# Coherent States of Coupled Oscillators

I am given the following coupled Hamiltonian $$H=\hbar\omega[a_x^\dagger a_x+a_y^\dagger a_y+1+K(a_x^\dagger a_y+a_y^\dagger a_x)]$$ Which I can decouple into two different oscillators $$H_u=\frac 12 \hbar \omega(1+K)(a_u^\dagger a_u+1)$$ $$H_v=\frac 12 \hbar \omega(1-K)(a_v^\dagger a_v+1)$$ I am then given that the system is prepared in a state $|\psi(0)\rangle=|0\rangle_x|\alpha(0)\rangle_y$ is a coherent state such that $$\langle\psi(0)|y|\psi(0)\rangle=y_0$$ $$\langle\psi(0)|p_y|\psi(0)\rangle=0$$ I find that $$\alpha(0)=y_0\sqrt{\frac{m\omega}{2\hbar}}$$ Finally, I am asked to show that each of the oscillators is in a coherent state for all time and that $\langle x(t)\rangle$ and $\langle y(t)\rangle$ act as coupled classical oscillators. I am not sure how to show that they act as coherent states. Showing the expectation values seems simple enough once I can find x and y as functions of time.

1. Show that the initial state is a (tensor product of) coherent state for the new oscillator modes i.e., it is also an eigen vector of both $a_u$ and $a_v$.
2. Next, show that any coherent state continues to remain a coherent state under time evolution, possibly with a time-dependent $\alpha$, where $\alpha$ is the parameter associated with the coherent state.
• And the initial state was that $|0\rangle_x|\alpha(0)\rangle_y$. So I act the u and v operators to make sure I get them back? – yankeefan11 Feb 28 '14 at 3:59
• Yes! You should show that $a_u |\psi(0)\rangle = \alpha_u|\psi(0)\rangle$ and $a_v |\psi(0)\rangle = \alpha_v|\psi(0)\rangle$ for some constants $\alpha_u$ and $\alpha_v$. – suresh Feb 28 '14 at 7:18
• So I have been trying this. I have that $a_u=(a_x+a_y)$ so $a_u|\psi(0)\rangle=a_x|0\rangle_x|\alpha(0)\rangle_y + a_y|0\rangle_x|\alpha(0)\rangle_y$. I believe that the $a_x$ will eliminate the zero state, but can I figure what $a_y|\alpha(0)\rangle_y$ is? – yankeefan11 Mar 2 '14 at 1:34
• Ah. So you don't know that coherent states are eigenstates of the annihilation operator. Well, they can be defined as follows: $a\ |\alpha\rangle = \alpha\ |\alpha\rangle$, where $\alpha$ is any complex number. $\alpha=0$ corresponds to the ground state of the SHO. – suresh Mar 2 '14 at 2:18
• I know that from your answer. I am just wondering if I can say that $a_y$ on the x state will be ignored. – yankeefan11 Mar 2 '14 at 2:21