1
$\begingroup$

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

EDIT 0

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?

EDIT 1

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*}

$\endgroup$
6
  • 2
    $\begingroup$ What have you tried so far? Have you expanded the first exponential in powers of the double integral? $\endgroup$ Mar 12 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
    Mar 12 at 10:03
  • $\begingroup$ Yes, normally you just have to see that $\frac{\delta}{\delta \phi(x)}\int d^d z\left( -\frac{\lambda}{3!}\phi^3+J\phi \right)=-\frac{\lambda}{2}\phi^2(x)+J(x)$. $\endgroup$ Mar 12 at 10:14
  • $\begingroup$ Oh, I didn't notice your edit, you only have to expand the first exponential. $\endgroup$ Mar 12 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). $\endgroup$ Mar 12 at 10:46
2
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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
    Mar 12 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 $\phi=0$. $\endgroup$ Mar 12 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
    Mar 12 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.