Write each free field as creation and annihilation parts, $\phi_i=\phi_i^-+\phi_i^+$. Moving every annihilation part to the right produces commutators $[\phi_i^+,\phi_j^-]$, which are precisely the contractions. Applying the $n=2$ identity once and then normal ordering the remaining field gives
$$
\boxed{
T\{\phi_1\phi_2\phi_3\}
=:\!\phi_1\phi_2\phi_3\!:
+D_F(x_1-x_2):\!\phi_3\!:
+D_F(x_1-x_3):\!\phi_2\!:
+D_F(x_2-x_3):\!\phi_1\!: .}
$$
No double contraction is possible for three fields. This is exactly <Wick theorem> at $n=3$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 4 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-301/4/ii/solution