Quantum codes do not increase fidelity against isotropic errors
Given an $m-$qubit $Φ_0$ and an $(n,m)-$quantum code $\mathcal{C}$, let $Φ$ be the $n-$qubit that results from the $\mathcal{C}-$encoding of $Φ_0$. Suppose that the state $Φ$ is affected by an isotropic error (decoherence), becoming $Ψ$, and that the corrector circuit of $\mathcal{C}$ is applied to $Ψ$, obtaining the quantum state $\tildeΦ$. Alternatively, we analyze the effect of the isotropic error without using the quantum code $\mathcal{C}$. In this case the error transforms $Φ_0$ into $Ψ_0$. Assuming that the correction circuit does not introduce new errors and that it does not increase the execution time, we compare the fidelity of $Ψ$, $\tildeΦ$ and $Ψ_0$ with the aim of analyzing the power of quantum codes to control isotropic errors. We prove that $F(Ψ_0) \geq F(\tildeΦ) \geq F(Ψ)$. Therefore the best option to optimize fidelity against isotropic errors is not to use quantum codes.