Comments on Wick Rotation

In this article I’ll discuss a few of the ways we can “evaluate” integral expressions for the free propagator.

1. The Propagator

The propagator is the (unique tempered) fundamental solution for the Klein-Gordon equation. That is, it is a function {D: {\mathbb R}^{4} \rightarrow {\mathbb C}} satisfies

\displaystyle -(\partial^2 + m^2)D(x) = \delta^{(4)}(x)


  • {\partial^2} is the “Minkowski space Laplacian”: {\partial^2 f = \frac{\partial^2 f}{\partial x_{0}^{2}} - \frac{\partial^2 f}{\partial x_{1}^{2}} - \frac{\partial^2 f}{\partial x_{2}^{2}} - \frac{\partial^2 f}{\partial x_{3}^{2}} }
  • {\delta^{(4)}(x)} is the Dirac delta distribution on 4-dimensional Minkowski space.

{D(x)} is the amplitude for a particle to propagate from {0} to the point {x} in space-time.

We can take the Fourier Transform of both sides to turn this into an algebraic equation

\displaystyle -(-k^2 + m^2)\hat{D}(k) = 1

Where we define the Fourier transform as

\displaystyle \hat{f}(k) = \int_{{\mathbb R}^4} d^4 x e^{-i <k,x>} f(x)

And the inner products and “squares” are all using the Minkowski metric:

\displaystyle \begin{array}{rcl} <k,x> &=& k_0 x_0 - k_1 x_1 - k_2 x_2 - k_3 x_3 \\ k^2 &=& <k,k> \end{array}

This lets us write

\displaystyle \hat{D}(k) = \frac{1}{(k^2 - m^2)} \ \ \ \ \ (1)

Inverting the Fourier Transform, we now find

\displaystyle D(x) =\frac{1}{(2\pi)^{4}} \int_{{\mathbb R}^4} d^4 k\ \frac{ e^{i<k,x>}}{k^2 - m^2} \ \ \ \ \ (2)

When computing the effective mass of a particle in {\phi^4} theory, we have to sum over all chains of self-interactions. Each of these self-interactions will involve a particle propagating from a point in space time back to itself — propagating from 0 to 0 — and will add a factor of {D(0)} to a term in our sum.

So we need to compute {D(0)}

2. Propagating to Zero

We want to compute the amplitude of propagating to zero:

\displaystyle D(0) = \frac{1}{(2\pi)^{4}} \int_{{\mathbb R}^4} \frac{d^4k}{k^2 - m^2} \ \ \ \ \ (3)

We will do this in three ways, one not terribly rigorous and two that are solid:

  1. Integrate over sheets where {k^2} is constant.
  2. Use calculus of residues to evaluate the integral in the time variable first.
  3. Use a Wick Rotation to convert this to a Euclidean integral.

2.1. Contour Selection and the meaning of that {i\epsilon}

Before proceeding, note that for all fixed values of the space variables {k_1, k_2, k_3} the integrand in the time variable {k_0} has two poles, one at each of {\pm\omega_k} where {\omega_k = \sqrt{k_1^2 + k_2^2 + k_3^2 + m^2}}. To define the integral we must specify whether we go around these poles to the left or to the right. There are, of course, four possible choices. We will follow Feynman and go below the negative pole and above the positive one:

In the literature, you will often see the integral formula for the propagator expressed as

\displaystyle D(x) =\frac{1}{(2\pi)^{4}} \int_{{\mathbb R}^4} d^4 k\ \frac{ e^{i<k,x>}}{k^2 - m^2 + i\epsilon} \ \ \ \ \ (4)

The addition of {i\epsilon} moves the positive pole down a little bit and moves the negative pole up a little bit, allowing us to avoid the poles with a simple contour, like this:

As {\epsilon \rightarrow 0} this is equivalent to our first choice of contour. All the “{i\epsilon} term” is doing is telling us which of the four contours to pick when evaluating Equation 2.

2.2. Approach 1: a first attempt at symmetry

For our first attempt at evaluating Equation 3, notice that the integrand is left unchanged by Minkowski rotations, so it is constant on Minkowski spheres. Say we could compute the surface area of the Minkowski sphere of radius {r}, call it {S(r)}. Then we can use our symmetry to change our 4D integral into a 1D integral in {r}. On the {r}-sphere, the integrand takes the value {\frac{1}{r^2 - m^2}}, so our integral will now be

\displaystyle D(0) = \frac{1}{(2\pi)^{4}} \int_{{\mathbb R}^4} \frac{d^4k}{k^2 - m^2} = \frac{1}{(2\pi)^{4}} \int_{{\mathbb R}} \frac{S(r)dr}{r^2 - m^2}

But the surface area of the sphere in 4D space scales like {r^3}, so this is

\displaystyle D(0)= \frac{1}{(2\pi)^{4}} \int_{{\mathbb R}} \frac{S(1)r^3 dr}{r^2 - m^2}

And it appears that we are integrating a function that grows linearly in {r}. Yes, cancellation may helps is, we need to calculate residues, etc. but none of that really matters because {S(1)} is infinite. Using this symmetry was a good idea, but it needs some work.

2.3. Approach 2: Integrate in time first

Going for the throat with the symmetry didn’t work, so let’s try straightforward calculation. Looking at the contour in the diagram above, we can imagine forming it into a closed loop by adding an arc along the semicircle of radius {R} in the upper half plane. For {R} large, the integrand will be {O(R^{-2})} along this arc and the arc has length {O(R)}, so the contribution of the arc to the integrand is {O(R^{-1})}; it vanishes.

Thus our integral will just be the limit of our closed contour integrals as we let {R \rightarrow \infty}. What is the contour integral? There is only one pole inside the contour, at {-\omega_k}, and it has residue {-\frac{1}{2\omega_k}}. So our integral is {\frac{-i\pi}{\omega_k}}.

To evaluate {D(0)} we now integrate this in the space variables used to define {\omega_k}.

\displaystyle D(0) = \frac{-i\pi}{(2\pi)^{4}} \int_{{\mathbb R}^3} \frac{d^3k}{\sqrt{k^2 + m^2}}

Where now {k^2} represents the square of the ordinary Euclidean norm in {{\mathbb R}^3}. We still have rotational symmetry here so the integrand is constant on spheres. The surface area of the {r}-sphere is {4\pi r^2} so

\displaystyle D(0) = \frac{-i}{4\pi^2} \int_0^\infty \frac{r^2 dr}{\sqrt{r^2 + m^2}} \ \ \ \ \ (5)

Of course this integral still diverges, this is QFT after all. Notice that the integrand grows linearly, so we have a quadratic divergence. We could perform this full integration in {k_0}, then employ cutoffs for renormalization in the space variables. I expect we would get the same results, but that is not what is done. Instead, we use the Wick Rotation.

2.4. Approach 3: Wick Rotation

Instead of integrating all the way out in {k_0}, consider the path integral along the following contour:

This contour encloses no poles, so the integral along it is zero.  Because the integrand is an even function, the integral along the two large arcs will cancel perfectly.  Thus the integral along the two “straight lines” must also cancel perfectly. This gives the equation

\displaystyle D(0) = \frac{1}{(2\pi)^{4}} \int_{{\mathbb R}^4} \frac{d^4k}{k^2 - m^2} = \frac{1}{(2\pi)^{4}} \int_{{\mathbb R}^4} \frac{d^4x}{-\|x\|^2 - m^2}

where {x_0 = ik_0, x_1 = k_1, x_2 = k_2, x_3 = k_3} and {\|x\|} is the ordinary Euclidean norm on {{\mathbb R}^4}. Now we’ve changed our Minkowski rotational symmetry into a Euclidean rotational symmetry, and unlike the Minkowski sphere, the Euclidean sphere has finite surface area. This means we can play those games we tried in our first approach. Integrating over the radius of the sphere (which has 3-area {S(1) = }), we get

\displaystyle D(0) = \frac{-S(1)}{(2\pi)^{4}} \int_0^\infty \frac{r^3 dr}{r^2 + m^2} \ \ \ \ \ (6)

No poles, the integrand grows linearly in {r}, and we have quadratic divergence. When we apply cutoffs in QFT, this is the integral we cutoff. Notice that it is a bit different from cutting things off in the original {k_0} variable but in the end we still get the quadratic divergence we observed with approach 2.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s