The cartesian form of contracted gaussian is something like $$\phi^{CGF}=\sum_{n} s_nx^iy^jz^k \exp(-\alpha_nr^2)$$ and these is also a pure or spherical harmonic version (which I don't know the details about). Are the coefficients and exponents same between the two, different for $\ge d$ orbitals, or different for all orbitals.


They are the same.

Your equation can be rewritten as

$$ \phi_\mathbf l^\text{CGF}(r,\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l}\sum_ns_nr^l\exp(-\alpha_nr^2)\equiv\Phi_\mathbf l(\Omega)R_l(r) $$

where $l=l_x+l_y+l_z$ is the total angular momentum, $\Phi_\mathbf l(\Omega)$ is the angular part, and $R_l(r)$ is the radial part, and they are separable not only in mathematical form, but also for practical purposes. (This discussion omits the norm, which is not separable to radial and angular contributions.) When specifying a Gaussian basis set in quantum chemistry, one talks only about the radial part, independent of the angular part. The angular part is really considered an implementation detail, because its form usually doesn't have any significant effect on the results.

For completeness, two most common angular forms are Cartesian:

$$ \Phi_\mathbf l(\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l} $$

and spherical harmonic:

$$ \Phi_\mathbf l(\Omega)=Y_l^m(\theta,\varphi) $$

| improve this answer | |
  • $\begingroup$ I presume normalization constant for >p orbitals will be different? $\endgroup$ – packet0 Apr 26 '17 at 0:57
  • 1
    $\begingroup$ @packet0 Yes, the norm is not separable to the radial and angular part. I'll make this clear in the answer. $\endgroup$ – jhrmnn Apr 26 '17 at 9:41

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