I have a question about an equation in Condensed Matter Field Theory (2nd edition) by Altland/Simons.

On page 201, there is an expression for a general pertubative expansion, which was introduced before (eq. 5.14). In the following the "first non trivial term" is deduced from it, given as equation 5.15.

I have no idea, how this result (5.15) was calculated. I thought, that I just have to set n=1 in eq. 5.14 and deal with this order, but then I cannot possbily get two terms, from which one has a different sign.

My second idea was, that I have to use som form of partial fraction decomposition from terms with higher n, but then I would have much more terms in 5.15.

Somehow, I think to remember that someone said, that eq. 5.14 cannot be taken at face value and is somehow more symbologically. But I simply cannot guess how.

So what do I miss?

Edit: In this example: $X[\phi]=\langle \phi(x) \phi(x') \rangle$

equation 5.14: $\langle X[\phi] \rangle =\frac{\sum_n \frac{(-g)^n}{n!}\langle X[\phi] \left(\int dx \ \phi^4 \right)^n \rangle_0 }{\sum_n \frac{(-g)^n}{n!}\langle \left(\int dx \ \phi^4 \right)^n \rangle_0} \approx \sum_{n=0} X^{(n)}$

equation 5.15: (first non-trivial term:) $G^{(1)}(x,x')=-g\left( \langle \phi(x) \int dy \ \phi(y)^4 \phi(x') \rangle_0 -\langle \phi(x) \phi(x')\rangle_0 \langle \int dy \ \phi(y)^4 \rangle_0 \right)$

Edit2: Later on in the book, it is mentioned that the mixed term stems from the denominator. But I do not understand how a contribution of the denominator appears without an $()^{-1}$

  • 1
    $\begingroup$ Please include all the relevant equations here. Thank you for your collaboration. $\endgroup$ Commented Mar 29, 2017 at 13:51

1 Answer 1


So this is a very tricky result because the heart is very simple but the notation is very complicated. Suppose we have the expression $$f(x) = \frac{a + bx + c x^2}{p + qx + rx^2},$$ where $c = c(x)$ and $r = r(x)$ represent the "rest of the sums". How do we get the expansion of this for $x \approx 0$ in normal everyday calculus?

Well, we have $f(0) = a/p$ and then $$f'(x) = \frac{(b + 2 cx + c' x^2)(p + qx + r x^2) - (a + bx + cx^2)(q + 2r x + r' x^2)}{\left(p + qx + r x^2\right)^2},$$ by the division rule. This seems more complicated than it is because evaluating at $x=0$ we have $f'(0) = (bp - aq)/p^2.$ So the denominator does indeed "come up to the top", roughly by the expansion that $1/(1 + x) \approx 1 - x + x^2 - x^3 + \dots$ if that helps to better visualize where it's coming from.

  • $\begingroup$ Thanks for your help! So the trick is basically to insert the desired terms for wanted orders and then expand once again for small g? $\endgroup$
    – Djoser42
    Commented Mar 29, 2017 at 15:32
  • $\begingroup$ Great, I was able to reproduce the result from above! Thank you very much. $\endgroup$
    – Djoser42
    Commented Mar 29, 2017 at 15:42

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