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?


1 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.

  • $\begingroup$ awesome! thanks a lot @DaftWullie... $\endgroup$ Commented Nov 25, 2020 at 22:02
  • $\begingroup$ Hello @DaftWullie, could you please comment on how this formulation of depolarizing channel is equivalent to applying the Kraus operator representation of depolarizing channel? Thanks... $\endgroup$ Commented Nov 26, 2020 at 19:02
  • $\begingroup$ Also, the diagonal elements no longer adds up to 1, don't they? Do I have to renormalize the final density operator? $\endgroup$ Commented Nov 26, 2020 at 19:10
  • 1
    $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Commented Nov 27, 2020 at 7:51

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