Wick Rotation — What happens to the integration limits?

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This is a genuine question that arises in the minds of students in a class where Wick rotation is discussed carelessly. It is a very simple observation.

Consider a definite integral over the real line of a real-valued function $f: k^0 \mapsto f(k^0)\in \mathbb R$.

\[I := \int_{\mathbb R} dk^0\ f(k^0) \equiv \lim_{a\to \infty} \int_{-a}^a dk^0\ f(k^0)\]

Now, the teacher asks you to make an innocent substitution (change of variables), $k^0\to k_E^0 := -ik^0$ which substitution he/she christens as 'Wick rotation' and asks you to compute the integral with the new so-called Euclidean variable. Naively, you think that this is something you have done since kindergarten, and observe that as $k^0 \to \pm \infty$, the new variable $k_E^0 \to \mp i\infty$, and therefore you write,

\[ I = i \int_{i\infty}^{-i\infty}dk_E^0\ f(ik_E^0) \equiv -i \lim_{a\to\infty}\int_{-ia}^{ia}dk_E^0\ f(ik_E^0)\,.\]

But then you notice, assuming you are smart and careful, that the teacher slyly and quietly changes the integration limits while computing it. He/she writes

\[ I = \color{red}{+}i \lim_{a\to\infty} \int_{\color{red}{-a}}^{\color{red}{a}}dk_E^0\ f(ik_E^0) \,, \]

which you know is apriori wrong. You revolt in class, but you are silenced by the teacher with a slurry of hand-wavy arguments that are simply wrong, or by the voice of the majority who play nods with the teacher and pretend to understand this silly kindergarten substitution.

But you see, the truth is you are onto something. This substitution is not as innocent as it seems. Look! You have rotated in the complex plane and you are no longer in the real line. The new variable is a complex one, and you need to learn a thing or two about complex integration before setting out to compute it. After all, given your intelligence, you should ask if complex integration is even possible, or if our intuitions about real integration hold true even in the complex case. The answers are non-trivial and I would like to remind you about them.

For one, it is true that one can define well-behaved complex integration over curves in the complex plane, known as contour integration, and I am sure you have come across them several times. Since we are familiar with integration over real variables only, we define contour integration in terms of real variables as well. We always choose a real parametrization $z: \mathcal I \subset \mathbb R \to \mathcal C, t \mapsto z(t) \in \mathcal C$ of a contour $\mathcal C$ in the complex plane, over which we wish to integrate a complex function $f:z\mapsto f(z)$. The contour integration is then defined as follows.

\[ \int_{\mathcal C} dz\ f(z) := \int_{\mathcal I} dt\ f(z(t)) z'(t)\,. \]

Note that one never has complex integration limits, except perhaps for a understood figurative for the contour. The actual integration is always done over real variables.

Secondly, you ought to digest that the fundamental theorem of calculus does not necessarily hold true for complex integrals. In general, it is not true that 

\[ \int_{\mathcal C} dz\ f'(z) = f(z)\Big|_{\partial\mathcal C} \,.\]

If you fix two end points, the integral of a function along a path joining these endpoints depends on which path the integration was done, unless you make sure that the function happened to be analytic. Most of our intuition from calculus hold true for analytic functions only, otherwise not!

Now, going back to our discussion of Wick rotation, first let me tell you that a Wick-rotated integral is not trivially the same as the original integral. The kindergarten "substitution" that supposedly achieves Wick rotation is ill-defined, and to be honest, indicative of a misunderstanding. To go the right way, first of all one needs to analytically extend the domain of the variable $k^0$ by complexifying it. The second step is the transformation of the integration contour achieved by a right-angled rotation, say anticlockwise. Understand that we are making no substitution business here. Our variable $k^0 \in \mathbb C$ has been complexified and our original integral $I$ can now be seen as a contour integral where the contour is the real line. Now, we consider a different integral which has the same integrand as before, however we integrate it over a different contour, the Wick-rotated contour. This new integral is called the Wick-rotated integral.

\[ I_E := \int_{\mathcal C}dk^0\ f(k^0)\,, \]

where the contour $\mathcal C$ is the imaginary line. Note that there is apriori no reason to believe that $I=I_E$. A rigorous proof of the equivalence between a theory and its Wick-rotated version is a matter of extraordinary non-triviality and is the subject of the Osterwalder-Schrader reconstruction theorem in Wightman QFT.

In order to evaluate $I_E$, we must parametrize the contour $\mathcal C$ using a real variable. There is a natural choice of parametrization for $\mathcal C$, namely its coordinate: $k^0(t)=it$ where $t\in\mathbb R$. Then, according to our definition of contour integrals, we have that
\[ I_E = \int_{\mathbb R} dt\ f(it) \frac{d}{dt}(it) = i \int_{-\infty}^{\infty} dt\ f(it)\,. \]
Note that your teacher merely had a different name for the parameter $t$, namely $t\equiv k_E^0 \in \mathbb R$. Keep in mind that $k_E^0$ is a real parameter, and therefore, the integration limits used by your teacher is right.
\[ I_E = i  \int_{-\infty}^{\infty}dk_E^0\ f(ik_E^0) \stackrel{?}{=} I \]
Now you understand the correct description of Wick rotation. It's simple, no? Next time someone gives you a load of bollocks, break their bones. Cheers! Until next time!