The Gaussian orbital for $p_x$ is defined as:

$$\psi_{p_x}=C(x-x_A)e^{-(\vec r - \vec R_A)^2}$$

Where $C$ is a constant.

Now if we multiply two such orbitals with two different centers, then

$$C_1(x-x_A)e^{-(\vec r - \vec R_A)^2} \cdot C_2(x-x_B)e^{-(\vec r - \vec R_B)^2}=C_3(x-x_A)(x-x_B)e^{-(\vec r - \vec R_A)^2} \cdot e^{-(\vec r - \vec R_B)^2}$$

I understand that the product of the exponential parts above can be transferred to a third center. But I am not sure about $(x-x_A)(x-x_B)$ part. This part, as far as I know, can't be transferred to $(x-x_C)=(x-x_A)(x-x_B)$. I don't think such $x_C$ exists unless $x=0$.

So my question is, usage of Gaussian orbitals only helps when the integral has all $s$ type orbitals. Whenever $p$, $d$ type orbitals get involved I can not transfer four center integrals to two center integrals.

Is my understanding above correct?


1 Answer 1


No, your understanding is not correct. Let the third centre at which the gaussian product is centred be $R_P$. Now we can write


so defining $x_{PA}$. Similarly





$$(x-x_A)e^{-(\vec r - \vec R_A)^2} * (x-x_B)e^{-(\vec r - \vec R_B)^2}=D(x-x_P)^2 e^{-(\vec r - \vec R_P)^2}-D(x_{PA}+x_{PB})(x-x_P) e^{-(\vec r - \vec R_P)^2}+Dx_{PA}x_{PB}e^{-(\vec r - \vec R_P)^2}$$

The first term in this is a d type function, the second a p type function, and the third an s type function, all centred at $R_P$. Thus your product gets transformed to a linear combination of functions all centred at the same site, and thus the four centre integral can be reduced to a two centre one.

This trick of recentring all to the same site is common, and can be extended to any angular momentum.


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