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I'm currently trying to implement the depolarizing channel on qiskit. But, as I see in my calculation it doesn't match with the qiskit aer_noise.

So, for the Depolarizing Channel we got : $$ \mathcal{E}(\rho) = (1 - p) \rho + p \frac{I}{2} $$

Ant the choi matrix equation is: $$ \Lambda_\mathcal{E} = (\mathcal{I} \otimes \mathcal{E})(|\Phi\rangle \langle \Phi|) $$ So, this is the tricky part where i get lost, I know that the Choi matrix got the Channel acting in every ketbra on the base, so in this case, for a one qubit i got: $$ \mathcal{E}(|0\rangle \langle 1|) = (1-p) |0\rangle \langle 1| + p/2 (|0\rangle \langle 0| + |1\rangle \langle 1|) $$ $$ \mathcal{E}(|1\rangle \langle 0|) = (1-p) |1\rangle \langle 0| + p/2 (|0\rangle \langle 0| + |1\rangle \langle 1|) $$ $$ \mathcal{E}(|0\rangle \langle 0|) = (1-p/2) (|0\rangle \langle 0|) + p/2 (|1\rangle \langle 1|) $$ $$ \mathcal{E}(|1\rangle \langle 1|) = (1-p/2) (|1\rangle \langle 1|) + p/2 (|0\rangle \langle 0|) $$

Sho, the Choi matrix for this channel would be $$ \Lambda_\mathcal{E}= \begin{pmatrix} 1 - \frac{p}{2} & 0 & \frac{p}{2} & 1 - p \\ 0 & \frac{p}{2} & 0 & \frac{p}{2} \\ \frac{p}{2} & 0 & \frac{p}{2} & 0 \\ 1 - p & \frac{p}{2} & 0 & 1 - \frac{p}{2} \end{pmatrix} $$. However, if I use the qiskit funciton aer.get_noise and later use the Choi(noise) i get the form : $$ \Lambda_\mathcal{E}= \begin{pmatrix} 1 - \frac{p}{2} & 0 & 0 & 1 - p \\ 0 & \frac{p}{2} & 0 & 0\\ 0 & 0 & \frac{p}{2} & 0 \\ 1 - p & 0 & 0 & 1 - \frac{p}{2} \end{pmatrix} $$.

Can someone help me to understand why is the second form? Thanks.

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Your definition of the depolarizing channel is slightly off, it should read $$ \mathcal E(\rho)=(1-p)\rho+p\,\boxed{{\rm tr}(\rho)}\frac I2\,. $$ This extra term ${\rm tr}(\rho)$ is necessary for $\mathcal E$ to be a linear map (without it, the $0$ matrix would not be mapped to the $0$ matrix which would contradict linearity).

Of course if $\rho$ is a state, then ${\rm tr}(\rho)=1$ which recovers your definition. What changes—and what constitutes your extra terms causing the mismatch—is that the off-diagonal blocks of the Choi matrix read $$ \mathcal E(|0\rangle\langle 1|)=(1-p)|0\rangle\langle 1|+p\,\underbrace{{\rm tr}(|0\rangle\langle 1|)}_{=\langle 1|0\rangle=0}\frac I2=\begin{pmatrix}0&1-p\\0&0\end{pmatrix} $$ so there are no diagonal terms, as output by qiskit.

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  • $\begingroup$ Omg I completely miseed that part. Now it works as intended. Thanks for helping me! This term applies to any QC that is CPTP? $\endgroup$
    – felipechoy
    May 16 at 4:10
  • $\begingroup$ You're welcome! And yes, any (seemingly) constant term in the definition of any channel necessarily needs a factor ${\rm tr}(\rho)$ for the sake of linearity $\endgroup$ May 16 at 4:12
  • $\begingroup$ Thank you again! Is there any book or paper that mentions it? I see its kinda obvious but, It will help me for my article. $\endgroup$
    – felipechoy
    May 16 at 6:08
  • $\begingroup$ I don't know of any work that references this fact in general for it is too vague, but if you're okay with a book where just the depolarizing channel is defined rigorously (i.e. with this trace factor) check, e.g., Definition 6.36 in Holevo's book "Quantum Systems, Channels, Information" (2012) $\endgroup$ May 16 at 12:52
  • $\begingroup$ Thank you! Sorry to ask again but. Revisiting some lectures, there are some papers who mention that the Choi matrix is the way i mention here. But, there is the factor 1/d. Why qiskit and well, this matrix doesn't use the 1/d factor? $\frac{1}{d} \sum |i \geq \leq j| \otimes \mathcal{E} ( |i\geq \leq j |)$ $\endgroup$
    – felipechoy
    May 17 at 6:29

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