Constructing molecular integrals with derivative on the r_inv operator by partial integration

In quantum chemstry integrals of the following form are sometimes needed:

\[F_x = \left<\phi_i\left|\nabla^x r^{-1}\right|\phi_j\right>\]

But these integrals are often not directly available.

What instead is available is integrals of the form:

\[f_x = \left<\nabla^x\phi_i\left|r^{-1}\right|\phi_j\right>\] \[f_{x,y} = \left<\nabla^x\phi_i\left|r^{-1}\right|\nabla^y\phi_j\right>\]

Let us therefore try to formulate \(F_x\) in terms of \(f_x\) and \(f_{x,y}\). Now switching the integral notation to the more “regular” notation:

\[F_x = \int_\Omega \phi_i \phi_j \nabla^x r^{-1} \mathrm{d}r\]

Doing integration by parts:

\[F_x = \left[ \phi_i \phi_j \nabla^{x-1} r^{-1} \right]_{\partial\Omega} - \int_\Omega \nabla \left(\phi_i \phi_j \right) \nabla^{x-1} r^{-1} \mathrm{d}r\]

The wave function goes to zero at infinity, therefore the first term is zero:

\[F_x = - \int_\Omega \nabla \phi_i \phi_j \nabla^{x-1} r^{-1} \mathrm{d}r - \int_\Omega \phi_i \nabla \phi_j \nabla^{x-1} r^{-1} \mathrm{d}r\]

Doing integration by parts again:

\[F_x = -\left(\left[ \nabla \phi_i \phi_j\nabla^{x-2} r^{-1} \right]_{\partial\Omega} - \int_\Omega \nabla \left(\nabla \phi_i \phi_j\right) \nabla^{x-2} r^{-1} \mathrm{d}r \right) -\left(\left[ \phi_i \nabla \phi_j\nabla^{x-2} r^{-1} \right]_{\partial\Omega} - \int_\Omega \nabla \left( \phi_i \nabla \phi_j \right) \nabla^{x-2} r^{-1} \mathrm{d}r \right)\]

The wave function goes to zero at infinity, therefore the first term is zero:

\[F_x = -\left(-\int_\Omega \nabla^2 \phi_i \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r - \int_\Omega \nabla \phi_i \nabla \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r\right) -\left(-\int_\Omega \nabla \phi_i \nabla \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r - \int_\Omega \phi_i \nabla^2 \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r \right)\]

Simplifying:

\[F_x = \int_\Omega \nabla^2 \phi_i \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r + \int_\Omega \phi_i \nabla^2 \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r + 2\int_\Omega \nabla \phi_i \nabla \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r\]

We therefore now have:

\[F_1 = \left<\phi_i\left|\nabla r^{-1}\right|\phi_j\right> = - \int_\Omega \nabla \phi_i \phi_j r^{-1} \mathrm{d}r - \int_\Omega \phi_i \nabla \phi_j r^{-1} \mathrm{d}r\]

and,

\[F_2 = \left<\phi_i\left|\nabla^2 r^{-1}\right|\phi_j\right> = \int_\Omega \nabla^2 \phi_i \phi_j r^{-1} \mathrm{d}r + \int_\Omega \phi_i \nabla^2 \phi_j r^{-1} \mathrm{d}r + 2\int_\Omega \nabla \phi_i \nabla \phi_j r^{-1} \mathrm{d}r\]

I.e. the integrals have been constructed without taking the derivative of \(r^{-1}\). If hihger derivatives are needed, simply just do partial integration again.

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