Timeline for Can a Kraus representation act as the identity on any operator?
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| Jul 18, 2020 at 0:53 | history | edited | keisuke.akira | CC BY-SA 4.0 |
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| Jul 18, 2020 at 0:52 | comment | added | keisuke.akira | One way to obtain the result in $d$ dimensions is to use the $d^2$ Heisenberg-Weyl operators (or the finite dimensional representation of the Heisenberg-Weyl algebra). If $X(i)Z(j)$ is the $(i,j)$th operator then, we have, $\frac{1}{d^{2}} \sum_{i, j=0}^{d-1} X(i) Z(j) \rho Z^{\dagger}(j) X^{\dagger}(i)=\frac{\mathbb{I}}{d}$. See, for example, Page 176 of this book. | |
| Jul 17, 2020 at 8:16 | comment | added | Amplituhedron | Are you assuming the state $\rho$ is written in terms of the Pauli matrices? For a higher dimension, I found it very difficult to compute the commutation relation with unitaries and Hermitian matrices given in a quantum tomography for $\rho$. (I tried with the U(n) basis.) Nevertheless in some dimensions like 2^n dimensions, we can construct such a basis with tensor products of Pauli matrices. | |
| Jul 15, 2020 at 3:06 | history | answered | keisuke.akira | CC BY-SA 4.0 |