## 2 Laser Control of Ring Currents in a Molecule Having a Single Aromatic Ring
### Laser-Polarization Effects on Ring Currents within the Frozen-Nuclei Approximation
#### 2.2.3 Case of Degenerate Excited States  
Before proceeding to the solution of Equation (16) for a quasi-degenerate system, let us consider the degenerate case of \(\Delta\omega\) = 0 for highly symmetric aromatic molecules such as benzene and Mg porphyrin. In this case, Equation (16) becomes simple, namely:  
\[\frac{d\mathbf{D}(t)}{dt}=i\frac{\overline{\Omega}}{2}f(t)\left( \begin{array}[ ]{ccc}0&1&0\\ 1&0&0\\ 0&0&0\end{array} \right)\mathbf{D}(t). \tag{17}\]  
The system practically consists of the two states \(|G\rangle\) and \(|U_{+}\rangle\). It is easy to integrate Equation (17):  
\[\mathbf{D}(t)=\left( \begin{array}{c}\cos\Big{[}\frac{\overline{\Omega}}{2}F (t)\Big{]}\\ i\sin\Big{[}\frac{\overline{\Omega}}{2}F(t)\Big{]}\\ 0\end{array} \right), \tag{18}\]  
where \(F(t)\equiv\int_{0}^{t}dt^{\prime}f(t^{\prime})\). This indicates that the so-called \(\pi\) pulse [59], which satisfies \(\overline{\Omega}F\Big{(}t_{f}\Big{)}=\pi\), achieves complete population inversion from \(|G\rangle\) to \(|U_{+}\rangle\). Inserting Equation (18) into Equation (14) yields the following:  
\[|\Psi(t)\rangle=\cos\Bigg{[}\frac{\overline{\Omega}}{2}F(t)\Bigg{]}|G\rangle+i \frac{\Omega_{X}}{\overline{\Omega}}\sin\Bigg{[}\frac{\overline{\Omega}}{2}F( t)\Bigg{]}e^{-i(\omega t+\varphi)}\Big{(}|X\rangle+e^{i\theta}|Y\rangle\Big{)}. \tag{19}\]  
The relative quantum phase between \(|X\rangle\) and \(|Y\rangle\) coincides with that between the Rabi frequencies \(\Omega_{X}\) and \(\Omega_{Y}\), i.e., \(\theta\), which is independent of time and controllable by the ellipticity angle \(\beta\) and orientation angle \(\delta\). A desired superposition of the degenerate states can be created by tuning the polarization of incident light properly. The populations of \(|+\rangle\) and \(|-\rangle\), which are the exact eigenstates of \(\hat{l}_{z}\) in this case, are given by:  
\[P_{\pm}(t)\equiv|\langle\pm|\Psi(t)\rangle|^{2}=\frac{1}{2}\sin^{2}\Bigg{[} \frac{\overline{\Omega}}{2}F(t)\Bigg{]}(1\pm\sin\theta) \tag{20}\]  
and the expectation value of \(\hat{l}_{z}\) is:  
\[l_{z}(t)\equiv\Big{\langle}\Psi(t)\Big{|}\hat{l}_{z}\Big{|}\Psi(t)\Big{\rangle} =h\sin^{2}\Bigg{[}\frac{\overline{\Omega}}{2}F(t)\Bigg{]}\sin\theta. \tag{21}\]  
The sign of \(l_{z}(t)\) (in other words, the direction of ring currents) depends on the relative quantum phase \(\theta\) and remains unchanged throughout the time evolution: Ring current flows in a sole direction even after a laser pulse vanishes at \(t=t_{f}\). We revert to Equation (12), which connects the relative quantum phase \(\theta\) to the ellipticity angle \(\beta\) and orientation angle \(\delta\). As stated for benzene, where \(\left|G\right\rangle=\left|1^{1}\text{A}_{1g}\right\rangle\), \(\left|X\right\rangle=\left|1^{1}\text{E}_{1 ux}\right\rangle\), and \(\left|Y\right\rangle=\left|1^{1}\text{E}_{1 uw}\right\rangle\), in Section 2.1, \(\mu_{X}\) and \(\mu_{Y}\) of highly symmetric aromatic molecules have the same magnitude and are perpendicular to each other: \(\left|\ \mu_{X}\ \right|\ =\ \left|\ \mu_{Y}\ \right|\ \mid\ \text{and}\ \mu_{X}\cdot\mu_{Y}=0\). The \(x\) and \(y\) axes can thus be chosen to be parallel to \(\mu_{X}\) and \(\mu_{Y}\), respectively. As a result, Equation (12) is reduced to:  
\[e^{i2\delta}\tan\beta=-\frac{e^{i\theta}-i}{e^{i\theta}+i}=e^{i\pi/2}\tan\left[ \frac{1}{2}\left(\frac{\pi}{2}-\theta\right)\right], \tag{22}\]  
which involves neither transition moments nor spherical unit vectors and is purely imaginary. Given a value of \(\beta\), there exist two solutions of Equation (22):  
\[\delta=\pm\frac{\pi}{4}\text{ and }\theta=\frac{\pi}{2}\mp 2\beta. \tag{23}\]  
The major axis of the polarization ellipse with \(\delta=\pi/4\) bisects the angle between \(\mu_{X}\) and \(\mu_{Y}\); that with \(\delta=-\pi/4\) is perpendicular to it. In both cases, the value of \(l_{z}(t_{f})\) for the \(\pi\)-pulse excitation is \(h\text{sin}\theta=h\text{cos}2\beta\).