计谱原理（二）|作业11

1

\[ \begin{align} \bra{g} \eta \ket{n} ^{*}\times \bra{g} \eta \ket{n} &=\bra{g} (\cos kr+i\sin kr)\nabla\ket{n} ^{*}\times\bra{g} (\cos kr+i\sin kr)\nabla \ket{n}) \\ &=\bra{g}\cos kr\nabla \ket{n} \times \bra{g} \cos kr\nabla \ket{n} +\bra{g} \sin kr\nabla \ket{n} \times \bra{g}\sin kr\nabla\ket{n}+i\bra{g} \cos kr\nabla \ket{n} \times \bra{g} \sin kr\nabla \ket{n} -i\bra{g} \sin kr\nabla \ket{n} \times \bra{g} \cos kr\nabla \ket{n} \\ &=2i\bra{g} \cos kr\nabla \ket{n} \times \bra{g} \sin kr\nabla \ket{n}\\ A^{2}_{0}i\hat{\epsilon_{3}}\cdot[\bra{g} \eta \ket{n} ^*\times \bra{g} \eta \ket{n} ]&=A^2_{0}i\hat{\epsilon_{3}}\cdot 2i\bra{g} \cos kr\nabla \ket{n} \times \bra{g} \sin kr\nabla \ket{n} \\ &=-2A_{0}^{2}\mathrm{Im}\bra{g} e^{ -ikr }\nabla_{\hat{\epsilon}}\ket{n} \bra{g} e^{ ikr }\nabla_{\hat{\epsilon}}\ket{g}\\ &=\Delta ECD \end{align} \]

2

\[ \begin{align} \mathcal{H}'_{2} (t)&=\frac{1}{2}(\hat{A_{1}}\cdot\hat{A_{1}}+\hat{A_{2}}\cdot\hat{A_{2}}+\hat{A_{2}}\cdot\hat{A_{1}}+\hat{A_{1}}\cdot\hat{A_{2}}) \\ \end{align} \] 其中， 只有 \(\hat{A}_{1}\cdot \hat{A}_{2}+\hat{A}_{2}\hat{A}_{1}\) 有意义 \[ \begin{align} A_{1}=\sqrt{ \frac{2\pi}{\omega_{1} V} }\hat{b_{1}}e^{ -2\omega_{1} t } \\ A_{2}=\sqrt{ \frac{2\pi}{\omega_{2} V}}\hat{b_{2}} e^{ -2\omega_{2} t } \end{align} \] \[ \begin{align} \mathcal{H'_{(2)}}(t)&=\frac{1}{2}(A_{1}\cdot A_{2}+A_{2} \cdot A_{1} ) \\ & =\frac{2\pi}{ \omega_{1} V}(\hat{b_{1}^2e^{ -4\omega_{1}t }}) \end{align} \] \[ \begin{align} \mathcal{H'}_{(2)}(t) & =\hat{A_{1}}\cdot \hat{p}+\hat{A_{2}}\cdot \hat{p} \\ & =2\sqrt{ \frac{2\pi}{V\omega_{1} } }(\hat{p} \hat{b_{1}}e^{ -i\omega t}) \end{align} \]