= Solution
The <Heisenberg picture> evolution of the <annihilation operator> follows from $[H,a_{\mathbf p}]=-E_{\mathbf p}a_{\mathbf p}$. With $p^0=E_{\mathbf p}$, this gives the <mode expansion of a free field>
$$
\phi(x)=\int_{\mathbf p}\frac1{\sqrt{2E_{\mathbf p}}}
\left(a_{\mathbf p}e^{-ip\cdot x}+a^\dagger_{\mathbf p}e^{ip\cdot x}\right).
$$
Define $\Delta^+(z)=\int_{\mathbf p}e^{-ip\cdot z}/(2E_{\mathbf p})$. The <time-ordered product> then gives the <Feynman propagator>
$$
\Delta_F(z)=\theta(z^0)\Delta^+(z)+\theta(-z^0)\Delta^+(-z)
=\int\frac{d^4p}{(2\pi)^4}\frac{i\,e^{-ip\cdot z}}{p^2-m^2+i0}.
$$
The last expression follows by closing the $p^0$ <contour integral> below for $z^0>0$ and above for $z^0<0$; the <scalar Feynman propagator pole prescription> supplies the required positive- and negative-frequency terms.
For $z^0\ne0$ every term obeys the homogeneous <Klein-Gordon equation>. At $z^0=0$, the equal-time <commutator> of the <fields> vanishes, so the <propagator> is continuous and there is no derivative of a <Dirac delta function>. Its first time derivative has the jump
$$
\left[\partial_{z^0}\Delta_F\right]_{0^-}^{0^+}
=\int_{\mathbf p}(-i)e^{i\mathbf p\cdot\mathbf z}
=-i\delta^3(\mathbf z).
$$
The second time derivative therefore supplies precisely the source:
$$
\boxed{(\Box_x+m^2)\Delta_F(x-y)=-i\delta^4(x-y).}
$$
Thus the defined <Feynman propagator> is a <Green's function> with source normalization $-i$; $i\Delta_F$ has source normalization $+1$. Keeping this factor avoids confusing the <vacuum expectation value> convention with conventions which define the <propagator> after dividing by $i$.<Quantum Field Theory by David Tong>[Sections 2.7.1–2.7.2, time ordering, pole prescription and the inhomogeneous Klein-Gordon equation]{ref}
Back to article page
