What is the rank of a quantum channel?

Question

I read the following sentence in a paper:

We consider a quantum channel $\mathcal{E}_{\omega}(\rho)=\sum_{i=1}^{r} K_{i} \rho K_{i}^{\dagger}$ where $r$ is the rank of the channel.

I didn't find the definition of the rank of the quantum channel online. So I guess the meaning is that the rank of the quantum channel is the number of Kraus operators we used to describe the quantum channel? If it is right, is there the maximum number of the Kraus operators to describe a quantum channel?

Answer 1

Every quantum channel has many Kraus representations that may differ in the number of Kraus operators. For example, for any positive integer $n$ and numbers $p_i$ with $i=1,\dots,n$ and $\sum_{i=1}^np_i=1$ the matrices $E_i=\sqrt{p_i}I$ form a valid, if impractical, Kraus representation of the identity channel with $n$ Kraus operators. This example also shows that there is no maximum on the number of non-zero Kraus operators one can use to describe a quantum channel.

However, there is a minimum. The minimum is called Kraus rank or Choi rank of the channel. The letter term is justified by the fact that Choi rank is equal to the rank - in the usual linear algebraic sense - of the channel's Choi matrix. This implies that the Choi rank of a channel $\mathcal{E}: L(\mathcal{X})\to L(\mathcal{Y})$ is an integer between $1$ and $\dim\mathcal{X} \cdot \dim\mathcal{Y}$.

Occasionally, a channel's Choi rank is also simply called its "rank". However, this can be misleading as every channel is also a linear map and therefore has the regular, linear algebraic rank which may be different$^1$ than its Choi rank. In a context where one uses the term "rank" to refer to the number of operators in a Kraus representation - as it is done in the paper in question - it is likely that this unfortunate convention is being employed.

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^{$^1$ Consider the amplitude damping channel with $K_0=\begin{bmatrix}1&0 \\ 0&0\end{bmatrix}$ and $K_1=\begin{bmatrix}0&1 \\ 0&0\end{bmatrix}$.}