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Split the free scalar field into its annihilation part ${\varphi }^{(+)}$ and creation part ${\varphi }^{(-)}$. Normal ordering moves every creation operator to the left of every annihilation operator, without adding the commutators produced by that rearrangement. For two fields,Its vacuum expectation is zero. Time ordering instead places the field at the later time on the left:The equal-time convention is immaterial away from coincident singularities; these expressions are understood as operator-valued distributions.

The only nonzero vacuum contraction uses $a_{\mathbf{p}}{a}_{\mathbf{q}}^{†}$. The oscillator commutator therefore gives the Wightman functionPut $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. Combining the two time orders, and reversing $\mathbf{p}$ in the second when necessary, givesTo recover the four-dimensional Fourier transform, use the Feynman i-epsilon prescription in the energy variable:The positive-energy pole lies just below the real axis, and the negative-energy pole just above it. For $\tau >0$, close below clockwise: the residue at $E_{\mathbf{p}}-i0$ is $i{e}^{-i{E}_{\mathbf{p}}\tau }/(2E_{\mathbf{p}})$, and the clockwise factor $-i$ after dividing by $2\pi$ gives the required positive coefficient. For $\tau <0$, close above counterclockwise: the residue at $-E_{\mathbf{p}}+i0$ is $-i{e}^{i{E}_{\mathbf{p}}\tau }/(2E_{\mathbf{p}})$, and the factor $+i$ gives the other time order. Equivalently the energy contour passes above the positive-energy pole and below the negative-energy pole. HenceThe limit is distributional. The scalar Feynman propagator pole prescription is essential; omitting it from the denominator must be accompanied by the specified contour.

Finally, moving the annihilation part of $\varphi (x)$ past the creation part of $\varphi (y)$ givesThe normal-ordered two-field product is symmetric in $x,y$, because the two creation parts commute and the two annihilation parts commute. Applying the two time orders thus proves the two-field scalar Wick identity