I am reading Quantum Field Theory in a Nutshell by A. Zee.

Zee introduces the rationale/machinery behind Feynman diagrams in three steps: Baby -> Child -> "Real".

The baby problem generates diagrams by expanding the following one-dimensional integral to a double series with respect to $e^{- \frac{\lambda}{4!}q^4}$ and $e^{Jq}$: $$Z(J) = \int_{- \infty}^{\infty} dq e^{-\frac{1}{2}m^2q^2 - \frac{\lambda}{4!}q^4+Jq}$$
So far so good.

The child problem promotes the above integral into a multiple integral with the "substitutions" $m^2 \to A$ ($n \times n$ symmetric matrix) and $J,q$ to $n$ vectors: $$Z(J) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \ldots \int_{- \infty}^{\infty} dq_1 dq_1 \ldots dq_n e^{-\frac{1}{2} q \cdot A \cdot q - \frac{\lambda}{4!} \sum_{i=1}^{n}q_i^4 +J \cdot q}=$$ $$= Z(0,0) \sum_{s=0}^{\infty} \sum_{i_1=1}^{n} \ldots \sum_{i_s=1}^{n} \frac{1}{s!} J_{i_1} \ldots J_{i_s} G^{(s)}_{i_1 \ldots i_s}$$

For $s=2$ I manage to get the right answer (Because there's no possibility of permutations..).
for $s=4$ I fail to understand where the sum on permutations is coming from, so, I reach back to the appendix.
To make things as simple as possible I choose $n=4$ and obtain: $$\langle x_1 x_2 \rangle = \frac{\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} dx_1 dx_2 dx_3 dx_4 e ^{-\frac{1}{2} x \cdot A \cdot x} x_1 x_2}{\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} dx_1 dx_2 dx_3 dx_4 e ^{-\frac{1}{2} x \cdot A \cdot x} } = A_{12}^{-1}$$

But when I try to do $\langle x_1 x_2 x_3 x_4\rangle$, I get only $A_{12}^{-1} A_{34}^{-1}$, and not a sum on permutations. I get this by applying $\frac{d^4}{dJ_4 dJ_3 dJ_2 dJ_1}$ on $e^{\frac{1}{2} J \cdot A^{-1} \cdot J}$ (This is the solution of the 4-d gaussian with the extra $J \cdot x$ term in the exponential) and then setting $J=0$.

I must have misunderstood the instructions: "Differentiate $p$ times with respect to $J_i, J_j, \ldots , J_k,$and $J_l$, and then set $J=0$."

Please tell me if my question lacks some details that might shead light on what I'm doing wrong (Or what I don't understand)...


1 Answer 1


Right, one is only supposed to put the sources $J=0$ to zero after the very last $J$-differentiation has been performed. Figuratively speaking, short of writing out the calculation in full detail: Some of the $J$s downstairs can "couple" to the $J$s upstairs in the exponential.

  • $\begingroup$ I (think I) know. And I took that into account. But I do now notice that changing the order of differentiation on the integral supposedly changes nothing. But, changing the order on the "result" does change it, in the 3 possible ways. So, is this the reason for summing the permutations? $\endgroup$
    – user76568
    Commented Oct 5, 2014 at 12:25
  • $\begingroup$ Hint: Keeping the differentiations in the standard order, try to trace how a contribution $A^{-1}_{13}A^{-1}_{24}$ could possibly arise. $\endgroup$
    – Qmechanic
    Commented Oct 5, 2014 at 12:28
  • $\begingroup$ Got it. Guess I didn't take it into account like I thought I did. Thank you! $\endgroup$
    – user76568
    Commented Oct 5, 2014 at 12:41

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