= Solution
Use $\eta_{ab}=\operatorname{diag}(1,-1,-1,-1)$ and $\hbar=1$. The free <real scalar field> has <action>
$$
S=\frac12\int d^4x\,[\partial_a\phi\partial^a\phi-m^2\phi^2].
$$
The <canonical momentum> is $\pi=\partial\mathcal L/\partial\dot\phi=\dot\phi$, and the <Hamiltonian operator> is obtained by the <Legendre transform> of the density:
$$
H=\frac12\int d^3x\,[\pi^2+(\boldsymbol\nabla\phi)^2+m^2\phi^2].
$$
<Canonical quantization> promotes $\phi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time <canonical commutation relations>
$$
[\phi(t,\mathbf x),\pi(t,\mathbf y)]=i\delta^3(\mathbf x-\mathbf y),\qquad
[\phi(t,\mathbf x),\phi(t,\mathbf y)]=[\pi(t,\mathbf x),\pi(t,\mathbf y)]=0.
$$
Their <Heisenberg equation of motion> gives $\dot\phi=\pi$, $\dot\pi=\boldsymbol\nabla^2\phi-m^2\phi$, hence the <Klein-Gordon equation> $(\Box+m^2)\phi=0$.
Let $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$ and $p\cdot x=E_{\mathbf p}t-\mathbf p\cdot\mathbf x$. The Hermitian field mode expansion is
$$
\phi(x)=\int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}
\left[a(\mathbf p)e^{-ip\cdot x}+a^\dagger(\mathbf p)e^{ip\cdot x}\right].
$$
The <scalar field oscillator inversion> extracts
$$
a(\mathbf p)=e^{iE_{\mathbf p}t}\int d^3x\,e^{-i\mathbf p\cdot\mathbf x}
\left[\sqrt{\frac{E_{\mathbf p}}2}\,\phi(t,\mathbf x)+\frac{i}{\sqrt{2E_{\mathbf p}}}\,\pi(t,\mathbf x)\right],
$$
and its adjoint obtained by <Hermitian conjugation> extracts $a^\dagger$. Substitute these expressions into the equal-time <canonical commutation relation>. The two mixed field-momentum terms give
$$
\begin{aligned}
[a(\mathbf p),a^\dagger(\mathbf q)]
&=e^{i(E_{\mathbf p}-E_{\mathbf q})t}\frac12
\left(\sqrt{\frac{E_{\mathbf p}}{E_{\mathbf q}}}+\sqrt{\frac{E_{\mathbf q}}{E_{\mathbf p}}}\right)
(2\pi)^3\delta^3(\mathbf p-\mathbf q)\\
&=\boxed{(2\pi)^3\delta^3(\mathbf p-\mathbf q)}.
\end{aligned}
$$
For $[a(\mathbf p),a(\mathbf q)]$, the corresponding coefficient is a difference of those square roots and multiplies $\delta^3(\mathbf p+\mathbf q)$; it vanishes because $E_{\mathbf p}=E_{-\mathbf p}$. Taking the adjoint gives $[a^\dagger,a^\dagger]=0$. Thus these are bosonic <annihilation operators> and <creation operators>. The vacuum satisfies $a(\mathbf p)|0\rangle=0$, and <normal ordering> gives $:H:=\int d^3p\,E_{\mathbf p}a^\dagger(\mathbf p)a(\mathbf p)/(2\pi)^3$, after removing the constant zero-point energy. A one-particle excitation has <energy> $E_{\mathbf p}$ and spin zero.
The <Feynman propagator> $G_F(x-y)=\langle0|T\phi(x)\phi(y)|0\rangle$ is the <vacuum expectation value> of the <time-ordered product> of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^\dagger$ contraction survives, so with $t=x^0-y^0$ and $\mathbf r=\mathbf x-\mathbf y$,
$$
\begin{aligned}
G_F(t,\mathbf r)
&=\int\frac{d^3p}{(2\pi)^3,2E_{\mathbf p}}
\left[\theta(t)e^{-iE_{\mathbf p}t+i\mathbf p\cdot\mathbf r}
+\theta(-t)e^{iE_{\mathbf p}t-i\mathbf p\cdot\mathbf r}\right]\\
&=\int\frac{d^3p}{(2\pi)^3,2E_{\mathbf p}}
e^{i\mathbf p\cdot\mathbf r}e^{-iE_{\mathbf p}|t|}.
\end{aligned}
$$
Here $\theta$ is the <Heaviside step function>, and changing $\mathbf p$ to $-\mathbf p$ in the second term gives the last line.
For the <Fourier transform> convention $\widetilde G_F(\omega,\mathbf p)=\int dt\,d^3r\,e^{i\omega t-i\mathbf p\cdot\mathbf r}G_F(t,\mathbf r)$, insert a positive damping factor $e^{-\epsilon|t|}$ and integrate the positive and negative half-lines separately:
$$
\widetilde G_F(\omega,\mathbf p)
=\lim_{\epsilon\downarrow0}\frac{i}{2E_{\mathbf p}}
\left[\frac1{\omega-E_{\mathbf p}+i\epsilon}-\frac1{\omega+E_{\mathbf p}-i\epsilon}\right]
=\boxed{\frac{i}{\omega^2-\mathbf p^2-m^2+i0}}.
$$
The last equality is a <distribution> identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The <Feynman i-epsilon prescription> means that the positive-frequency pole is just below the real <energy> axis and the negative-frequency pole is just above it. Closing the <contour integral> below for $t>0$, and above for $t<0$, reproduces the oscillator result by the <residue theorem>. This prescription fixes which homogeneous solutions are added to the <Green function>. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, so
$$
(\Box+m^2)G_F(x)=-i\delta^4(x).
$$
It is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.
The vacuum is a centered free Gaussian state. The <Wick theorem> expresses the four-field <time-ordered product> as the sum of all pair <Wick contractions> plus terms containing a <normal-ordered product>. The latter terms have zero <vacuum expectation value>. There are exactly three complete pairings, with no <fermionic signs> for this bosonic field. Thus the <free scalar four-point function> is
$$
\boxed{\begin{aligned}
\langle0|T\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)|0\rangle
={}&G_F(x_1-x_2)G_F(x_3-x_4)\\
&+G_F(x_1-x_3)G_F(x_2-x_4)\\
&+G_F(x_1-x_4)G_F(x_2-x_3).
\end{aligned}}
$$
These formulas describe <canonical quantization of a real scalar field>; in particular, a <real scalar field> uses a single oscillator family rather than independent charged-particle and antiparticle families.
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