Prove that the tensor product of two CPTP maps is also CPTP We can take the depolarization map $\mathcal{E}$ as an example, which is defined as:
$$
\mathcal{E}(\rho) = (1-p)\rho + p Tr(\rho)\frac{\mathbb{I}}{2},
$$
for an arbitrary density operator $\rho$ in two dimensions and probability p. It can be proved that $\mathcal{E}$ is completely positive and trace preserving (CPTP). I want to prove that the map $\mathcal{E} \otimes \mathcal{E}$ is also CPTP. How to proceed? Would Choi-Jamiołkowski representation help? TIA.
 A: The useful thing in you context is the Kraus Decomposition. If $\mathcal E=\sum_jE_j\cdot E_j^*$, then
$$
(\mathcal E\otimes\mathcal E)(\rho\otimes\eta)=\sum_{k,j}E_k\rho E_k^*\otimes E_j\eta E_j^*
=\sum_{k,j}(E_k\otimes E_j)(\rho\otimes\eta)(E_k\otimes E_j)^*. 
$$
By linearity this holds for all elements in $M_n(\mathbb C)\otimes M_n(\mathbb C)$.
The trace property is obvious if you know that the trace on the tesor product is the product of the traces. If you don't, take $\{e_k\}$ an orthonormal basis of $H$. Then $\{e_k\otimes e_j\}_{k,j}$ is an orthonormal basis of $H\otimes H$. So
\begin{align}
\operatorname{Tr}((\mathcal E\otimes\mathcal E)(\rho\otimes\eta))
&=\sum_{k,j}\langle (\mathcal E(\rho)\otimes\mathcal E(\eta))(e_k\otimes e_j),e_k\otimes e_j\rangle\\[0.3cm]
&=\sum_{k,j}\langle \mathcal E(\rho)e_k\otimes\mathcal E(\eta)e_j,e_k\otimes e_j\rangle\\[0.3cm]
&=\sum_{k,j}\langle \mathcal E(\rho)e_k,e_k\rangle\,\langle\mathcal E(\eta)e_j,e_j\rangle\\[0.3cm]
&=\Big(\sum_k\langle \mathcal E(\rho)e_k,e_k\rangle\Big)\,\Big(\sum_k\langle \mathcal E(\eta)e_j,e_j\rangle\Big)\\[0.3cm]
&=\operatorname{Tr}(\mathcal E(\rho))\operatorname{Tr}(\mathcal E(\eta))\\[0.3cm]
&=\operatorname{Tr}(\rho)\operatorname{Tr}(\eta)\\[0.3cm]
&=\operatorname{Tr}(\rho\otimes\eta),
\end{align}
where the last equality is obtained by undoing the previous steps (without $\mathcal E$).
