I have a few issues with making the transition between these:
$\phi(x)=\int{\frac{d^3p}{2\pi^3}\frac{1}{\sqrt{2\omega_{\vec{p}}}}(a_{\vec{p}}e^{i \vec{p} \vec{x}}+ a^{\dagger}_{\vec{p}}e^{-i \vec{p} \vec{x}}})$
$\pi(x)=\int{\frac{d^3p}{2\pi^3}(-i)\sqrt{\frac{\omega_{\vec{p}}}{2}}(a_{\vec{p}}e^{i \vec{p} \vec{x}}- a^{\dagger}_{\vec{p}}e^{-i \vec{p} \vec{x}}})$.
So from the Lagrangian of the real scalar field, we get
$\pi(x)=\partial_0 \phi(x)$
which just means differentiating our $\phi(x)$ w.r.t time. But where does the $-i$ come from (as in the minus sign)? Also aren't the exponentials independent of time as $\vec{p} \vec{x}$ isn't a four vector? So now I'm wondering where the $\omega_{\vec{p}}$ has come from as well!
Answer
The answer depends on whether you represent the operators in the Schrödinger or Heisenberg picture. In the Schrödinger picture, there is no explicit time-dependence of the operators. Hence, the conjugate momentum is not given by the time derivative of the field. The concrete form is rather given by analogy to the position and momentum operators of a harmonic oscillator:
$$x=\frac{1}{\sqrt{2\omega}}(a+a^\dagger)$$ $$p=-i\sqrt{\frac{\omega}{2}}(a-a^\dagger).$$
In the Heisenberg picture, the operators acquire an explicit time-dependence, i.e.
$$\phi(x)\rightarrow e^{iHt}\phi(x) e^{-iHt}.$$
In this case, taking the derivative with respect to time makes sense, and the relation between the field and its conjugate momentum can be shown to hold.
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