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How to show $$\displaystyle\int\int\int f(p,p')e^{ip\cdot x-ip'\cdot x}d^3pd^3p'd^3x=(2\pi)^3\int f(p,p)d^3p$$ ?

I have $p\cdot x=Et-\bf p\cdot x$

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up vote 3 down vote accepted

So, following the suggestion of Olaf and Vladimir, assume that the momenta are on-shell, so that $E = E(p)$. Then we first do the position integral to get a delta function which lets us perform one of the momentum integrals: $$\begin{align} \int d^3p\,d^3p'\,d^3x\ f(p,p')e^{ip\cdot x-ip'\cdot x} &=\int d^3p\,d^3p'\,d^3x\ f(p,p')e^{i(E(p)-E(p'))t - i(\vec p - \vec p')\cdot\vec x}\\ &=(2\pi)^3\int d^3p\,d^3p'\ f(p,p')e^{i(E(p)-E(p'))t}\delta^{(3)}(\vec p - \vec p')\\ &=(2\pi)^3\int d^3p \ f(p,p) \end{align}$$

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Are you sure that $x\cdot p$ is not just the ordinary 3D dot product? T

Because in this case you can use the property of the delta function,

$$\int e^{i(p-p')\cdot x} d^3x = (2\pi)^3 \delta^{(3)}(p-p')$$

which you can use to integrate out $p'$.

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Olaf, $p\cdot x$ can be $Et-\vec{p}\vec{x}$: after integrating over $d^3x$ the energies $E'(\vec{p}')=E(\vec{p})$, so their difference vanishes. –  Vladimir Kalitvianski Nov 22 '11 at 18:38
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