  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.
