Quantum Liouville-Propagator Operator Trace Identity

Question

How can I prove:

$Tr(Ae^{Lt}B)=Tr(Be^{-Lt}A)$

where $A$ and $B$ are observables in the Schrödinger picture and $L$ is the quantum Liouville super-operator defined by:

$LA={ i \over \hbar} [H,A]$

so defining the Liouville propagator as: $A(t)= e^{Lt} A(0)$

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user pindakaas

Comments

$\mathrm{tr}(X^\dagger) = \mathrm{tr}(X)$ and $(XYZ)^\dagger = Z^\dagger Y^\dagger X^\dagger$.

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user ACuriousMind

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As pointed out above, it follows from standard properties of the trace... Voting to close.

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user Danu

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Oh of course taking the adjoint leaves the trace invariant aswell I totally overlooked that. Thanks^^ In fact I am not sure I even ever knew that.

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user pindakaas

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actually how is this true? $Tr(A)=Tr(A^T)$ shure but how is a sum of numbers invariant under complex conjugation?

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user pindakaas

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Okay It is true for self adjoint operators... But the Liouville operator is anti self adjoint right? @Danu how is that a standard property of the trace?

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user pindakaas

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In the same way that cyclicity is.

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user Danu

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I still don't get it $\mathrm{tr}(X^\dagger) = \mathrm{tr}(X)$ would only hold if $X$ was real or self adjoint or am I missing something? So how Does it work if the Liouville propagator is anti-self adjoint?

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user pindakaas

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$tr(X)=\sum_n \langle n| X |n \rangle =\sum_n \langle n| X^\dagger |n \rangle= tr(X^\dagger)$ if $X=X^\dagger$ but $(Ae^{Lt}B)\neq (Ae^{Lt}B)^\dagger$ since ${e^{Lt}}^\dagger = e^{-Lt}$ rigth? So how does it work?

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user pindakaas

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Meta question: meta.physics.stackexchange.com/questions/7002/….

This post imported from StackExchange Physics at 2015-08-27 17:28 (UTC), posted by SE-user HDE 226868