Two qubit state + Depolarizing channel = Bell diagonal state?

Question

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?

Answer 1

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.