Zee says in Section I.3 of QFT in a nutshell:

The functional integral $$Z = \int D \varphi e^{i \int d^4 x [\frac{1}{2} (\partial \varphi)^2 - V(\varphi) + J(x) \varphi (x)]} \tag{11} $$ is impossible to do except when $$\mathcal L (\varphi) = \frac{1}{2} [(\partial \varphi)^2 - m^2 \varphi ^ 2].\tag{12}$$
The corresponding theory is called the free or Gaussian theory.

This restriction gives sudden birth to the Klein-Gordon equation and also practically allows the entire part I of the book to proceed as it is.

So, two and a half questions:

  1. How can I understand the necessity of this restriction? And, what does "impossible to do" mean here?
  2. As part I unfolds, Zee explains that the above is not so much "meant" to be solved, but rather, it is a generating functional. So, why is the restriction necessary in the first place?
  • 1
    $\begingroup$ I was about asking you what "impossible" should mean here! I do not think it means anything with any rigorous meaning. The general notion of Lorentzian path integral is not mathematically well defined as an integral since it is not based upon a true measure. It can be defined as a "limit" notion, for instance, relying upon well defined mathematical tools (Euclidean path integral for example). Maybe the author would mean that these procedures do not work if the action is not the standard KG one. But I do not think it is generally true. $\endgroup$ – Valter Moretti Oct 5 '14 at 16:39
  • $\begingroup$ I see your correction, impossibly "to do", i.e. to explicitly compute? Indeed the other cases are usually handled just pertubatively. $\endgroup$ – Valter Moretti Oct 5 '14 at 16:42
  • $\begingroup$ @Qmechanic Thanks for the edits. I'll raise my standard for my future questions :) $\endgroup$ – user76568 Oct 5 '14 at 16:48

First of all, let me say the following: If anyone (perhaps you, @V.Moretti?) would be able to provide a more mathematically oriented perspective on this question, I think that would make a valuable complement to this answer, which can be characterized as pragmatic (or handwavey, depending who you'd ask!), rather than deep.

That being said, I will now answer both subquestions:

  1. I think there is nothing all too deep behind this statement by Zee. In particular, I don't think he means to make a rigorous statement on the well-definedness of the path integral, and the way this may (or may not) depend on the particular form of $\mathcal L$. He simply means that the free field theory allows us to - sweeping all serious mathematical issues involved in defining the path integral measure etc. etc. under the rug - to explicitly perform this integration, as it all boils down to nothing more than (a generalized version of) the standard Gaussian integral $$ \int_\mathbb R e^{-x^2}\mathrm{d} x$$ In any serious model of the interactions between fundamental particles we have to (rather unsurprisingly) consider interaction terms. These spoil the - once again, ignoring many mathematical issues - simplicity of the integral, and are usually attacked by means of a perturbative expansion, as opposed to solving exactly.

  2. First considering a free field theory is instructive physically, because we usually expand around the free field theory (in the sense that we consider small/weak interactions). Furthermore, the mathematical treatment of interacting theories is quite similar, so the free field Lagrangian can be considered as an excellent warm-up exercise in this aspect as well, allowing the beginning student to build up some intuition and familiarity with common techniques.

  • $\begingroup$ I'm guessing the measure issue is caused by the possible 0 and negative valued (squared) length of segments..? Where can I read about this? $\endgroup$ – user76568 Oct 6 '14 at 10:48

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