# Quantum Liouville-Propagator Operator Trace Identity

+ 2 like - 0 dislike
84 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
retagged Aug 27, 2015
$\mathrm{tr}(X^\dagger) = \mathrm{tr}(X)$ 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
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
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
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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.