Molecules

We can study simple molecules to understand the physical phenomena of molecules in general. The simplest molecule we can work with is the $\mathrm {H}_2^+$ ion. It has two nuclei (A and B) sharing one electron (1).

\begin{displaymath}\bgroup\color{black}H_0 = {p^2_e\over {2m}} - {e^2\over{r_{1A}}} - {e^2\over{r_{1B}}} + {e^2\over{R_{AB}}}\egroup\end{displaymath}

\bgroup\color{black}$R_{AB}$\egroup is the distance between the two nuclei. We calculate the ground state energy using the Hydrogen states as a basis.

The lowest energy wavefunction can be thought of as a (anti)symmetric linear combination of an electron in the ground state near nucleus A and the ground state near nucleus B

\begin{displaymath}\bgroup\color{black}\psi_\pm\left(\vec{r},\vec{R}\right)=C_\pm(R)\left[\psi_A\pm\psi_B\right]\egroup\end{displaymath}

where \bgroup\color{black}$\psi_A=\sqrt{1\over{\pi a^3_0}}e^{-r_{\scriptscriptstyle 1A}/a_0}$\egroup is g.s. around nucleus A. \bgroup\color{black}$\psi_A$\egroup and \bgroup\color{black}$\psi_B$\egroup are not orthogonal; there is overlap. The symmetric (bonding) state has a large probability for the electron to be found between nuclei. The antisymmetric (antibonding) state has a small probability there, and hence, a much larger energy. Remember, this symmetry is that of the wavefunction of one electron around the two nuclei.

The $\mathrm {H}_2$ molecule is also simple and its energy can be computed with the help of the previous calculation. The space symmetric state will be the ground state.

\begin{displaymath}\bgroup\color{black}\left< \psi\vert H\vert\psi\right>=2E_{H_...
...si\left\vert {e^2\over{r_{12}}} \right\vert \psi\right> \egroup\end{displaymath}

The molecule can vibrate in the potential created when the shared electron binds the atoms together, giving rise to a harmonic oscillator energy spectrum.

Molecules can rotate like classical rigid bodies subject to the constraint that angular momentum is quantized in units of \bgroup\color{black}$\hbar$\egroup.

\begin{displaymath}\bgroup\color{black}E_{rot}={1\over 2}{L^2\over I}={\ell(\ell...
...c^2\over 2}\approx{m\over M}E
\approx {1\over {1000}} eV\egroup\end{displaymath}

Jim Branson 2013-04-22