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For a centered free real scalar field, its vacuum time-ordered product has expectation $⟨T{\varphi }_1{\varphi }_2{\varphi }_3{\varphi }_4⟩=G_{12}{G}_{34}+G_{13}{G}_{24}+G_{14}{G}_{23}$, where $G_{ij}$ is the Feynman propagator between insertions $i,j$. These are the three complete pairings in the Wick theorem. All remaining normal-ordered products have zero vacuum expectation. No fermionic interchange signs occur for a bosonic field.

A normal-ordered product of free-field oscillator operators places all creation operators to the left of all annihilation operators, including the fermionic sign required to reorder fermions. Its vacuum expectation vanishes if any nontrivial oscillator factors remain. The Wick theorem expresses a free-field time-ordered product through these products and all Wick contractions.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

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 isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, 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 $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ 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 $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,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 ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately: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 $t=0$, soIt 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 isThese 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.