Explicit computation of Choi matrix for a qubit channel

Question

I'm struggling with the explicit computation of the Choi matrix of a generic quantum channel ${\Phi}:\mathbb{C}^2\to\mathbb{C}^2$.

I know that I can write the channel in the Bloch representation as ${\Phi}(v_0 I+\mathbf{v}\cdot\vec{\sigma})=v_0I+(\mathbf{t}+T\mathbf{v})\cdot\vec{\sigma}$, with $\vec{\sigma}=(I,\sigma_x,\sigma_y,\sigma_z)$ the vector formed by the basis of the real vector space of Hermitian $2\times2$ matrices $\mathcal{H}(2,\mathbb{C})$, given by the 3 Pauli matrices plus the identity.

I am interested in the case in which $\mathbf{t}$ is a generic translation vector $(t_1,t_2,t_3)$ and $T$ is diagonal $T=\text{diag}(\mu_1,\mu_2,\mu_3)$, i.e. the cases in which $\Phi$ can be represented by the $4\times4$ matrix $$ \mathbb{T}=\begin{pmatrix} 1 & 0 & 0 & 0\\ t_1 & \mu_1 & 0 & 0\\ t_2 & 0 & \mu_2 & 0\\ t_3 & 0 & 0 & \mu_3 \end{pmatrix} $$

Since the Choi matrix $\Lambda_\Phi$ of $\Phi$ is $\Lambda_\Phi=\begin{pmatrix}\Phi(E_{11}) & \Phi(E_{12})\\ \Phi(E_{21}) & \Phi(E_{22})\end{pmatrix}$ (where each $E_{ij}$ is a $2\times2$ matrix with $1$ in the $(i,j)$-th position and $0$ elsewhere) in order to compute it I've been trying to use the definition of $\Phi$ above to calculate the four $\Phi(E_{ij})$s.

I wrote $E_{11}$ as a linear combination of the elements of the basis $\{ I,\sigma_x,\sigma_y,\sigma_z \}$ and got $E_{11}=\frac{1}{2}I+\frac{1}{2}\sigma_z$ and so $$\Phi(E_{11})=\frac{1}{2}\begin{pmatrix} 1+t_3+\mu_3 & t_1-it_2\\ t_1+it_2 & 1-t_3-\mu_3\end{pmatrix}.$$

Analogously $E_{22}=\frac{1}{2}I-\frac{1}{2}\sigma_z$ and so $$\Phi(E_{22})=\frac{1}{2}\begin{pmatrix} 1+t_3-\mu_3 & t_1-it_2\\ t_1+it_2 & 1-t_3+\mu_3\end{pmatrix}.$$

So far everything is alright and in accordance with what I should get.

I have problems in the calculations of $\Phi(E_{12})$ and $\Phi(E_{21})$.

In a first attempt I did the same reasoning and got $E_{12}=\frac{1}{2}\sigma_x+\frac{i}{2}\sigma_y$ and so $\Phi(E_{12})=\frac{1}{2}\begin{pmatrix}t_3 & t_1-it_2+\mu_1+\mu_2 \\ t_1+it_2+\mu_1-\mu_2 & -t_3\end{pmatrix}$. But this is wrong because I know that I should get $\frac{1}{2}\begin{pmatrix}0 & \mu_1+\mu_2 \\ \mu_1-\mu_2 & 0\end{pmatrix}$.

I think that my mistake is the following: I cannot write $E_{12}$ as a linear combination of $\{I,\sigma_x,\sigma_y,\sigma_z\}$ with complex coefficients, because $\mathcal{H}(2,\mathbb{C})$ is a real vector space. But since the two matrices $E_{12}$ and $E_{21}$ are not Hermitian (they are not symmetric) it is not possible to write them as a linear combination $\{I,\sigma_x,\sigma_y,\sigma_z\}$ with real coefficients, so I don't know how to proceed.

How can I compute $\Phi(E_{12})$ and $\Phi(E_{21})$?

Answer 1

Your general argument is correct, and the mistake you made is keeping an identity component where there shouldn't be one. Indeed using ($\mathbb C$-)linearity of $\Phi$ your calculation yields $$ \Phi(E_{12})=\frac{1}{2}\Phi(\sigma_x)+\frac{i}{2}\Phi(\sigma_y) =\frac{\mu_1}{2}\sigma_x+\frac{i\mu_2}{2}\sigma_y =\frac{1}{2}\begin{pmatrix}0& \mu_1+\mu_2 \\ \mu_1-\mu_2 & 0\end{pmatrix} $$ which is exactly what it should be. I assume that you accidentally carried over the extra summand $$ \frac12\begin{pmatrix} t_3&t_1-it_2\\t_1+it_2&-t_3 \end{pmatrix}=\Phi(I) $$ from your calculations of $\Phi(E_{11})$, $\Phi(E_{22})$ to $\Phi(E_{12})$.

As for your second question: I am not aware of the existence of something like a Choi matrix for linear maps $\Phi$ on real symmetric matrices. Actually, I doubt that such a thing exists -- or if it does, it does not resemble the Choi matrix as we know it. Recall that the Choi matrix works as it does because it embeds an orthonormal basis of the domain of $\Phi$ into a larger matrix $C$ which is positive semi-definite. Now if the Choi matrix of $\Phi$ (that is, $(\Phi\otimes\operatorname{id})(C)$, resp. $(\operatorname{id}\otimes\Phi)(C)$) is positive semi-definite, then it can be diagonalized into pairwise orthogonal vectors. Partitioning these vectors and re-arranging them into $n\times n$ matrices yields the Kraus operators of $\Phi$ -- and the Kraus form is readily verified to always be completely positive. The two big problems I see are the following:

• As you pointed out an orthonormal basis of $ \mathsf{Sym}_2:=\{A\in\mathbb R^{2\times 2}\,:\,A=A^T\} $ is given by $$ \Big\{\frac1{\sqrt2}\begin{pmatrix}1&0\\0&1\end{pmatrix},\frac1{\sqrt2}\begin{pmatrix}1&0\\0&-1\end{pmatrix},\frac1{\sqrt2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\Big\}\,, $$ for short $\{E_0,E_x,E_z\}$. While the usual Choi-matrix is given by $$ C=E_0\otimes E_0+E_x\otimes E_x+E_y\otimes E_y+E_z\otimes E_z=\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}\,,\tag{1} $$ we do not have access to $E_y$. But getting rid of the $E_y\otimes E_y$ term yields $$ E_0\otimes E_0+E_x\otimes E_x+E_z\otimes E_z=\frac12\begin{pmatrix}2&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&2\end{pmatrix} $$ which is not positive semi-definite anymore. Making it positive semi-definite (while keeping the coefficients equal; this is important as we will see later on) leads to $$ \frac12\begin{pmatrix}1&0&0&1\\0&1&1&0\\0&1&1&0\\1&0&0&1\end{pmatrix}=E_0\otimes E_0+E_x\otimes E_x $$ meaning it does not capture the action of $\Phi$ on $E_z$; hence it cannot characterize complete positivity.

• Of course every symmetric matrix can be made positive by adding a "sufficient amount" of the identity to it. So we can always play with the weights in (1) to obtain a positive (but "asymmetric" in the weights) larger matrix $$ C=2E_0\otimes E_0+E_x\otimes E_x+E_z\otimes E_z=\frac12\begin{pmatrix}3&0&0&1\\0&1&1&0\\0&1&1&0\\1&0&0&3\end{pmatrix}\,. $$ By assumption we know that $$ (\operatorname{id}\otimes\Phi)(C)=\frac12\begin{pmatrix} \Phi\begin{pmatrix} 3&0\\0&1\end{pmatrix}&\Phi\begin{pmatrix} 0&1\\1&0 \end{pmatrix}\\\Phi\begin{pmatrix} 0&1\\1&0 \end{pmatrix}&\Phi\begin{pmatrix} 1&0\\0&3 \end{pmatrix} \end{pmatrix}\geq 0 $$ so there exist pairwise orthogonal (not necessarily non-zero) vectors $g_1,\ldots,g_4\in\mathbb R^4$ such that $(\operatorname{id}\otimes\Phi)(C)=\sum_{i=1}^4g_ig_i^T$. The usual procedure turns these $g_i$ into Kraus matrices $K_i:=\operatorname{vec}^{-1}(g_i)$ where $\operatorname{vec}^{-1}$ is the inverse of usual vectorization. A straightforward computation shows that \begin{align*} \sum_{i=1}^4K_i\begin{pmatrix}a&b\\b&c\end{pmatrix}K_i^T&=\frac{a}2\Phi\begin{pmatrix} 3&0\\0&1 \end{pmatrix}+\frac{b}2\Phi\begin{pmatrix} 0&1\\1&0 \end{pmatrix}+\frac{b}2\Phi\begin{pmatrix} 0&1\\1&0 \end{pmatrix}+\frac{c}2\Phi\begin{pmatrix} 1&0\\0&3 \end{pmatrix}\\ &=\Phi\begin{pmatrix}\frac32a+\frac12c&b\\b&\frac32a+\frac12c\end{pmatrix} \end{align*} for all $a,b,c\in\mathbb R$, meaning that not $\Phi$, but rather $$ \Phi_S\begin{pmatrix}a&b\\b&c\end{pmatrix}:=\Phi\begin{pmatrix}\frac32a+\frac12c&b\\b&\frac12a+\frac32c\end{pmatrix} $$ is completely positive. This may seem promising as all we have to do now is "un-mix" the diagonals -- but doing so is not (completely) positive, and applying a non-positive map to a positive one can yield anything. Indeed, $\Phi=\Phi_S\circ S^{-1}$ where $$ S^{-1}\begin{pmatrix}a&b\\b&c\end{pmatrix}:=\begin{pmatrix}\frac34(a-\frac13c)&b\\b& \frac34(c-\frac13a)\end{pmatrix}\,, $$ but $S^{-1}(E_{11})=\operatorname{diag}(\frac34,-\frac14)\not\geq 0$ so complete positive of $\Phi_S$ does not imply anything about $\Phi$.