- #1
matteo137
- 43
- 9
- TL;DR Summary
- What is the state evolution under a P*P entangling Hamiltonian
Consider two harmonic oscillators, described by annihilation operators [itex]a[/itex] and [itex]b[/itex], both initially in the vacuum state. Let us imagine that there is a coupling mechanism governed by the Hamiltonian [itex]H=P_A P_B[/itex], where [itex]P_i[/itex] is the momentum operator for the oscillator [itex]i[/itex]. For example [itex] P_A = (a-a^\dagger)/(i\sqrt{2}) [/itex].
I would like to derive the time evolution
[tex]
\vert \psi \rangle = e^{-i t P_A P_B} \vert 0,0 \rangle
[/tex]
but it is not very clear to me how to proceed. I can decompose the exponential as [itex]e^{i \mu P_A P_B} = e^{i t P_A (b-b^\dagger)}[/itex], and using [itex][P_A b, - P_A b^\dagger]=-P_A^2 [/itex] in the Baker-Campbell-Hausdorff (BCH) formula I obtain [itex] 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} [/itex]. If needed the BCH can be applied once more to [itex]P_A[/itex]. However, I'm not sure if this is useful, or how to continue from there.
I would like to derive the time evolution
[tex]
\vert \psi \rangle = e^{-i t P_A P_B} \vert 0,0 \rangle
[/tex]
but it is not very clear to me how to proceed. I can decompose the exponential as [itex]e^{i \mu P_A P_B} = e^{i t P_A (b-b^\dagger)}[/itex], and using [itex][P_A b, - P_A b^\dagger]=-P_A^2 [/itex] in the Baker-Campbell-Hausdorff (BCH) formula I obtain [itex] 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} [/itex]. If needed the BCH can be applied once more to [itex]P_A[/itex]. However, I'm not sure if this is useful, or how to continue from there.