Exponential of momenta to entangle harmonic oscillators

Click For Summary
SUMMARY

The discussion focuses on deriving the time evolution of two coupled harmonic oscillators using the Hamiltonian \( H = P_A P_B \), where \( P_A \) and \( P_B \) are the momentum operators defined as \( P_A = \frac{(a - a^\dagger)}{i\sqrt{2}} \) and \( P_B \) similarly for the second oscillator. The time evolution is expressed as \( \vert \psi \rangle = e^{-i t P_A P_B} \vert 0,0 \rangle \). The Baker-Campbell-Hausdorff (BCH) formula is utilized to decompose the exponential, leading to the expression \( e^{i t P_A (b - b^\dagger)} = e^{i t P_A b} e^{-i t P_A b^\dagger} e^{-\frac{t^2}{2} P_A^2} \). The complete Hamiltonian also includes the free non-interacting harmonic oscillator terms \( \omega_a a^\dagger a + \omega_b b^\dagger b \), which must be incorporated for a full solution.

PREREQUISITES
  • Understanding of quantum mechanics and harmonic oscillators
  • Familiarity with annihilation and creation operators
  • Knowledge of the Baker-Campbell-Hausdorff (BCH) formula
  • Basic concepts of Hamiltonian mechanics in quantum systems
NEXT STEPS
  • Study the Baker-Campbell-Hausdorff (BCH) formula in detail
  • Learn about the time evolution operator in quantum mechanics
  • Explore diagonalization techniques for Hamiltonians involving multiple operators
  • Investigate the implications of coupling in quantum harmonic oscillators
USEFUL FOR

Quantum physicists, researchers in quantum mechanics, and students studying harmonic oscillators and their interactions will benefit from this discussion.

matteo137
Messages
42
Reaction score
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.
 
Physics news on Phys.org
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!
 
You could probably diagnaolize that Hamiltonian with a linear combination of ##a##s and ##b##s.
 
  • Like
Likes   Reactions: vanhees71

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K