Consider the Hamiltonian $$ \hat{H} = \frac{E_C}{2} \hat{k}^2 - E_J f(\hat{\Phi}), $$ where $\hat{k}$ and $\hat{\Phi}$ are canonically conjugated variables satisfying the commutation relations $[\hat{\Phi}, \hbar\hat{k}] = i\hbar$ and $f(\hat{\Phi})$ is a regular function, say for example $f(\hat{\Phi}) = \hat{\Phi}^2$ or $f(\hat{\Phi}) = \cos{\hat{\Phi}}$. The physical background is not really important for the question, but the latter is essentially a way to describe an atomic Josephson junction beyond semi-classical approximation, see e.g. [1].
In this paper the authors claim that: "An accurate expression for the lowest excitation frequency is provided by the ratio $\hbar^2\omega^2_P = m_3/m_1$", where $$ m_3 = \langle [ [\hat{k}, \hat{H}], [\hat{H}, [\hat{H}, \hat{k}]]] \rangle \;\;\;\;\;\;\; m_1 = \langle [ \hat{k}, [\hat{H}, \hat{k}] ] \rangle $$ and $\langle \cdot \rangle$ denotes the bra-ket with respect to the ground state of $\hat{H}$. From a quick research, I would say that $m_1$ and $m_3$ are moments of the structure factor associated to the operator $\hat{k}$ as a consequence of sum rules.
I have two questions about this result:
- what does $\hbar\omega_P$ represent? As far as I understand, it is the energy of the first excited state of $\hat{H}$, but I am not completely sure.
- where does the result $\hbar^2\omega^2_P = m_3/m_1$ come from? How can it be proved (a sketchy proof is also fine)? I understand that this has to do with linear response theory, but I can't really find this result in standard references on the topic.
