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An explicit example avoids an unspecified correlation time. For each coordinate axis ${\mathbf{e}}_j$, choose transverse unit vectors ${\mathbf{a}}_j,{\mathbf{b}}_j$ with ${\mathbf{e}}_j\times {\mathbf{a}}_j={\mathbf{b}}_j$, and take the three real circular polarizationsThe propagation directions are mutually perpendicular and all three waves have the same handedness. Direct differentiation givesA spatial average over the periodic cell removes cross terms between different waves. Hence the mean kinetic helicity of their sum isThe sum has isotropic second moments, $⟨u_i{u}_j⟩=U^2{\delta }_{ij}$.

For a locally uniform test magnetic field $\overline{\mathbf{B}}$, the periodic first-order smoothing response obeysSet $a=\eta k^2>0$. Since ${\partial }_t=-\Omega {\partial }_{\varphi }$ and ${\partial }_{\varphi }^2{\mathbf{u}}_j=-{\mathbf{u}}_j$, direct substitution gives the response after transients decay:The term proportional to ${\mathbf{u}}_j$ contributes no cross product with that same wave. Cross terms between distinct waves average to zero. Thereforeand the isotropic alpha effect of three helical traveling waves isThis is the alpha-helicity relation with the exact response time ${\tau }_{\mathrm{eff}}=a/(a^2+{\Omega }^2)$ for these monochromatic waves. In the quasistatic limit $\Omega =0$, $\alpha =-\mathcal{H}/(3\eta k^2)$. A single wave would give an anisotropic response along its propagation direction; the three equal waves make the response tensor a scalar multiple of the identity. The calculation retains the validity assumptions of first-order smoothing, rather than extrapolating the formula to arbitrary fluctuation amplitude.