# Two qubit state + Depolarizing channel = Bell diagonal state?

In multiple sources, e.g. RGK, KGR, it is stated (without proof) that if you take any two qubit state and send it through a depolarizing channel, the resulting state would be a Bell-diagonal state. I understand that a bipartite Bell-diagonal state $$\rho_{AB}$$ has the form:

$$\rho_{AB} = \lambda_1 |\Psi^+\rangle\langle \Psi^+| + \lambda_2 |\Psi^-\rangle\langle \Psi^-| +\lambda_3 |\Phi^+\rangle\langle \Phi^+| +\lambda_4 |\Phi^-\rangle\langle \Phi^-|,$$ where $$|\Psi^+\rangle, |\Psi^-\rangle, |\Phi^+\rangle, |\Phi^-\rangle$$ are the usual Bell states. The action of a depolarizing channel $$\mathcal{E}$$ on two qubits is defined as:

$$\mathcal{E}(\rho_{AB}) = \sum_i (E_i \otimes E_i) \rho_{AB} (E_i \otimes E_i)^\dagger,$$ where $$E_i \in \{\mathbb{I}, \sigma_x, \sigma_y, \sigma_z\}$$ are the Pauli operators. However, I don't see why ANY bipartite density operator would be transformed into a Bell-diagonal state. Is there any proof of this claim?

Firstly, note that every Bell state $$|\psi_{ij}\rangle=(|0i\rangle+(-1)^j|1\bar i\rangle)/\sqrt{2}$$ is an eigenstate of $$E_i\otimes E_i$$ for all $$i$$ (the eigenvalues are either $$\pm 1$$). Hence, a Bell-diagonal state remains Bell-diagonal under the action of the map. This already suggests that a Bell-diagonal state is likely to be the ultimate destination of the map, but let us prove that.
Consider an arbitrary state $$|\Psi\rangle$$. This can be decomposed in the Bell basis, $$|\Psi\rangle=\sum_{i,j}a_{ij}|\psi_{ij}\rangle.$$ We have $$XX|\psi_{i1}\rangle=-|\psi_{i,1}\rangle$$ and $$XX|\psi_{i0}\rangle=|\psi_{i,0}\rangle$$. So, for example, if I calculate $$|\Psi\rangle\langle\Psi|+XX|\Psi\rangle\langle\Psi|XX,$$ then this knocks out any cross terms such as $$|\psi_{i0}\rangle\langle\psi_{j1}|$$
Similarly, $$ZZ|\psi_{0i}\rangle=|\psi_{0,i}\rangle$$ and $$ZZ|\psi_{1i}\rangle=-|\psi_{1i}\rangle$$, so terms such as $$|\psi_{0i}\rangle\langle\psi_{1j}|$$ will also be knocked out. Ultimately, the only terms that are left are $$|\psi_{ij}\rangle\langle\psi_{ij}|$$, i.e. the state is Bell diagonal.
Strictly, to put all this together carefully, you want to say $$\rho_x=\rho+XX\rho XX$$ and $$\mathcal{E}(\rho)=\rho_x+ZZ\rho_xZZ$$ two see how the two separate steps that I've made fit together.
• The Krauss operators are $\sigma_i\otimes\sigma_i/2$. It's that factor of 2 that makes things properly normalised, so you don't need a separate step at the end. – DaftWullie Nov 27 '20 at 7:51