Time evolution of a Gaussian State

Question

I consider two coupled harmonic oscillators with annihilation operators $\hat{a}$ and $\hat{b}$ described by the gaussian Hamiltonian: $$ \hat{H} = g(\hat{a}\hat{b}^\dagger + \hat{a}^\dagger \hat{b}) + (\epsilon(t)\hat{a}^\dagger + \epsilon(t)^*\hat{a}) $$ What is the time evolution of a coherent state $|\psi(t=0)\rangle = |\alpha\rangle \otimes |\beta \rangle$ ?

Since there are only gaussian operations, the system is classical and I guess the state at time $t$ should stay coherent, i.e. $|\psi(t)\rangle = |\alpha(t)\rangle \otimes |\beta(t) \rangle$. With only the first term in the Hamiltonian, we would see a beamsplitter operation which interchanges the initial coherent states $|\alpha\rangle$ and $|\beta\rangle$ at rate $g$. With only the second term, the $|\alpha\rangle$ coherent state would be displaced to $|\alpha(t)\rangle = |\alpha - i\int \epsilon(t)\mathrm{d}t\rangle$. But I can't seem to be able to solve the full system. Any idea on how to proceed?

Thanks.

Answer 1

We can choose normal modes for the beam splitter portion, of the form (you can choose your normalization constants) $$ \hat{a}=\hat{c}-\hat{d},\qquad \hat{b}=\hat{c}+\hat{d} $$ such that the Hamiltonian transforms to $$ H=2g(\hat{c}^\dagger\hat{c}-\hat{d}^\dagger\hat{d})+\left[\epsilon(t)\hat{c}+\epsilon(t)^*\hat{c}^\dagger\right] -\left[\epsilon(t)\hat{d}+\epsilon(t)^*\hat{d}^\dagger\right]. $$ Now this is much more manageable - we go to the rotating frame to get rid of the first term, then we simply have a time-dependent displacement on each of the normal modes, which you already explained how to solve! At the end you can go back to the original modes if you like, and everything will indeed remain coherent.