I use $\phi^4$ theory as an example here, but similar things happen to other theories as well.
In $\phi^4$ theory, we can easily use the Lagrangian here to write down the interaction Hamiltonian (in interaction picture, canonical quantization): $$ H_I=\frac{g}{4!}\int d^3 x \phi(x)^4, $$ My first question: must I impose normal ordering on $H_I$ and write down $$ H_I=\frac{g}{4!}\int d^3 x :\phi(x)^4: $$ instead? I have seen at different places the interaction Hamiltonian written without the normal ordering colons $::$, and I am confused by this. Does that mean normal ordering can be implied implicitly? After all, the Hamiltonian of free theory without interaction is normal ordered.
The reason why I think we should have normal ordering is this: to compute the scattering amplitude, we use Wick's theorem, and express time ordering in terms of normal orderings. If we write out the first order term of the S-matrix (in Dyson series): $$ S-\text{Id} \sim \frac{-ig}{4!} \mathcal T \int d^4 x \phi(x)^4, $$ we see that we are time ordering ($\mathcal T$) four operators that are at the same time and same place: $$ \mathcal T (\phi(x)\phi(x)\phi(x)\phi(x)) $$ Of course, now the time are all the same, so it makes little sense to order this in time. But I include $\mathcal T$ here anyway because we are going to apply wick's theorem: $$ \mathcal T (\phi_1(x_1)\phi_2(x_2)\phi_3(x_3)\phi_4(x_4))=:\phi_1(x_1)\phi_2(x_2)\phi_3(x_3)\phi_4(x_4): + \text{contraction}(\phi_1(x_1)\phi_2(x_2)):\phi_3(x_3)\phi_4(x_4): \\+ \text{all other contractions} $$ Applying this formula to $\mathcal T (\phi(x)\phi(x)\phi(x)\phi(x))$ means that we are expressing $\phi(x)\phi(x)\phi(x)\phi(x)$ in terms of its normal orderings. But the Feynman propagator $\text{contraction}(\phi(x)\phi(x))=\Delta_F(0)\sim \delta(0)=\infty$, so if $\phi(x)\phi(x)\phi(x)\phi(x)$ is NOT treated as normal ordered, then the extra contraction terms will ALL be infinite. So we need to ignore all these terms to give a useful scattering amplitude.
In short, it seems that in a theory with two fields $\phi_1, \phi_2$, we should treat all fields at the same point in the spacetime as if they are already normal ordered, or in other words, stipulate that their contraction $\text{contraction}(\phi_1(x)\phi_2(x))$ is ZERO.
Is my summary accurate?
