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

  1. Aug 10, 2011 #1
    in QF, every dirac delta function is accompanied by [itex]2\pi[/itex],i.e.[itex](2\pi)\delta(p-p_0)[/itex] or [itex](2\pi)^3\delta(\vec{p}-\vec{p_0})[/itex]

    the intergral element in QF is [itex]\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_P}[/itex], it comes from the integral element [itex]\int\frac{d^4p}{(2\pi)^4}(2\pi)\delta(p^2-m^2)[/itex],I want to know why there is [itex]2\pi[/itex]in the second equation. Is it a convention? Is so where does the [itex]2\pi[/itex]comes from originally.
  2. jcsd
  3. Aug 10, 2011 #2


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  4. Aug 10, 2011 #3


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    It's a convention, where you shift all the [itex]2 \pi[/itex] 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 [itex]2\pi[/itex]'s to the momentum-space integrals, i.e., you define Green's functions etc. as

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

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

    [tex]p \cdot x=p_{\mu} x^{\mu}=p^0 x^0-\vec{p} \cdot \vec{x}.[/tex]

    Thus, the inverse reads

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

    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)

    [tex]\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}}},[/tex]

    with [itex]E_{\vec{p}}=\sqrt{m^2+\vec{p}^2}[/itex].

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


    But for this latter issue you have nearly as many conventions as textbooks :-(.
  5. Aug 10, 2011 #4


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    One representation of the delta function is
    [tex]\delta(x)=\frac{1}{2\pi}\int dk e^{ikx}.[/tex]
    If the integral appears in some equation without the [tex]\frac{1}{2\pi}[/tex],
    you get [tex]2\pi\delta[/tex].
    Last edited: Aug 10, 2011
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