Quantum Channels are generalizations of Quantum Operations. Since, in general our operations maybe noisy (inherently) and the system we are dealing with maybe open so Quantum Channels us the study of noisy openQuantum Operations.

$$\rho \xrightarrow{U}\rho'$$

The above only holds for closed quantum systems. Thus, in case of noisy open quantum systems we have the following.

$$\rho \xrightarrow{\mathcal E} \rho'$$

Here, $\mathcal E$ is a combination of:

unitaries $(U)$

adding systems $(\rho \to \rho \otimes \sigma)$

subtracting systems or partial tracing $(\rho_{AB} \to \text{tr}_B \rho_{AB} = \rho_A)$

Review of Density Matrices

Moreover, to deal with noisy open systems we also need to review density matrices which are the general way of representing noisy quantum states.

$$\rho \equiv \text{noisy quantum state}\\
\mathcal E \equiv \text{noisy open quantum channel}$$

Now since, $\rho$ is positive semi-definite thus, $\frac{1 + ||\vec{a}||}{2} \geq 0 \implies ||\vec{a}|| \leq 1$.

This results in the Bloch Ball representation for general mixed state qubits, which serves as a generalization of the Bloch Sphere that serves as representation for pure state qubits. Pure state qubits have eigenvalues of $0$ or $1$, $\lambda(\rho) = 0\ \text{or}\ 1$.

The eigenvalue of the maximally mixed state is ${I}/{2}$ where the state is actually given by the following.

This can be generalized for noisy open channels as combinations of unitaries $(U)$, adding systems together $(\rho \to \rho \otimes \sigma)$ and subtracting systems $(\rho_{AB} \to \text{tr}_B(\rho_{AB}) = \rho_A)$.

But why is partial tracing allowed? Doesn't this "delete" information?