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 $O\in L(W)$.

  • $\begingroup$ I apologize if I'm missing something obvious, but can you clarify what exactly $d$ is? I feel like it should be something like a dimension of an ancilla space, but I can't quite pin it down since I don't see it actually in the definition of $\rho_{2,d}$ anywhere. Is $d = n$? $\endgroup$ May 26, 2022 at 1:49

1 Answer 1


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 $A\in L(V)$.



$$=\Gamma(RA_1S,\dots,RA_rS;RB_1S,\dots,RB_rS)(X)=\sum_{k=1}^{r}RA_kSX(RB_kS)^*$$ $$=\sum_{k=1}^rRA_kSXS^*B^*_kR^* =R\big(\Gamma(A_1,\dots,A_r;B_1,\dots,B_r)(SXS^*)\big)R^*.$$


$$\mho(\Gamma(A_1,\dots,A_r;B_1,\dots,B_r),1_{L(V)},T_{R,S})(X) =\Gamma(A_1,\dots,A_r;RB_1S,\dots,RB_rS)(X)$$

$$=\sum_{k=1}^rA_kX(RB_kS)^*=\sum_{k=1}^rA_kXS^*B_k^*R^* =\big(\Gamma(A_1,\dots,A_r;B_1,\dots,B_r)(XS^*)\big)R^*.$$

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 $\rho(\mho_{R,S}^-(\mathcal{E}))^2>\rho(\mho_{R,S}^+(\mathcal{E}))\cdot\rho(\mathcal{E})$.


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