]^3} 1/|\vec{x}| \mathrm{d}^3x$ is convergent, which can be seen after using spherical coordinates. The integrals above (in the question) are also convergent, which can be seen by using the multi-dimensional substitution-rule. Thanks again for your answer! $\endgroup$ – Antihero May 9 at 3:34

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