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?

  • $\begingroup$ Related:quantumcomputing.stackexchange.com/questions/28104/… Mind the statement in the comment. $\endgroup$
    – narip
    Sep 28, 2022 at 6:57
  • $\begingroup$ is this question actually related to the Choi? It seems to me like you're just asking whether/why it makes sense to consider the action of a generic channel on an operator that does not represent a physical state $\endgroup$
    – glS
    Sep 28, 2022 at 7:20
  • $\begingroup$ @glS, you are right, and in particular it I am only concerned with this issue when it comes to Choi matrix that we want to calculate for a map which is, by definition, defined for valid a density matrix. $\endgroup$
    – User101
    Sep 28, 2022 at 7:59

1 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 $\Phi(A) = \sum_i \mu_i \Phi(\rho_i)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.