Solving to get the kinetic energy integral for restricted closed shell hartree fock calculations

Question

The Kinetic energy integral given in the book Modern Quantum Chemistry by Attila Szabo and Neil Ostlund is given by (Appendix A eqn A.10) $$ \left(A\left|-\frac{1}{2} \nabla^{2}\right| B\right)=\int d \mathbf{r}_{1} \tilde{g}_{1 s}\left(\mathbf{r}_{1}-\mathbf{R}_{A}\right)\left(-\frac{1}{2} \nabla_{1}^{2}\right) \tilde{g}_{1 s}\left(\mathbf{r}_{1}-\mathbf{R}_{B}\right) $$

where the g1s functions represent gaussians centered at Ra and Rb. They are unnormalised wave function g1s can be written as $\phi_{1 s}^{\mathrm{GF}}\left(\alpha, \mathbf{r}-\mathbf{R}_{A}\right)=e^{-\alpha\left|\mathbf{r}-\mathbf{R}_{A}\right|^{2}}$ While solving this integral I am getting stuck at $$ \beta\int e^{-\alpha\left|\mathbf{r}-\mathbf{R}_{A}\right|^{2}}[3-2\beta|\mathbf{r}-\mathbf{R}_{B}|^2]e^{-\beta\left|\mathbf{r}-\mathbf{R}_{B}\right|^{2}}dr $$ Now I know that the two gaussians with combine to form $$ K\beta\int e^{-p\left|\mathbf{r}-\mathbf{R}_{p}\right|^{2}}[3-2\beta|\mathbf{r}-\mathbf{R}_{B}|^2]dr $$ Where $p=\alpha+\beta$ and $\mathbf{R}_{P}=\left(\alpha \mathbf{R}_{A}+\beta \mathbf{R}_{B}\right) /(\alpha+\beta)$

To calculate the overlap integral where the del square term wasnt present we could just substitute $|r-Rp|=r_1$ and $dr=dr_1$ and solve the integral. But what substitution do I make here. In the book the final given equation is as follows. (eqn A.11) $$\begin{aligned}\left(A\left|-\frac{1}{2} \nabla^{2}\right| B\right)=& \alpha \beta /(\alpha+\beta)\left[3-2 \alpha \beta /(\alpha+\beta)\left|\mathbf{R}_{A}-\mathbf{R}_{B}\right|^{2}\right][\pi /(\alpha+\beta)]^{3 / 2} \\ & \times \exp \left[-\alpha \beta /(\alpha+\beta)\left|\mathbf{R}_{A}-\mathbf{R}_{B}\right|^{2}\right] \end{aligned}$$

Answer 1

Probably the easiest way to do this integral is to realize that \begin{equation} |{\bf r}-{\bf R}_B|^2 e^{-\beta|\bf{ r}-\bf{R}_B|^2} = -\frac{\partial}{\partial \beta} e^{-\beta|\bf{r}-\bf{R_B}|^2} \,. \end{equation} Then if you do the easier gaussian integral \begin{equation} I(\beta) \equiv \int d{\bf r} e^{-\alpha|\bf{r}-\bf{R_A}|^2} e^{-\beta|\bf{r}-\bf{R_B}|^2} \,, \end{equation} Your integral is \begin{equation} \beta\left [3I(\beta)+2\beta \frac{\partial I(\beta)}{\partial \beta} \right ] \end{equation} which should give the desired result.

Here's some Maple that shows the result

I0 := a*b/(a+b)*(3-2*a*b/(a+b)*R2)*(Pi/(a+b))^(3/2)*exp(-a*b/(a+b)*R2);
I1 :=(Pi/(a+b))^(3/2)*exp(-a*b/(a+b)*R2);
I2 := simplify(b*(3*I1+2*b*diff(I1,b)));
simplify(I2-I0);

The first line is your desired result, the second is the gaussian integral for $I(\beta)$, the third is the derivative expression I gave, and the fourth line simplifies this showing the two expressions are equal. Here is the output

> I0 := a*b/(a+b)*(3-2*a*b/(a+b)*R2)*(Pi/(a+b))^(3/2)*exp(-a*b/(a+b)*R2);
                         /    2 a b R2\ / Pi  \3/2       a b R2
                     a b |3 - --------| |-----|    exp(- ------)
                         \     a + b  / \a + b/          a + b
               I0 := -------------------------------------------
                                        a + b

> I1 :=(Pi/(a+b))^(3/2)*exp(-a*b/(a+b)*R2);
                              / Pi  \(3/2)       a b R2
                        I1 := |-----|      exp(- ------)
                              \a + b/            a + b

> I2 := simplify(b*(3*I1+2*b*diff(I1,b)));
                                          3/2 /  1  \1/2       a b R2
               (2 R2 a b - 3 a - 3 b) b Pi    |-----|    exp(- ------) a
                                              \a + b/          a + b
       I2 := - ---------------------------------------------------------
                                              3
                                       (a + b)
> simplify(I2-I0);
memory used=3.5MB, alloc=32.3MB, time=0.12
                                       0




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