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I want to calculate the first order perturbative corrected ground state energy $E^{(1)} = \langle H'\rangle$ with the perturbing Hamilton $H' = \frac{e^2}{|\vec r_1 - \vec r_2 |}$ using the spatial part of the wave function $\psi_0 = \psi_{100} (r_1) \cdot \psi_{100} (r_2)$ with $\psi_{100} = \sqrt{\frac{4Z^3}{a^3}} \text e^{-\frac{Zr}a} \frac 1{\sqrt{4\pi}}$.

So far I represented it in spherical coordinates $$ E^{(1)} = \int d\vec r_1 \int d\vec r_2 \psi^*_0 \frac{e^2}{|\vec r_1 - \vec r_2 |} \psi_0 \\ = \frac{64 e^2}{\pi^2 a^6} \int^\infty_0 dr_1\int^\frac \pi 2 _{-\frac \pi 2} d\theta_1 \int^{2\pi}_0 d\varphi_1 \int^\infty_0 dr_2\int^\frac \pi 2 _{-\frac \pi 2} d\theta_2 \int^{2\pi}_0 d\varphi_2 ~r_1^2 ~r_2^2 ~\text e^{-\frac 4a (r_1 + r_2)} ~\frac 1{|\vec r_1 - \vec r_2 |} ~\sin \theta_1 \sin \theta_2 $$

but I don't get, how to use the given tip $$ \int^1_{-1} dx \frac 1{\sqrt{a^2 + b^2 - 2 abx}} = \frac{a+b-|a-b|}{ab} ~. $$ I see, that I can get something similar for $$ \frac 1{|\vec r_1 - \vec r_2 |} = \frac 1{\sqrt{ (\vec r_1 - \vec r_2 )^2}} = \frac 1{\sqrt{r_1^2 + r_2^2 - 2 r_1 ~r_2}} ~. $$

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  • $\begingroup$ Your final equality is incorrect. (I didn’t check all of them.) $\endgroup$ – G. Smith Jan 20 at 3:18
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The correct identity is $|\vec{r}_{1} - \vec{r}_{2}|^{2} = r_{1}^{2} + r_{2}^{2} - 2r_{1}r_{2}\cos\theta\,$, where $\theta$ is the relative angle between them. The integral that you need to perform is (only keeping the relevant terms):

$$E^{(1)} = \int d^{3}r_{1} \int d^{3}r_{2} \,\, \frac{e^{-(r_{1}+r_{2})}}{|\vec{r}_{1} - \vec{r}_{2}|}$$

You can now split the integral over $r_{2}$ as

$$E^{(1)} = \int d^{3}r_{1} \int_{0}^{\infty} dr_{2}\,r_{2}^{2}\int_{-1}^{1}d(\cos \theta)\int_{0}^{2\pi} d\phi \, \frac{e^{-(r_{1}+r_{2})}}{\sqrt{r_{1}^{2} + r_{2}^{2} - 2r_{1}r_{2}\cos\theta}} $$

Now, you can use the identity that you have mentioned.

$$E^{(1)} = 2\pi \int d^{3}r_{1} \int_{0}^{\infty} dr_{2}\,r_{2}^{2} e^{-(r_{1}+r_{2})}\frac{r_{1}+r_{2}-|r_{1}-r_{2}|}{r_{1}r_{2}} \\ \\ = 2\pi \int d^{3}r_{1} \int_{0}^{\infty} dr_{2}\,\frac{r_{2}}{r_{1}}(r_{1}+r_{2}-|r_{1}-r_{2}|) \ e^{-(r_{1}+r_{2})}$$

The final integral can be done easily by splitting the $r_{1}$ from $0<r_{1}<r_{2}$ and $r_{2}<r_{1}$.

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