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  Quantum Liouville-Propagator Operator Trace Identity

+ 2 like - 0 dislike
1477 views

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

asked Aug 22, 2015 in Theoretical Physics by pindakaas (10 points) [ revision history ]
retagged Aug 27, 2015
Most voted comments show all comments
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
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
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
$tr(X)=\sum_n \langle n| X |n \rangle =\sum_n \langle n| X^\dagger |n \rangle= tr(X^\dagger)$ 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
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
Most recent comments show all comments
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
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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