# Divergent path integral

What does it mean to have a divergent path integral in a QFT?

More specifically, if $$\int e^{i S[\phi]/\hbar} D\phi (t)=\infty$$ What does this mean for the QFT of the field $\phi$?

The field $\phi$ has action $$S[\phi]=\int\left(\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-V(\phi)\right)\mbox{d}vol$$ where we use Minkowski signature $(+,-,-,-)$.

• What's the action? – Alex Nelson Mar 2 '14 at 0:16
• The reason I ask about the action is: (a) you could be dealing with a "trivial infinity" from the volume of a gauge orbit; (b) it could be a legitimate infinity [e.g., the action could be unbounded from below after Wick rotating], where you would need to use Lattice approximations or effective field theory; (c) sometimes it could be "ghosts of departed quantities" when taking the continuum limit of finite differences. Can't tell which unless we see the action... – Alex Nelson Mar 3 '14 at 1:50
• Comment to the question (v3): What is the potential $V$? – Qmechanic Mar 3 '14 at 12:42
• Presumably, it's any interaction $V(\phi)~\mathcal{O}(\phi^3)$. I'm at work, and cannot answer (perhaps @Qmechanic or someone else will give a coherent/non-tweet-lengthed response). Basically, you consider $\int\exp(S_{0}[\phi]+\int V(\phi)\,\mathrm{d}^{n}x)\, D\phi = \int\sum_{k=0}(\int V(\phi)\,\mathrm{d}^{n}x)^{k}\exp(S_{0}[\phi])\,D\phi$ then truncate the series after a couple terms, otherwise it diverges and you get infinities. – Alex Nelson Mar 3 '14 at 15:35
• @AlexNelson I understand that. However, what does it mean physically for a divergent path integral? – user28355 Mar 3 '14 at 23:40

If the path integral itself diverges, it means that the v.e.v. diverges. That by itself is bad, because then any arbitrary $n$-point function vanishes. Recall that to compute correlation functions, we append a $J(x)\phi(x)$ to the action and calculate $$\frac{\delta^n}{\delta J(x_1)\ldots \delta J(x_n)} \int e^{i S[\phi]/\hbar + J(x)\phi(x)}\mathcal{D}\phi = \langle\phi(x_1)\ldots \phi(x_n)\rangle$$ which is normalized by the v.e.v.. Thus, you wouldn't be able to calculate anything sensible.