计谱原理（三）|作业12

1

其中，\(\hat{p}=i[\mathcal{H}_{0},\hat{\mu_{2}}]\)，则（以第一项为例）： \[ \begin{align} \bra{f}\epsilon_{2} \cdot\hat{p}\ket{r}\bra{r}\epsilon_{1} \cdot \hat{\mu_{1}}\ket{i} & =i\bra{f}\epsilon_{2} \cdot(\hat{\mathcal{H}_{0}}\hat{\mu_{2}}-\hat{\mu_{2}}\hat{\mathcal{H_{0}}})\ket{r}\bra{r}\epsilon_{1} \cdot \hat{\mu_{1}}\ket{i} \\ & =i(\omega_{f}-\omega_{r})\bra{f} \epsilon_{2}\hat{\cdot}\mu_{2} \ket{r} \bra{r }\epsilon_{1}\hat{\cdot}\mu_{1}\ket{i} \\ & =i\omega_{fi}\bra{f} \epsilon_{2}\hat{\cdot}\mu_{2} \ket{r} \bra{r }\epsilon_{1}\hat{\cdot}\mu_{1}\ket{i} \\ & =i(\omega_{1}-\omega_{2}-\omega_{ri})\bra{f} \epsilon_{2}\hat{\cdot}\mu_{2} \ket{r} \bra{r }\epsilon_{1}\hat{\cdot}\mu_{1}\ket{i} \end{align} \] \[ \begin{align} LEFT & =\sum_{r}\left[ \frac{(\omega_{1}-\omega_{2}-\omega_{ri})\bra{f} \epsilon_{2}\cdot\hat{\mu_{2}}\ket{r} \bra{r} \epsilon_{1}\cdot\hat{\mu_{1}}\ket{g} }{\omega_{ri}-\omega_{1} }+\frac{(\omega_{1}-\omega_{2}-\omega_{ri})\bra{f} \epsilon_{1}\cdot\hat{\mu_{1}}\ket{r} \bra{r} \epsilon_{2}\cdot\hat{\mu_{2}}\ket{g} }{\omega_{ri}+\omega_{2} } \right] \\ & =RIGHT \end{align} \]

2

\[ \begin{align} \alpha_{pq}=\frac{\partial\alpha^e_{pq}}{\partial Q_{k}}\bra{\nu^f} Q_{k}\ket{\nu^{i}} \end{align} \] $$ \[\begin{aligned} \frac{I_{S}}{I_{aS}} &=\frac{e^{ \hbar \omega_{k}/k_{B}T }\cdot{(\omega_{1}-\omega_{k})^{4}}}{(\omega_{1}+\omega_{k})^{4}}\cdot \frac{c^{4}}{c^{4}} \\ &=(\frac{\omega_{1}-\omega_{k}}{\omega_{1}+\omega_{k}})^{4}e^{ \hbar \omega_{k}/k_{B}T } \end{aligned}\]

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