Semiclassic limit of a QFT in Zinn-Justin

Question

I am reading the Zinn-Justin book "Quantum Field Theory and Critical Phenomena" and i have come across a perplexing point. Given the partition functional, in Euclidean QFT:

$$Z[J, \hbar] = \int d^N \phi e^{-\frac{1}{\hbar}(S[\phi] - J\phi)}\tag{7.86}$$

Where integration over $x$ is intended in the exponent, we find the saddle point by means of:

$$\frac{\delta}{\delta \phi}(S[\phi]-J\phi)|_{\phi_c} = 0.\tag{7.112}$$

We then expand around the classical saddle point as:

$$\phi = \phi_c + \chi\sqrt{\hbar}.\tag{7.113}$$

The action becomes

$$S= \frac{S[\phi_c]}{\hbar} + \frac{1}{2}S^{(2)}_{ij}(\phi_c)\chi_i \chi_j + O(\hbar^{1/2})\tag{7.114}$$

With $$\frac{\delta^2S[\phi]}{\delta \phi_i \delta \phi_j}|_{\phi_c}=S^{(2)}_{ij}(\phi_c).\tag{7.115}$$

Also we should get: $$J\phi = J\phi_c + J\chi\sqrt{\hbar}$$

At first order we have then:

$$Z[J, \hbar]_0 = \mathcal{N} e^{-\frac{1}{\hbar}(S[\phi_c] - J\phi_c)}$$

The next order, according to Zinn-Justin, is given by:

$$Z[J, \hbar]_1 = Z[J, \hbar]_0\int[d\chi]e^{-\frac{1}{2}\int dx_1dx_2S^{(2)}_{12}(\phi_c)\chi_1 \chi_2}.$$

My question is: why is he omitting the term $e^{ \frac{ \sqrt{\hbar} }{\hbar} J\chi}$ from the integral?

Answer 1

Perhaps it should be stressed that a stationary field configuration $\phi_c[J]$ depends on the external source $J$.

In particular the total action is $$S_J[\phi]~=~S[\phi]- J \phi,$$ and its expansion around a stationary solution $\phi_c[J]$ does not contain any linear terms in the fluctuation $\chi$, because by definition $$\left.\frac{\delta S_J[\phi]}{\delta\phi}\right|_{\phi=\phi_c[J]}~=~0,$$ cf. above comment by naturallyInconsistent.

For the proof of the semiclassical expansion, see e.g. this related Phys.SE post, where the external source $J$ is implicit in the notation.