Consider an arbitrary quantum operation defined by a series of Kraus operators $\sum_j K_j\rho K_j^\dagger$ over the density matrix of the system $\rho$. The operation might or might not be unitary, but the Kraus operators satisfy the completely positive trace preserving condition $\sum_j K_j^\dagger K_j= I$. Is there a way to represent such operations in a generic matrix $A$ such that $A\rho A^\dagger=\sum_j K_j\rho K_j^\dagger$? Is this in any way related to Choi's theorem and the Choi matrix, or does anybody have a good reference to look this up?


1 Answer 1


No, this is not possible in general.

To see it, consider for example what happens taking the trace of that expression. You'd get: $$\operatorname{tr}(A^\dagger A\rho)=\operatorname{tr}\left(\sum_j K_j^\dagger K_j \rho\right)$$ for all $\rho$. This implies $A^\dagger A=\sum_j K_j^\dagger K_j=I$, with the last identity from the definition of Kraus operators of a trace-preserving operation. Thus $A$ would need to be an isometry, and the channel be an isometric channel. But clearly not all channels are isometric.

You can also see it from the Choi. The Choi representation of a quantum map $\Phi(X)=\sum_j A_j X A_j^\dagger$ is $J(\Phi)=\sum_j \operatorname{vec}(A_j)\operatorname{vec}(A_j)^\dagger$. So the identity translates in terms of the Chois as the identity $$\sum_j \operatorname{vec}(K_j)\operatorname{vec}(K_j)^\dagger = \operatorname{vec}(A)\operatorname{vec}(A)^\dagger.$$ But this would imply that the LHS has rank 1, which clearly isn't always the case (think as a counterexample something like the depolarizing channel: you'll readily verify that the matrix you get has more than one nonzero eigenvalue).


Your Answer

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

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