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Commutation of Lindblad Operators

  1. Dec 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Let operator $$\mathcal{L}_{AD}(\rho)$$ and $$\mathcal{L}_{PD}(\rho)$$ is defined as


    and $$\mathcal{L}_{PD}(\rho)=2a^{\dagger}a\rho{a^{\dagger}}a-(a^{\dagger}a)^2\rho-\rho(a^{\dagger}a)^2$$

    As we could see it is easy to show that $$\mathcal{L}_{AD}(\rho)$$ and $$\mathcal{L}_{PD}(\rho)$$ commute with each other. But if I do explicit calculation to show they commute with each other,
    i.e, $$[\mathcal{L}_{AD}(\rho),\mathcal{L}_{PD}(\rho)]=0$$

    2. Relevant equations

    3. The attempt at a solution
    on explicit calculation I got,
    $$ [a{\rho}{a}^{\dagger} , a^{\dagger}a{\rho}a^{\dagger}a] - 2[a^{\dagger}a{\rho} , a^{\dagger}a{\rho}a^{\dagger}a] - 2[\rho{a}^{\dagger}a , a^{dagger}a{\rho}a^{dagger}a] - 2[a{\rho}a^{\dagger} , (a^{\dagger}a)^2{\rho}] + [a^{\dagger}a{\rho} , (a^{\dagger}a)^2{\rho}] + [{\rho}{a}^{\dagger}a , (a^{\dagger}a)^2{\rho}] - 2[a{\rho}a^{\dagger , {\rho}(a^{\dagger}a)^2}] + [a^{\dagger}a{\rho} , {\rho}(a^{\dagger}a)^2] + [{\rho}{a^{\dagger}a , {\rho}(a^{\dagger}a)^2] $$

    But from here I don't know how to cancel the terms in these commutators, I don't how to proceed further.

    Attached Files:

    Last edited: Dec 21, 2016
  2. jcsd
  3. Dec 27, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
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