# Euclidean QFT definition

I have a question on Euclidean field theories and their relationship with QFT defined on a Minkowski spacetime.

In order to compute the generating function $Z$, one has to compute the integral

$$Z = \int \mathcal D \phi \exp{\left(i\int \mathrm {d}^4x \mathcal L\left[\phi \right]\right)}$$

where for the sake of simplicity I assumed only a scalar field.

Now, quoting my teachers, the convergence of this integral is bad because we have an imaginary exponent, thus an oscillating integral (BAD!).

A possible solution to this is to analytically continue the lagrangian and then rotate the axis of the $\mathrm{dx^0}$ integration in the exponent.

$$\left( x^0, \vec{x}\right) \rightarrow \left( -i\tilde{x}, \vec{\tilde{x}}\right)$$

where this means that i just changed the axis of integration by a $\pi/2$ clockwise rotation.

(This works because we assume that when we rotate the axis of integration, we do not cross any pole ; otherwise the integral, computed via residue theorem, would give us a different result I suppose)

So, after this step, we have to redefine the integral variable to the euclidean variable.

$$\left(\tilde{x}^0, \vec{\tilde{x}}\right) = \left(-ix^4_E, \vec{x}_E\right)$$

Next step is to introduce euclidean fields, defined such that

$$iS\left[ \phi \right] = -S_E\left[ \phi_E \right]$$

After this, one should be ready to compute the integral without problems, but something doesn't fully convince me about this procedure.

1) How can I be sure about the fact that the euclidean action (evaluated on euclidean fields) is definite positive? If this is not the case, the whole "pass in Euclidean because of convergence reason" is useless.

2)Is there any convincing result about the absence of poles that you cross whilde doing the Wick rotation?

• Ok, I curbed out some of the subquestions. I will be more careful next time – otillaf May 23 '18 at 20:26
• – AccidentalFourierTransform May 23 '18 at 20:39