Why there is 2$\pi$ in every dirac delta function

1. Aug 10, 2011

in QF, every dirac delta function is accompanied by $2\pi$,i.e.$(2\pi)\delta(p-p_0)$ or $(2\pi)^3\delta(\vec{p}-\vec{p_0})$

the intergral element in QF is $\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_P}$, it comes from the integral element $\int\frac{d^4p}{(2\pi)^4}(2\pi)\delta(p^2-m^2)$,I want to know why there is $2\pi$in the second equation. Is it a convention? Is so where does the $2\pi$comes from originally.

2. Aug 10, 2011

dextercioby

3. Aug 10, 2011

vanhees71

It's a convention, where you shift all the $2 \pi$ factors around in quantum mechanics or quantum field theory and where to put the - sign in the exponential. In the HEP community the most common convention is to put all the $2\pi$'s to the momentum-space integrals, i.e., you define Green's functions etc. as

$$G(x)=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 p}{(2 \pi)^4} \tilde{G}(p) \exp(-\mathrm{i} p \cdot x),$$

where I used the west-coast-metric convention, i.e.,

$$p \cdot x=p_{\mu} x^{\mu}=p^0 x^0-\vec{p} \cdot \vec{x}.$$

$$\tilde{G}(p)=\int_{\mathbb{R}^4} \mathrm{d}^4 x G(x) \exp(+\mathrm{i} p \cdot x).$$

The only exception are the field operators themselves. Their Fourier representation in terms of creation and annihilation operators read (here for the scalar Klein-Gordon field)

$$\hat{\phi}(x)=\int_{\mathrm{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 E_{\vec{p}}}} \left [\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x) + \hat{b}^{\dagger}(\vec{p}) \exp(+\mathrm{i} p \cdot x) \right]_{p^0=E_{\vec{p}}},$$

with $E_{\vec{p}}=\sqrt{m^2+\vec{p}^2}$.

The reason for this is to get simply normalized commutation relations for the creation and annihilation ops, e.g.,

$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=\delta^{(3)}(\vec{p}-\vec{p}').$$

But for this latter issue you have nearly as many conventions as textbooks :-(.

4. Aug 10, 2011

clem

One representation of the delta function is
$$\delta(x)=\frac{1}{2\pi}\int dk e^{ikx}.$$
If the integral appears in some equation without the $$\frac{1}{2\pi}$$,
you get $$2\pi\delta$$.

Last edited: Aug 10, 2011