PHY.K02UF Molecular and Solid State Physics
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PHY.K02UF Molecular and Solid State Physics | ||||
The total quantum state of a molecule is described by a many-electron wavefunction. In the standard approximation, the many-electron wavefunction can be expressed as a product of single-electron wavefunctions called molecular orbitals. Molecular orbitals are used very much like the hydrogen wavefunctions are used to construct the many-electron wavefunctions of atoms. In the simplest approximation, the molecular orbitals are the wavefunction of a single electron moving in a potential created by all of the positively charged nuclei in a molecule. The molecular orbital Hamiltonian can be written,
\[ \begin{equation} H_{\text{mo}}= - \frac{\hbar^2}{2m_e}\nabla^2 -\sum\limits_{a} \frac{Z_ae^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_a|}. \end{equation} \]Here $\vec{r}_a$ are the positions of the nuclei in the molecule, $\vec{r}$ is the position of the single electron, and $a$ sums over the atoms in the molecule. The molecular orbital Hamiltonian is often solved by a method called the Linear Combination of Atomic Orbitals (LCAO). It is assumed that the wave function can be written in terms of hydrogen atomic orbitals that are centered around the nuclei,
\begin{equation} \psi_{\text{mo}}(\vec{r})= \sum\limits_{a} \sum\limits_{ao} c_{ao,a}\phi^{Z_a}_{ao}\left(\vec{r}-\vec{r}_a\right). \end{equation}where $ao$ labels the atomic orbitals ($ao=1$: 1s, $ao=2$: 2s, $ao=3$: 2px, $\cdots$). The number of molecular orbitals that we calculate will be equal to the number of unknown coefficients. We call this number $N$. We have to decide how many atomic orbitals should be included for each atom. A reasonable choice is to take all of the occupied atomic orbitals of the isolated atoms. For instance, for water one might use the 1s, 2s, and 2p orbitals of oxygen and the 1s orbitals of the two hydrogen atoms. In that case, there would be $N=7$ terms in the wave function for the molecular orbital. There is no strict rule as to which atomic orbitals to include. Including more atomic orbitals leads to a higher accuracy but makes the numerical calculation more difficult.
At this point it is convenient to relabel the atomic orbitals used in the wave function with integers $p=1\cdots N$. For water we might choose $\phi_1 = \phi^{Z=1}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{H1}}\right)$, $\phi_2 = \phi^{Z=1}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{H2}}\right)$, $\phi_3 = \phi^{Z=8}_{\text{1s}}\left(\vec{r}-\vec{r}_{\text{O}}\right)$, $\phi_4 = \phi^{Z=8}_{\text{2s}}\left(\vec{r}-\vec{r}_{\text{O}}\right)$, etc. The trial wavefunction can then be written more compactly as,
\begin{equation} \psi_{\text{mo}} = \sum\limits_{p=1}^N c_p\phi_p. \end{equation}The time independent Schrödinger equation is,
\[ \begin{equation} H_{\text{mo}}\psi_{\text{mo}}=E\psi_{\text{mo}} . \end{equation} \]Multiply the Schrödinger equation from the left by each of the atomic orbitals and integrate over all space. This results in a set of $N$ algebraic equations called the Roothaan equations.
\[ \begin{equation} \begin{matrix} \langle \phi_1 | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_1|\psi_{\text{mo}} \rangle \\ \langle \phi_2 | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_2|\psi_{\text{mo}} \rangle\\ \vdots \\ \langle \phi_N | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi_N|\psi_{\text{mo}} \rangle \end{matrix} \end{equation} \]By substituting in the form for $\psi_{\text{mo}}$ from above, the Roothaan equations can be written in matrix form,
\[ \begin{equation} \left[ \begin{matrix} H_{11} & H_{12} & \cdots & H_{1N} \\ H_{21} & H_{22} & \cdots & H_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ H_{N1} & H_{N2} & \cdots & H_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right] = E\left[ \begin{matrix} S_{11} & S_{12} & \cdots & S_{1N} \\ S_{21} & S_{22} & \cdots & S_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ S_{N1} & S_{N2} & \cdots & S_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right]. \end{equation} \]Here the elements of the Hamiltonian matrix and the overlap matrix are,
\[ \begin{equation} H_{pq}= \langle\phi_{p}|H_{\text{mo}}|\phi_{q}\rangle \hspace{1cm}\text{and}\hspace{1cm} S_{pq}= \langle\phi_{p}|\phi_{q}\rangle . \end{equation} \]The Roothaan equations can be solved numerically to find $N$ solutions for the energy $E$ along with the corresponding coefficients that describe the $N$ wave functions which are the molecular orbitals. Multiplying from the left with the inverse of the overlap matrix results in an eigenvalue problem,
$$\textbf{S}^{-1}\textbf{H}\,\vec{c}=E\,\vec{c}.$$Often when the are calculating the molecular orbitals of a molecule with LCAO, it turns out that the resulting Hamitonian matrix and the overlap matrix have nearly block diagonal form. For instance, consider a carbon monoxide molecule. For a trial wavefuntion of the form,
$$\psi_{\text{mo}} = c_1\phi^{\text{C}}_{\text{2s}}+c_2\phi^{\text{O}}_{\text{2s}}+c_3\phi^{\text{C}}_{\text{2px}}+c_4\phi^{\text{O}}_{\text{2px}}+ c_5\phi^{\text{C}}_{\text{2py}}+ c_6\phi^{\text{O}}_{\text{2py}}+ c_7\phi^{\text{C}}_{\text{2pz}}+ c_8\phi^{\text{O}}_{\text{2pz}},$$Evaluated in the molecular orbital Hamiltonian,
$$ H_{\text{mo}}= - \frac{\hbar^2}{2m_e}\nabla^2 - \frac{Z^C_{\text{eff}}e^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_C|}- \frac{Z^O_{\text{eff}}e^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_O|},$$with $Z^C_{\text{eff}}=3.25$ and $Z^O_{\text{eff}}=4.55$, the Hamiltonian and overlap matrices are,
\begin{equation} H= \left[ \begin{matrix} -90.63 & -52.64 & 20.73 & -32.35 & 0.00 & 0.00 & 0.00 & 0.00\\ -52.64 & -111.57 & 59.54 & -12.40 & 0.00 & 0.00 & 0.00 & 0.00\\ 20.73 & 59.54 & -102.59 & 43.97 & 0.00 & 0.00 & 0.00 & 0.00\\ -32.35 & -12.40 & 43.97 & -116.66 & 0.00 & 0.00 & 0.00 & 0.00\\ 0.00 & 0.00 & 0.00 & 0.00 & -87.01 & -31.20 & 0.00 & 0.00\\ 0.00 & 0.00 & 0.00 & 0.00 & -31.20 & -109.27 & 0.00 & 0.00\\ 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & -87.01 & -31.20\\ 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & -31.20 & -109.27\\ \end{matrix} \right], \qquad S = \left[ \begin{matrix} 1.00 & 0.47 & 0.00 & 0.26 & 0.00 & 0.00 & 0.00 & 0.00\\ 0.47 & 1.00 & -0.49 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00\\ 0.00 & -0.49 & 1.00 & -0.30 & 0.00 &0.00 & 0.00 &0.00\\ 0.26 & 0.00 & -0.30 & 1.00 & 0.00 & 0.00 & 0.00 & 0.00\\ 0.00 & 0.00 & 0.00 & 0.00 & 1.00 & 0.26 & 0.00 & 0.00\\ 0.00 & 0.00 &0.00 & 0.00 & 0.26 & 1.00 & 0.00 & 0.00\\ 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 0.00 & 1.00 & 0.26\\ 0.00 & 0.00 &0.00 & 0.00 & 0.00 & 0.00 & 0.26 & 1.00\\ \end{matrix} \right]. \end{equation}Notice that there is a $4\times 4$ block of the $2s$ and $2p_x$ orbitals, a $2\times 2$ block of the $2p_y$ orbitals, and another $2\times 2$ block of the $2p_z$ orbitals. When the Hamiltonian matrix and the overlap matrix have block diagonal form, it is possible to solve each of the blocks separately. People will often refer to molecular orbitals by stating which atomic orbitals contribute to the relevant block.