# Wick rotation: still trouble in getting how it works

I'm preparing my second exam in QFT and I still have trouble in getting the Wick rotation and its analytic continuation. I know that this topic have been discussed a lot in previous threads, but I hope that someone could help me in detail here. I'll use Greiner (Field Quantization) and Peskin-Schroeder (An Introduction to Quantum Field Theory) as references. In both cases, in order to get the transition amplitude $$\langle q',t'|T[q(t_1)q(t_2)]|q,t\rangle$$ in terms of the vacuum expectation value $$\langle 0|T[q(t_1)q(t_2)]|0\rangle$$, one analytically extend the time variable $$t$$ in the complex plane, setting $$t=\exp(i\delta)\tau$$, where $$\tau$$ and $$\delta$$ are real. Setting for example $$\delta=-\pi/2$$, one can get: $$\begin{equation} \langle q',t'|T[q(t_1)q(t_2)]|q,t\rangle= \langle q',-i\tau'|T[q(t_1)q(t_2)]|q,-i\tau\rangle \end{equation}$$ At this point one takes the limit for $$t'\rightarrow\infty$$ and $$t\rightarrow -\infty$$, or equivalently $$\begin{equation} \lim_{t'\rightarrow\infty,t\rightarrow-\infty}\langle q',t'|T[q(t_1)q(t_2)]|q,t\rangle=\lim_{\tau'\rightarrow\ i\infty,\tau\rightarrow -i\infty} \langle q',-i\tau'|T[q(t_1)q(t_2)]|q,-i\tau\rangle \end{equation}$$ Then Greiner sets: $$\begin{equation} \lim_{\tau'\rightarrow\ i\infty,\tau\rightarrow -i\infty} \langle q',-i\tau'|T[q(t_1)q(t_2)]|q,-i\tau\rangle\leftrightarrow\lim_{\tau'\rightarrow\ \infty,\tau\rightarrow -\infty} \langle q',-i\tau'|T[q(t_1)q(t_2)]|q,-i\tau\rangle \end{equation}$$ saying that "This [step] will only be admissible mathematically if the matrix element is an analytic function in the variables $$t$$ and $$t'$$". However, I still can't get what's going on here. How can the result be independent on $$\delta$$? We're doing this cheap trick of Wick rotation because, for $$t\rightarrow\infty$$ we have oscillatory terms which do not admit limit. So, here we're taking complex values of $$t$$ (and they cannot be purely real), setting their limit for $$t\rightarrow\infty$$ in order to get a finite quantity, than going back to real values in order to have a finite number, differently from before, when we hadn't? How can that be rigorous? I noticed that Ryder keeps the limit to be complex, and then reverse in the real time adding, however, a $$i\varepsilon q$$ term in the lagrangian. I do prefer this approach, but are they equivalent? Thank you very much for your answers!

• It's just basic loop integral on the complex plane, which totals to zero if the enclosed area is analytic (no residual/singular points). This fact establishes the equivalence of integrals along the real and imaginary axis. Jul 24, 2019 at 15:02
• Feynman's $i\epsilon$ prescription is to ensure the correct locals of the aforementioned non-analytic (singular) points, so that causality is preserved. Jul 24, 2019 at 15:11
• @MadMax, but here I don't have integrals, right? In physics we often have an integral on the real axis, so we estend the integral on the complex plane, and kill (or absorb) some terms evaluated in complex curves, but the real axis still stays the same. Here we're rotating the axis itself, and then we're re-rotating it back ignoring the terms that was previously not defined (the oscillating terms at infinity). Jul 25, 2019 at 14:47

A Wick rotation is a mathematical trick which requires a contour integral in a complex plane $$z$$. It exploits the residue theorem which states the integral of a function along a closed curve as $$2 \pi i$$ times the sum of the residues of the function inside the curve.

If you have an integral along the real axis from $$-\infty$$ to $$+\infty$$ you can, via a semicircle, extend the integral to a closed curve covering half of the complex plane. If the integrand vanishes fast enough as $$\vert z \vert \to \infty$$ the contour integral outcome is the same as the integral along the real axis.

You can now rotate the contour by 90° provided that it does not pass over any pole to guaranty that the outcome does not change. This is possible if the poles are situated all in the left side or all in the right side of the half complex plane. Now the contour integral outcome is given by the the integral along the imaginary axis.

The procedure is formalized as $$t$$ replaced by $$i \tau$$.

Note: You do not rotate the real axis, but rotate the contour integral.

• Not having poles shd not be enough to guarantee the same result after rotation. The functon has to be holomorphic doesnt it? So to allow for a holomorphic extension of the function on the real line, we have to know it is analytic. Big question: how do you prove in reality that your function you want to wick rotate is analytic. Aug 3, 2019 at 6:04
• The residue theorem requires the function to be analytic along the closed curve, not inside. Inside the loop you may have poles. What matters is that the rotated contour contains the same poles as the non rotated contour. Aug 3, 2019 at 6:38
• Having poles is independent of beiing analytic. Just hsving the same poles doesnt imply you can rotate. Aug 3, 2019 at 8:00
• Where is the contour integral in OP's question? The "Wick rotation" OP is asking about is on the states not the time argument in the fields. Aug 4, 2019 at 0:35
• @lalala. The probability amplitude, via the path integral, in the ground state as both the intial and final state can be simplified if you add $-i \epsilon$ to the Lagrangian, where $\epsilon$ is a positive infinitesimal. Keeping track of the $\epsilon$ implies that the poles of the propagator are not encountered along the real axis and so that the function is analytic along the integration curve. Aug 4, 2019 at 15:08

It's not really about being analytic or contour integrals (at least I don't think that's the best way to understand it). It's about the simple fact that the vacuum has the lowest energy. If you have a state, $$e^{-iHT}|q\rangle$$ as long as $$T$$ has some small negative imaginary part $$T=|T|(1-i\epsilon)$$, then because of the $$e^{-\epsilon |T| H}$$ factor, the state with the lowest energy will dominate as $$|T|\rightarrow \infty$$. It doesn't matter how big the negative imaginary part is, or in other words it doesn't depend on $$\delta$$. As long as $$\delta$$ is in the right quadrant the vacuum will dominate in the limit.

Now you might argue that it seems unjustified to give $$T$$ a small imaginary part at all, but this is actually the same $$i\epsilon$$ that appears in the denominator of the propagator. If you keep track of it all throughout you can verify this. It has to be there or the theory doesn't make sense.