# How can the depolarizing channel be a quantum operation?

In Quantum Computing: From Linear Algebra to Physical Realizations it states that

A quantum operation maps a density matrix to another density matrix linearly

But let $$\rho\in M_2$$ be a density matrix and consider the Depolarizing Channel $$\varepsilon(\rho) = (1-p)\rho + p\frac{I}{2}$$ How is this linear? We have that $$\varepsilon(\rho_1 + \rho_2) = (1-p)(\rho_1 + \rho_2) + p\frac{I}{2} \neq (1-p)(\rho_1 + \rho_2) + 2p\frac{I}{2} = \varepsilon(\rho_1) + \varepsilon(\rho_2).$$ I'm trying to calculate the Choi Matrix, which my professor defines in the following way:

$$C(\varepsilon) = (C_{ij})$$ where $$C_{ij} = \varepsilon(E_{ij})$$ where $$\{E_{ij} = |i\rangle\langle j|: i, j \in \{0,1 \}\}$$ is the standard basis for $$M_2$$.

I do not understand the definition for the Choi matrix $$C(\varepsilon)$$; how can we plug in $$E_{01}$$ into $$\varepsilon$$ when $$E_{01}$$ is not a density matrix? My professor says it is first necessary to decompose $$E_{01}$$ as a linear combination of density matrices, but if $$\varepsilon$$ is linear, is that not pointless? We would get the same result. It only makes sense to talk about the Choi Matrix of a linear map (right?), so how can I be asked to calculate the Choi Matrix for the Depolarizing Channel?

• @FDGod I added an explanation as to what I mean by $E_{12}$. I have the $\neq$ because the coefficients of the last term in the LHS and RHS don't match. Commented Apr 17 at 2:27
• @JohnHippisley I have not seen this definition before, if I use the one in Eq. (1.20) of Gottesman's book (cs.umd.edu/class/spring2024/cmsc858G/QECCbook-2024-ch1-13.pdf), the depolarizing channel is indeed linear. Commented Apr 17 at 2:48
• possible duplicate of quantumcomputing.stackexchange.com/q/28445/55
– glS
Commented Apr 17 at 8:03
• This kind of problem seems to arise quite regularly ... maybe we should have a canonical answer (phrased broadly enough) and close as duplicates. Commented Apr 17 at 9:44
• Does this answer your question? Depolarizing channel for $n$ qubits: why is there a trace term? Commented Apr 17 at 9:44

The correct linear form of the depolarizing channel is $$\varepsilon(\rho) = (1-p)\rho + p\frac{I}{2}{\rm Tr}(\rho).$$ For density matrices $${\rm Tr}(\rho)=1$$, so you can usually see the form without the $${\rm Tr}$$ part.
Since $$\varepsilon$$ is just a linear map between matrices you can use any matrix as an input, not just density matrices.
E.g. $$\varepsilon(E_{01}) = (1-p)E_{01}.$$
$$E_{01} = (X + iY)/2 = (X+I)/2 + i(Y+I)/2 - (i+1)I/2.$$
• I see, so the $Tr(\rho)$ guarantees that $\varepsilon$ is linear when we consider any matrix as an input. Commented Apr 18 at 16:59