I am trying to derive the form of the product of two Gaussian functions of the form $\chi = e^{-\alpha|r - A|^2}$, where $\alpha$ is a positive constant and $A$ is a 3D vector. These are used in computational chemistry to represent the orbitals, so I know what the result should be, but I don't quite get that result and can't find any detailed proof of this.
Let $\chi_P$ be the product of two Gaussian functions of that form: $$ \chi_p = e^{-\alpha_a|r - A|^2}e^{-\alpha_b|r - B|^2} = e^{-\alpha_a(r² -2Ar + A²) -\alpha_b(r² -2Br + B²)} $$ Let us consider only the exponent ($\kappa$) for now. $$ \kappa = -((\alpha_a + \alpha_b)r² - 2*(A\alpha_a + B\alpha_b)r + ( \alpha_aA² + \alpha_bB²)) $$ By completing the square, one arrives at: $$ \kappa = -\left(r² - 2\frac{(A\alpha_a + B\alpha_b)r}{\alpha_a + \alpha_b} + \frac{(A\alpha_a + B\alpha_b)²}{(\alpha_a + \alpha_b)² } - \frac{(A\alpha_a + B\alpha_b)²}{(\alpha_a + \alpha_b)² } + \frac{\alpha_aA² +\alpha_bB²}{\alpha_a + \alpha_b}\right) = -\left(\left(r - \frac{A\alpha_a + B\alpha_b}{\alpha_a + \alpha_b}\right)^2 - \frac{(A\alpha_a + B\alpha_b)²}{(\alpha_a + \alpha_b)² } + \frac{\alpha_aA² +\alpha_bB²}{\alpha_a + \alpha_b}\right) = -\left(\left(r - \frac{A\alpha_a + B\alpha_b}{\alpha_a + \alpha_b}\right)^2 + \frac{(\alpha_a\alpha_b)*(A-B)^2}{(\alpha_a + \alpha_b)^2}\right) $$ Accordingly, $\chi_P$ is: $$ e^{-\left(r - \frac{A\alpha_a + B\alpha_b}{\alpha_a + \alpha_b}\right)^2}*e^{-\frac{(\alpha_a\alpha_b)*(A-B)^2}{(\alpha_a + \alpha_b)^2}} $$
But actually, the result should be: $$ e^{-(\alpha_a + \alpha_b)\left(r - \frac{A\alpha_a + B\alpha_b}{\alpha_a + \alpha_b}\right)^2}*e^{-\frac{(\alpha_a\alpha_b)*(A-B)^2}{\alpha_a + \alpha_b}} $$
I don't quite see how to get this form or what mistake I made.
My mistake was on the third line.
$$ \kappa = -\left(r^2 - 2\frac{(A\alpha_a + B\alpha_b)r}{\alpha_a + \alpha_b} + \frac{(A\alpha_a + B\alpha_b)^2}{(\alpha_a + \alpha_b)^2 }\right) $$ Should instead be $$ \kappa = -(\alpha_a + \alpha_b)\left(r^2 - 2\frac{(A\alpha_a + B\alpha_b)r}{\alpha_a + \alpha_b} + \frac{(A\alpha_a + B\alpha_b)^2}{(\alpha_a + \alpha_b)^2 }\right) $$
Following the same steps, we arrive at: $$ \kappa = -(\alpha_a + \alpha_b)\left(\left(r - \frac{A\alpha_a + B\alpha_b}{\alpha_a + \alpha_b}\right)^2 + \frac{(\alpha_a\alpha_b)*(A-B)^2}{(\alpha_a + \alpha_b)^2}\right) $$
And thus: $$ \chi_P = e^{-(\alpha_a + \alpha_b)\left(r - \frac{A\alpha_a + B\alpha_b}{\alpha_a + \alpha_b}\right)^2}*e^{-(\alpha_a + \alpha_b)\frac{(\alpha_a\alpha_b)*(A-B)^2}{(\alpha_a + \alpha_b)^2}} = e^{-(\alpha_a + \alpha_b)\left(r - \frac{A\alpha_a + B\alpha_b}{\alpha_a + \alpha_b}\right)^2}*e^{-\frac{(\alpha_a\alpha_b)*(A-B)^2}{(\alpha_a + \alpha_b)}} $$