附錄：微擾理論下的能階修正計算
利用微擾理論，當 \( \mathcal{H} = H_0 + \Delta H \)， \( |E_0^m\rangle \to |E^m\rangle \)、 \( E_0^m \to E^m = E_0^m + \delta E^m \)，我們知道對能階 \( E_0^m \) 的修正 \( \delta E^m \) 為：
\[ \delta E^m = \langle E_0^m | \Delta H | E_0^m \rangle = \langle E_0^m | \left[-\frac{\hbar^2}{96C} \left(\frac{\pi}{\Phi_0}\right)^2 (a^\dagger + a)^4\right] | E_0^m \rangle \]
完整推導過程
\[ \langle E_0^m | (a^\dagger + a)^4 | E_0^m \rangle = \langle E_0^m | (aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger)^2 | E_0^m \rangle \]
展開為：
\[ \langle E_0^m | aa(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) + aa^\dagger(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) \] \[ + a^\dagger a(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) + a^\dagger a^\dagger(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) | E_0^m \rangle \]
定義能階差：
\[ \Delta E^m_{m-1} = \frac{\hbar}{\sqrt{LC}} - \frac{\hbar^2}{96C} \left(\frac{\pi}{\Phi_0}\right)^2 \left\{6[m^2 - (m-1)^2] + 6\right\} \]
\[ = \frac{\hbar}{\sqrt{LC}} - \frac{\hbar^2 \pi^2}{96C} \frac{e^2}{\pi^2 \hbar^2} \left[6(2m - 1) + 6\right] \]
最終得到：
\[ \Delta E^m_{m-1} = \frac{\hbar}{\sqrt{LC}} - m\frac{e^2}{8C} \]
王培儒