= Solution
The normalized <Fock vacuum> obeys $a_{\mathbf p}|0\rangle=0$ for every momentum and $\langle0|0\rangle=1$. Taking adjoints gives $\langle0|a_{\mathbf p}^\dagger=0$. Of the four products in the mode expansion of $\phi(x)\phi(y)$, only $a_{\mathbf p}a_{\mathbf q}^\dagger$ has a nonzero <vacuum expectation value>. The <commutator> implies
$$
\langle0|a_{\mathbf p}a_{\mathbf q}^\dagger|0\rangle=(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q).
$$
Consequently
$$
\begin{aligned}
\langle0|\phi(x)\phi(y)|0\rangle
&=\int\frac{d^3p\,d^3q}{(2\pi)^6\sqrt{2E_{\mathbf p}\,2E_{\mathbf q}}}(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q)e^{-ipx+iqy}\\
&=\boxed{\int\frac{d^3p}{(2\pi)^3\,2E_{\mathbf p}}e^{-ip\cdot(x-y)}}.
\end{aligned}
$$
This unordered <Wightman function> differs from the time-ordered <Feynman propagator> when the temporal ordering is reversed.
Back to article page
