Prove that the depolarizing channel is completely positive

Question

In two dimensions, for a density operator $\rho$ and probability $\lambda$, a depolarizing channel can be written as:

$$\mathcal{E}(\rho) = (1-\lambda) \frac{\mathbb{I}}{2} + \lambda\rho$$

In wikipedia page about Quantum depolarizing channel, it is mentioned that this channel can be viewed as a completely positive trace preserving (CPTP) map. But there is no reference about how to prove this statement? It only says that the range of $\lambda$ should be: $$ -\frac{1}{d^2 - 1} \le \lambda \le 1 $$ Any pointers about how to prove that a depolarizing channel is CPTP?

Answer 1

Ok, so starting with Trace Preserving, since it's easier:

$$Tr(I/2) = 1$$ $$Tr(\rho) = 1$$ $$Tr((1-\lambda)I/2 + \lambda\rho) = (1-\lambda)Tr(I/2) + \lambda Tr(\rho) = 1$$

Now for a map to be completely positive, it must take positive elements to positive elements. So since depolarizing noise is essentially just adding in a bit of the identity operator (who's eigenvalues are all 1, which is positive), and all vectors are eigenvectors of the identity operator, any vector which previously had some eigenvalue $\Lambda_0$ now has eigenvalue $\lambda\Lambda_0 + \frac{(1-\lambda)}{2}$ which will also be positive.

So since it preserves positivity and trace, its CPTP.

Answer 2

Here's an alternative proof: first note that any quantum map, $\Phi(\rho) \mapsto \sigma$ that can be written in the Kraus form, that is, as $\Phi(\cdot) = \sum_{j} K_j (\cdot) K_j^\dagger$, with, $K_j^\dagger K_j \geq 0, \sum_j K_j^\dagger K_j = \mathbb{I}$ is a CP map (see for example, Nielsen and Chuang, or Page 26 of https://arxiv.org/abs/1902.00967). This is also the "usual" way to prove CP-ness: find a set of Kraus operators for the map $\Phi$ that satisfy the above condition. Also, note that the TP part is straightforward since you can just take the trace of $\mathcal{E}(\rho)$ and show that is it $1$.

Now, note that $\rho+X \rho X+Y \rho Y+Z \rho Z=2 I$, therefore, $$ \Phi(\rho) = \left( 1- \lambda \right) \frac{1}{4} \left( \rho+X \rho X+Y \rho Y+Z \rho Z \right) + \lambda \rho = \frac{1}{4} \left(1 + 3\lambda \right) \rho + \frac{(1- \lambda)}{4} \left(X \rho X+Y \rho Y+Z \rho Z \right). $$ Then, we can see that the Kraus operators are $K_{0} = \frac{1}{2} \sqrt{1 + 3 \lambda} \mathbb{I}$ and $K_{i} = \frac{1}{2} \sqrt{(1- \lambda)} \sigma_{i}$, $i=1,2,3$, where $\{ \sigma_{i} \}$ are the sigma matrices. Hence this map is CP because it has a Kraus representation.

Note: The Kraus operator form also reveals why in $d=2$, the limit for CP is $-1/3 \leq \lambda \leq 1$ (note the square roots in the Kraus representation). This can be generalized to $d$-dimensions.