# Andrew Quantum Vibrator

The Hamiltonian is modeled in the supplemental material: ${\hat {H}}=-h{\frac {f_{q}}{2}}\sigma _{z}+hf_{r}a_{-}a_{+}-{\frac {ih\Omega }{2}}(\sigma _{-}a_{+}-\sigma _{+}a_{-})$,

where $\sigma$ represents the quibit and $a$ represents the resonator. The subscripts represent the z and raising and lowering operators. The first two terms are energy terms (look at the Harmonic oscillator in chapter 2), the third term is the interaction term

This Hamiltonian is the coupled variety so it produces a coupled wave function: $\Psi ={\frac {1}{{\sqrt {2}}}}\left(|q_{0}\;r_{1}\rangle e^{{{\frac {-iE_{{01}}t}{\hbar }}}}+|q_{1}\;r_{0}\rangle e^{{{\frac {-iE_{{10}}t}{\hbar }}}}\right)$

assuming the cubit and resonator only has 2 states (they go on to say, in actuality there are probably 5 or 6 states, but you can use this for simplicity)

then $\langle V_{{int}}\rangle =\langle \Psi |{\hat {V}}_{{int}}|\Psi \rangle >$

where ${\hat {V}}_{{int}}$ is the third term of the hamiltonian

If you assume $\sigma _{-}a_{+}|q_{0}\;r_{1}\rangle =0$ $\sigma _{-}a_{+}|q_{1}\;r_{0}\rangle =|q_{0}\;r_{1}\rangle$ $\sigma _{+}a_{-}|q_{0}\;r_{1}\rangle =|q_{1}\;r_{0}\rangle$ $\sigma _{+}a_{-}|q_{1}\;r_{0}\rangle =0$,

which is true for a coupled two state system.

If you do ${\hat {V}}_{{int}}$ you will get the expectation for an interaction to occur as a function of time with a sinusoidal relationship. If you can show that then that is your quantitative piece.