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I'm working in the $\varphi^4$ QFT where

$$S(\varphi)=\frac{1}{2} \mu \varphi^{2}+\frac{1}{4 !} \lambda_{4} \varphi^{4}$$

and the text says that we can expand (assuming small $\lambda_4$) as

$$\exp (-S(\varphi))=\exp \left(-\frac{1}{2} \mu \varphi^{2}\right) \sum_{k \geq 0} \frac{1}{k !}\left(-\frac{\lambda_{4}}{24}\right)^{k} \varphi^{4 k}$$

which makes sense, but then it says that we interchange the series expansion in $\lambda_4$ with an integration over $\varphi$ and arrive at the following Green's functions

$$G_{2 n}=H_{2 n} / H_{0}$$

where

$$H_{2 n}=\frac{1}{\mu^{n}} \sum_{k \geq 0} \frac{(4 k+2 n) !}{2^{2 k+n}(2 k+n) ! k !}\left(-\frac{\lambda_{4}}{24 \mu^{2}}\right)^{k}$$

and for the life of me I can't figure out how they are getting to this.

I understand how it worked for the free theory, where

$$\begin{aligned} Z(J) &=N \int \exp \left(-\frac{1}{2} \mu \varphi^{2}+J \varphi\right) d \varphi \\ &=N \int \exp \left(-\frac{1}{2} \mu\left(\varphi-\frac{J}{\mu}\right)^{2}+\frac{J^{2}}{2 \mu}\right) d \varphi=\exp \left(\frac{J^{2}}{2 \mu}\right) \end{aligned}$$ and we can use the taylor expansion of $\exp \left(\frac{J^{2}}{2 \mu}\right)$ to get $$G_{2 n}=\frac{(2 n) !}{2^{n} n !} \frac{1}{\mu^{n}}$$

But I'm miffed how I can't get to the result stated for the $\varphi^4$ theory. If I were to follow the logic directly from the free theory case, then I wouldn't even taylor expand to get $$\exp (-S(\varphi))=\exp \left(-\frac{1}{2} \mu \varphi^{2}\right) \sum_{k \geq 0} \frac{1}{k !}\left(-\frac{\lambda_{4}}{24}\right)^{k} \varphi^{4 k}$$ but instead complete the square on this expression.

Any insights would be greatly appreciated!

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  • $\begingroup$ $G_{2n}$ is the result but it has $H_{2n}$ in it. So definition $\endgroup$
    – Некто
    Commented Aug 13, 2021 at 23:40
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    $\begingroup$ Maybe it's worth clarifying in the post that this is a 0-dimensional field theory, ie an integral over just a single real number $\varphi$. I was a bit confused trying to read this initially. Also it would be helpful to state which text you're following. $\endgroup$
    – Zack
    Commented Aug 14, 2021 at 1:27
  • $\begingroup$ @Zack yes you're right... that's a good clarification to make, this is 0-dimensional $\endgroup$
    – Некто
    Commented Aug 16, 2021 at 15:51

1 Answer 1

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Here are some tips (to do in order):

  1. Swap the order of integration and summation (truncate the series first since we cannot do this with an infinite series and we are just forming an asymptotic expansion anyway).
  2. Change the integration variable to $x=\frac{\mu^2 \phi^2}{2}$.
  3. Use the definition of the gamma function.
  4. Use Stirling's approximation.

(just google the definitions of 3 and 4 if you need to).

Let me know if you need any more guidance!

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