Lorentz invariant integration measure and Heaviside step function I'm currently studying Klein-Gordon fields and I ran onto the concept of the Lorentz invariant integration measure, namely:
\begin{equation}
\frac{d^3k}{(2\pi)^32E_k}
\end{equation}
where $E_k=\sqrt{\boldsymbol{k}^2+m^2}$. I see from Lorentz Invariant Integration Measure that in my integral I should include 
$\theta(k_0)\delta(k^{\mu}k_{\mu}-m^2)$. I get the reason for the presence of the delta (I want on-shell relativistic particles) but I don't get why I want to select particles with positive energy and neglect the negative solutions.
On my notes, I find that $k_0=\pm E_k$ and "Negative energies can't be neglected. They will eventually be interpreted as antiparticles with positive energy in QFT". So why, integrating, I select only particles with $k_0>0$?
 A: It is just mathematics, forget about the physics for a moment. What we want to do is to find an integration measure $\mathrm d\mu(\boldsymbol k)$ that is invariant under Lorentz transformations. It should be clear that
$$
\mathrm d\mu(\boldsymbol k)=\delta(k^2-m^2)\Theta(k^0)
$$
does the trick, irrespective of what the factors represent. You may wonder why we choose this form for $\mathrm d\mu$ instead of some other possibility; the reason is that this measure is rather natural in this context. You can find a possible motivation for $\mathrm d\mu$ in this post of mine.
By using $\Theta(k^0)$ you are not selecting particles. After all, you are solving the Klein-Gordon equation, at which point there are no particles yet. On the other hand, if you are using $\mathrm d\mu(\boldsymbol k)$ to derive a formula for the cross-section, the particles are already there! You cannot change the spectrum of the Hamiltonian by choosing some integration measure or other. The physics of a problem are independent of how you choose to solve the equations. Choosing this specific form for $\mathrm d\mu(\boldsymbol k)$ is convenient. Choosing other forms won't remove, nor add, particles from your system.
