Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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

share|improve this question

2 Answers 2

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: $$\begin{align} \int d^3p\,d^3p'\,d^3x\ f(p,p')e^{ip\cdot x-ip'\cdot x} &=\int d^3p\,d^3p'\,d^3x\ f(p,p')e^{i(E(p)-E(p'))t - i(\vec p - \vec p')\cdot\vec x}\\ &=(2\pi)^3\int d^3p\,d^3p'\ f(p,p')e^{i(E(p)-E(p'))t}\delta^{(3)}(\vec p - \vec p')\\ &=(2\pi)^3\int d^3p \ f(p,p) \end{align}$$

share|improve this answer

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 $p'$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.