How to represent a Liouville projection superoperator in Hilbert space? Is there a general way to represent a Liouville projection operator in Hilbert space, or can they take on any form so long as they satisfy the required properties of a projector?
e.g. The thermal projection superoperator is represented in Hilbert space as:
$$ \mathcal{P}\hat{\rho}=\hat{\rho}^{(B)}_{eq}\otimes \mathrm{Tr}_B[\hat{\rho}] $$ 
Where $\hat{\rho}$ is the density matrix, $\mathrm{Tr}_B$ is a partial trace over "bath" states and $\hat{\rho}^{(B)}_{eq}$ is an equilibrium density matrix in the bath subspace.
This is the only projection superoperator I've come across and I was wondering if they can always be written in a form similar to this, any information on projection superoperators would be appreciated.  
 A: A simple way to construct projection superoperators is to follow the direct analogy with the usual Hilbert space. Let $\mathscr H$ be the Hilbert space and $\mathscr L(\mathscr H)$ the space of linear operators on $\mathscr H$ equipped with the trace scalar product, $(A|B) = Tr(A^\dagger B)$. Then ${\mathcal P}:{\mathscr L}({\mathscr H}) \rightarrow {\mathscr L}({\mathscr H})$ is a projector provided ${\mathcal P} = {\mathcal P}^2$. 
If we ask that in addition ${\mathcal P} = {\mathcal P}^\dagger$, then by the usual argument nonvanishing eigenvalues of ${\mathcal P}$ are all unit, and if $\lbrace \hat \alpha \rbrace_\alpha$, $(\hat\alpha|\hat\alpha') = Tr(\hat\alpha^\dagger \hat\alpha') = \delta_{\alpha'\alpha}$ is its orthonormal eigenbasis in $\mathscr L(\mathscr H)$, we have 
$$
{\mathcal P} = \sum_\alpha{|\hat\alpha)(\hat\alpha|}\\
{\mathcal P}(\hat \chi) = \sum_\alpha{Tr(\hat\alpha^\dagger \hat\chi) \hat\alpha}\\
{\mathcal P}^2(\hat \chi) = {\mathcal P}\left(\sum_\alpha{Tr(\hat\alpha^\dagger \hat\chi) \hat\alpha}\right) = \sum_\alpha{Tr(\hat\alpha^\dagger \hat\chi) {\mathcal P}(\hat\alpha)} =\\
= \sum_\alpha{Tr(\hat\alpha^\dagger \hat\chi) \sum_\beta{Tr(\hat\beta^\dagger \alpha) \hat\beta}} = \sum_\alpha{Tr(\hat\alpha^\dagger \hat\chi) \sum_\beta{\delta_{\alpha\beta} \hat\beta}} = {\mathcal P}(\hat \chi)
$$ 
But the thermal projector is not self-adjoint. It reads instead
$$
{\mathcal P}_{th} = |\hat\rho_{eq})(\hat I|\\
{\mathcal P}_{th}(\hat \chi) = \hat\rho_{eq} Tr(\hat\chi)
$$
or for a bipartite system AB, 
$$
{\mathcal P}^{(B)}_{th} \equiv {\mathcal I}^{(A)}\otimes{\mathcal P}^{(B)}_{th} = {\mathcal I}^{(A)}\otimes|\hat\rho^{(B)}_{eq})_B(\hat I^{(B)}|\\
{\mathcal P}^{(B)}_{th}(\hat\chi) = Tr_B(\hat\chi)\otimes\hat\rho^{(B)}_{eq} 
$$
Fortunately other non-selfadjoint superoperator projectors are not difficult to construct. For instance any
$$
{\mathcal P} = |\hat\eta)(\hat\xi|\;\;\;\text{with} \;\;Tr(\hat\xi^\dagger\hat\eta) = 1\\
{\mathcal P}(\hat\chi) = Tr(\hat\xi^\dagger\hat\chi) \hat\eta
$$
Or take $\lbrace \hat \alpha \rbrace_\alpha$ as a nonorthogonal basis, $(\hat\alpha|\hat\alpha') = Tr(\hat\alpha^\dagger \hat\alpha') \neq \delta_{\alpha'\alpha}$, and let $\lbrace \hat {\bar \alpha} \rbrace_\alpha$ be the biorthogonal dual, $(\hat{\bar\alpha}|\hat\alpha') = Tr(\hat{\bar\alpha}^\dagger \hat\alpha') = \delta_{\alpha'\alpha}$. Then for any subset $S$ of labels $\alpha$ we can define 
$$
{\mathcal P} = \sum_{\alpha\in S}{|\hat\alpha)(\hat{\bar\alpha}|}\\
{\mathcal P}(\hat \chi) = \sum_{\alpha\in S}{Tr(\hat{\bar\alpha}^\dagger \hat\chi) \hat\alpha}
$$
and verify that ${\mathcal P} = {\mathcal P}^2$. 
Note that superoperator projectors may have features without analog for regular projectors $\hat P\in\mathscr L(\mathscr H)$. For instance there are superoperator projectors that map any self-adjoint operator into another self-adjoint operator, the thermal projector being a trivial example, in the sense that it maps any trace class operator into a self-adjoint operator.
However, here are the more familiar examples: Let $\hat P\in\mathscr L(\mathscr H)$, $\hat P^2 = \hat P$ be a regular projector on $\mathscr H$, and let $\hat Q = \hat I - \hat P$ be its complement. Then all of the following define superoperator projectors on $L(\mathscr H)$: 
$$
{\mathcal P}(\hat\chi) = \hat P \hat\chi,\;\;\;\tilde{\mathcal P}(\hat\chi) = \hat\chi\hat P \\
{\mathcal Q}(\hat\chi) = \hat Q \hat\chi,\;\;\;\tilde{\mathcal Q}(\hat\chi) = \hat\chi\hat Q \\
{\mathcal P}\tilde{\mathcal P}(\hat\chi) = \hat P\hat \chi\hat P,\;\;\;{\mathcal Q}\tilde{\mathcal Q}(\hat\chi) = \hat Q\hat \chi\hat Q\\
{\mathcal P}\tilde{\mathcal Q}(\hat\chi) = \hat P\hat \chi\hat Q,\;\;\;{\mathcal Q}\tilde{\mathcal P}(\hat\chi) = \hat Q\hat \chi\hat P
$$
and assorted combinations involving $\hat P^\dagger$, $\hat Q^\dagger$, etc. In particular, we can check that ${\mathcal P}^\dagger(\hat\chi) = \hat P^\dagger\hat\chi$ and that ${\mathcal P}^\dagger\tilde{\mathcal P}$ maps any self-adjoint operator into a self-adjoint operator, etc. 
