How to show causality for a Klein-Gordon field in 1+1 dimensions using field commutators?

Question

For a non-interacting massive scalar field $\phi$ in an $n+1$ dimensional minkowskian spacetime,
the field commutator between two event points is

$$ [\phi(x),\phi(y)] = \int \frac{\mathrm{d}^n p}{(2\pi)^n} \frac{1}{2\sqrt{p^2+m^2}} \left( \mathrm{e}^{-\mathrm{i}p\cdot(x-y)} -\mathrm{e}^{-\mathrm{i}p\cdot(y-x)} \right) =: D(x-y) - D(y-x) $$

For the normal $3+1$ D case, since for any space-like separation $(x-y)$,
there is a Lorentz transform $\cal{L}$ continuous to the identity (i.e., ${\cal{L}}\in \mathrm{SO}^+(1,n)$), such that ${\cal{L}}(y-x) = (x-y)$,
it follows that $D(y-x)=D(x-y)$ and the commutator vanishes,
as illustrated by the $2+1$ D example in the following figure:

The argument however cannot be made for the $1+1$ D case; is there some other way to show causality for this case?

Some posts I've encountered:

• In QFT, why does a vanishing commutator ensure causality?
• Does the commutator of vanish for $m^2<0$?

Answer 1

Consider the commutator in $n+1$ space-time dimensions, $$[\phi(x),\phi(0)]= \int d\mu(p) \left( e^{-ip\cdot x}-e^{+ip\cdot x}\right) \tag{1} \label{1}$$ with $$d\mu(p)= \frac{d^n p}{(2\pi)^n \, 2 \, \omega(\vec{p})}, \qquad \omega(\vec{p})=\sqrt{\vec{p}^2+m^2}, \qquad p\cdot x= \omega(\vec{p}) \, x^0-\vec{p} \cdot \vec{x}.\tag{2} \label{2} $$ As $$\int d\mu(p)\, e^{\pm ip\cdot x}= \int \frac{d^{n+1} p}{(2\pi)^n}\, \theta(p^0)\, \delta(p^2-m^2)\, e^{\pm ip\cdot x} \tag{3} \label{3},$$ we see that \eqref{1} is invariant under proper orthochronous Lorentz transformations. Note that in \eqref{3}, $p^0$ has now become an integration variable with $p^2=(p^0)^2-\vec{p}^2$ and $p\cdot x = p^0 x^0-\vec{p}\cdot \vec{x}$!

For spacelike $x$ (i.e. $x^2 = (x^0)^2-\vec{x}^2\lt 0$) one can always find a proper orthochronous Lorentz tansformation $x^\prime = L x$ with $x^{\prime \, 0}=0$. Computing \eqref{1} in this reference frame, we obtain $$\int \frac{d^n p}{(2\pi)^n \, 2 \, \omega(\vec{p})}\left(e^{i \vec{p} \cdot \vec{x}} -e^{-i \vec{p} \cdot \vec{x}}\right) =0, \tag{4} \label{4}$$ where the transformation of variables $\vec{p} \to -\vec{p}$ was performed on the second term in the last step. This shows that indeed $$[\phi(x), \phi(0)]= 0 \quad \text{for} \quad x^2\lt 0 \tag{5} \label{5}, $$ independent of the space dimension $n$.

Remark: The general relation $$[\phi(x), \phi(y)] = 0 \quad \text{for} \quad (x-y)^2 \lt 0 \tag{6} \label{6}$$ is an immediate consequence of translation invariance.