Finding the irreducible representation of Kraus operators of a dephasing channel

I would like to understand an example of finding a noiseless subsystem of a quantum channel from the irreducible representation of its Kraus operators.

Assume we have $$2$$ dephasing channels acting on two qubits, then $$\mathcal{E}(\rho_{AB})= \sum_{i=1}^4 K_i \rho_{AB} K_i^\dagger,$$ where the Kraus operators are $$K_1=\mathbf{1}_{AB}/2, \qquad K_2= \frac{1}{2}\mathbf{1}_{A}\otimes Z_B, \qquad K_3=\frac{1}{2}Z_A\otimes \mathbf{1}_{B}, \qquad K_4=\frac{1}{2}Z_{A}\otimes Z_B.$$ How do I decompose this as $$\bigoplus_j \mathbf{1}_{n_j}\otimes M(d_j)$$? Will each Kraus operator be expressible in this decomposition?

I can see they are diagonal, and this channel has three decoherence-free subspaces; one spanned by $$|00\rangle$$, one spanned by $$|11\rangle$$,and one spanned by $$\{|01\rangle,|10\rangle\}$$. As mentioned in this book, decoherence free subspaces are a special case of noiseless subsystems, where $$M(d_j)$$ is just a constant, and $$n_j$$ is the dimension of the decoherence-free subspace. How do I extract these $$n_j$$'s from the Kraus operators or their irreps? Thanks in advance.

• In the form you write these are not Kraus operators. Also, what is the "irreducible representation of Kraus operators" - it would be good if you could define this. Aug 22 '20 at 20:23
• I normalized them, so I believe now they should be correct? What I mean is that, technically the 'algebra of Kraus operators is isomorphic to a direct sum of irreducible representations', which I think means each of the Kraus operators can be expressed by a block diagonal matrix of the form $\bigoplus \mathbf{1}_{n_j} \otimes M(d_j)$ , where $M(d_j)$ is some complex matrix. Aug 23 '20 at 6:54
• The formula for E(rho) you write is not a Kraus form. Also, I'd say that a map with these Kraus operators does not have dec. free subspaces with dimension >1. Aug 23 '20 at 11:52
• I see, may be I am confused between the dephasing channel and the collective dephasing where $|{0}\rangle$ is unaffected, and $|{1}\rangle$ is mapped to $e^{i\phi} |1\rangle$. Thanks. Aug 24 '20 at 13:21
• BTW, I think it is very confusing that your tensor product is not always ordered the same way (i.e. A x B). Aug 24 '20 at 14:21

To clear up my own confusion, these DFSs exist for the channel that maps $$|0\rangle$$ to $$|0\rangle$$ and $$|1\rangle$$ to $$e^{j\phi}|1\rangle$$. This is not the channel discussed in the question.