$\newcommand{\ld}{\mathcal{L}}\newcommand{\adj}[1]{#1^\dagger}\newcommand{\dc}[1]{\overline{#1}}\newcommand{\Psi}{\varPsi}\newcommand{\dd}{\mathrm{d}}$Take a typical Lagrangian density defined over Minkowski spacetime with metric $\eta=\operatorname{diag}(-1,1,1,1)$, say the scalar field one \begin{equation} \ld= -\frac12\partial^\mu\phi\partial_\mu\phi-\frac12 m^2\phi^2. \end{equation} In order to switch to an Euclidean field theory, I have to perform a Wick rotation. I imagine that a rigorous approach to the "rotation of the integration domain" goes like this: consider, firstly, the closed path $\Gamma_R$ given by the maps

  • $t\mapsto -R+2Rt$,
  • $t\mapsto Re^{i\frac{\pi}{2}t}$,
  • $t\mapsto iR-2iRt$,
  • $t\mapsto -iRe^{-i\frac{\pi}{2}t}$,

in succession, all with $t\in[0,1]$. Now I have to consider the analytic continuation of $\ld$ as a function of $x^0$ into the complex plane: assuming that the involved function are analytic where needed, by Cauchy's theorem the integral in $x^0$ along $\Gamma_R$ is zero for every $R$. Moreover, for $R\to+\infty$, assuming that the fields fall off at infinity rapidly enough for the integral over the two arcs to go to zero, I obtain that \begin{equation} \int_{-\infty}^{+\infty}\ld\,\dd x^0= -\int_{i\infty}^{-i\infty}\ld\,\dd x^0= \int_{-i\infty}^{i\infty}\ld\,\dd x^0 \end{equation} and then renaming $x^0=ix^4$ \begin{equation} \int_{-\infty}^{+\infty}\ld(x^0,\dotsc)\,\dd x^0= i\int_{-\infty}^{+\infty}\ld(ix^4,\dotsc)\,\dd x^4. \end{equation} Therefore, in the path integral weight $e^{iS}$, the $i$ factor together with the one I get from the Wick rotation gives a $-1$, so $e^{iS}$ becomes $e^{S'}$ with \begin{equation} S'= \frac12\int_{\mathbb{R}^4}(\partial^i\phi\partial_i\phi+m^2\phi^2)\,\dd x^1\,\dd x^2\,\dd x^3\,\dd x^4 \end{equation} that is the opposite of what I wanted, that is an exponentially decaying function $e^{-S}$ with $S$ a positive definite quadratic form in the fields.

If I close, instead, the integration contour in the other two quadrants of the complex plane, I obtain \begin{equation} \int_{-\infty}^{+\infty}\ld\,\dd x^0= -\int_{-i\infty}^{i\infty}\ld\,\dd x^0 \end{equation} and this minus sign gets me what I want in the end.

I feel that the first procedure should have some kind of obstruction. After all, if both are possible, that means that $\ld$ is an entire function of $x^0$, and since we assume that the fields (and their derivatives) go to zero when $\lvert x^0\rvert\to +\infty$, as required by the above calculations to work, by Liouville's theorem $\ld=0$, if I'm correct.

Now, assuming that the above procedures are correct: does $\ld$ have some singularities in the complex plane that prevent from going both ways? If so, can I determine them? Or is there some error in my "proof" of the Wick rotation?

  • $\begingroup$ Hint: For the Euclidean path integral to be convergent, the Boltzmann factor should be an exponentially decaying function of the field $\phi$. This in turn dictates that the direction of the Wick rotation is $t_E=it_M$, and not the opposite. Related: physics.stackexchange.com/q/231328/2451 and links therein. $\endgroup$ – Qmechanic Jul 3 '18 at 17:47
  • $\begingroup$ Well, ok, I have to discard the wrong one. But it seems that it's still possible, even if it's useless, and there remain the doubts about the analiticity properties of the functions, at least in the "famous simple cases" such as the scalar, Weyl, Dirac Lagrangian... $\endgroup$ – yellowquark Jul 3 '18 at 19:49

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