# Scalar Field Theory with Interaction¶

This section introduces the self-interaction theory of scalar field. As usual, the interaction is implemented as an additional interaction term $$\mathcal{L}_{\text{int}}$$ in Lagrangian. The interaction discussed here is assumed to satisfy the following condition

• It is the self-interaction, i.e. $$\mathcal{L}_{\text{int}} = \mathcal{L}_{\text{int}}[\varphi]$$
• It has the form of polynomials.

Similarly, the construction starts from the generating functional. The generating functional of an interaction theory is

$\begin{split}W[J] &= \int\mathcal{D}\varphi\exp\{\mathrm iS\} = \int\mathcal{D}\varphi\exp\{\mathrm i\!\!\int\!\!\mathcal{L}\! + \!\mathcal{L}_{\text{int}}[\varphi]\}\\ &= \int\mathcal{D}\varphi\exp\{\mathrm i\!\!\int\!\!\mathcal{L}_{\text{int}}[\varphi]\}\exp\{\mathrm i\!\!\int\!\!\mathcal{L}\}\end{split}$

Next, we expand the first exponential (the one with interaction) according to definition

$W[J] = \int\mathcal{D}\varphi\sum_{n=0}^\infty\frac{1}{n!}\left[\mathrm i\!\!\int\!\!\mathcal{L}_{\text{int}}[\varphi]\}\right]^n\exp\{\mathrm i\!\!\int\!\!\mathcal{L}\}$

Now, recall our first identity

$\frac{1}{\mathrm{i}}\frac{\delta}{\delta J(x)}W_0[J] = \int\mathcal{D}\varphi\ \varphi(x)\exp\{\mathrm{i}S\}$

Therefore, each $$\varphi(x)$$ in $$\mathcal{L}_{\text{int}}[\varphi]$$ can be replaced with $$\delta/\mathrm{i}\delta J(x)$$, as long as $$\mathcal{L}_{\text{int}}[\varphi]$$ is a pure polynomial of $$\varphi$$. In this case, we can write

$\begin{split}W[J] = \int\mathcal{D}\varphi\sum_{n=0}^\infty\left[\mathrm i\!\!\int\!\!\mathcal{L}_{\text{int}}\!\!\left[\frac{1}{\mathrm{i}}\!\frac{\delta}{\delta J}\right]\}\right]^n\exp\{\mathrm i\!\!\int\!\!\mathcal{L}\}\\ = \sum_{n=0}^\infty\frac{1}{n!}\left[\mathrm i\!\!\int\!\!\mathcal{L}_{\text{int}}\!\!\left[\frac{1}{\mathrm{i}}\!\frac{\delta}{\delta J}\right]\}\right]^n\int\mathcal{D}\varphi\exp\{\mathrm i\!\!\int\!\!\mathcal{L}\}\end{split}$

where expression $$\mathcal{L}_{\text{int}}\!\!\left[\delta/\mathrm{i}\delta J\right]$$ represents the the expression where all $$\varphi$$ in $$\mathcal{L}_{\text{int}}$$ is replaced with $$\delta/\mathrm{i}\delta J(x)$$. The terms after the replacement no longer contain $$\varphi$$ and thus can be moved out of the path integral. And we find now that the path integral gives the generating functional $$W_0[J]$$ of free field. Transform back into the exponential form and we get

$W[J] = \exp\{-\mathrm{i}\int\mathcal{L}_{\text{int}}\!\!\left[\frac{1}{\mathrm{i}}\!\frac{\delta}{\delta J}\right]\} W_0[J]$

which is the result of second stage of our construction. The interaction of other kinds of fields (except for vector field in gauge theories) can be performed through similar manner.