Why there is 2[itex]\pi[/itex] in every dirac delta function

[nadia8999]

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\piin the second equation. Is it a convention? Is so where does the 2\picomes from originally.

 

[dextercioby]

I think the wiki article (especially the section on <Other Conventions>) http://en.wikipedia.org/wiki/Fourier_transform pretty much explains it. Essentially the 2\pi comes from the Euler's e^i2\pi = -1.

 

[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}.

Thus, the inverse reads

\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}&#039;)]=\delta^{(3)}(\vec{p}-\vec{p}&#039;).

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

 

[Meir Achuz]

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.