I am attempting to understand the normalization of one-particle state $|\textbf{p}\rangle \propto a_{\textbf{p}}^\dagger$ in the context of Klein-Gordon field quantization from Peskin & Schroder's QFT textbook (page no. 22-23).
They compute $\langle\textbf{p}|\textbf{q}\rangle = (2\pi)^3 \delta(\textbf{p}-\textbf{q})$ and demonstrate that it is not Lorentz invariant by considering the effect of a boost in the 3-direction. Under such a boost we have $p_3' = \gamma(p_3 + \beta E)$ and $E' = \gamma (E+\beta p_3)$, where the symbols have their usual meanings in the discussion of Lorentz transformation. They recommend using the following Dirac delta identity:
$$\delta(f(x) – f(x_0)) = \frac{\delta(x-x_0)}{\vert f'(x_0) \vert}$$
and with that they show that
$$\delta(\textbf{p}-\textbf{q}) = \delta(\textbf{p}'-\textbf{q}') \frac{E'}{E}.$$ (I put their calculation in the Appendix.)
To reproduce the result, I started in the following way.
$$
\delta(\textbf{p}'-\textbf{q}') = \delta(p'_1-q'_1)\delta(p'_2-q'_2)\delta(p'_3-q'_3) \\
= \delta(p_1-q_1) \, \delta(p_2-q_2) \, \delta[\gamma(p_3 + \beta E_p) – \gamma(q_3 + \beta E_q)],
$$
where $E_p = \sqrt{|\textbf{p}|^2 + m^2} = \sqrt{p_1^2 + p_2^2 + p_3^2 + m^2}$ and $E_q = \sqrt{|\textbf{q}|^2 + m^2} = \sqrt{q_1^2 + q_2^2 + q_3^2 + m^2}$ and $m$ is the mass of the particle. Here, I have dropped the primes in the argument first two Dirac delta functions as $p_1, q_1$ and $p_2, q_2$ do not change due to boost considered.
My Question
From here I am not sure how to proceed or use the Dirac Delta identity to get to the final result. Could you please help me here a little bit?
Appendix
Best Answer
The $f$ you want to use the identity on is $$f(x) = \gamma (x+ \beta E(x)),$$ where
$$E(x) = \sqrt{p_1^2+p_2^2+ x^2 +m^2}.$$
Then taking the derivative $$f'(x) = \gamma(1+ \beta \frac{dE}{dx}(x)).$$
So the identity gives
$$\delta(p_3'-q_3')=\delta(p_3-q_3)\frac{1}{\gamma(1+\beta \frac{dE}{dx}(q_3))}$$
and one can calculate
$$\frac{dE}{dx}(p_3) = \frac{p_3}{E(p_3)}.$$
Now just multiply the denominator over
$$\begin{align}\delta(p_3-q_3) &= \delta(p_3'-q_3')\gamma(1+\beta \frac{dE}{dx}(p_3))\\ &=\delta(p_3'-q_3')\frac{\gamma}{E(p_3)}(E(p_3)+\beta p_3)\\ &=\delta(p_3'-q_3')\gamma\frac{E'(p_3)}{E(p_3)}\end{align}$$