Approximations of the spectral radii of completely positive superoperators Let $V$ be a finite dimensional complex Hilbert space. Let $L(V)$ denote the collection of all linear operators from $V$ to $V$. An operator $\mathcal{E}:L(V)\rightarrow L(V)$ is said to be positive if whenever $A\geq 0$, we have $\mathcal{E}(A)\geq 0$ as well. We say that $\mathcal{E}$ is completely positive if $\mathcal{E}\otimes 1_{L(W)}:L(V\otimes W)\rightarrow L(V\otimes W)$ is positive for each finite dimensional complex Hilbert space $W$. A linear mapping $\mathcal{E}:L(V)\rightarrow L(V)$ is said to be trace preserving if $\text{Tr}(\mathcal{E}(A))=\text{Tr}(A)$ whenever $A\in L(V)$. A channel is a completely positive trace preserving map $\mathcal{E}:L(V)\rightarrow L(V)$. A unital channel is a channel where $\mathcal{E}(1_V)=1_V$.
If $A_1,\dots,A_r\in L(V)$, then define a mapping $\Phi(A_1,\dots,A_r):L(V)\rightarrow L(V)$ by $\Phi(A_1,\dots,A_r)(X)=A_1XA_1^*+\dots A_rXA_r^*$. Then the mapping $\Phi(A_1,\dots,A_r)$ is a completely positive mapping, and every completely positive mapping $\mathcal{E}:L(V)\rightarrow L(V)$ is of this form.
Define $$\rho_{2}(A_1,\dots,A_r)=\rho(\Phi(A_1,\dots,A_r))^{1/2}.$$
The Cauchy-Schwarz inequality holds for $\rho_{2}$: $$\rho(A_1\otimes B_1+\dots+A_r\otimes B_r)\leq\rho_2(A_1,\dots,A_r)\rho_2(B_1,\dots,B_r).$$
Define $$\rho_{2,d}(A_1,\dots,A_r)$$
$$=\sup\{\frac{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)}{\rho_{2}(X_1,\dots,X_r)}\mid X_1,\dots,X_r\in M_d(\mathbb{C}),\rho_{2}(X_1,\dots,X_r)\neq 0\}.$$ Observe that $\rho_{2,d}(A_1,\dots,A_r)\leq\rho_{2,g}(A_1,\dots,A_r)\leq\rho_2(A_1,\dots,A_r)$ whenever $d\leq g$, and  $\rho_{2,d}(A_1,\dots,A_r)=\rho_2(A_1,\dots,A_r)$ whenever $d\geq\dim(V).$
Theorem: $\Phi(A_1,\dots,A_r)=\Phi(B_1,\dots,B_r)$ if and only if there is an $r\times r$ unitary matrix $(u_{i,j})_{i,j}$ where $B_{i}=\sum_{j=1}^{r}u_{i,j}A_{j}$ for $1\leq i\leq r$.
The $\leftarrow$ direction is easy to prove, and a proof of the direction $\rightarrow$ can be found in the book The Theory of Quantum Information by John Watrous.
Lemma: Suppose that $A_1,\dots,A_r,B_1,\dots,B_r,X_1,\dots,X_r,Y_1,\dots,Y_r$ are matrices over the same field and whose dimensions are proper so that $A_1\otimes X_1+\dots A_r\otimes X_r,B_1\otimes Y_1+\dots B_r\otimes Y_r$ both make sense and have the same dimension. Suppose that $(u_{i,j})_{i,j},(v_{i,j})_{i,j}$ are $r\times r$-matrices over the field $K$ and
$(u_{i,j})_{i,j}^{-1}=(v_{i,j})_{i,j}^{T}$. Furthermore, suppose that
$A_i=\sum_{j=1}^{r}u_{i,j}B_j,X_i=\sum_{j=1}^{r}v_{i,j}Y_j$ for $1\leq i\leq r$. Then
$$A_1\otimes X_1+\dots A_r\otimes X_r=B_1\otimes Y_1+\dots B_r\otimes Y_r.$$
I was able to prove the following fact (it is not too hard to verify that this fact is correct using computer calculations).
Theorem: If $\Phi(A_1,\dots,A_r)=\Phi(B_1,\dots,B_r)$, then $\rho_{2,d}(A_1,\dots,A_r)=\rho_{2,d}(B_1,\dots,B_r)$.
Proof: If $\Phi(A_1,\dots,A_r)=\Phi(B_1,\dots,B_r)$, then there is a unitary map $(u_{i,j})_{i,j}$ where $A_i=\sum_{j=1}^{r}u_{i,j}\cdot B_j$ for $1\leq i\leq r$. Therefore, set $(v_{i,j})_{i,j}=((u_{i,j})_{i,j}^{-1})^{T}$. Then whenever $X_1,\dots,X_r\in M_n(\mathbb{C})$, and $Y_i=\sum_{j=1}^{r}v_{i,j}X_{j}$, we have $\rho_{2}(X_1,\dots,X_r)=\rho_{2}(Y_1,\dots,Y_r)$, and $\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)=\rho(B_1\otimes Y_1+\dots+B_r\otimes Y_r)$. Therefore, $$\frac{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)}{\rho_2(X_1,\dots,X_r)}=\frac{\rho(B_1\otimes Y_1+\dots+B_r\otimes Y_r)}{\rho_2(Y_1,\dots,Y_r)}$$
whenever $\rho_2(Y_1,\dots,Y_r)\neq 0$, so $\rho_{2,d}(B_1,\dots,B_r)\leq\rho_{2,d}(A_1,\dots,A_r)$. The reverse inequality is established in a similar manner. Q.E.D.
Therefore, if $\mathcal{E}:L(V)\rightarrow L(V)$ is a completely positive mapping, then we can define $\rho_{2,d}(\mathcal{E})$ by letting $\rho_{2,d}(\mathcal{E})=\rho_{2,d}(A_1,\dots,A_r)^{2}$ where $\mathcal{E}=\Phi(A_1,\dots,A_r)$.
If $d\geq\dim(V)$, then $\rho_{2,d}(\mathcal{E})=\rho(\mathcal{E})$.
If $1\leq d<\dim(V)$, then is there a characterization of $\rho_{2,d}(\mathcal{E})$ that does not require us to decompose $\mathcal{E}$ as $\Phi(A_1,\dots,A_r)$? Is there such a characterization of $\rho_{2,d}(\mathcal{E})$ in the special case when $\mathcal{E}$ is a channel? What about when $\mathcal{E}$ is a unital channel or a mixed unitary channel? Can $\rho_{2,d}(\mathcal{E})$ be generalized to the case when $\mathcal{E}$ is no longer necessarily completely positive?
It would be great if there were a quantum algorithm that often efficiently computes $\rho_{2,d}(\mathcal{E})$ when there is a quantum computer that sends the mixed state $D$ to the mixed state $\mathcal{E}(D)$, but perhaps this is too much to ask for.
Added 5/27/2022
Suppose that $(e_{a}\mid a\in\Sigma)$ is an orthonormal basis for $W$. Let
$A:V\rightarrow V\otimes W$ be a linear operator. Suppose that
$A=\sum_{a\in\Sigma}A_a\otimes e_a$. Then the mapping $\mathcal{E}_A:L(V)\rightarrow L(V)$ defined by letting $\mathcal{E}_A(X)=\text{Tr}_W(AXA^*)$ ($\text{Tr}_W$ denotes the partial trace) is a completely positive mapping, and
$\text{Tr}_W(AXA^*)=\sum_{a\in\Sigma}A_aXA_a^*$, so every completely positive mapping is of the form $\mathcal{E}_A$ for some $A$.
Furthermore, if $B\in L(U,W\otimes U)$, and $B=\sum_{b\in\Sigma}e_b\otimes B_b$, then $$\text{Tr}_W(A\otimes B^*)=\sum_{a\in\Sigma}A_a\otimes B_b^*.$$
Therefore, $$\rho_{2,d}(\mathcal{E})^{1/2}=
\sup\{\frac{\rho(\text{Tr}_W(A\otimes B^*))}{\rho(\mathcal{E}_{B})^{1/2}}\mid B\in L(U,W\otimes U)\}$$
whenever $\dim(U)=d$. This characterization of $\rho_{2,d}(\mathcal{E})^{1/2}$ depends on the choice of $A$ and is not much different than the other definition of $\rho_{2,d}(\mathcal{E})^{1/2}$. It is known that
if $\mathcal{E}_{A_1}=\mathcal{E}_{A_2}$, then
$A_1=(1_{V}\otimes O)A_2$ for some unitary map $O\in L(W)$.
 A: Yes. We can characterize $\rho_{2,d}(\mathcal{E})$ whenever $\mathcal{E}$ is completely positive without needing to first decompose $\mathcal{E}$ as $\Phi(A)$ or $\Phi(A_1,\dots,A_r).$ As a consequence, we can define $\rho_{2,d}(\mathcal{E})$ for all linear operators $\mathcal{E}:L(V)\rightarrow L(V)$, but if $\mathcal{E}$ is not completely positive, then $\rho_{2,d}(\mathcal{E})$ is usually infinite, so $\rho_{2,d}(\mathcal{E})$ it not well behaved when we do not assume that $\mathcal{E}$ is completely positive.
Let $U_1,U_2,V_1,V_2,U_1^\sharp,U_2^\sharp,V_1^\sharp,V_2^\sharp,U,V$ be finite dimensional complex inner product spaces.
If $A_1,\dots,A_r:U_2\rightarrow V_2,B_1,\dots,B_r:U_1\rightarrow V_1$ are linear, then define a mapping $\Gamma(A_1,\dots,A_r;B_1,\dots,B_r):L(U_1,U_2)\rightarrow L(V_1,V_2)$ by letting $$\Gamma(A_1,\dots,A_r;B_1,\dots,B_r)(X)=\sum_{k=1}^rA_kXB_k^*.$$
Suppose now that $A_1,\dots,A_r:U_2\rightarrow V_2,B_1,\dots,B_r:U_1\rightarrow V_1$. Let $S:L(U_2,V_2)\rightarrow L(U_2^\sharp,V_2^\sharp)$ be linear and let
$T:L(U_1,V_1)\rightarrow L(U_1^\sharp,V_1^\sharp)$ also be linear. Then define
$\mho(\Gamma(A_1,\dots,A_r;B_1,\dots,B_r),S,T):L(U_1^\sharp,U_2^\sharp)\rightarrow L(V_1^\sharp,V_2^\sharp)$ by letting
$$\mho(\Gamma(A_1,\dots,A_r;B_1,\dots,B_r),S,T)=\Gamma(S(A_1),\dots,S(A_r);T(B_1),\dots,T(B_r)).$$ The mapping $\mho$ is well-defined; i.e. it depends on $\Gamma(A_1,\dots,A_r;B_1,\dots,B_r)$ rather than the particular choice of $A_1,\dots,A_r;B_1,\dots,B_r$.
$\rho_{2,d}(\mathcal{E})$ is the maximum value of $$\frac{\rho(\mho(\mathcal{E},1_{L(V)},T))^2}{\rho(\mho(\mathcal{E},T,T))}$$
where $T:L(V)\rightarrow L(U)$ is a linear operator and $\dim(U)=d$, and this definition generalizes to all linear operators $\mathcal{E}:L(V)\rightarrow L(V)$ when we replace the word 'maximum' with 'supremum'.
Suppose now that $\mathcal{E}=\Gamma(A_1,\dots,A_r;B_1,\dots,B_r)$ where $A_1,\dots,A_r,B_1,\dots,B_r$ are linearly independent and $\Phi(A_1,\dots,A_r)$ is not nilpotent. Let $T:L(V)\rightarrow L(V)$ be a linear mapping where $T(A_j)=A_j,T(B_j)=\alpha\cdot A_j$ for $1\leq j\leq r$.
Then
$\mho(\mathcal{E},1_{L(V)},T)=\mho(\mathcal{E},T,T)=\alpha\cdot\Phi(A_1,\dots,A_r)$.
Therefore, $$\rho_{2,d}(\mathcal{E})\geq\frac{\rho(\alpha\cdot\Phi(A_1,\dots,A_r))^2}{\rho(\alpha\cdot\Phi(A_1,\dots,A_r))}=\alpha\cdot\rho(\Phi(A_1,\dots,A_r)).$$
Since $\alpha$ can be made arbitrarily large, we have $\rho_{2,d}(\mathcal{E})=\infty$ in this case.
A modest generalization: added 8/15/2022
There is another way to generalize $\rho_{2,d}(\mathcal{E})$ to some operators that are not completely positive, but this generalization is based on a conjecture.
Conjecture: Suppose that $\mathcal{E}:L(V)\rightarrow L(V)$ is completely positive and $\alpha\geq 0$. Then $\rho_{2,d}(\mathcal{E}+\alpha\cdot 1_{L(V)})=\rho_{2,d}(\mathcal{E})+\alpha$.
From this conjecture, we can define $\rho_{2,d}(\mathcal{E})$ whenever there is some $\alpha\geq 0$ where $\mathcal{E}+\alpha\cdot 1_{L(V)}$ is completely positive by letting $\rho_{2,d}(\mathcal{E})=\rho_{2,d}(\mathcal{E}+\alpha\cdot 1_{L(V)})-\alpha$. Here, we can have negative values of $\rho_{2,d}(\mathcal{E})$, so $\rho_{2,d}$ more closely resembles the maximum value of a Hermitian operator than the spectral radius.
