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Suppose we have an initial ensemble described by a density matrix $\rho$ and any given member of the ensemble scatters from one of some set of scattering matrices $\{S_g \equiv O_g S O_g^\dagger : g \in G \}$ where $S$ is some unitary matrix and the $O_g$ form a unitary representation of some symmetry group $G$. Suppose further that each $g$ is equally likely so that the ensemble $\rho'$ after scattering is given by \begin{equation} \rho' = \frac{1}{|G|}\sum_g S_g \rho S^\dagger_g \tag{1}\label{1} \end{equation} The transformation taking $\rho \to \rho'$ is a "super-operator" (operator on operators) which we will define as $\tilde{S}$ so that we can rewrite equation \eqref{1} as $$ \rho' = \tilde{S} \rho $$ We can also transform the representation $O_g$ into a "super-representation" by defining the super-operators $\tilde{O}_g$: $$ \tilde{O}_g \rho := O_g \rho O^\dagger_g $$ It is easy to see that, while $S$ may have no special symmetry with respect to the $O_g$, $\tilde{S}$ turns out to be totally symmetric with respect to the $\tilde{O}_g$, i.e. for any $g \in G$ we have $$ \tilde{O}_g \tilde{S} \tilde{O}^\dagger_g = \tilde{S} $$ Suppose the $S$ act on states in hilbert space of dimension $N$, so that it takes $N^2$ real numbers to define an arbitrary (unitary) scattering matrix. My question is

How many real numbers does it take to describe a general $\tilde{S}$?

It seems reasonable that in general it would be less than $N^2$ in light of the symmetric nature of the $\tilde{S}$.

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I will talk of linear maps or operators instead of super-operators, but in the context it should be clear.

Assume pure states are parametrized by a Hilbert space $V$. Then density operators are i.p. bounded operators on $V$, which is denoted by $B(V)$. What you call superoperators are elements of $B(B(V))$, bounded operators on the bounded operators.

You have outlined how an unitary representation of a group $G$ on $V$ may be lifted to a representation of $G$ on $B(B(V))$, and you ask what is the most general operator symmetric under this symmetry.

The answer is: it's complicated and there is no general answer. First of all, if the representation of $G$ on $B(B(V))$ is irreducible, the most general operator that is invariant under it is proportional to the identity by Schur's Lemma, that is, one real number (an angle in the complex plane) suffices to parametrize your Scattering operator.

If the representation on $B(B(V))$ is reducible, one may reduce it to its blocks, and any symmetric operator will be constant on each block.

So to proceed you'll have to think of the kind of $G$ you want to consider.

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  • $\begingroup$ I believe I am asking a different question, since there may be totally symmetric super-operators which are not in the image of the map $S \to \tilde{S}$. Also I agree that the answer will depend on the representation $O_g$. $\endgroup$ – creillyucla Sep 20 at 13:49
  • $\begingroup$ It's true: you consider the completely positive maps. This implies that if the representation of the symmetry group is irreducible, there is, in fact, no free parameter. The problem remains that there is nothing general that can be said without further knowledge of the representation at hand. $\endgroup$ – Lorenz Mayer Sep 20 at 13:54
  • $\begingroup$ While Choi's theorem gaurantees that any completely positive map $\Phi$ admits a decomposition $\Phi = \sum_i V_i (\cdot) V_i^\dagger$ I don't see why it is that we can always find a $V$ such that the $V_i$ should be of the form $O_g V O^\dagger_g$. Also I agree that the number of free parameters depends on the representation but I would like to know how it depends, i.e. some procedure for producing coordinates on the space of possible $\tilde{S}$. $\endgroup$ – creillyucla Sep 20 at 14:38
  • $\begingroup$ The above comment should read "...any completely positive and totally symmetric map $\Phi \in B(B(V))$ admits..." $\endgroup$ – creillyucla Sep 20 at 14:52
  • $\begingroup$ The problem is that what the conjugacy class of a given element is depends strongly on the representation - and might be the whole space (for example if G is the group of unitaries). You may be right in that there are some constraints by requiring to sum over all elements of a given class, but on the other hand this might be equivalent to the cp map defined by them being symmetric. $\endgroup$ – Lorenz Mayer Sep 20 at 15:19

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