Quantum Liouville-Propagator Operator Trace Identity How can i proove:
$Tr(Ae^{Lt}B)=Tr(Be^{-Lt}A)$
were $A$ 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: $A(t)= e^{Lt} A(0)$
 A: The trace identity you are looking for is actually 
$$
Tr(X^\dagger) = \left( Tr(X) \right)^*
$$
It follows from the definition of the adjoint, $<m|X^\dagger|n> = <n|X|m>^* $, as applied to the diagonal elements in the trace. 
However, the original identity you ask about derives from the definition of the adjoint super-operator on the space of linear operators using the trace inner product. So, if X, Y are arbitrary linear operators, not necessarily self-adjoint observables, their inner product is
$$
(X|Y) = Tr(X^\dagger Y) = (Y|X)^*
$$
and the adjoint of a super-operator $\mathfrak{O}$ is defined as usual by
$$
(X|{\mathfrak O}^\dagger|Y) = (Y|{\mathfrak O}|X)^*\\
\text{or} \;\;Tr\left( X^\dagger {\mathfrak O}^\dagger(Y) \right) = Tr^*\left( Y^\dagger {\mathfrak O}(X) \right)
$$
Now look at your identity. For $A=A^\dagger$, $B=B^\dagger$, ${\mathfrak L}A = \frac{i}{\hbar}[H,A]$, we have
$$
Tr\left( A e^{{\mathfrak L}t} B \right) = (A|e^{{\mathfrak L}t}|B)\\
Tr\left( B e^{-{\mathfrak L}t} A \right) = (B|e^{-{\mathfrak L}t}|A)
$$
Taking into account that ${\mathfrak L}^\dagger = -{\mathfrak L}$, we also have
$$
(B|e^{-{\mathfrak L}t}|A) = (B|e^{{\mathfrak L}^\dagger t}|A) = (B|\left(e^{{\mathfrak L} t}\right)^\dagger|A) = (A|e^{{\mathfrak L} t}|B)^*
$$
This is what we need but for the complex conjugate. To resolve the latter, note that the action of ${\mathfrak L}$ on a self-adjoint operator produces a self-adjoint operator,
$$
\left( {\mathfrak L}A \right)^\dagger = \left( \frac{i}{\hbar}[H,A] \right)^\dagger = \frac{i}{\hbar}[H,A] = {\mathfrak L}A
$$
and so the action of the propagator $e^{{\mathfrak L}t}$ on a self-adjoint operator also produces a self-adjoint operator,
$$
\left( e^{{\mathfrak L}t}A \right)^\dagger = e^{{\mathfrak L}t}A
$$
Now take into account the above and the cyclic property of trace to obtain
$$
(A|e^{{\mathfrak L} t}|B)^* = (A|\left(e^{{\mathfrak L} t}B\right))^* = ( \left(e^{{\mathfrak L} t}B\right)|A) = Tr\left( \left(e^{{\mathfrak L} t}B\right) A\right) = (A|e^{{\mathfrak L} t}|B) 
$$
We can conclude that 
$$
(B|e^{-{\mathfrak L}t}|A) = (A|e^{{\mathfrak L} t}|B)\\
\text{or} \;\; Tr\left( A e^{{\mathfrak L}t} B \right) = Tr\left( B e^{-{\mathfrak L}t} A \right)
$$
