Assume we have a quantum channel $\Phi$. The single qubit Pauli basis is $\sigma_0, \sigma _1, \sigma_2, \sigma_3$. Now we apply $\Phi$ to Pauli basis and get $\gamma_0=\Phi(\sigma_0), \gamma_1 = \Phi(\sigma_1), \gamma_2=\Phi(\sigma_2), \gamma_3 = \Phi(\sigma_3)$. My question is whether $\gamma_0, \gamma_1, \gamma_2, \gamma_3 $ forms a basis.
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2 Answers
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No, not necessarily. For example, the channel $\Phi(\rho) = \operatorname{Tr}(\rho) \vert 0 \rangle \langle 0 \vert$ makes $\{\Phi(\sigma_0), \ldots, \Phi(\sigma_3)\}$ linearly dependent. (In fact, the three outputs corresponding to the traceless Pauli matrices are the all-zero matrices in this case.)
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As a more general statement: a linear operator always sends linearly independent vectors in its support into linearly independent vectors in its image.
In other words, if $\{\mathbf v_k\}$ is a basis for some finite-dimensional space $V$, and $A$ is an operator on $V$, then $\{A\mathbf v_k\}$ is a basis iff $A$ is injective, i.e. $\ker(A)=\{0\}$.
So in the case of channels, you have to look at whether any of the Pauli matrices is in the ker of $\Phi$. If none are, then $\Phi$ is injective (as an operator on the space of Hermitian operators), and thus sends the Pauli matrices (or any other operatorial basis of Hermitian operators) into another basis.