In the usual Fourier expansion of Schrodinger fields \begin{align} \Psi(\vec{x}) = \frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3 k \hat{a}_k e^{-i (wt-\vec{k}\cdot \vec{x})}, \quad \Psi^{*}(\vec{x}) = \frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3 k \hat{a}^{\dagger}_k e^{i (wt-\vec{k}\cdot \vec{x})} \end{align} while Foruier expansion of Klein-Gordon fields are followings
\begin{align} \Psi(\vec{x}) = \int \frac{d^3 k}{(2\pi)^{3}2w_k} \left[\hat{a}_k e^{-i (wt-\vec{k}\cdot \vec{x})}+\hat{a}^{\dagger}_k e^{i(wt-\vec{k}\cdot \vec{x})}\right] \end{align}
Here i wonder why the expansion of schrodinger field and klein gordon fields are different. First in schrodinger field $\Psi$ is wrttien in one variable($a$), while klein gordon field is written in two variables. ($a, a^{\dagger}$). Second the factor of $2w_k$ comes out. I guess this comes from the Lorentz invariance of measure $\frac{d^3k}{(2\pi)^3 2w_k}$. hmm, in this sense, i guess Lorentz invariance is key difference between two theories.
Can anyone gives some reliable explanation about the difference between Schrodinger fields and Klein-Gordon fields?