In this video, Artur Ekert shows that for a single qubit, 4 Kraus operators can be chosen such that the action on state $\rho$ is given as $\rho \rightarrow \sum_m p_m A_m \rho A_m^\dagger$. We can choose $A_m$ to be the Pauli matrices $I,X,Y,Z$. Then, how does one get the off-diagonal terms such as $X\rho Y$ and so on, which are said to be removed using the Pauli twirling approximation (PTA)? In other words, what exactly is the definition of the PTA if one can always express the action of a single qubit channel as mentioned above? In this reference, they consider a single qubit channel in eqn 8 which doesn't automatically reduce to a diagonal form in eqn 16 before applying the PTA to get the diagonal form in eqn 17. Is it just that the matrix of coefficients ($d_{ij}$ in the video and $\tau_{m,a}$ in the reference is not always diagonalizable?

  • $\begingroup$ I don't quite follow the question. Where are you saying the "off-diagonal terms" should come from? $\endgroup$
    – glS
    Jul 23, 2022 at 9:58
  • $\begingroup$ In general, there can be off-diagonal terms in a single qubit noise channel. You can check Eqn 16 in the reference I linked in the question and also Eqn 6 of this reference arxiv.org/pdf/1701.03708.pdf. My question is the definition of PTA because in the video lecture I linked, it seems you can always reduce to a diagonal form assuming that the matrix of coefficients $d_ij$ can be diagonalized. So my question is- does the PTA make sense to use as an approximation when $d_ij$ cannot be diagonalized? $\endgroup$
    – user111
    Jul 23, 2022 at 21:07
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    $\begingroup$ @user111 The Kraus operator decomposition, which allows to describe a completely positive trace preserving operation $\mathcal{E}$ can be written as $\mathcal{E}=(\rho)\sum_m K_m \rho K_m^{\dagger}$, the $K_m$ are the Kraus operators which are not necessarily Pauli operators, i.e. if you redecompose this sum, in general you will have off-diagonal terms as you suggested. I believe that the Twirling is a trick that you do in your quantum circuit (by considering applying some random maps before and after each gate) that kills these extra diagonal terms. $\endgroup$ Aug 17, 2022 at 14:07
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    $\begingroup$ Hence, we cannot always represent the channel without the extra diagonal terms, but by applying some trick in the circuit (twirling) we can remove these. Was that your question? Note that I don't know Twirling very well so you should double check my suggestion. $\endgroup$ Aug 17, 2022 at 14:08
  • $\begingroup$ @MarcoFellous-Asiani thanks, that makes sense $\endgroup$
    – user111
    Aug 20, 2022 at 3:11


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