Could this map be a quantum channel?

Question

A quantum channel is defined as a completely positive, trace-preserving (CPTP) linear map between spaces of operators.

Suppose we have the following operation $$\mathcal{E}(\rho) = \sum_{j=1}^n K_j \rho K_j^{\dagger}, \tag{1}$$ where $\rho$ is a density matrix.

From what I know, since $\mathcal{E}(\cdot)$ has the Kraus operator representation, this operation is completely positive.

However, this operation is not trace-preserving because $$ \sum_{j=1}^n K_j^{\dagger}K_j = \begin{pmatrix} 1 + 2k^2dt && - 2 k \alpha dt\\ - 2 k \alpha dt && 1 + 2k^2dt \end{pmatrix} = I + O(dt).\tag{2} $$ In the above, $dt \rightarrow 0$, $k \in (0,1)$, $\alpha \in \mathbb{R}$. Here, we have $dt$ infinitesimally small but non-zero.

The eigenvalues of this matrix are very close to being $1$ given that $\alpha$ is not too large: \begin{align} \lambda_1 &= 1 + 2(-\alpha k + k^2)dt \\ \lambda_2 &= 1 + 2(\alpha k + k^2)dt \end{align}

For example, for a fixed small $dt$, a numerical representation of these eigenvalues could be $\lambda_1 = 1.0002$ and $\lambda_2 = 0.9998$. Analytically, since $dt$ tends to zero, eigenvalues should be tending to $1$.

My questions are:

1. Could $\mathcal{E}(\cdot)$ be viewed as a quantum channel for $dt>0$ and $dt \rightarrow 0$?
2. What if I normalize $\mathcal{E}(\rho)$ by $Tr(\mathcal{E}(\rho))$? Would that be a channel, albeit non-linear, in $\rho$.
3. Perhaps, more generally, what is the meaning of an operation that is CP but not TP?

Answer 1

As the third questions seems to have been addressed in the comments already let me focus on the other two:

1. Because of trace preservation your $\mathcal E$ is not a channel for any $dt>0$, only for $dt\to 0$.

2. If you normalize by dividing $\mathcal E(\rho)$ by ${\rm tr}(\mathcal E(\rho))$, then—as you correctly pointed out—$\mathcal E$ is not linear anymore. Hence $\mathcal E$ is not a channel, it is merely a map; a channel is defined as linear, completely positive, trace preserving.

However, there is a way of turning (almost) any CP map into a channel and it is to skew the input before applying $\mathcal E$: If $\sum_jK_j^\dagger K_j$ is positive definite (i.e. has no zero eigenvalues), then $X:= (\sum_jK_j^\dagger K_j)^{-1/2}$ (inverse of square root) is well-defined. With this the modified map $\mathcal E'(\rho):=\mathcal E(X\rho X^\dagger)$ is linear, completely positive (with Kraus operators $\{K_jX\}_j$), and trace preserving because \begin{align*} {\rm tr}(\mathcal E(\rho))&=\sum_j{\rm tr}(K_jX\rho X^\dagger K_j^\dagger)\\ &=\sum_j{\rm tr}(X^\dagger K_j^\dagger K_jX\rho)\\ &={\rm tr}\Big(X^\dagger\Big(\sum_j K_j^\dagger K_j\Big)X\rho \Big)\\ &={\rm tr}\Big(\Big(\sum_j K_j^\dagger K_j\Big)^{-1/2}\Big(\sum_j K_j^\dagger K_j\Big)\Big(\sum_j K_j^\dagger K_j\Big)^{-1/2}\rho \Big)={\rm tr}(\rho)\,. \end{align*}