0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Prove your answer in two steps.

  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.

You are done.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – yankeefan11 Feb 28 '14 at 3:59
  • $\begingroup$ 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 $\alpha_v$. $\endgroup$ – suresh Feb 28 '14 at 7:18
  • $\begingroup$ 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? $\endgroup$ – yankeefan11 Mar 2 '14 at 1:34
  • $\begingroup$ 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. $\endgroup$ – suresh Mar 2 '14 at 2:18
  • $\begingroup$ I know that from your answer. I am just wondering if I can say that $a_y$ on the x state will be ignored. $\endgroup$ – yankeefan11 Mar 2 '14 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.