Hooke's atom has the following wave function:

$$\Psi( \boldsymbol{r_1}, \boldsymbol{r_2}) \propto \left( 1 + \frac{|\boldsymbol{r_1} - \boldsymbol{r_2}|}{2} \right) \exp\left(-\frac{(r_1^2+r_2^2)}{4}\right).$$

Question: How to calculate electronic density $\rho (\boldsymbol{r_1})$ using this wave function?

The answer, which should contain Error function, is on Wikipedia but not method.

I tried $\rho (\boldsymbol{r_1}) = \int |\Psi (( \boldsymbol{r_1}, \boldsymbol{r_2}))|^2 d\boldsymbol{r_2}$ but got wrong answer. In spherical coordinates:

$\rho (\boldsymbol{r_1}) = \int |\Psi (( \boldsymbol{r_1}, \boldsymbol{r_2}))|^2 d\boldsymbol{r_2} = \int_0^{2 \pi} \int_0^{\pi} \int_0^{\infty} |\Psi (( \boldsymbol{r_1}, \boldsymbol{r_2}))|^2 r_2^2 d{r_2} \sin(\theta_2) d\theta_2 d\phi_2$

$\rho (\boldsymbol{r_1}) \propto \int_0^{2 \pi} \int_0^{\pi} \int_0^{\infty} \left( 1 + \tfrac{1}{2} \sqrt{r_1^2 + r_2^2 -2 \boldsymbol{r_1} \cdot \boldsymbol{r_2} } \right)^2 \exp\left(-\frac{(r_1^2+r_2^2)}{2}\right) r_2^2 d{r_2} \sin(\theta_2) d\theta_2 d\phi_2$

where $\boldsymbol{r_1} \cdot \boldsymbol{r_2} = r_1 r_2 \left[ \sin(\theta_1) \sin(\theta_2) \cos(\phi_1) \cos(\phi_2) + \sin(\theta_1) \sin(\theta_2) \sin(\phi_1) \sin(\phi_2) + \cos(\theta_1) \cos(\theta_2) \right]$

Aso according to the article linked on wikipedia answer should contain Error function. However Mathematica does not yield this result upon carrying out above integral. Where is my mistake?

(Note: crosspost from physics.se).


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