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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.
     
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  3. Aug 10, 2011 #2

    dextercioby

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  4. Aug 10, 2011 #3

    vanhees71

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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.,

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

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

    clem

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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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