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1. Dec 21, 2016

### Muthumanimaran

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

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

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.

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Last edited: Dec 21, 2016
2. Dec 27, 2016