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 $O\in L(W)$.



Approximations of the spectral radii of completely positive superoperators
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.


**Projective mappings: added 3/21/2023**


Define $\rho\_{2,d}^P(A\_1,\dots,A\_r)$ to be the supremum of all values of the form $$\frac{\rho(A\_1\otimes\overline{RA\_1S}+\dots+A\_r\otimes\overline{RA\_rS})}{\rho(RA\_1S\otimes\overline{RA\_1S}+\dots+RA\_rS\otimes\overline{RA\_rS})^{1/2}}$$
where $R\in M\_{d,n}(\mathbb{C}),S\in M\_{n,d}(\mathbb{C})$. I conjecture that $\rho\_{2,d}^P(A\_1,\dots,A\_r)=\rho\_{2,d}(A\_1,\dots,A\_r)$ (computer calculations support this conjecture at least sometimes), but it seems like $\rho\_{2,d}^P(A\_1,\dots,A\_r)$ is easier to work with than $\rho\_{2,d}(A\_1,\dots,A\_r)$.


If $S:U\rightarrow V,R:V\rightarrow U$ are linear maps, then define a mapping $T\_{R,S}:L(V)\rightarrow L(U)$ by setting $T\_{R,S}(A)=RAS$ for $A\in L(V)$.


$$\mho(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r),T\_{R,S},T\_{R,S})(X)$$


$$=\Gamma(T\_{R,S}(A\_1),\dots,T\_{R,S}(A\_r);T\_{R,S}(B\_1),\dots,T\_{R,S}(B\_r))(X)$$


$$=\Gamma(RA\_1S,\dots,RA\_rS;RB\_1S,\dots,RB\_rS)(X)=\sum\_{k=1}^{r}RA\_kSX(RB\_kS)^\*$$
$$=\sum\_{k=1}^rRA\_kSXS^\*B^\*\_kR^\*
=R\big(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r)(SXS^\*)\big)R^\*.$$


Similarly,


$$\mho(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r),1\_{L(V)},T\_{R,S})(X)
=\Gamma(A\_1,\dots,A\_r;RB\_1S,\dots,RB\_rS)(X)$$


$$=\sum\_{k=1}^rA\_kX(RB\_kS)^\*=\sum\_{k=1}^rA\_kXS^\*B\_k^\*R^\*
=\big(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r)(XS^\*)\big)R^\*.$$


Therefore, $\mho(\mathcal{E},T\_{R,S},T\_{R,S})(X)=R\big(\mathcal{E}(SXS^\*)\big)R^\*$ and $\mho(\mathcal{E},1\_{L(V)},T\_{R,S})(X)=\big(\mathcal{E}(XS^\*)\big)R^\*$ whenever $\mathcal{E}:L(V)\rightarrow L(V)$ and $R,S,X$ are suitable linear operators. Therefore, if $\mathcal{E}:L(V)\rightarrow L(V)$, then define $\mho\_{R,S}^-(\mathcal{E})(X)=\big(\mathcal{E}(XS^\*)\big)R^\*$ and
$\mho\_{R,S}^+(\mathcal{E})(X)=R\big(\mathcal{E}(SXS^\*)\big)R^\*$.


$\rho\_{2,d}^P(A\_1,\dots,A\_r)$ is the supremum of all values of the form
$$\frac{\rho(\mho\_{R,S}^-(\Phi(A\_1,\dots,A\_r)))}{\rho(\mho\_{R,S}^+(\Phi(A\_1,\dots,A\_r))^{1/2}}.$$ Therefore, we may define $\rho\_{2,d}^P(\mathcal{E})$ for all completely positive operators $\mathcal{E}:L(V)\rightarrow L(V)$ by setting
$\rho\_{2,d}^P(\mathcal{E})=\rho\_{2,d}^P(A\_1,\dots,A\_r)^2$ whenever $\mathcal{E}=\Phi(A\_1,\dots,A\_r)$. Then $\rho\_{2,d}^P(\mathcal{E})$ is the supremum of all values of the form $\frac{\rho(\mho\_{R,S}^-(\mathcal{E}))^2}{\rho(\mho\_{R,S}^+(\mathcal{E}))}$ where $R\in L(U,V),S\in L(V,U)$ and $U$ is a complex inner product space $\dim(U)=d$. This definition of $\rho\_{2,d}^P(\mathcal{E})$ makes sense even when $\mathcal{E}:L(V)\rightarrow L(V)$ and $V$ is an infinite dimensional complex Hilbert space (we will probably need to use Stinespring's dilation theorem to generalize $\rho\_{2,d}$ to infinite dimensional spaces).


Example: $\rho\_{2,d}^P(\mathcal{E})$ can be infinite for superoperators which are positive but not completely positive. For example, the transpose map $T:M\_n(\mathbb{C})\rightarrow M\_n(\mathbb{C})$ defined by $T(X)=X^T$ is positive but not completely positive whenever $n>1$, and we shall prove that $T$ is not completely positive by showing that $\rho\_{2,d}^P(T)=\infty$. A straightforward calculation yields $\rho\_{2,d}^P(T)=\sup\_{R\in M\_{d,n}(\mathbb{C}),S\in M\_{n,d}(\mathbb{C})}\frac{\rho(SR)^2}{\rho(R\overline{S}\overline{R}S)}.$


If $R,S$ are rank-1 matrices, then we can set $R=uv^\*,S=wx^\*$ for vectors $u,v,w,x$. In this case,
$\frac{\rho(SR)^2}{\rho(R\overline{S}\overline{R}S)}=|\frac{v^\*w}{v^Tw}|^2$. If $v=w=[1,i]^T$, then $\frac{\rho(SR)^2}{\rho(R\overline{S}\overline{R}S)}=(2/0)^2=+\infty.$


It seems like a good way of showing that a superoperator $\mathcal{E}:L(V)\rightarrow L(V)$ is not completely positive is to show that $\rho\_{2,d}^P(\mathcal{E})>\rho(\mathcal{E})$ by finding matrices $R,S$ where $\rho(\mho\_{R,S}^-(\mathcal{E}))^2>\rho(\mho\_{R,S}^+(\mathcal{E}))\cdot\rho(\mathcal{E})$.


