# Why is the Choi matrix different from the analytic form for a depolarizing channel?

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.

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.
• 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 Commented May 16 at 4:12
• 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? $\frac{1}{d} \sum |i \geq \leq j| \otimes \mathcal{E} ( |i\geq \leq j |)$ Commented May 17 at 6:29