For the free Klein-Gordon theory: If I change the definition of time-ordering such that:
$\mathcal{T} A(x)B(y) = A(x) B(y) \Theta( x^{0} - y^{0} ) - B(y) A(x) \Theta(y^{0}-x^{0})$
In the above $\Theta$ is the Heaviside step function (and normally, it would be + sign instead of a minus sign).
If I know that $<0| \phi(x) \phi(y) |0> = \int \frac{d^{3}p}{(2\pi)^{3}2E_{\mathbf{p}}}e^{-i p \cdot (x-y)}$, I'm supposed to show that $< 0 | \mathcal{T} \phi(x) \phi(y) | 0 >$ is $\mathbf{NOT}$ Lorentz invariant.
I can write the above in a piece-wise manner so that:
$< 0 | \mathcal{T} \phi(x) \phi(y) | 0 > = \begin{cases} \int \frac{d^{3}p}{(2\pi)^{3}2E_{\mathbf{p}}}e^{-i p \cdot (x-y)} \ \ \ & \ , \ x^{0}>y^{0} \\ -\int \frac{d^{3}p}{(2\pi)^{3}2E_{\mathbf{p}}}e^{i p \cdot (x-y)} \ \ \ & \ , \ y^{0}>x^{0} \end{cases}$
I'm just totally lost as to how to show this is not lorentz invariant...
I know that $\Theta$, the measure $\frac{d^{3}p}{(2\pi)^{3}2E_{\mathbf{p}}}$, and the four dot-product $p \cdot (x-y)$ are all Lorentz invariant....so why wouldn't the entire thing above be also?
I have no idea where to start and would appreciate some direction...
EDIT: while I understand the proofs for invariance of the measure and dot product, I don't totally grasp why $\Theta(x^{0})$ is Lorentz invariant. Due to this post Lorentz-invariance of step function I'm going to assume through that this is indeed the case.
