Lindblad superoperator
\[We\,\,ignore\,\,time\,\,in\,\,\rho , i.e., \rho \,\,is\,\,shorthand\,\,for\,\,\rho \left( t \right)
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\dot{\rho}=-i\left[ H,\rho \right]
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we\,\,define\,\,a\,\,linear\,\,map\,\,\mathcal{L} \rho \equiv -i\left[ H,\rho \right]
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so\,\,we\,\,have\,\,\dot{\rho}=\mathcal{L} \rho
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the\,\,solution\,\,of\,\,it\,\,is\,\,e^{\mathcal{L} t}\rho \left( 0 \right) , proof\,\,as\,\,follows
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\partial _te^{\mathcal{L} t}\rho \left( 0 \right) =\partial _t\left[ I+\mathcal{L} t+\frac{\left( \mathcal{L} t \right) ^2}{2!}+... \right] \rho \left( 0 \right)
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=\left[ \mathcal{L} +\frac{\left( \mathcal{L} t \right) \mathcal{L}}{1}+... \right] \rho \left( 0 \right)
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=\mathcal{L} \left( e^{\mathcal{L} t}\rho \left( 0 \right) \right)
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hence\,\,e^{\mathcal{L} t}\rho \left( 0 \right) \,\,satisfy\,\,the\,\,equation, a\,\,solution\]
We now transfer the solution into a more familiar form.
\[e^{\mathcal{L} t}\rho \left( 0 \right)
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=\left[ I+\mathcal{L} t+\frac{\left( \mathcal{L} t \right) ^2}{2!}+... \right] \rho \left( 0 \right)
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=\rho \left( 0 \right) +\left( -it \right) \left[ H,\rho \left( 0 \right) \right] +\frac{\left( -it \right) ^2\left[ H,\left[ H,\rho \left( 0 \right) \right] \right]}{2!}+...
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=e^{-iHt}\rho \left( 0 \right) e^{iHt}
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by\,\,formula:e^ABe^{-A}=B+\left[ A,B \right] +\frac{1}{2!}\left[ A,\left[ A,B \right] \right] +...\]