PHY.K02UF Molecular and Solid State Physics

Atomic Orbitals

The atomic orbitals are solutions to the equation,

\begin{equation} - \frac{\hbar^2}{2m}\nabla^2\phi(\vec{r}) - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}|}\phi(\vec{r}) = E\phi(\vec{r}). \end{equation}

In spherical coordinates the Laplacian operator is,

\[\begin{equation} \nabla ^2 \phi = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \phi}{\partial r}\right)+\frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}\left(\sin \theta~\frac{\partial \phi}{\partial \theta}\right)+\frac{1}{r^2 \sin^2\theta} \frac{\partial^2 \phi}{\partial \varphi^2} \end{equation}\]

Often the atomic orbitials are used to evaluate the matrix elements of the form, $\langle \phi_m |H|\phi_n\rangle$. To perform this calculation it is necessary to calculate $\nabla^2 \phi_n$. Since numerical differentiation can be imprecise, it is useful to calculate the Laplacian analytically. To facilitate such calculations, a table of atomic orbitals and the Laplacian of the orbitals is given below.

\[\begin{equation} \begin{array}{c,c} \phi_{1s}=\sqrt{\frac{Z^3}{\pi a^3_0}}\exp\left(-\frac{Zr}{a_0}\right)& \nabla ^2\phi_{1s}=\sqrt{\frac{Z^5}{\pi a^5_0}} \left(\frac{Z}{a_0}-\frac{2}{r}\right)\exp\left(-\frac{Zr}{a_0}\right)\\ \phi_{2s}=\frac{1}{4} \sqrt{\frac{Z^3}{2\pi a_0^3}}\left(2-\frac{Zr}{a_0}\right)\exp\left(-\frac{Zr}{2a_0}\right)& \nabla ^2\phi_{2s}=\frac{1}{4}\sqrt{\frac{Z^5}{2\pi a^5_0}} \left(-\frac{Z^2r}{4a_0^2}+\frac{5Z}{2a_0}-\frac{4}{r}\right)\exp\left(-\frac{Zr}{2a_0}\right)\\ \phi_{2p_x}=\frac{1}{4} \sqrt{\frac{Z^5}{2\pi a_0^5}}r \exp\left(-\frac{Zr}{2a_0}\right) \sin\theta \cos\varphi & \nabla^2\phi_{2p_x}=\sqrt{\frac{Z^7}{\pi 2^5 a_0^7}}\left(\frac{Zr}{4a_0}-2\right)\sin\theta\cos\varphi \exp\left(-\frac{Zr}{2a_0}\right) \\ \phi_{2p_y}=\frac{1}{4} \sqrt{\frac{Z^5}{2\pi a_0^5}}r \exp\left(-\frac{Zr}{2a_0}\right) \sin\theta \sin\varphi & \nabla^2\phi_{2p_y}=\sqrt{\frac{Z^7}{\pi 2^5 a_0^7}}\left(\frac{Zr}{4a_0}-2\right)\sin\theta\sin\varphi \exp\left(-\frac{Zr}{2a_0}\right)\\ \phi_{2p_z}=\frac{1}{4} \sqrt{\frac{Z^5}{2\pi a_0^5}}r \exp\left(-\frac{Zr}{2a_0}\right) \cos\theta& \nabla^2\phi_{2p_z}=\sqrt{\frac{Z^7}{\pi 2^5 a_0^7}}\left(\frac{Zr}{4a_0}-2\right)\cos\theta \exp\left(-\frac{Zr}{2a_0}\right)\\ \phi_{3s} = \frac{1}{81}\sqrt{\frac{Z^3}{3\pi a_0^3}} \left(27-\frac{18Zr}{a_0}+\frac{2Z^2r^2}{a_0^2}\right) \exp\left(\frac{-Zr}{3a_0}\right) & \nabla^2\phi_{3s} = \frac{1}{81}\sqrt{\frac{Z^3}{3\pi a_0^3}} \left(-\frac{54 Z}{a_0 r}+\frac{39Z^2}{a_0^2}-\frac{6Z^3r}{a_0^3}+\frac{2Z^4r^2}{9a_0^4}\right) \exp\left(\frac{-Zr}{3a_0}\right) \\ \phi_{3p_x} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(6-\frac{Zr}{a_0}\right)\frac{Zr}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) \sin\theta\cos\varphi & \nabla^2\phi_{3p_x} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(-\frac{12Z}{a_0r}+\frac{8Z^2}{3a_0^2}-\frac{Z^3r}{9a_0^3}\right) \frac{Zr}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) \sin\theta\cos\varphi \\ \phi_{3p_y} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(6-\frac{Zr}{a_0}\right)\frac{Zr}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) \sin\theta\sin\varphi & \nabla^2\phi_{3p_y} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(-\frac{12Z^2}{a_0^2}+\frac{8Z^3r}{3a_0^3}-\frac{Z^4r^2}{9a_0^4}\right) \exp\left(\frac{-Zr}{3a_0}\right) \sin\theta\sin\varphi \\ \phi_{3p_z} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(6-\frac{Zr}{a_0}\right)\frac{Zr}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) \cos\theta & \nabla^2\phi_{3p_z} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(-\frac{12Z^2}{a_0^2}+\frac{8Z^3r}{3a_0^3}-\frac{Z^4r^2}{9a_0^4}\right) \exp\left(\frac{-Zr}{3a_0}\right) \cos\theta \\ \phi_{3d_{z^2}} = \frac{1}{81}\sqrt{\frac{Z^7}{6\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\left(3\cos^2\left(\theta\right)-1\right) & \nabla^2\phi_{3d_{z^2}} = \frac{1}{243}\sqrt{\frac{Z^9}{6\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\left(3\text{cos}^2\left(\theta\right)-1\right)\\ \phi_{3d_{xz}} = \frac{1}{81}\sqrt{\frac{2Z^7}{\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\sin\left(\theta\right)\cos\left(\theta\right)\cos\left(\phi\right) & \nabla^2\phi_{3d_{xz}} = \frac{1}{243}\sqrt{\frac{2Z^9}{\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\sin\left(\theta\right)\cos\left(\theta\right)\cos\left(\phi\right)\\ \phi_{3d_{yz}} = \frac{1}{81}\sqrt{\frac{2Z^7}{\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\sin\left(\theta\right)\cos\left(\theta\right)\sin\left(\phi\right) & \nabla^2\phi_{3d_{yz}} = \frac{1}{243}\sqrt{\frac{2Z^9}{\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\sin\left(\theta\right)\cos\left(\theta\right)\sin\left(\phi\right)\\ \phi_{3d_{xy}} = \frac{1}{81}\sqrt{\frac{2Z^7}{\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\sin^2\left(\theta\right)\sin\left(\phi\right)\cos\left(\phi\right) & \nabla^2\phi_{3d_{xy}} = \frac{1}{243}\sqrt{\frac{2Z^9}{\pi a_0^9}}~r\left(\frac{Zr}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\sin^2\left(\theta\right)\sin\left(\phi\right)\cos\left(\phi\right)\\ \phi_{3d_{x^2-y^2}} = \frac{1}{81}\sqrt{\frac{Z^7}{2\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\sin^2\left(\theta\right)\text{cos}\left(2\phi\right) & \nabla^2\phi_{3d_{x^2-y^2}} = \frac{1}{243}\sqrt{\frac{Z^9}{2\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\sin^2\left(\theta\right)\cos\left(2\phi\right)\\ \end{array} \end{equation}\]

A project has been established to collect some code for calculating with atomic orbitals. There are programs to check that the atomic orbitals are defined correctly for the 1s and 2pz atomic orbitals. In the code block below, the atomic orbitals $\phi (r,\theta ,\varphi )$ are defined and then 10 random positions are chosen for $r$, $\theta$, and $\varphi$. The energy $E=H\phi/\phi$ is calculated at these positions. If the wave functions are properly programmed, the energy should be the same at all of these positions.

If the JavaScript code on the left is modified, pressing the Execute button displays the results of executing the modified code.

If an atom is located at position $(x_i,y_i,z_i)$, sometimes it is easier to work in Cartesan coordinates where $r=\sqrt{(x-x_i)^2+(y-y_i)^2+(z-z_i)^2}$.

\[\begin{equation} \begin{array}{c,c} \phi_{1s}=\sqrt{\frac{Z^3}{\pi a^3_0}}\exp\left(-\frac{Zr}{a_0}\right)& \nabla^2\phi_{1s}=\sqrt{\frac{Z^5}{\pi a^5_0}} \left(\frac{Z}{a_0}-\frac{2}{r}\right)\exp\left(-\frac{Zr}{a_0}\right)\\ \phi_{2s}=\frac{1}{4} \sqrt{\frac{Z^3}{2\pi a_0^3}}\left(2-\frac{Zr}{a_0}\right)\exp\left(-\frac{Zr}{2a_0}\right)& \nabla^2\phi_{2s}=\frac{1}{4}\sqrt{\frac{Z^5}{2\pi a^5_0}} \left(-\frac{Z^2r}{4a_0^2}+\frac{5Z}{2a_0}-\frac{4}{r}\right)\exp\left(-\frac{Zr}{2a_0}\right)\\ \phi_{2p_x}=\frac{1}{4} \sqrt{\frac{Z^5}{2\pi a_0^5}}(x-x_i) \exp\left(-\frac{Zr}{2a_0}\right) & \nabla^2\phi_{2p_x}=\sqrt{\frac{Z^7}{\pi 2^5 a_0^7}}\left(\frac{Zr}{4a_0}-2\right)\frac{x-x_i}{r} \exp\left(-\frac{Zr}{2a_0}\right),\\ \phi_{2p_y}=\frac{1}{4} \sqrt{\frac{Z^5}{2\pi a_0^5}}(y-y_i) \exp\left(-\frac{Zr}{2a_0}\right) & \nabla^2\phi_{2p_y}=\sqrt{\frac{Z^7}{\pi 2^5 a_0^7}}\left(\frac{Zr}{4a_0}-2\right)\frac{y-y_i}{r} \exp\left(-\frac{Zr}{2a_0}\right),\\ \phi_{2p_z}=\frac{1}{4} \sqrt{\frac{Z^5}{2\pi a_0^5}}(z-z_i) \exp\left(-\frac{Zr}{2a_0}\right) & \nabla^2\phi_{2p_z}=\sqrt{\frac{Z^7}{\pi 2^5 a_0^7}}\left(\frac{Zr}{4a_0}-2\right)\frac{z-z_i}{r} \exp\left(-\frac{Zr}{2a_0}\right),\\ \phi_{3s} = \frac{1}{81}\sqrt{\frac{Z^3}{3\pi a_0^3}} \left(27-\frac{18Zr}{a_0}+\frac{2Z^2r^2}{a_0^2}\right) \exp\left(\frac{-Zr}{3a_0}\right) & \nabla^2\phi_{3s} = \frac{1}{81}\sqrt{\frac{Z^3}{3\pi a_0^3}} \left(-\frac{54 Z}{a_0 r}+\frac{39Z^2}{a_0^2}-\frac{6Z^3r}{a_0^3}+\frac{2Z^4r^2}{9a_0^4}\right) \exp\left(\frac{-Zr}{3a_0}\right) \\ \phi_{3p_x} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(6-\frac{Zr}{a_0}\right)\frac{Z(x-x_i)}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) & \nabla^2\phi_{3p_x} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(-\frac{12Z}{a_0 r}+\frac{8Z^2}{3a_0^2}-\frac{Z^3 r}{9a_0^3}\right) \frac{Z(x-x_i)}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) \\ \phi_{3p_y} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(6-\frac{Zr}{a_0}\right)\frac{Z(y-y_i)}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) & \nabla^2\phi_{3p_y} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(-\frac{12Z}{a_0 r}+\frac{8Z^2}{3a_0^2}-\frac{Z^3r}{9a_0^3}\right) \frac{Z(y-y_i)}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) \\ \phi_{3p_z} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(6-\frac{Zr}{a_0}\right)\frac{Z(z-z_i)}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) & \nabla^2\phi_{3p_z} = \frac{1}{81}\sqrt{\frac{2Z^3}{\pi a_0^3}} \left(-\frac{12Z}{a_0 r}+\frac{8Z^2}{3a_0^2}-\frac{Z^3 r}{9a_0^3}\right) \frac{Z(z-z_i)}{a_0} \exp\left(\frac{-Zr}{3a_0}\right) \\ \phi_{3d_{z^2}} = \frac{1}{81}\sqrt{\frac{Z^7}{6\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\left(3\frac{(z-z_i)^2}{r^2}-1\right) & \nabla^2\phi_{3d_{z^2}} = \frac{1}{243}\sqrt{\frac{Z^9}{6\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\left(3\frac{(z-z_i)^2}{r^2}-1\right)\\ \phi_{3d_{xz}} = \frac{1}{81}\sqrt{\frac{2Z^7}{\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\frac{(x-x_i)(z-z_i)}{r^2} & \nabla^2\phi_{3d_{xz}} = \frac{1}{243}\sqrt{\frac{2Z^9}{\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\frac{(x-x_i)(z-z_i)}{r^2}\\ \phi_{3d_{yz}} = \frac{1}{81}\sqrt{\frac{2Z^7}{\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\frac{(y-y_i)(z-z_i)}{r^2} & \nabla^2\phi_{3d_{yz}} = \frac{1}{243}\sqrt{\frac{2Z^9}{\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\frac{(y-y_i)(z-z_i)}{r^2}\\ \phi_{3d_{xy}} = \frac{1}{81}\sqrt{\frac{2Z^7}{\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\frac{(x-x_i)(y-y_i)}{r^2} & \nabla^2\phi_{3d_{xy}} = \frac{1}{243}\sqrt{\frac{2Z^9}{\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\frac{(x-x_i)(y-y_i)}{r^2}\\ \phi_{3d_{x^2-y^2}} = \frac{1}{81}\sqrt{\frac{Z^7}{2\pi a_0^7}}~r^2\exp\left(\frac{-Zr}{3a_0}\right)\frac{(x-x_i)^2-(y-y_i)^2}{r^2} & \nabla^2\phi_{3d_{x^2-y^2}} = \frac{1}{243}\sqrt{\frac{Z^9}{2\pi a_0^9}}~r\left(\frac{Z}{3a_0}r-6\right)\exp\left(\frac{-Zr}{3a_0}\right)\frac{(x-x_i)^2-(y-y_i)^2}{r^2}\\ \end{array} \end{equation}\]

The code below defines the atomic orbitals in Cartesian coordinates centered at position $(x_i,y_i,z_i)$.

The following code checks the energies and the normalizations of the atomic orbitals up to 3d.