Normalization of momentum eigenstates in QFT

Question

Inspired by a previous question, I'd like to ask about the normalization of one-particle states in QFT.

The most common normalization seems to be the covariant one: $$ \langle \vec p'|\vec p\rangle = (2\pi)^3 2E(\vec p)\delta^{(3)}(\vec p-\vec p') \quad\leftrightarrow\quad 1\!\!1 = \int\frac{\text{d}^3\vec p}{(2\pi)^3}\frac{1}{2E(\vec p)} |\vec p\rangle\langle \vec p| \tag{1} $$ but other choices seem to be possible, e.g. $$ \langle \vec p'|\vec p\rangle = (2\pi)^3 \delta^{(3)}(\vec p-\vec p') \quad\leftrightarrow\quad 1\!\!1 = \int\frac{\text{d}^3\vec p}{(2\pi)^3} |\vec p\rangle\langle \vec p| \tag{2} $$

1. What's the main advantage of the normalization (1) (as most textbooks seem to use it)?
2. What freedom do we have in choosing a(nother) normalization in QFT?

Answer 1

If you're prepared to keep track of extra factors in the LSZ formula and such, it seems to me that you can change the normalization however you want. But the reason why the zeroth component appears in the standard measure is because it allows you to write the integral over 3-momenta as a Lorentz invariant integral over 4-momenta. \begin{equation} \int \frac{d^3 \vec{p}}{(2\pi)^3} \frac{1}{2E(\vec{p})} = \int \frac{d^4 p}{(2\pi)^3} \delta(p^2 + m^2) \theta(p^0) \end{equation} More precisely, this is invariant under the proper orthochronous Lorentz group since $\theta(p^0)$ is there to pick out a single zero from the mass shell condition.