# Is the scalar propagator an even function?

The scalar propagator for the Klein-Gordon Lagrangian is given by:

$$D(x-y)=\int \frac{d^{4} k}{(2 \pi)^{4}} \frac{e^{i k(x-y)}}{k^{2}-m^{2}+i \varepsilon}$$

I need to know if it is an even function, i.e.:

$$D(x-y) = D(-(x-y)) = \int \frac{d^{4} k}{(2 \pi)^{4}} \frac{e^{i k(x-y)}}{k^{2}-m^{2}+i \varepsilon}\tag{1}$$

Since $$k^2=k_0^2-\vec{k}^2$$ and $$k_0^2$$ and $$\vec{k}^2$$ are both positive.

Also

$$\int_{-\infty}^{\infty} d^4k e^{ik(x-y)}=\int_{-\infty}^{\infty} d^4k e^{-ik(x-y)}=\int_{-\infty}^{\infty} d^4k e^{ik(y-x)}$$

So can we conclude that equation 1 is indeed true?

edit: rephrased the question hope it makes sense now

• Do you mean $D(x-y) = D(y-x)$ ? – Señor O Sep 29 '19 at 17:50
• I think you mean, when asking if it is even, if $D(x-y) = D(y-x)$ since what you've written cancels to $D(x-y)=D(x-y)$ which is obviously true. – JamalS Sep 29 '19 at 17:52
• Yes, it is true that $D(x-y) = D(-(y-x))$. – knzhou Sep 29 '19 at 18:03

The fact that $$D(x) = D(-x)$$ follows immediately from the fact that the both the integrand and the integration measure are invariant under Lorentz transformations, and recalling that $$x \to -x$$ (parity transformation) is indeed a Lorentz transformation.
Alternatively, it's not difficult to prove it explicitly. \begin{align} D(-x) &= \int_{R^4}\frac{d^4k}{(2\pi)^4}\frac{e^{ik\cdot (-x)}}{k^2-m^2+i\varepsilon} \\ &= \int_{R^4}\frac{d^4k}{(2\pi)^4}\frac{e^{i(-k)\cdot x}}{k^2-m^2+i\varepsilon}. \end{align} Now make the substitution $$k' = -k$$. Since $$|\det(\partial k'/\partial k)|=|-1|=1$$ and under this substitution $$R^4$$ gets mapped into itself, we have, by substitution for multiple variables, \begin{align} \phantom{D(-x)} &= \int_{R^4}\frac{d^4k'}{(2\pi)^4}\frac{e^{ik'\cdot x}}{(-k')^2-m^2+i\varepsilon} \\ &= \int_{R^4}\frac{d^4k'}{(2\pi)^4}\frac{e^{ik'\cdot x}}{k'^2-m^2+i\varepsilon} \\ &= D(x) \end{align} (since $$k'$$ is just a dummy variable; we also used the fact that $$(-k)^2 = k^2$$).