# Klein-Gordon Propagator for spatial separation $x - y = (0, r)$

On pg. 27 of Peskin and Schroeder I would like to know how we get the first equality when deriving the Klein-Gordon propagator for $$x^0 - y^0 = 0, \vec{x} - \vec{y} = \vec{r}$$:

$$D(x-y) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_p} e^{i \vec{p} \cdot \vec{r}} = \frac{2\pi}{(2\pi)^3} \int_{0}^{\infty} dp \frac{p^2}{2 E_p} \frac{e^{ipr} - e^{-ipr}}{ipr}$$

The subsequent contour integral (2.52) makes sense but I'm a little confused about the derivation to get there from (2.50).

Why we are integrating only the imaginary part of $$e^{ipr}$$ and why the factor $$ipr$$ in the denominator?

• Just take the integral over $\cos\theta$. Commented Jan 12, 2021 at 2:39

The easiest way to proceed is to notice that the first integral can only depend on $$r := |\vec{r}|$$. To convince yourself of this, you can rotate $$\vec{r}$$ by any rotation $$R \in SO(3)$$, and you notice that you can always "unwind" this rotation by a redefinition of the momentum variable being integrated (if you take $$\vec{r}\to R\vec{r}$$ then change variables to $$R^{T} \vec{p} := \vec{p}'$$, where the Jacobian is just $$1$$). Since this works for any $$R$$ you know this is only a function of $$r$$.
Because of this fact, you can make your life easier by picking $$\vec{r} = ( 0, 0, r )$$ (notice that $$|( 0, 0, r )| = r$$ is still true). If you plug this into the first integral you get $$\int \frac{d^3 \vec{p}}{(2\pi)^3 } \frac{1}{2 E_p} e^{- i \vec{p} \cdot \vec{r}} \ = \ \int \frac{d^3 \vec{p}}{(2\pi)^3 } \frac{1}{2 E_p} e^{- i p_3 r}$$ in spherical coordinates this becomes $$\cdots \ = \ \int \frac{dp \; d\theta \; d\phi}{(2\pi)^3 } \frac{p^2 \sin\theta}{2 E_p} e^{- i r p \cos\theta } \ .$$ I leave it up to you to simplify this to your expression ($$\phi$$ integrates easily, and use the coordinate change $$\mu = \cos\theta$$).
• Hi, I have a query here. Conceptually, the vector space to which $\vec{r}$ and $\vec{p}$ belong are different. So how can we "unwind" the rotation that we did on $\vec{r}$ via a reverse transformation on $\vec{p}$? Commented Mar 17, 2021 at 16:15
• I like to think of functions $f(\mathbf{p})$ in momentum space, or their Fourier transform $\tilde{f}(\mathbf{x}) = \int \frac{d^3\mathbf{p}}{(2\pi)^3} e^{i \mathbf{p} \cdot \mathbf{x}} f(\mathbf{p})$ in position space, as expressions of the same function $f$ but being expressed in different "bases". The fundamental thing is the function (expressed in whatever momentum or position "basis"): in this case, the function is rotationally invariant (in whatever basis you express it in) Commented Mar 17, 2021 at 20:58