# 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 $k_0>0$?

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.
• Ok I get it. I thought there was a specific physical reason and this was the origin of my confusion. Thank you for linking your answer, it's very detailed and it will be useful. So I can as well use $\delta(k^{\mu}k_{\mu}-m^2)$? (as you did in your answer) – Luthien Dec 15 '16 at 17:10
• Im glad I could help. What do you mean by "So I can as well use $\delta(k^2-m^2)$?" – AccidentalFourierTransform Dec 15 '16 at 17:12
• As a measure. I saw from your answer that you didn't bother to ass the heaviside but, elegantly, you simply integrated over $dk_0$. The Lorentz invariance is preserved but your measure was simply $d^4k\delta(k^2-m^2)$. (I admit I'm very weak with the concept of "measure" so I may say really stupid things) – Luthien Dec 15 '16 at 17:17