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I am currently studying Quantum Field Theory from the textbook Overview of Quantum Field Theory and I am confused by two formulae presented in chapter 2 (2.39) and (2.40). The first is

$$(1)_{1-particle} = \int \frac{d^3p}{(2\pi)^3}|p\rangle\frac{1}{2E_p}\langle p|.\tag{2.39}$$

The textbook says this is the completeness relation. I sort of see it but I do not understand where the $\frac{1}{2E_p}$ comes from. Can anyone explain?

The second equation is also confusing; can anyone explain why this is true? My guess that it comes from the above completeness relation but I am not sure.

$$\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p} = \int \frac{d^4 p}{(2\pi)^4}(2\pi)\delta(p^2 - m^2).\tag{2.40}$$

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Equation (2.39) is written the way it is because one has chosen the covariant normalization. In this normalization the momentum eigenstates are normalized as

$$\langle p'|p\rangle = (2\pi)^3 (2E_p) \delta^3(\mathbf{p}-\mathbf{p'}).\tag{1}$$

In that case observe that if we decompose a state as $$|\phi\rangle=\int d^3p\ \phi(p) |p\rangle\tag{2},$$

then you can find out by using (1) that $\phi(p)$ is given by $$\phi(p) = \dfrac{1}{(2\pi)^3}\dfrac{1}{2E_p} \langle p|\phi\rangle\tag{3}.$$

This means that (2) is actually the statement that

$$|\phi\rangle=\int \dfrac{d^3p}{(2\pi)^3 2E_p}|p\rangle \langle p|\phi\rangle\tag{4},$$

from which you can read off the resolution of the identity in the form of your equation (2.39).

As for (2.40) you can easily derive it from the right-hand side by recalling that $$\delta(f(x))=\sum_{x_i} \dfrac{\delta(x-x_i)}{|f'(x_i)|}\tag{5},$$

where $x_i$ are the zeroes of $f(x)$. View $p^2-m^2$ as a function of $p^0$ and in the end use the above result to eliminate the integration over $p^0$ to understand the relation with the left-hand side. That's a good exercise.

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