# 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.

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 $$O\in L(W)$$.

• 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 $d = n$? May 26, 2022 at 1:49

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.

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.

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