Action of a channel on an "unphysical" state

Question

Suppose we are given a rule $\Phi$ which is completely positive and trace preserving operation takes an input qubit state $\rho$ to an output qubit state $\rho^\prime$ (as an example of such rule see equation (11) of this article). In other words, its input as well as output are valid density matrix (with unit trace). The rule does not tell us how what happens if input is not a valid density matrix.

One often encounters situations, for example, the Choi matrix

$$ \chi = \left( \mathbb{1} \otimes \Phi \right) |\tilde{\psi} \rangle \langle \tilde{\psi} | = \sum_{ij} |i\rangle \langle j| \otimes \Phi(|i\rangle \langle j|), $$ where $|\tilde{\psi} \rangle = |00\rangle + |11\rangle$ is maximally entangled state Bell state apart from a normalization factor (assuming $\Phi$ is a qubit operation).

Now $\Phi(|0\rangle \langle 0|)$ and $\Phi(|1\rangle \langle 1|)$ make perfect sense since both $|0\rangle \langle 0|$ and $|1\rangle \langle 1|$ are valid density matrices. But how about $\Phi(|i\rangle \langle j|)$ with $i\ne j$? Since $|i\rangle \langle j|$ ($i\ne j$) is not a valide density matrix, we do not how $\Phi$ acts on it (remember we only have the rule $\Phi$ for valid density a matrix). Then how can one construct such quantities?

Answer 1

A quantum channel sends states to states, which are positive semidefinite matrices. But it is a linear map. There's no problem in extending its action to all matrices by linearity, since every matrix is a linear combination of pure states with complex coefficients. It's even simpler to treat quantum channel as a map from all matrices to matrices.

Any matrix $A$ can be written as $$ A = \frac{1}{2}(A+A^\dagger) + \frac{1}{2}(A - A^\dagger) = \frac{1}{2}(A+A^\dagger) - \frac{i}{2}(i(A - A^\dagger)). $$

Matrices $(A+A^\dagger)$ and $i(A - A^\dagger)$ are Hermitian (here $A - A^\dagger$ is skew Hermitian). Thus they are linear combinations of pure states by spectral theorem. Hence $A$ is also a linear combination of pure states, say $A = \sum_i \mu_i \rho_i$, $\mu_i \in \mathbb{C}$. Then $\Phi(A) = \sum_i \mu_i \Phi(\rho_i)$.