Appendix: Energy Level Corrections Using Perturbation Theory
Using perturbation theory, when \( \mathcal{H} = H_0 + \Delta H \), and the unperturbed eigenstate and eigenvalue are \( |E_0^m\rangle \) and \( E_0^m \), respectively, the perturbed eigenstate and eigenvalue are \( |E^m\rangle \) and \( E^m = E_0^m + \delta E^m \). The energy correction \(\delta E^m\) is given by:
\[ \delta E^m = \langle E_0^m | \Delta H | E_0^m \rangle = \langle E_0^m | \left[-\frac{\hbar^2}{96C} \left(\frac{\pi}{\Phi_0}\right)^2 (a^\dagger + a)^4\right] | E_0^m \rangle. \]
Complete Derivation
\[ \langle E_0^m | (a^\dagger + a)^4 | E_0^m \rangle = \langle E_0^m | (aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger)^2 | E_0^m \rangle. \]
Expanding this:
\[ \langle E_0^m | aa(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) + aa^\dagger(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) \] \[ + a^\dagger a(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) + a^\dagger a^\dagger(aa + aa^\dagger + a^\dagger a + a^\dagger a^\dagger) | E_0^m \rangle. \]
Taking the expectation value leaves only terms where \( a \) and \( a^\dagger \) appear in equal powers:
\[ \langle E_0^m | aaa^\dagger a^\dagger + aa^\dagger aa^\dagger + aa^\dagger a^\dagger a + a^\dagger aaa^\dagger + a^\dagger aa^\dagger a + a^\dagger a^\dagger aa | E_0^m \rangle. \]
Simplifying further:
\[ \langle E_0^m | a(a^\dagger a + 1)a^\dagger + aa^\dagger aa^\dagger + (a^\dagger a + 1)a^\dagger a + a^\dagger a(a^\dagger a + 1) + a^\dagger aa^\dagger a + a^\dagger (aa^\dagger - 1)a | E_0^m \rangle. \]
After further simplifications:
\[ \langle E_0^m | 2aa^\dagger aa^\dagger + aa^\dagger + 4a^\dagger aa^\dagger a + a^\dagger a | E_0^m \rangle. \]
The final result is:
\[ \langle E_0^m | 6a^\dagger aa^\dagger a + 6a^\dagger a + 3 | E_0^m \rangle = 6m^2 + 6m + 3. \]
The energy difference between levels \(m\) and \(m-1\) is:
\[ \Delta E^m_{m-1} = \frac{\hbar}{\sqrt{LC}} - \frac{\hbar^2}{96C} \left(\frac{\pi}{\Phi_0}\right)^2 \left\{6[m^2 - (m-1)^2] + 6\right\}. \]
\[ = \frac{\hbar}{\sqrt{LC}} - \frac{\hbar^2 \pi^2}{96C} \frac{e^2}{\pi^2 \hbar^2} \left[6(2m - 1) + 6\right]. \]
Finally, we find:
\[ \Delta E^m_{m-1} = \frac{\hbar}{\sqrt{LC}} - m\frac{e^2}{8C}. \]
Originally written in Chinese by the author, these articles are translated into English to invite cross-language resonance.
Peir-Ru Wang