I am studying interacting scalar fields (from Osborn) using the path integral approach.

We define the functional integral

\begin{equation*} Z[J] := \int d[\phi] e^{iS[\phi] + i\int d^d x J(x) \phi(x)} \tag{1} \end{equation*}

The idea is to differentiate $Z[J]$ with respect to $J$ and end up defining correlation functions. We can define this integral by a perturbation expansion. This can be expressed in terms of Feynman diagrams, and for each diagram there is an amplitude given by the Feynman rules.

I see that, formally, $(1)$ takes the form

\begin{align*} Z[J] &= \exp\left(\frac{i}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \exp\left( i \int d^d x (-V(\phi(x))+J(x)\phi(x))\right) \Big|_{\phi=0} \end{align*}

Then Osborn states to "expand this integral to get the perturbation expansion" and then he goes straight to explain the Feynman rules.

My issue is that I do not see how this perturbative expansion leads to write down the Feynman rules.

Could you please explain me the easiest case I could find, $V(\phi(x))=\lambda \frac{\phi^3}{3!}$? In other words, how to expand the following integral perturbatively

\begin{align*} Z[J] &= \exp\left(\frac{i}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \exp\left( i \int d^d x (-\lambda \frac{\phi^3}{3!}+J(x)\phi(x))\right) \Big|_{\phi=0} \tag{2} \end{align*}

So that I see how to establish the Feynman rules (i.e. to find out what the factor of (…) for a $\phi^3$ is etc), what kind of integrals (at least 1-loop) are associated to which diagrams and find all connected one- and two-loop graphs which contribute to $\langle \phi(x_1) \phi(x_2) \rangle$ and $\langle \phi(x_1) \phi(x_2) \phi(x_3)\rangle$

Please note that , to do so, we should only do an inspection of $(2)$ and not the explicit full expansion.

Please note that this is not a homework exercise: I am just looking for a particular solved example so that I can understand how the whole machinery works. You could explain it based on another potential if you wish.

Source: Osborn notes, section 2.2. Interacting Scalar Field Theories


By expanding $(2)$ as $e^x e^y = 1 + xy + (xy)^2/4 + ...$ i.e.

\begin{equation*} Z[J] = 1+ \left(-\frac{1}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \left( \int d^d x (-\lambda \frac{\phi^3}{3!}+J(x)\phi(x))\right) + ...\end{equation*}

I get what to me looks like a messy expression. What am I missing?


Alright, so based on your comment I would say we get

\begin{align*} Z[J] &= 1+ \left(-\frac{1}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \left( \int d^d x (-\lambda \frac{\phi^3}{3!}+J(x)\phi(x))\right) + ... \\ &= 1+ \left(- \int d^d y \frac{\delta}{\delta \phi(y)} \Delta_F (x-y) \right) \times \\ &\times \left( (-\lambda \phi^2/2+J(x)\right) + ... &= \end{align*}

  • 2
    $\begingroup$ What have you tried so far? Have you expanded the first exponential in powers of the double integral? $\endgroup$ Commented Mar 12, 2021 at 9:52
  • $\begingroup$ @JeanbaptisteRoux I naively tried to expand the exponentials but I get a messy expresion... Is this what you meant? $\endgroup$
    – JD_PM
    Commented Mar 12, 2021 at 10:03
  • $\begingroup$ Yes, normally you just have to see that $\frac{\delta}{\delta \phi(x)}\int d^d z\left( -\frac{\lambda}{3!}\phi^3+J\phi \right)=-\frac{\lambda}{2}\phi^2(x)+J(x)$. $\endgroup$ Commented Mar 12, 2021 at 10:14
  • $\begingroup$ Oh, I didn't notice your edit, you only have to expand the first exponential. $\endgroup$ Commented Mar 12, 2021 at 10:19
  • 2
    $\begingroup$ For me, it is OK following your method (I think you forgot a 1/2 somewhere), but I insist, it should be more doable expanding only the first exponential. You have to do all the functional derivatives, then all the integrals, and at the end only take $\phi=0$, I will post an "answer" showing how to proceed for the propagator only (so 0 loops). $\endgroup$ Commented Mar 12, 2021 at 10:46

1 Answer 1


So, this is not a full answer but I'll do the calculations up to the first-order in the perturbative expansion (so no loops, see the comments) : \begin{align*} &\left.e^{\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}}e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \\ &=\left[1+\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}+\cdots \right]\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \\ &=\left[1+\frac{i}{2}\int d^d x\int d^d y \frac{\delta}{\delta\phi(x)}\Delta_F(x-y)\left( -\frac{\lambda}{2}i\phi^2(y)+iJ(y)\right) +\cdots\right]\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \\ &=\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}+\frac{i}{2}\int d^d x\int d^d y\,\Delta_F(x-y)\left( -\lambda i\phi(y)\delta(x-y)\right)\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \\ &\hphantom{=}+\frac{i}{2}\int d^d x \int d^d y\,\Delta_F (x-y)\left( -\frac{\lambda}{2}i\phi^2(y)+iJ(y)\right)\left( -\frac{\lambda}{2}i\phi^2(x)+iJ(x)\right)\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} \\ &\hphantom{=}+\cdots \\ &=1+\frac{i}{2}\times 0+\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y))+\cdots \end{align*} So at the first order the propagator term is given by $\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y))$, more precisely the propagator is $\Delta(x-y)$ and the external lignes are $iJ$.

Hope this helps a bit

  • $\begingroup$ Thank you very much, your answer helps a lot! Let me ask you some naïve questions: why does the delta function appear in the third equality? why $i/2 \int d^d x \int d^d y \Delta_F (x-y)\left( -\frac{\lambda}{2}i\phi^2(y)+iJ(y)\right)\left( -\frac{\lambda}{2}i\phi^2(x)+iJ(x)\right)\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0} = \frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y))$? i.e. what happened to the contributions coming from the $-\frac{\lambda}{2}i \phi^2$ terms? $\endgroup$
    – JD_PM
    Commented Mar 12, 2021 at 11:58
  • 2
    $\begingroup$ A Dirac delta appear because you make a derivative with respect to $\phi(x)$, but on a term that depends on $\phi(y)$, remember that $\frac{\delta \phi(y)}{\delta \phi(x)}=\delta(x-y)$. The contributions coming from the $-\frac{\lambda}{2}i\phi^2$ terms are erazed since you take the value for $\phi=0$. $\endgroup$ Commented Mar 12, 2021 at 12:08
  • $\begingroup$ Ahhh I see it now, thanks for you patience! I will perform the analogous computation for second order. If I get stuck, I will post another question. Hope to see you around :) $\endgroup$
    – JD_PM
    Commented Mar 12, 2021 at 12:14

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