Quantum codes do not increase fidelity against isotropic errors

Abstract

Given an $m-$qubit ${\Phi }_0$ and an $(n,m)-$quantum code $\mathcal{C}$, let $\Phi$ be the $n-$qubit that results from the $\mathcal{C}-$encoding of ${\Phi }_0$. Suppose that the state $\Phi$ is affected by an isotropic error (decoherence), becoming $\Psi$, and that the corrector circuit of $\mathcal{C}$ is applied to $\Psi$, obtaining the quantum state $\tilde{\Phi }$. Alternatively, we analyze the effect of the isotropic error without using the quantum code $\mathcal{C}$. In this case the error transforms ${\Phi }_0$ into ${\Psi }_0$. Assuming that the correction circuit does not introduce new errors and that it does not increase the execution time, we compare the fidelity of $\Psi$, $\tilde{\Phi }$ and ${\Psi }_0$ with the aim of analyzing the power of quantum codes to control isotropic errors. We prove that $F({\Psi }_0)\ge F(\tilde{\Phi })\ge F(\Psi )$. Therefore the best option to optimize fidelity against isotropic errors is not to use quantum codes.