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Consider the path integral expression for the (unrenormalised) propagator in $\phi^4$ theory with a real scalar field $\phi$ on 4D Minkowski space and with a complex mass term:

$$ G[f,g] = \int\mathcal{D}\phi\,{\textstyle(\int f\phi)(\int g\phi)\exp(i\int[\frac12(\partial_\mu\phi)(\partial^\mu\phi) - \frac12(m^2 - i\epsilon)\phi^2 - \frac1{4!}\lambda\phi^4]}) $$

where $f$ and $g$ are smooth, square-integrable functions on Minkowski space. In the limit $\epsilon\to0$ this is just the propagator of $\phi^4$ theory which I know is UV divergent. My question is whether this expression is also divergent for finite $\epsilon>0$.

I have one argument which shows that $G[f,g]$ is finite and one which shows that it is infinite and I can't figure out which one is wrong:

Argument 1: I know that $\mathcal{D}\phi$ is just a sloppy physicist notation for an ill-defined functional integration measure, but if my understanding of Gaussian functional integration measures (which is somewhat limited and comes mostly from math-ph/0510087) is correct then

$$ d\mu_\epsilon(\phi)\equiv\mathcal{D}\phi{\textstyle\exp(-\frac12\epsilon\int\phi^2)} $$

can be regarded as a well-defined functional integration measure which satisfies

$$ \int d\mu_\epsilon(\phi) = 1 \quad,\quad \int d\mu_\epsilon(\phi){\textstyle(\int f\phi)(\int g\phi)} = \frac1\epsilon{\textstyle\int fg} $$

If this is true we may write

$$ G[f, g] = \int d\mu_\epsilon(\phi){\textstyle(\int f\phi)(\int g\phi)\exp(i\int[\frac12(\partial_\mu\phi)(\partial^\mu\phi) - \frac12m^2\phi^2 - \frac1{4!}\lambda\phi^4]}) $$

and since the expression in the exponent is now purely imaginary the exponential has magnitude 1 and we have

$$ |G[f,g]| \leq \left|\int d\mu_\epsilon(\phi){\textstyle(\int f\phi)(\int g\phi)}\right| = \frac1\epsilon|{\textstyle\int fg}| < \infty $$

Argument 2: The integrals we encounter when we calculate $G[f, g]$ in perturbation theory have the same UV divergences irrespective of whether $\epsilon$ is zero or not. For example, the one-loop 'tadpole' integral is

$$ \int d^4l \frac1{l^2 - m^2 + i\epsilon} = -i\int_0^\infty dr\frac{2\pi^2r^3}{r^2 + m^2 - i\epsilon} $$

where we obtained the right-hand side by Wick rotating and integrating out the angular variables. The UV divergence is associated with the behaviour of the integrand for $r\to\infty$ and for this the value of $\epsilon$ is irrelevant. Thus, if $G[f, g]$ is UV divergent for $\epsilon\to0$ it should be UV divergent for all $\epsilon>0$.

My instinct is to trust the non-perturbative argument 1 over the perturbative argument 2, but this would mean that a complex mass term can serve as a UV regulator and I've never heard of anyone use or even discuss such a regularisation.

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  • $\begingroup$ Ah, nevermind. I found the answer with the help of these lecture notes. (Example 2.6, to be specific.) It's Argument 1 which is flawed. The expression is UV divergent. Since this got upvoted I'll post a proper answer later. $\endgroup$ – Martin Wiebusch Jun 4 '18 at 11:19
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Argument 1 is incorrect because (without additional regularisations) the functional integral is still ill-defined. The reason is that the the derivatives in the integrand are only defined on a dense subspace of $L^2(\mathbb{R}^4)$ (the space of square fields on Minkowski space) but we are trying to integrate over the whole space $L^2(\mathbb{R}^4)$. The issue is discussed in detail in Example 2.6 of these lecture notes.

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