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?


1 Answer 1


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. $$ \varepsilon(E_{01}) = (1-p)E_{01}. $$

BTW, any matrix is indeed a linear combination of density matrices, but with complex coefficients.

$$ E_{01} = (X + iY)/2 = (X+I)/2 + i(Y+I)/2 - (i+1)I/2. $$

  • 1
    $\begingroup$ I see, so the $Tr(\rho)$ guarantees that $\varepsilon$ is linear when we consider any matrix as an input. $\endgroup$ Commented Apr 18 at 16:59

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