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 $F(\Psi_0) \geq F(\tilde\Phi) \geq F(\Psi)$. Therefore the best option to optimize fidelity against isotropic errors is not to use quantum codes.