I am trying find the definition of twirled (super)operator. One such is Definition 2.3.16 on p. 33 of Christoph Dankert, Efficient Simulation of Random Quantum States and Operators. However, the notations puzzles me.

  1. What does $\Lambda(V^\dagger XV)$ mean?
  2. How do the two integrals apparently both involving $\Lambda_T$ relate to each other?

1 Answer 1

  1. $X$ is a density matrix. $V^\dagger X V$ is another density matrix. $\Lambda$ is a channel that takes one density matrix to another density matrix. So $\Lambda(V^\dagger X V)$ means take the density matrix $V^\dagger X V$ and apply the channel $\Lambda$ to it.

  2. The two integrals are two ways to write the same formula. In the first, they use the Liouville operators $\hat V:=V\otimes V^*$ and $\hat\Lambda$ to define the Liouville operator version of the twirled channel $\hat\Lambda_T$. In the second, they use the channel notation $\Lambda(\cdot)$ to define the twirled channel $\Lambda_T(\cdot)$. If you just follow the definition of the Liouville operator, you can see that the Liouville operator of the thing they define as $\Lambda_T$ is indeed what they define as $\hat \Lambda_T$.

  • $\begingroup$ 1) How is $\hat\Lambda$ defined? Is it the following? Given $V\in U(d), \hat\Lambda(X):=\Lambda(VXV^\dagger), \forall$ density operator $X$. Note in this setting the $\hat\cdot$ operator is dependent on $V$. 2) How does the tensor product definition of $\hat V$ reconcile with the second definition of $\Lambda_T(X)$ which is defined in the non-tensor space? $\endgroup$
    – Hans
    Oct 30, 2022 at 20:17
  • $\begingroup$ @Hans That is not how $\hat\Lambda$ is defined. $\hat\Lambda$ is a linear superoperator matrix, or Liouville matrix. The basic idea is that we know $\Lambda$ is a linear operator on the $d^2$-dimensional space of density matrices, so if you pick a basis for that space you can write $\Lambda$ as a $d^2\times d^2$ matrix. That matrix is $\hat\Lambda$. See arxiv.org/pdf/0911.2539.pdf, in particular III.B $\endgroup$ Oct 31, 2022 at 1:19
  • $\begingroup$ @Hans Although honestly, you could do better by just reading this answer, I think it's clearer than the paper I linked quantumcomputing.stackexchange.com/a/6646/9438 $\endgroup$ Oct 31, 2022 at 1:20

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