   PHY.K02UF Molecular and Solid State Physics

## 2pz Atomic Orbital

The 2pz atomic orbital is,

$\begin{equation} \phi^Z_{2p_z}= \sqrt{\frac{Z^3}{\pi 2^5 a_0^3}}\frac{Zz}{a_0} \exp\left(-\frac{Zr}{2a_0}\right). \end{equation}$

To account for the screening due to the core electrons, the effective nuclear charge $Z$ should be determined by Slater's rules. For carbon, the effective nucelar charge is $Z=3.25$.

 $\left(\frac{a_0}{Z}\right)^{3/2}\phi_{2p_z}$ $\frac{Zz}{a_0}$ The $2p_z$ orbital plotted along the $z$-axis. The maximum is at $z=\frac{2a_0}{Z}$.

To check that the 2pz orbital solves the Schrödinger 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}

we let the Laplacian operator act on the orbital,

$\begin{equation} \nabla^2\phi^Z_{2p_z}=\sqrt{\frac{Z^7}{\pi 2^5 a_0^7}}\left(\frac{Zr}{4a_0}-2\right)\frac{z}{r} \exp\left(-\frac{Zr}{2a_0}\right), \end{equation}$

and then calculate the energy $E_2= - \frac{\hbar^2}{2m}\frac{\nabla^2\phi_{2p_z}(\vec{r})}{\phi_{2p_z}(\vec{r})} - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}|}$. The code below defines the 2pz orbital and its Laplacian in Cartesian coordinates centered at position $(x_i,y_i,z_i)$. It then chooses random numbers for $x$, $y$, $z$, $x_i$, $y_i$, and $y_i$ and calculates the energy. If the orbital has been programmed properly, the energy should be -35.9 eV at every position for $Z=3.25$.

Consider the overlap integral of two 2pz orbitals located at positions $\vec{r}_1$ and $\vec{r}_2$,

\begin{equation} S_{12}=\langle \phi_{2p_z}^Z(\vec{r}-\vec{r}_1)|\phi_{2p_z}^Z(\vec{r}-\vec{r}_2)\rangle . \end{equation}

For a ethene molecule, $Z_{eff}=3.25$, $\vec{r}_1=-0.67\,\hat{x}$ Å and $\vec{r}_2=0.67\,\hat{x}$ Å. Below $\phi_{2p_z}^Z(\vec{r}-\vec{r}_1)\phi_{2p_z}^Z(\vec{r}-\vec{r}_2)$ is plotted along the $x$-direction at $z=\frac{2a_0}{Z}$ and along the $z$-direction through one of the atoms.

 $\left(\frac{a_0}{Z}\right)^3\phi_{2p_z}^Z(\vec{r}-\vec{r}_1)\phi_{2p_z}^Z(\vec{r}-\vec{r}_2)$ $\frac{Zx}{a_0}\,@\,y=0,\,z=\frac{2a_0}{Z}$

The code below uses a Monte-Carlo method to integrate $\phi_{2p_z}^C(\vec{r}-\vec{r}_1)\phi_{2p_z}^C(\vec{r}-\vec{r}_2)$ and calculate $S_{12}$. For carbon, $Z_{eff}=3.25$.



Press the 'Execute' button a few times and notice that the answer keeps changing. By doing this you can estimate the error in the calculation. The error should decrease like $1/\sqrt{N}$ where $N$ is the number of random numbers chosen. By setting $x_1=x_2=0$ in the code and pressing 'Execute', you calculate $\langle \phi_{2p_z}(\vec{r})|\phi_{2p_z}(\vec{r})\rangle$ which should equal to 1 if the wave function is properly normalized.

The Hamiltonian matrix element $H_{11}$ for ethene with two 2pz orbitals is,

\begin{equation} H_{11}=\Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{2p_z}^C(\vec{r}-\vec{r}_1) \Big \rangle . \end{equation}

This can be broken into two terms,

\begin{equation} H_{11}=\Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}\right|\phi_{2p_z}^C(\vec{r}-\vec{r}_1) \Big \rangle + \Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left|- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{2p_z}^C(\vec{r}-\vec{r}_1) \Big \rangle . \end{equation}

The wave function $\phi_{2p_z}^C(\vec{r}-\vec{r}_1)$ is an eigenfunction of the atomic orbital Hamiltonian in the first term $H\phi_{2p_z}^C(\vec{r}-\vec{r}_1) = E_2 \phi_{2p_z}^C(\vec{r}-\vec{r}_2)$, so the first term is easily evaluated,

\begin{equation} H_{11}=E_2 - \Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left| \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{2p_z}^C(\vec{r}-\vec{r}_1) \Big \rangle . \end{equation}

The code below uses a Monte-Carlo method to and calculate $H_{11}$.



The Hamiltonian matrix element $H_{12}$ for ethene with two 2pz orbitals located at positions $\vec{r}_1$ and $\vec{r}_2$ is,

\begin{equation} H_{12}=\Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{2p_z}^C(\vec{r}-\vec{r}_2) \Big \rangle . \end{equation}

This can be broken into two terms,

\begin{equation} H_{12}=\Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|}\right|\phi_{2p_z}^C(\vec{r}-\vec{r}_2) \Big \rangle + \Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left|- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|} \right|\phi_{2p_z}^C(\vec{r}-\vec{r}_2) \Big \rangle . \end{equation}

The wave function $\phi_{2p_z}^C(\vec{r}-\vec{r}_2)$ is an eigenfunction of the atomic orbital Hamiltonian in the first term $H\phi_{2p_z}^C(\vec{r}-\vec{r}_2) = E_2 \phi_{2p_z}^C(\vec{r}-\vec{r}_2)$, so the first term is easily evaluated,

\begin{equation} H_{12}=E_2S_{12} - \Big \langle \phi_{2p_z}^C(\vec{r}-\vec{r}_1)\left| \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|} \right|\phi_{2p_z}^C(\vec{r}-\vec{r}_2) \Big \rangle . \end{equation}

Below is a plot of the integrand of the second term. Even though there is a factor of $\frac{1}{|\vec{r}-\vec{r}_1|}$, there is no singularity because the $\phi_{2p_z}^C(\vec{r}-\vec{r}_1)$ wave function goes to zero at that point.

 $-\frac{a_0^3}{Z^3}\frac{\phi_{2p_z}^Z(\vec{r}-\vec{r}_1)\phi_{2p_z}^Z(\vec{r}-\vec{r}_2)}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}$ $\frac{Zz}{a_0}$ The integrand of the matrix element plotted along the $z$-direction through the atom at $\vec{r}_1$.

The code below uses a Monte-Carlo method to and calculate $H_{12}$.



A Matlab script to calculate the overlap matrix and the Hamiltonian matrix is calcHamiltonian.m. This script uses the atomic orbitals defined in atorbit.m.