计谱原理（二）|作业11

1

$$\begin{array}{rl}⟨g|\eta {|n⟩}^{\ast }\times ⟨g|\eta |n⟩& =⟨g|(\cos kr+i\sin kr)\mathrm{\nabla }{|n⟩}^{\ast }\times ⟨g|(\cos kr+i\sin kr)\mathrm{\nabla }|n⟩)\\ & =⟨g|\cos kr\mathrm{\nabla }|n⟩\times ⟨g|\cos kr\mathrm{\nabla }|n⟩+⟨g|\sin kr\mathrm{\nabla }|n⟩\times ⟨g|\sin kr\mathrm{\nabla }|n⟩+i⟨g|\cos kr\mathrm{\nabla }|n⟩\times ⟨g|\sin kr\mathrm{\nabla }|n⟩-i⟨g|\sin kr\mathrm{\nabla }|n⟩\times ⟨g|\cos kr\mathrm{\nabla }|n⟩\\ & =2i⟨g|\cos kr\mathrm{\nabla }|n⟩\times ⟨g|\sin kr\mathrm{\nabla }|n⟩\\ A_0^{2}i\widehat{{ϵ}_3}\cdot [⟨g|\eta {|n⟩}^{\ast }\times ⟨g|\eta |n⟩]& =A_0^{2}i\widehat{{ϵ}_3}\cdot 2i⟨g|\cos kr\mathrm{\nabla }|n⟩\times ⟨g|\sin kr\mathrm{\nabla }|n⟩\\ & =-2A_0^2\mathrm{Im}⟨g|e^{-ikr}{\mathrm{\nabla }}_{\widehat{ϵ}}|n⟩⟨g|e^{ikr}{\mathrm{\nabla }}_{\widehat{ϵ}}|g⟩\\ & =\mathrm{\Delta }ECD\end{array}$$

2

$$\begin{array}{rl}{\mathcal{H}}_2^{\prime }(t)& =\frac{1}{2}(\widehat{A_1}\cdot \widehat{A_1}+\widehat{A_2}\cdot \widehat{A_2}+\widehat{A_2}\cdot \widehat{A_1}+\widehat{A_1}\cdot \widehat{A_2})\end{array}$$ 其中， 只有 ${\hat{A}}_1\cdot {\hat{A}}_2+{\hat{A}}_2{\hat{A}}_1$ 有意义 $$\begin{array}{r}{A}_1=\sqrt{\frac{2\pi }{{\omega }_{1}V}}\widehat{b_1}{e}^{-2{\omega }_{1}t}\\ A_2=\sqrt{\frac{2\pi }{{\omega }_{2}V}}\widehat{b_2}{e}^{-2{\omega }_{2}t}\end{array}$$ $$\begin{array}{rl}{\mathcal{H}}_{\mathcal{(}\mathcal{2}\mathcal{)}}^{\prime }(t)& =\frac{1}{2}(A_1\cdot A_2+A_2\cdot A_1)\\ & =\frac{2\pi }{{\omega }_{1}V}(\widehat{b_1^2{e}^{-4{\omega }_{1}t}})\end{array}$$ $$\begin{array}{rl}{{\mathcal{H}}^{\prime }}_{(2)}(t)& =\widehat{A_1}\cdot \hat{p}+\widehat{A_2}\cdot \hat{p}\\ & =2\sqrt{\frac{2\pi }{V{\omega }_1}}(\hat{p}\widehat{b_1}{e}^{-i\omega t})\end{array}$$