Gaussian type orbitals and p-type orbital

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?




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


$$(x-x\_A)=((x-x\_P)-(x\_A-x\_P))=(x-x\_P)-x\_{PA}$$


so defining $x\_{PA}$. Similarly


$$(x-x\_B)=((x-x\_P)-(x\_B-x\_P))=(x-x\_P)-x\_{PB}$$


So


$$(x-x\_A)(x-x\_B)=(x-x\_P)^2-(x\_{PA}+x\_{PB})(x-x\_P)+x\_{PA}x\_{PB}$$


Thus


$$(x-x\_A)e^{-(\vec r - \vec R\_A)^2} \* (x-x\_B)e^{-(\vec r - \vec R\_B)^2}=D(x-x\_P)^2 e^{-(\vec r - \vec R\_P)^2}-D(x\_{PA}+x\_{PB})(x-x\_P) e^{-(\vec r - \vec R\_P)^2}+Dx\_{PA}x\_{PB}e^{-(\vec r - \vec R\_P)^2}$$


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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