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if we adopt the convention, [itex] a^{\dagger}_\textbf{p} |0\rangle = |\textbf{p}\rangle [/itex]

then we get a normalization that is not Lorentz invariant, i.e. [itex] \langle \textbf{p} | \textbf{q} \rangle = (2\pi)^3 \delta^{(3)}(\textbf{p} - \textbf{q}) [/itex].

How do I explicitly show that this delta function (of 3-vectors) is not lorentz invariant, say for a boost in the p^3 direction ? What is [itex] \langle \textbf{p'} | \textbf{q'} \rangle [/itex] where the primed frame is the boosted frame?

I thought since [itex] \int d^{3}\textbf{p} \hspace{2mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = 1 [/itex] is lorentz invariant (integral over all space, so includesq)

and as [itex] \hspace{3mm} d^3 \textbf{p} = \gamma^{-1} d^3 \textbf{p'} [/itex], (primed frame is boosted frame)

so [itex] \hspace{5mm} \delta^{(3)}(\textbf{p} - \textbf{q}) = \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'}) [/itex].

so that [itex] \int \gamma\,\delta^{(3)}(\textbf{p'} - \textbf{q'}) \gamma^{-1} d^3 \textbf{p'} = 1 [/itex] stays Lorentz invariant.

However, Peskin gets for the same calculation (p.22-23, eqn.2.34) the result

[itex] \delta^{(3)}(\textbf{p'} - \textbf{q'}) = \delta^{(3)}(\textbf{p} - \textbf{q}) (\frac{E}{E'}) [/itex] using a method that I don't really understand well.

Have I made a mistake somewhere?

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# Normalization of free scalar field states

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