The field in terms of creation and annihilation operators:
$$\phi(x)=\frac{1}{(2\pi)^3}\int\frac{d^3p}{\sqrt{2{E_p}}}\left(\hat{a}_{p}e^{-i({E_p}t-\vec{p}\cdot\vec{x})}+\hat{a}_{p}^{\dagger}e^{i({E_p}t-\vec{p}\cdot\vec{x})}\right)$$
$$\pi(x)=\dot{\phi}(x)=\frac{-i}{(2\pi)^3}\int d^3p\sqrt{\frac{{E_p}}{2}}\left(\hat{a}_{p}e^{-i({E_p}t-\vec{p}\cdot\vec{x})}-\hat{a}_{p}^{\dagger}e^{i({E_p}t-\vec{p}\cdot\vec{x})}\right)$$
$$\left(\partial_{\mu}\partial^{\mu}+m^2\right)\phi(x)=0$$

where \({E_p}=\sqrt{p^2+m^2}\). The commutation relations:
$$\left[\phi(t,\vec{x}),\pi(t,\vec{y})\right]=i\delta^{(3)}(\vec{x}-\vec{y})$$
$$\left[\hat{a}_{p},\hat{a}_{p'}^{\dagger}\right]=(2\pi)^3\delta^{(3)}(\vec{p}-\vec{p}')$$
The Klein-Gordon Hamiltonian
$$H=\int d^3x\mathcal{H}=\int d^3x\left(\left|\partial_{0}\phi\right|^2+\left|\nabla\phi\right|^2+m^2\left|\phi\right|^2\right)=\frac{1}{2}\int\frac{d^3p}{(2\pi)^3}\,{E_p}\left(\hat{a}_{p}^{\dagger}\hat{a}_{p}+\hat{a}_{p}\hat{a}_{p}^{\dagger}\right)$$
$$\left[H,\hat{a}_{p}\right]={E_p}\,\hat{a}_{p},\quad\quad\left[H,\hat{a}_{p}^{\dagger}\right]=-{E_p}\,\hat{a}_{p}^{\dagger}$$

$$\left|p\right.\rangle=\sqrt{2{E_p}}\,\hat{a}_{p}^{\dagger}\left|0\right.\rangle,\quad\langle p\left|p'\right.\rangle=(2\pi)^32{E_p}\,\delta^{(3)}(\vec{p}-\vec{p}')$$
$$\phi(x)\left|0\right.\rangle=\frac{1}{(2\pi)^3}\int d^3p\,e^{i({E_p}t-\vec{p}\cdot\vec{x})}\left|p\right.\rangle,\quad\langle p\left|\phi(x)\right|0\rangle=e^{i({E_p}t-\vec{p}\cdot\vec{x})}$$

The amplitude for a particle to propagate \(x\rightarrow y\):
$$\langle0\left|\phi(y)\phi(x)\right|0\rangle=\frac{1}{(2\pi)^3}\int\frac{d^3p}{2{E_p}}e^{i\left({E_p}(x_0-y_0)-\vec{p}\cdot(\vec{x}-\vec{y})\right)}$$

Klein-Gordon propagator
$$D_F(y-x)=\theta(y_0-x_0)\langle0\left|\phi(y)\phi(x)\right|0\rangle\,+\,\theta(x_0-y_0)\langle0\left|\phi(x)\phi(y)\right|0\rangle=$$
$$=\frac{1}{(2\pi)^3}\int\frac{d^3p}{2{E_p}}\left(\theta(y_0-x_0)\,e^{i\left({E_p}(x_0-y_0)-\vec{p}\cdot(\vec{x}-\vec{y})\right)}-\theta(x_0-y_0)\,e^{-i\left({E_p}(x_0-y_0)-\vec{p}\cdot(\vec{x}-\vec{y})\right)}\right)$$
$$=\frac{i}{(2\pi)^4}\int\frac{d^4p}{p^2-m^2-i\epsilon}e^{i\left({E_p}(x_0-y_0)-\vec{p}\cdot(\vec{x}-\vec{y})\right)}$$