Let us consider Benzene ring \(\displaystyle \mathrm{C}_{6}\mathrm{H}_{6}\) as an example. We will demonstrate quantum chemical calculations of the molecular structure using a rough approximation model (the annular well potential approximation). The \(\pi\) electrons inside Benzene ring \(\displaystyle \mathrm{C}_{6}\mathrm{H}_{6}\) have a radius of 140[pm] under the rough approximation with electrons moving freely on the ring. Under this approximation, the \(\pi\) electrons can be described by the Schrödinger equation with the periodic boundary condition.
There are six \(\pi\) electrons inside Benzene ring. Since their energy levels are occupied by the \(\pi\) electrons two by two in order from the ground state energy level, The energy levels of the magnetic quantum numbers \(m=0\) and \(m=\pm 1\) are occupied by the electrons. Therefore, the highest occupied energy level (HOMO) is \(m=\pm 1\) and the levels immediately above these levels are \(m=\pm 2\). The energy required to excite an electron is described as follows:
\(\Delta E = E_{\pm 2} – E_{\pm 1} = \dfrac{\hbar^{2}}{2mr^{2}}\left((\pm 2)^{2} – (\pm 1)^{2}\right) = 9.34\times 10^{-19} \hspace{3pt}[\mathrm{J}]\).
Therefore, the wavelength of the electromagnetic wave with this energy \(\Delta E\) is written as
\(\lambda = \dfrac{hc}{\Delta E} = 213 \hspace{3pt}[\mathrm{nm}]\).
Note that the energy of an electromagnetic wave is \(\Delta E=hc/\lambda\).
ベンゼン環\(\displaystyle \mathrm{C}_{6}\mathrm{H}_{6}\)を例として，大胆な近似モデル（環状井戸型ポテンシャル近似）により分子構造の量子化学計算を示してみよう．ベンゼン環\(\displaystyle \mathrm{C}_{6}\mathrm{H}_{6}\)内の\(\pi\)電子は，半径140\([\mathrm{pm}]\)の円環上を自由に動くと大胆な近似を行うことにする．この近似により，その\(\pi\)電子は周期境界条件を持つシュレディンガー方程式で記述することができる．
ベンゼン環に含まれる\(\pi\)電子は6個ある．これらのエネルギー準位は基底状態のエネルギー準位から順に下から2個ずつエネルギー準位を電子が占有していくので，磁気量子数\(m=0\)と\(m=\pm 1\)のエネルギー準位までが電子で占有されることになる．最も高い占有されたエネルギー準位（HOMO）は\(m=\pm 1\)であり，これら準位のすぐ上の準位\(m=\pm 2\)に励起するために必要なエネルギーは
\(\Delta E = E_{\pm 2} – E_{\pm 1} = \dfrac{\hbar^{2}}{2mr^{2}}\left((\pm 2)^{2} – (\pm 1)^{2}\right) = 9.34\times 10^{-19} \hspace{3pt}[\mathrm{J}]\)
となる．これが電磁波のエネルギー\(\Delta E=hc/\lambda\)に等しいのだから，このエネルギーを持つ電磁波の波長は
\(\lambda = \dfrac{hc}{\Delta E} = 213 \hspace{3pt}[\mathrm{nm}]\)
となる．
