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.

  • $\begingroup$ The vacuum bubbles you drew there contribute, if you like, to the zero-point function $\langle 0 | 0 \rangle$ at second order. The normalisation of any n-point correlator is such that these are removed order by order in perturbation theory. $\endgroup$
    – CAF
    Jun 11, 2018 at 16:18

2 Answers 2


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.

  • $\begingroup$ And how does the $\frac{\delta}{\delta J(x_2)}$ act on $\exp\left[-\frac{i}{2}\int d^4 z_1 d^4 z_2 J(z_1)G(z_1-z_2)J(z_2) \right]$ ? I'm not used to functional derivatives. I know the relation $\frac{\delta}{\delta J(x_1)} \int d^4 y J(y)\phi(y)=\phi(x_1)$ does this help for my evaluation? $\endgroup$ Jun 12, 2018 at 8:54
  • 1
    $\begingroup$ @Kennethv.B. You use the product rule, together with e.g $\delta J(z_1)/\delta J(x_2) = \delta(z_1-x_2)$ and $\int d^4 z_1 \delta(z_1-x_2) \phi(z_1) = \phi(x_2)$. Is it clear? Yes that relation helps: it is $$\int d^4 y \frac{\delta J(y)}{ \delta J (x_1)} \phi(y) = \int d^4 y \delta(y-x_1)\phi(y) = \phi(x_1)$$ $\endgroup$
    – CAF
    Jun 12, 2018 at 10:08

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 $J=0$).

  • 1
    $\begingroup$ So for order $g^0$ i have to calculate $$-\left[\frac{1}{Z(0)}\frac{\delta}{\delta J(x_1)}\frac{\delta}{\delta J(x_2)} \exp\left[-\frac{i}{2} \int d^4 z_1 d^4z_2 \; J(z_1) G(z_1-z_2)J(z_2)\right]\right]_{J=0} $$ How do i apply the derivative to the J's? Im not very familiar with functional derivatives and how they act on the G's $\endgroup$ Jun 11, 2018 at 17:56
  • $\begingroup$ @Kennethv.B. indeed ! $\endgroup$
    – picop
    Jun 11, 2018 at 17:57
  • $\begingroup$ I edited my first comment (no notification i guess). How do i apply the derivative to the J's? Im not very familiar with functional derivatives and how they act on the J's in the Integral $\endgroup$ Jun 11, 2018 at 18:17

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