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?


  • $\begingroup$ Is $\hat{H}$ the full Hamiltonian of the system or is it an "interaction" Hamiltonian and there are additional terms $\hbar \omega_a a^\dagger a$ and $\hbar \omega_b b^\dagger b$ in the full Hamiltonian. (Or are you working in the interaction picture and the operators carry the "free" time-dependency.) $\endgroup$ Jun 30, 2021 at 15:22
  • $\begingroup$ But if you know the time-evolution of the individual terms and your ansatz is correct you can reduce the time-evolution to infinitesimal steps to obtain an ODE for $\alpha$ and $\beta$. (And in the end you can check whether your ansatz was correct by checking that your solution fulfils the Schrödinger equation.) $\endgroup$ Jun 30, 2021 at 15:25
  • $\begingroup$ I am working in the rotating frame, so there are no $\hat{a}^\dagger \hat{a}$ and $\hat{b}^\dagger \hat{b}$ terms. So, what would be the infinitesimal effect of the beamsplitter term? Is it $\dot{\alpha}(t) = g (\beta(t)-\alpha(t))$ and $\dot{\beta}(t) = g (\alpha(t)-\beta(t))$? $\endgroup$
    – Ronan
    Jun 30, 2021 at 15:31

1 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.


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