Atomic orbitals (AOs) and Basis sets#
Atomic orbitals for the pairing function and the Jastrow factor#
One of the most common choices for atomic orbitals in QMC is atom‑centered Gaussian‑type orbitals (GTOs). A primitive GTO \(\psi_{l,m,\alpha}(\mathbf{r})\) can be constructed using either solid harmonics or Cartesian polynomial basis functions.
Gaussian‑Type Orbitals with Solid‑Harmonic Basis#
Primitive orbitals with regular solid harmonics are given by:
\[\psi_{l,m,\alpha}(\mathbf{r}) = \mathcal{N}^{\rm solid}_{l,m,\alpha}\;\mathcal{R}_{\alpha}(\mathbf{r})\;\mathcal{S}_{l,m,\alpha}(\mathbf{r})\]
where the radial part is
\[\mathcal{R}_{\alpha}(\mathbf{r}) = e^{-Z_\alpha\,|\mathbf{r}-\mathbf{R}_\alpha|^2}\]
and the solid harmonics are
\[\mathcal{S}_{l,m,\alpha}(\mathbf{r}) = \sqrt{\frac{2l+1}{4\pi}}\;|\mathbf{r}-\mathbf{R}_\alpha|^l\;\mathcal{Y}_{l,m}(\theta_\alpha,\phi_\alpha).\]
Here \(\mathcal{Y}_{l,m}(\theta,\phi)\) are real spherical harmonics, and the Racah normalization ensures
\[\int_0^\pi\!\int_0^{2\pi} \sin\theta\,\mathcal{S}_{l,m,\alpha}^*(\mathbf{r})\,\mathcal{S}_{l,m,\alpha}(\mathbf{r})\,\mathrm{d}\phi\,\mathrm{d}\theta = \frac{4\pi}{2l+1}\;r^{2l}.\]
The normalization constant is
\[\mathcal{N}^{\rm solid}_{l,m,\alpha} = \sqrt{\frac{(2Z_\alpha/\pi)^{3/2}\,(4Z_\alpha)^l}{(2l-1)!!}}.\]
One may also write GTOs with spherical harmonics normalization
\[\psi_{l,m,\alpha}(\mathbf{r}) = \mathcal{N}^{\rm sphe}_{l,m,\alpha}\;|\mathbf{r}-\mathbf{R}_\alpha|^l\;\mathcal{R}_{\alpha}(\mathbf{r})\;\mathcal{Y}_{l,m}(\theta_\alpha,\phi_\alpha)\]
with
\[\mathcal{N}^{\rm sphe}_{l,m,\alpha} = \sqrt{\frac{2^{2l+3}(l+1)!(2Z_\alpha)^{l+3/2}}{(2l+2)!\,\sqrt{\pi}}}.\]
These two normalizations satisfy
\[\frac{\mathcal{N}^{\rm solid}_{l,m,\alpha}}{\mathcal{N}^{\rm sphe}_{l,m,\alpha}} = \sqrt{\frac{2l+1}{4\pi}}.\]
Gaussian‑Type Orbitals with Cartesian Basis#
Primitive orbitals in the Cartesian basis are
\[\psi_{n_x,n_y,n_z,\alpha}(\mathbf{r}) = \mathcal{N}^{\rm cart}_{n_x,n_y,n_z,\alpha}\;e^{-Z_\alpha\,|\mathbf{r}-\mathbf{R}_\alpha|^2}\;x^{n_x}y^{n_y}z^{n_z}.\]
The normalization is
\[\mathcal{N}^{\rm cart}_{n_x,n_y,n_z,\alpha} = \sqrt{\frac{(2Z_\alpha/\pi)^{3/2}\,(4Z_\alpha)^{n_x+n_y+n_z}}{(2n_x-1)!!\,(2n_y-1)!!\,(2n_z-1)!!}} = \sqrt{\frac{(2Z_\alpha/\pi)^{3/2}\,(8Z_\alpha)^{n_x+n_y+n_z}n_x!n_y!n_z!}{(2n_x)!(2n_y)!(2n_z)!}}.\]
We define the total angular momentum as \(l=n_x+n_y+n_z\). A basis of order \(l\) includes all monomials of that degree (e.g., \(l=2\) gives \(d_{xx},d_{xy},\dots,d_{zz}\)). Note that normalization constants differ among Cartesian orbitals with the same \(l\) (except for \(s\) and \(p\) shells). For example:
\[\mathcal{N}^{\rm cart}_{2,0,0,\alpha} = \sqrt{\frac{(2Z_\alpha/\pi)^{3/2}(4Z_\alpha)^2}{3\cdot1}},\]
\[\mathcal{N}^{\rm cart}_{1,1,0,\alpha} = \sqrt{\frac{(2Z_\alpha/\pi)^{3/2}(4Z_\alpha)^2}{1\cdot1}}.\]
Practical Tip#
In jQMC (JAX), Cartesian GTOs are computationally faster than spherical ones when using jit and grad, since they avoid branching logic by varying only polynomial exponents rather than basis‑function forms.
Real Spherical and Solid Harmonics#
The real spherical harmonics \(\mathcal{Y}_{l,m}(\theta,\phi)\) are built from the complex spherical harmonics \(Y_{l,m}(\theta,\phi)\) with the Condon–Shortley phase:
(1)#\[\begin{split}\mathcal{Y}_{l,m}(\theta,\phi) =
\begin{cases}
\displaystyle
\frac{1}{\sqrt{2}}\bigl(Y_{l,-|m|}(\theta,\phi) + (-1)^m\,Y_{l,|m|}(\theta,\phi)\bigr), & m>0,\\[1em]
Y_{l,0}(\theta,\phi), & m=0,\\[0.75em]
\displaystyle
\frac{i}{\sqrt{2}}\bigl(Y_{l,-|m|}(\theta,\phi) - (-1)^m\,Y_{l,|m|}(\theta,\phi)\bigr), & m<0.
\end{cases}\end{split}\]
Because the complex spherical harmonics satisfy
(2)#\[\int_{0}^{\pi}\!\!\int_{0}^{2\pi}
Y_{l',m'}^*(\theta,\phi)\,Y_{l,m}(\theta,\phi)\,
\sin\theta\,d\phi\,d\theta
= \delta_{l,l'}\,\delta_{m,m'},\]
the real ones are also orthonormal:
(3)#\[\int_{0}^{\pi}\!\!\int_{0}^{2\pi}
\mathcal{Y}_{l',m'}(\theta,\phi)\,\mathcal{Y}_{l,m}(\theta,\phi)\,
\sin\theta\,d\phi\,d\theta
= \delta_{l,l'}\,\delta_{m,m'}.\]
The spherical harmonics have singularities at the origin \((x,y,z)=(0,0,0)\). In practice one uses the regular solid harmonics centered at \(\mathbf{R}_\alpha\):
(4)#\[\mathcal{S}_{l,m,\alpha}(\mathbf{r})
= \sqrt{\frac{2l+1}{4\pi}}\;
|\mathbf{R}_\alpha-\mathbf{r}|^l\;
\mathcal{Y}_{l,m}(\theta_\alpha,\phi_\alpha).\]
Polynomial Form of Solid Harmonics#
One efficient way to compute \(\mathcal{S}_{l,m}\) is via polynomials in \((x,y,z)\). Define:
(5)#\[\begin{split}C_m(x,y) =
\begin{cases}
\displaystyle
\sum_{p=0}^{|m|}\binom{|m|}{p}x^p\,y^{|m|-p}\,\cos\!\bigl(\tfrac\pi2(|m|-p)\bigr),
& m\ge0,\\[0.75em]
\displaystyle
\sum_{p=0}^{|m|}\binom{|m|}{p}x^p\,y^{|m|-p}\,\sin\!\bigl(\tfrac\pi2(|m|-p)\bigr),
& m<0,
\end{cases}\end{split}\]
(6)#\[\gamma_{l,m,k}(z)
= (-1)^k\,2^{-l}
\,\binom{l}{k}\,\binom{2l-2k}{l}\,
\frac{(l-2k)!}{(l-2k-|m|)!},\]
(7)#\[\Gamma_{l,m}(z)
= \sum_{k=0}^{\lfloor (l-|m|)/2\rfloor}
\gamma_{l,m,k}(z)\;
|\mathbf{r}|^{2k}\;z^{\,l-2k-|m|}.\]
Then
(8)#\[\mathcal{S}_{l,m,\alpha}(\mathbf{r})
= \sqrt{\frac{(2-\delta_{m,0})(l-|m|)!}{(l+|m|)!}}\;
C_m(x-\!X_\alpha,\;y-\!Y_\alpha)\;
\Gamma_{l,m}(z-\!Z_\alpha).\]
Normalization of the Radial Part#
For a Gaussian radial factor \(\psi_{l,m,\alpha}(\mathbf{r})=\mathcal{N}_{l,\alpha}\,r^l\,e^{-Z_\alpha r^2}\,\mathcal{Y}_{l,m}(\theta,\phi)\), the overall normalization requires
(9)#\[\int_{\mathbb{R}^3}
\bigl|\psi_{l,m,\alpha}(\mathbf{r})\bigr|^2\,d\mathbf{r}
=1.\]
Separating radial and angular parts:
(10)#\[\mathcal{N}_{l,\alpha}^2
\int_{0}^{\infty}r^{2l+2}\,e^{-2Z_\alpha r^2}\,dr
=1
\;\Longrightarrow\;
\mathcal{N}_{l,\alpha}^2
\frac{(2l+2)!\,\sqrt{\pi}}{2^{2l+3}(l+1)!\,(2Z_\alpha)^{l+\frac{3}{2}}}
=1,\]
(11)#\[\int_{0}^{\pi}\!\!\int_{0}^{2\pi}
\bigl|\mathcal{Y}_{l,m}(\theta,\phi)\bigr|^2\,
\sin\theta\,d\phi\,d\theta
=1.\]
Hence the radial normalization constant is
(12)#\[\mathcal{N}_{l,\alpha}
= \sqrt{
\frac{2^{2l+3}(l+1)!\,(2Z_\alpha)^{\,l+\tfrac{3}{2}}}
{(2l+2)!\,\sqrt{\pi}}
}.\]
This factor is identical for all \((l,m)\) in the same shell.
Practical Implementation Note#
In JAX-based codes, Cartesian GTOs (polynomials in \(x,y,z\)) are often faster than spherical ones, because they avoid branching logic over \((l,m)\).
Tables of Real Solid Harmonics#
Below are the explicit real solid harmonics up to \(l=6\). Let \(r=\sqrt{x^2+y^2+z^2}\).
\(l=0\) (\(s\) orbital)#
(13)#\[Y_{0,0} = \frac{1}{2}\,\sqrt{\frac{1}{\pi}}.\]
\(l=1\) (\(p\) orbitals)#
(14)#\[\begin{split}\begin{aligned}
Y_{1,-1} &= i\sqrt{\tfrac12}\bigl(Y_1^{-1}+Y_1^1\bigr)
= \sqrt{\tfrac{3}{4\pi}}\;\frac{y}{r},\\
Y_{1,0} &= \;\;Y_1^0
= \sqrt{\tfrac{3}{4\pi}}\;\frac{z}{r},\\
Y_{1,1} &= \sqrt{\tfrac12}\bigl(Y_1^{-1}-Y_1^1\bigr)
= \sqrt{\tfrac{3}{4\pi}}\;\frac{x}{r}.
\end{aligned}\end{split}\]
\(l=2\) (\(d\) orbitals)#
(15)#\[\begin{split}\begin{aligned}
Y_{2,-2} &= i\sqrt{\tfrac12}\bigl(Y_2^{-2}-Y_2^2\bigr)
= \tfrac12\sqrt{\tfrac{15}{\pi}}\;\frac{xy}{r^2},\\
Y_{2,-1} &= i\sqrt{\tfrac12}\bigl(Y_2^{-1}+Y_2^1\bigr)
= \tfrac12\sqrt{\tfrac{15}{\pi}}\;\frac{yz}{r^2},\\
Y_{2,0} &= \;\;Y_2^0
= \tfrac14\sqrt{\tfrac{5}{\pi}}\;\frac{3z^2-r^2}{r^2},\\
Y_{2,1} &= \sqrt{\tfrac12}\bigl(Y_2^{-1}-Y_2^1\bigr)
= \tfrac12\sqrt{\tfrac{15}{\pi}}\;\frac{xz}{r^2},\\
Y_{2,2} &= \sqrt{\tfrac12}\bigl(Y_2^{-2}+Y_2^2\bigr)
= \tfrac14\sqrt{\tfrac{15}{\pi}}\;\frac{x^2-y^2}{r^2}.
\end{aligned}\end{split}\]
\(l=3\) (\(f\) orbitals)#
(16)#\[\begin{split}\begin{aligned}
Y_{3,-3} &= i\sqrt{\tfrac12}\bigl(Y_3^{-3}+Y_3^3\bigr)
= \tfrac14\sqrt{\tfrac{35}{2\pi}}\;\frac{y(3x^2-y^2)}{r^3},\\
Y_{3,-2} &= i\sqrt{\tfrac12}\bigl(Y_3^{-2}-Y_3^2\bigr)
= \tfrac12\sqrt{\tfrac{105}{\pi}}\;\frac{xyz}{r^3},\\
Y_{3,-1} &= i\sqrt{\tfrac12}\bigl(Y_3^{-1}+Y_3^1\bigr)
= \tfrac14\sqrt{\tfrac{21}{2\pi}}\;\frac{y(5z^2-r^2)}{r^3},\\
Y_{3,0} &= \;\;Y_3^0
= \tfrac14\sqrt{\tfrac{7}{\pi}}\;\frac{5z^3-3zr^2}{r^3},\\
Y_{3,1} &= \sqrt{\tfrac12}\bigl(Y_3^{-1}-Y_3^1\bigr)
= \tfrac14\sqrt{\tfrac{21}{2\pi}}\;\frac{x(5z^2-r^2)}{r^3},\\
Y_{3,2} &= \sqrt{\tfrac12}\bigl(Y_3^{-2}+Y_3^2\bigr)
= \tfrac14\sqrt{\tfrac{105}{\pi}}\;\frac{(x^2-y^2)z}{r^3},\\
Y_{3,3} &= \sqrt{\tfrac12}\bigl(Y_3^{-3}-Y_3^3\bigr)
= \tfrac14\sqrt{\tfrac{35}{2\pi}}\;\frac{x(x^2-3y^2)}{r^3}.
\end{aligned}\end{split}\]
\(l=4\) (\(g\) orbitals)#
(17)#\[\begin{split}\begin{aligned}
Y_{4,-4} &= i\sqrt{\tfrac12}\bigl(Y_4^{-4}-Y_4^4\bigr)
= \tfrac34\sqrt{\tfrac{35}{\pi}}\;\frac{xy(x^2-y^2)}{r^4},\\
Y_{4,-3} &= i\sqrt{\tfrac12}\bigl(Y_4^{-3}+Y_4^3\bigr)
= \tfrac34\sqrt{\tfrac{35}{2\pi}}\;\frac{y(3x^2-y^2)z}{r^4},\\
Y_{4,-2} &= i\sqrt{\tfrac12}\bigl(Y_4^{-2}-Y_4^2\bigr)
= \tfrac34\sqrt{\tfrac{5}{\pi}}\;\frac{xy(7z^2-r^2)}{r^4},\\
Y_{4,-1} &= i\sqrt{\tfrac12}\bigl(Y_4^{-1}+Y_4^1\bigr)
= \tfrac34\sqrt{\tfrac{5}{2\pi}}\;\frac{y(7z^3-3zr^2)}{r^4},\\
Y_{4,0} &= \;\;Y_4^0
= \tfrac{3}{16}\sqrt{\tfrac{1}{\pi}}\;\frac{35z^4-30z^2r^2+3r^4}{r^4},\\
Y_{4,1} &= \sqrt{\tfrac12}\bigl(Y_4^{-1}-Y_4^1\bigr)
= \tfrac34\sqrt{\tfrac{5}{2\pi}}\;\frac{x(7z^3-3zr^2)}{r^4},\\
Y_{4,2} &= \sqrt{\tfrac12}\bigl(Y_4^{-2}+Y_4^2\bigr)
= \tfrac38\sqrt{\tfrac{5}{\pi}}\;\frac{(x^2-y^2)(7z^2-r^2)}{r^4},\\
Y_{4,3} &= \sqrt{\tfrac12}\bigl(Y_4^{-3}-Y_4^3\bigr)
= \tfrac34\sqrt{\tfrac{35}{2\pi}}\;\frac{x(x^2-3y^2)z}{r^4},\\
Y_{4,4} &= \sqrt{\tfrac12}\bigl(Y_4^{-4}+Y_4^4\bigr)
= \tfrac{3}{16}\sqrt{\tfrac{35}{\pi}}\;\frac{x^2(x^2-3y^2)-y^2(3x^2-y^2)}{r^4}.
\end{aligned}\end{split}\]
\(l=5\) (\(h\) orbitals)#
(18)#\[\begin{split}\begin{aligned}
Y_{5,-5} &= i\sqrt{\tfrac12}\bigl(Y_5^{-5}+Y_5^5\bigr)
= \tfrac{3}{16}\sqrt{\tfrac{77}{2\pi}}\;\frac{5x^4y-10x^2y^3+y^5}{r^5},\\
Y_{5,-4} &= i\sqrt{\tfrac12}\bigl(Y_5^{-4}-Y_5^4\bigr)
= \tfrac{3}{16}\sqrt{\tfrac{385}{\pi}}\;\frac{4xyz(x^2-y^2)}{r^5},\\
Y_{5,-3} &= i\sqrt{\tfrac12}\bigl(Y_5^{-3}+Y_5^3\bigr)
= \tfrac{1}{16}\sqrt{\tfrac{385}{2\pi}}\;\frac{-(y^3-3x^2y)(9z^2-r^2)}{r^5},\\
Y_{5,-2} &= i\sqrt{\tfrac12}\bigl(Y_5^{-2}-Y_5^2\bigr)
= \tfrac{1}{8}\sqrt{\tfrac{1155}{\pi}}\;\frac{2xy(3z^3-zr^2)}{r^5},\\
Y_{5,-1} &= i\sqrt{\tfrac12}\bigl(Y_5^{-1}+Y_5^1\bigr)
= \tfrac{1}{16}\sqrt{\tfrac{165}{\pi}}\;\frac{y(21z^4-14z^2r^2+r^4)}{r^5},\\
Y_{5,0} &= \;\;Y_5^0
= \tfrac{1}{16}\sqrt{\tfrac{11}{\pi}}\;\frac{63z^5-70z^3r^2+15zr^4}{r^5},\\
Y_{5,1} &= \sqrt{\tfrac12}\bigl(Y_5^{-1}-Y_5^1\bigr)
= \tfrac{1}{16}\sqrt{\tfrac{165}{\pi}}\;\frac{x(21z^4-14z^2r^2+r^4)}{r^5},\\
Y_{5,2} &= \sqrt{\tfrac12}\bigl(Y_5^{-2}+Y_5^2\bigr)
= \tfrac{1}{8}\sqrt{\tfrac{1155}{\pi}}\;\frac{(x^2-y^2)(3z^3-zr^2)}{r^5},\\
Y_{5,3} &= \sqrt{\tfrac12}\bigl(Y_5^{-3}-Y_5^3\bigr)
= \tfrac{1}{16}\sqrt{\tfrac{385}{2\pi}}\;\frac{(x^3-3xy^2)(9z^2-r^2)}{r^5},\\
Y_{5,4} &= \sqrt{\tfrac12}\bigl(Y_5^{-4}+Y_5^4\bigr)
= \tfrac{3}{16}\sqrt{\tfrac{385}{\pi}}\;\frac{x^2z(x^2-3y^2)-y^2z(3x^2-y^2)}{r^5},\\
Y_{5,5} &= \sqrt{\tfrac12}\bigl(Y_5^{-5}-Y_5^5\bigr)
= \tfrac{3}{16}\sqrt{\tfrac{77}{2\pi}}\;\frac{x^5-10x^3y^2+5xy^4}{r^5}.
\end{aligned}\end{split}\]
\(l=6\) (\(i\) orbitals)#
(19)#\[\begin{split}\begin{aligned}
Y_{6,-6} &= i\sqrt{\tfrac12}\bigl(Y_6^{-6}-Y_6^6\bigr)
= \tfrac{1}{64}\sqrt{\tfrac{6006}{\pi}}\;\frac{6x^5y-20x^3y^3+6xy^5}{r^6},\\
Y_{6,-5} &= i\sqrt{\tfrac12}\bigl(Y_6^{-5}+Y_6^5\bigr)
= \tfrac{3}{32}\sqrt{\tfrac{2002}{\pi}}\;\frac{z(5x^4y-10x^2y^3+y^5)}{r^6},\\
Y_{6,-4} &= i\sqrt{\tfrac12}\bigl(Y_6^{-4}-Y_6^4\bigr)
= \tfrac{3}{32}\sqrt{\tfrac{91}{\pi}}\;\frac{4xy(11z^2-r^2)(x^2-y^2)}{r^6},\\
Y_{6,-3} &= i\sqrt{\tfrac12}\bigl(Y_6^{-3}+Y_6^3\bigr)
= \tfrac{1}{32}\sqrt{\tfrac{2730}{\pi}}\;\frac{-(11z^3-3zr^2)(y^3-3x^2y)}{r^6},\\
Y_{6,-2} &= i\sqrt{\tfrac12}\bigl(Y_6^{-2}-Y_6^2\bigr)
= \tfrac{1}{64}\sqrt{\tfrac{2730}{\pi}}\;\frac{2xy(33z^4-18z^2r^2+r^4)}{r^6},\\
Y_{6,-1} &= i\sqrt{\tfrac12}\bigl(Y_6^{-1}+Y_6^1\bigr)
= \tfrac{1}{16}\sqrt{\tfrac{273}{\pi}}\;\frac{y(33z^5-30z^3r^2+5zr^4)}{r^6},\\
Y_{6,0} &= \;\;Y_6^0
= \tfrac{1}{32}\sqrt{\tfrac{13}{\pi}}\;\frac{231z^6-315z^4r^2+105z^2r^4-5r^6}{r^6},\\
Y_{6,1} &= \sqrt{\tfrac12}\bigl(Y_6^{-1}-Y_6^1\bigr)
= \tfrac{1}{16}\sqrt{\tfrac{273}{\pi}}\;\frac{x(33z^5-30z^3r^2+5zr^4)}{r^6},\\
Y_{6,2} &= \sqrt{\tfrac12}\bigl(Y_6^{-2}+Y_6^2\bigr)
= \tfrac{1}{64}\sqrt{\tfrac{2730}{\pi}}\;\frac{(x^2-y^2)(33z^4-18z^2r^2+r^4)}{r^6},\\
Y_{6,3} &= \sqrt{\tfrac12}\bigl(Y_6^{-3}-Y_6^3\bigr)
= \tfrac{1}{32}\sqrt{\tfrac{2730}{\pi}}\;\frac{(11z^3-3zr^2)(x^3-3xy^2)}{r^6},\\
Y_{6,4} &= \sqrt{\tfrac12}\bigl(Y_6^{-4}+Y_6^4\bigr)
= \tfrac{3}{32}\sqrt{\tfrac{91}{\pi}}\;\frac{(11z^2-r^2)\bigl[x^2(x^2-3y^2)+y^2(y^2-3x^2)\bigr]}{r^6},\\
Y_{6,5} &= \sqrt{\tfrac12}\bigl(Y_6^{-5}-Y_6^5\bigr)
= \tfrac{3}{32}\sqrt{\tfrac{2002}{\pi}}\;\frac{z(x^5-10x^3y^2+5xy^4)}{r^6},\\
Y_{6,6} &= \sqrt{\tfrac12}\bigl(Y_6^{-6}+Y_6^6\bigr)
= \tfrac{1}{64}\sqrt{\tfrac{6006}{\pi}}\;\frac{x^6-15x^4y^2+15x^2y^4-y^6}{r^6}.
\end{aligned}\end{split}\]
\(l\ge7\) harmonics are rarely needed in practice.