In Quantum Field Theory in a Nutshell by A. Zee, the following integral

$$Z(J)=\int_{-\infty}^{+\infty} d q e^{-\frac{1}{2} m^{2} q^{2}-\frac{\lambda}{4!} q^{4}+J q}$$

is solved perturbatively by expansion of the $\lambda$ and the $J$ term.

For example expanding the $J$ term we obtain:

$$Z(J)=\displaystyle\sum_{s=0}^{\infty} \frac{1}{s !} J^{s} \int_{-\infty}^{+\infty} d q e^{-\frac{1}{2} m^{2} q^{2}-\left(\lambda / 4!) q^{4}\right.} q^{s} .\tag{1}$$

This method is extended to a multidimensional integral

$$Z(J)=\displaystyle\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \cdots \int_{-\infty}^{+\infty} d q_{1} d q_{2} \cdots d q_{N} e^{-\frac{1}{2} q \cdot A \cdot q-(\lambda / 4!) q^{4}+J \cdot q}$$

where $q^4\equiv\sum_i q_i^4$ and $A$ is an $N\times N$ matrix.

We can expand the $J$ term, obtaining the following according to A. Zee :

$$Z(J)=\displaystyle\sum_{s=0}^{\infty} \sum_{i_{1}=1}^{N} \cdots \sum_{i_{s}=1}^{N} \frac{1}{s !} J_{i_{1}} \cdots J_{i_{s}} \int_{-\infty}^{+\infty}\left(\prod_{l} d q_{l}\right) e^{-\frac{1}{2} q \cdot A \cdot q-(\lambda / 4 !) q^{4}} q_{i_{1}} \cdots q_{i_{s}}.\tag{2}$$

How is this a correct expansion of the $J$ term?

My guess would be straightforward using the exponential expansion, just like in equation $1$:

$$e^{J\cdot q}=e^{\displaystyle\sum_iJ_iq_i}=1+\sum_i J_iq_i+\frac{1}{2!}(\sum_i J_iq_i)^2\dots=\sum_{s=0}^{\infty}\dfrac{1}{s!}\Bigg(\sum_iJ_iq_i\Bigg)^s$$ Which results in an infinite power series in $J$ and $q$ while A. Zee's equation $2$, the powers of $J$ and $q$ are finite.

  • 1
    $\begingroup$ The powers are not finite, because $s$ runs all the way to $\infty$. Zee's expression can be obtained by plugging your expression in the definition of $Z(J)$. $\endgroup$ Sep 28, 2019 at 16:31

1 Answer 1


Yours and his expressions are equal. Remember that the sum over $i$ in your series run from $i=1$ to $i=N$, only. The other part of the combinatorics is just $$ e^{x+y}= e^xe^y $$ or $$\sum_s \frac 1{s!} (x+y)^s= \sum_{n=1}^\infty\sum_{m=1}^\infty \frac 1{n!} \frac 1{m!}x^n y^m $$


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.