matteo137
- 42
- 9
- 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
<br /> \vert \psi \rangle = e^{-i t P_A P_B} \vert 0,0 \rangle<br />
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.
I would like to derive the time evolution
<br /> \vert \psi \rangle = e^{-i t P_A P_B} \vert 0,0 \rangle<br />
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.