Exponential of momenta to entangle harmonic oscillators

[matteo137]

TL;DR

What is the state evolution under a P*P entangling Hamiltonian

Consider two harmonic oscillators, described by annihilation operators $a$ and $b$, both initially in the vacuum state. Let us imagine that there is a coupling mechanism governed by the Hamiltonian $H=P_A P_B$, where $P_i$ is the momentum operator for the oscillator $i$. For example $P_A = (a-a^\dagger)/(i\sqrt{2})$.

I would like to derive the time evolution
$$\vert \psi \rangle = e^{-i t P_A P_B} \vert 0,0 \rangle$$
but it is not very clear to me how to proceed. I can decompose the exponential as $e^{i \mu P_A P_B} = e^{i t P_A (b-b^\dagger)}$, and using $[P_A b, - P_A b^\dagger]=-P_A^2$ in the Baker-Campbell-Hausdorff (BCH) formula I obtain $e^{i t P_A (b-b^\dagger)}=e^{i t P_A b} e^{-i t P_A b^\dagger} e^{- \frac{t^2}{2} P_A^2}$. If needed the BCH can be applied once more to $P_A$. However, I'm not sure if this is useful, or how to continue from there.

 

[Paul Colby]

The hamiltonian you provide is only the interaction. To this one must add the free non-interacting HO hamiltonian $\omega_a a^\dagger a + \omega_b b^\dagger b$. Once this is added use the BHC relation to

$e^{itH}|0,0\rangle$

where

$H = \omega_a a^\dagger a + \omega_b b^\dagger b + P_AP_B$

good luck!

 

[HomogenousCow]

You could probably diagnaolize that Hamiltonian with a linear combination of $a$s and $b$s.