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.


1 Answer 1


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.

  • $\begingroup$ One should remember that density operators are elements of the trace class where the concept of orthogonality does not apply. Thus, if one wants to consider the problem in terms of the trace class topology the situation is somewhat different. $\endgroup$
    – Urgje
    Feb 29, 2016 at 11:45
  • $\begingroup$ @Urgje Superoperators should be able to act on the full matrix space, not only on the space of trace class operators. Otherwise, how would you define their action on part of a larger system? $\endgroup$ Feb 29, 2016 at 11:47
  • $\begingroup$ Given a superoperator acting on elements of the trace class one obtains it on the class of bounded operators (with the sup norm) by duality. What do you mean with full matrix space? See also en.wikipedia.org/wiki/Lindblad_superoperator for a general setting. $\endgroup$
    – Urgje
    Feb 29, 2016 at 11:57
  • $\begingroup$ @Urgje Your reference to the Lindblad generator suggests that what you have in mind is the convex set of positive definite operators whose trace class subset covers density matrices. Keep in mind though that the Heisenberg representation for a Lindblad generator, and with it the physics described, uses the entire ${\mathscr L}({\mathscr H})$ space. $\endgroup$
    – udrv
    Feb 29, 2016 at 12:32
  • $\begingroup$ @Urgje Regardless, superoperator projectors as a concept are completely independent of what happens on the set of trace class positive definite operators. Some map this set onto itself, see the last ${\mathcal P}^\dagger\tilde{\mathcal P}$ example in the answer, but most do not. If you are interested specifically in the former kind, that is another problem in itself. $\endgroup$
    – udrv
    Feb 29, 2016 at 12:33

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