# Lorentz Invariance of Propagator for Complex Scalar Field

## Homework Statement

Show that

$[\hat{\phi}(x_1),\hat{\phi}^\dagger(x_2)] = 0$
for $(x_1 - x_2)^2 < 0$

where $\phi$ is a complex scalar field

## Homework Equations

$\hat{\phi}=\int\frac{d^3 \mathbf{k}}{(2\pi)^3 \sqrt{2\omega}}[\hat{a}(k)e^{-ik\cdot x} + b^\dagger(k)e^{ik\cdot x}]$

with

$[\hat{a}(k),\hat{a}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})$
$[\hat{b}(k),\hat{b}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})$

and all other commutation relations vanish.

## The Attempt at a Solution

This is problem (7.3) in Aitchison and Hey's QFT book. I have a couple questions about the solutions that are posted online here. They get eventually to the difference of two integrals:

$\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{-ik\cdot (x_1-x_2)}-\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{ik\cdot (x_1-x_2)}$

So far so good. Then they argue that for space-like intervals, x1-x2 can be transformed to -(x1-x2) by a Lorentz transformation, and thus the second term cancels the first.

Now, I agree that each of the integrals must be Lorentz invariant, so you are allowed to evaluate each one in whichever frame you want, but I don't see how one is allowed to transform just the x1-x2 vector without also transforming the k.

And a related question: In a different problem (5.6), when we come to a similar integral, they suggest transforming to a frame where t1=t2, and since the resulting integral is an odd function of k, it must be zero. And since it's Lorentz invariant, it must be zero for all space-like intervals. It seems to me that this argument works for this problem, too, so I'm not sure why they bring up this other argument about transforming x1-x2 to -(x1-x2).

Anyone want to help explain what I'm missing?

Thanks

vanhees71
Gold Member
You are integrating over $\vec{k}$. Thanks to the invariance of $\mathrm{d}^3 \vec{k}/\omega$ where $\omega=\sqrt{m^2+\vec{k}^2}$ you know that the result is a Lorentz scalar $$F(x_1-x_2)$$. That means that under Lorentz transformations
$$F'(x')=F(x)=F(\Lambda^{-1} x).$$
Thus you only need to transform $x=x_1-x_2$.

It's only crucial to show that for a spacelike $x$ you can always find a Lorentz transformation such that $\Lambda x=-x$. Since you can orient your coordinate system always such that $x=(0,\xi,0,0)$ you just need to find a Lorentz transformation which makes out of this $\Lambda x=(0,-\xi,0,0)$.

• 1 person
Thanks for the reply. I guess I'm missing the last part of that equality $F(x) = F(\Lambda^{-1}x)$. As a test, if I define a function

$F(x)=k\cdot x$

where $k$ is some arbitrary 4-vector, then F(x) is a Lorentz scalar. If I have some transformation such that $x'=\Lambda x$, then

$F'(x')=k'\cdot x'=k\cdot x=F(x)\not=F(\Lambda^{-1} x)$

I understand that in the original problem we're integrating over $\vec{k}$, but it still seems to me we need to transform the $\vec{k}$.

vanhees71
$$F'(x')=F(x)=F(\Lambda^{-1} x'),$$
$$x'=\Lambda x \; \Leftrightarrow \; x=\Lambda^{-1} x.$$
Right after I posted that, I figured that's what you meant. And it finally struck me why we don't have to worry about the $\vec{k}$. Thanks for your help.