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) $$\begin{aligned}\left(A\left|-\frac{1}{2} \nabla^{2}\right| B\right)=& \alpha \beta /(\alpha+\beta)\left[3-2 \alpha \beta /(\alpha+\beta)\left|\mathbf{R}_{A}-\mathbf{R}_{B}\right|^{2}\right][\pi /(\alpha+\beta)]^{3 / 2} \\ & \times \exp \left[-\alpha \beta /(\alpha+\beta)\left|\mathbf{R}_{A}-\mathbf{R}_{B}\right|^{2}\right] \end{aligned}$$


1 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)));

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 := - ---------------------------------------------------------
                                       (a + b)
> simplify(I2-I0);
memory used=3.5MB, alloc=32.3MB, time=0.12

  • $\begingroup$ when we solve the integral $I(\beta)$ we get a term $[\pi /(\alpha+\beta)]^{3 / 2} \exp \left[-\alpha \beta /(\alpha+\beta)\left|\mathbf{R}_{A}-\mathbf{R}_{B}\right|^{2}\right]$ when differentiating this term with respect to beta a lot of terms emerge as we use the product rule. This does not give the correct result when the values of $I'(\bets)$ are put back. $\endgroup$ Dec 28, 2021 at 13:45
  • 1
    $\begingroup$ I added a little maple code to show that if you take the derivatives correctly you get your desired result. $\endgroup$
    – user200143
    Dec 28, 2021 at 17:33

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