Scattering in $g\phi^3$-Theory

Question

I have to consider the QFT with the Lagrangian

$$\mathscr{L}=\underbrace{\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{m^2}{2}\phi^2}_{=\mathscr{L}_0}\underbrace{-\frac{g}{6}\phi^3}_{\mathscr{L}_{\text{int}}}.$$

The question is to find the connected and time ordered two point function up to order $g^2$.

In general $n$-point functions are given by $$ G^{(n)}=(-i)^n\left[\frac{1}{Z(0)} \frac{\delta}{\delta J(x_1)\cdot}\cdot ...\cdot\frac{\delta}{\delta J(x_n)\cdot} Z(J)\right]_{J=0} $$

Since im only interested in the two point function i only have

$$ G^{(2)}=-\left[\frac{1}{Z(0)} \frac{\delta}{\delta J(x_1)\cdot}\frac{\delta}{\delta J(x_2)\cdot} Z(J)\right]_{J=0} .$$

This means i need the partition function Z(J): $$ Z(J)=\int \mathscr{D} \phi \exp\left[i\int d^4x \left(\mathscr{L}_0+\mathscr{L}_{\text{int}}-J(x)\phi(x)\right)\right]=Z_0(J)\cdot \exp\left[i\int d^4x\;\mathscr{L}_{\text{int}} \right] $$ $$ = Z_0(J)\cdot \exp\left[-i\frac{g}{3!}\int d^4x\;\phi^3(x) \right] $$

Now i can expand this up to the second order of the exponential funcion $$ \left(1-i\frac{g}{3!}\int d^4 x_1 \;\left(-i\frac{\delta}{\delta J(x_1)}\right)^3+\frac{(-i)^2}{2!}\left(\frac{g}{3!}\right)^2 \int d^4x_1 d^4 x_2 \;\left(-i\frac{\delta}{\delta J(x_1)}\right)^3\left(-i\frac{\delta}{\delta J(x_2)}\right)^3 \right) $$ $$ \cdot \exp\left[-\frac{i}{2} \int d^4 z_1 d^4z_2 \; J(z_1) G(z_1-z_2)J(z_2)\right] $$

This means at order $g^0$ we should have the free propagator $G^{(2)}_0(x_1,x_2)=G(x_1,x_2)$. The dots $x_1$and $x_2$connected by a line.

Now comes the part im stuck with:

1. The $\int d^4 x_1 \;\left(-i\frac{\delta}{\delta J(x_1)}\right)^3$ part should give us a vertex with 3 legs. If i connect $x_1$ to the vertex i have 3 options and then 2 with $x_2$ to the vertex. But then one connection is left. I don't know what to to with it, probably because im not sure what one of these diagrams means in terms of real particle interactions. Im not sure if this results in one of these tadpole diagrams i read about (not sure what is meant by "removing the source").

2. And another confusion are these propagators i found when i did some resarch on this theory

   

   Are they connected to the vacuum bubble or are they related to my question?

I appreciate any help to solve my confusion.

Answer 1

The three-leg vertex you mention arises in the three point function at $\mathcal O(g)$. All contributions to the two-point function up to $\mathcal O(g^2)$ will have, in a Feynman diagram fashion, two source points labelled $x_1$, $x_2$ with either zero, one or two interaction vertices, with coupling $g$.

In particular, by evaluating $$Z[J] \sim \exp \left(-i\frac{g}{3!} \int d^4 x\, \left(\frac{\delta}{\delta J(x)}\right)^3 \right) \exp \left(-\frac{i}{2} \int d^4 z_1 d^4 z_2 J(z_1) G(z_1-z_2) J(z_2)\right)$$ systematically through a perturbative expansion in $g$ you can obtain all the diagrams. At order $g^0$ indeed all you find is the propagator $G(x_1-x_2)$ between two source points. At order $g^0$, $$\langle 0 | T(\phi(x_1)\phi(x_2))|0\rangle = \frac{1}{Z[0]} \frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_2)} \exp \left(-\frac{i}{2} \int d^4 z_1 d^4 z_2 J(z_1) G(z_1-z_2) J(z_2)\right)|_{J=0}$$

At order $g$, you have three functional derivatives in $\mathcal L_{\text{int}}$ which when acting on $Z_0[J]$ vanish because of an excess of source terms after the differentiation which is just nullified due to the $J=0$ delimiter.

At $\mathcal O(g^2)$, you have six functional derivatives and there exists a non vanishing contribution when these act on the fourth order term of $Z_0[J]$ only. The interpretation is that three derivatives act at one vertex giving the three-prong $\phi^3$ interaction, another three act on another vertex and the two sources left over are identified with $x_1$ and $x_2$ after acting with $\delta/\delta J(x_1)$ and $\delta/\delta J(x_2)$.

As an exercise, convince yourself that the lower/higher order terms in $Z_0[J]$ may not contribute at $\mathcal O(g^2)$ to the two point function. They will of course in general contribute to other $n$-point functions.

Answer 2

You're maybe getting confused by the variables here. What you compute in the last part of your text are some vacuum graphs, because you have not included the derivatives that you write in $G^{(2)}$. With these, you'd have to hit some propagators to connect to the external points $x_1$ and $x_2$. You will find a one-loop correction, with two internal vertices but it will come from the term $\frac1{2!} L_{int}^2$, and indeed the term $L_{int}$ will not contribute to the two point function (all the terms it creates have an extra $J$ that vanishes after you set $J=0$).