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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.

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  • $\begingroup$ 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. $\endgroup$ – Norbert Schuch Aug 22 at 20:23
  • $\begingroup$ 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. $\endgroup$ – Dina Abdelhadi Aug 23 at 6:54
  • $\begingroup$ 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. $\endgroup$ – Norbert Schuch Aug 23 at 11:52
  • $\begingroup$ 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. $\endgroup$ – Dina Abdelhadi Aug 24 at 13:21
  • $\begingroup$ BTW, I think it is very confusing that your tensor product is not always ordered the same way (i.e. A x B). $\endgroup$ – Norbert Schuch Aug 24 at 14:21
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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.

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