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An identity

  1. Oct 5, 2008 #1
    [tex]\frac{1}{x-i\epsilon}[/tex]=[tex]\frac{x}{x²+a²}[/tex]+[tex]\frac{ia}{x²+a²}[/tex]= P [tex]\frac{1}{x}[/tex]+i pi [tex]\delta[/tex](x)

    P means the principal value, a is possibly infinitesimal (?), i is the imaginary unit

    Where does the pi, Dirac delta come from? What principal value?
    It is from a quantum field theory book.
     
    Last edited: Oct 5, 2008
  2. jcsd
  3. Oct 5, 2008 #2

    Mute

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    Homework Helper

    The expression contains a Dirac Delta, so that should be your clue to integrate against a test function, [itex]\varphi(x)[/itex] that goes to zero as [itex]x \rightarrow \pm \infty[/itex]. In particular, you want to consider a contour integral

    [tex]\int_\gamma dz \frac{\varphi(z)}{z - i\epsilon} [/tex]

    about some contour [itex]\gamma[/itex]. Note that the integrand has a pole at [itex]z = i\varepsilon[/itex]. To evaluate, consider the closed contour in the upper half plane that consists of a large semicircle of radius R with a small semicircle of radius b around the pole in the integrand at [itex]z = i\varepsilon[/itex]. Since the contour contains no poles, the contour integral is zero. Parameterize the contour:

    [tex]\int_{-R}^{-b}dx \frac{\varphi(x)}{x - i\epsilon} + \int_{b}^{R}dx \frac{\varphi(x)}{x - i\epsilon} + \int_0^{\pi}d(Re^{i\theta}) \frac{\varphi(Re^{i\theta})}{Re^{i\theta}-i\varepsilon} + \int_\pi^{0}d(be^{i\theta}) \frac{\varphi(be^{i\theta})}{be^{i\theta}-i\varepsilon} = 0[/tex]

    Now, take the limits as [itex]R \rightarrow \infty[/itex] and [itex]b \rightarrow 0[/itex], and use the definition of the principal part integral:

    [tex]\mathcal{P}\int_{-\infty}^{\infty}dx~f(x) = \lim_{a \rightarrow 0} \int_{-\infty}^{-a}dx~f(x) + \int_{a}^{\infty}dx~f(x)[/tex]

    and the half-residue theorem, that states that for a semi-circular arc about a pole, the contribution from that arc about the pole as the radius goes to zero is [itex]2\pi i[/itex] times half the residue about that pole. To get the delta function you note that at the pole the test function will give a contribution [itex]\varphi(0)[/itex], and so the action of the imaginary part of [itex](x - i\epsilon)^{-1}[/itex] is to pick out the value of the test function at the origin, which is just what a Dirac delta does. In the end you can set [itex]\epsilon = 0[/itex].
     
  4. Oct 5, 2008 #3
    Mute, thanks so much!!
     
  5. Oct 5, 2008 #4
    WAIT!!

    In my OP there was a mistake! I'm very sorry.

    It is x/(x²+a²)+ia/(x²+a²) = P 1/x + i pi delta(x) and a possibly infinitesimal.

    So RHS are two innocent algebraic terms, the LHS has something to two with integration.

    I still can't see how that works.
     
  6. Oct 5, 2008 #5

    Mute

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    The [itex]\epsilon[/tex] is an infinitesimal. It's there to enforce causality. You can set it to zero after you do your integrals over the left hand side because after that causality has been enforced and you don't need the infinitesimal anymore.

    The right hand side of the expression only makes sense inside an integral. What it says is that when you integrate

    [tex]\int_{-\infty}^{\infty}dx \frac{\varphi(x)}{x - i\epsilon}[/tex]

    it's equivalent to integrating

    [tex]\int_{-\infty}^{\infty}dx \left[\mathcal{P}\left(\frac{1}{x}\right) + i\pi \delta(x)\right]\varphi(x) = \mathcal{P}\left[\int_{-\infty}^{\infty}dx~\frac{\varphi(x)}{x}\right] + i\pi \varphi(0)[/tex].
     
  7. Oct 6, 2008 #6
    I really appreciate the effort Mute, but I can't see what you mean here.
     
  8. Oct 7, 2008 #7
    I wrote the author and he was so kind answering me. If anybody interested here is what he wrote.

    After comparing that with what Mute said, I think that I understand it now.
     
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