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What is the principal part in this identity?

  1. Mar 5, 2010 #1

    I'm reading a book at the moment in which the author states the identity:

    [tex]\frac{1}{x-i\epsilon}=\frac{x}{x^2+\epsilon^2}+\frac{i\epsilon}{x^2+\epsilon^2} [/tex]

    Which is fine, but then he goes on to state that this is equal to:
    [tex] P\frac{1}{x}+i\pi\delta(x) [/tex]

    Where P is the principal part. I think from googling the principle part means the sum of negative power terms in the Laurant expansion, yet I still have no clue as to how to get to this line.

    Thanks for any help at all
  2. jcsd
  3. Mar 6, 2010 #2
    Anyone? I cant find this any books, is this the wrong part of the forum for such a question?
  4. Mar 6, 2010 #3
    What is [itex]\delta[/itex]?
  5. Mar 7, 2010 #4
    I believe it's the Dirac delta function. Maybe the principal part isn't the sum of negative terms, more googling tells me it may instead be the normally called the "principal value", yet I still don't understand this identity.
  6. Mar 7, 2010 #5

    Ben Niehoff

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    One way to derive the identity is to use contour integration. You want to find

    [tex]\int_{-\infty}^{\infty} \frac{f(x) \; dx}{x - i\varepsilon}[/tex]

    for a generic function f(x). Then you can find what

    [tex]\frac{1}{x - i\varepsilon}[/tex]

    means, in a distribution sense. Look up distributions if you don't know what I'm talking about.

    Try to see if you can come up with a clever contour that allows you to compute this result.
  7. May 4, 2010 #6
    Sorry, I've just come back to this after seeing it again in another chapter of Srednicki, I never did get round to solving why this identity holds.

    My thoughts at the moment are, looking at:

    \int_{-\infty}^{\infty} \frac{f(x) \; dx}{x - i\varepsilon}
    This would normally have a pole at x=0, if it weren't for the [tex] i\epsilon[/tex]. Now the pole occurs offset from the real line, at [tex] x=+i\epsilon[/tex]. So assuming [tex]f(x)\rightarrow 0[/tex] fast enough, and Jordan's lemma holds, we could replace the whole integral by a semi circle in the upper half compex plane.

    Assuming this is the way to do it, then by residue theorem:
    \int_{-\infty}^{\infty} \frac{f(x) \; dx}{x - i\varepsilon}=2\pi i\times {f(i\epsilon)}

  8. May 4, 2010 #7
    I'd like to take a stab at that: I think it refers to the limit of the integral of that expression as [itex]\epsilon\to 0[/itex]:

    [itex]\lim_{\epsilon\to 0^{+}} \int_{-\infty}^{\infty} \frac{dx}{x-i\epsilon}[/itex]

    and it's not hard to use an analytic branch of log in the lower half-plane to show that limit goes to [itex]\pi i[/itex] and since the integral of the Dirac delta function in the interval is one, we could represent the limit using the delta function:

    [itex]\lim_{\epsilon\to 0^{+}} \int_{-\infty}^{\infty} \frac{dx}{x-i\epsilon}=P.V. \int_{-\infty}^{\infty}\left(\frac{1}{x}+\delta(x)\pi i\right)dx=\pi i[/itex]
  9. May 4, 2010 #8
    Thanks for the reply, but I'm afraid I don't understand your answer, could you maybe break it down for the uninitiated in all things to do with the principal value and distributions.
  10. May 4, 2010 #9
    I just tried to interpret your result. Not sure that's what you want. The integral is just a contour integral [itex]\mathop\int\limits_C \frac{1}{z}dz[/itex] along the real axis with a positive-oriented indentation around the origin. You'd have to understand what the Cauchy Principal-value means as opposed to the ordinary Riemann integral. Essentially, the part on the negative axis cancels the part on the positive axis with the integral over the indentation, as it's radius goes to zero, just being i times the angular distance around the contour. I just evaluated it using a limit over a horizontal contour that gradually rose up to the real axis. You can do that using an analytic branch of the antiderivative [itex]\log(x+i\epsilon)[/itex] to show the limit is just [itex]\pi i[/itex]. You really don't need the delta function. I just expressed it that way because you used it in your post and also recall the integral over the real axis of the Dirac delta function is one.
  11. May 17, 2010 #10
    In first place, sorry because my english is very bad. The second term
    [tex]\frac{1}{x-i\epsilon}=\frac{x}{x^2+\epsilon^2}+\frac{i\epsilo n}{x^2+\epsilon^2} [/tex]
    behaves like a delta(x) function, in the limit of epsilon going to zero. It is easy to check this with mathematica. If you integrate it with respect x from minus to plus infinity, you obtain
    [itex]i \pi[/itex]. On the other hand, if you integrate this expression plus a certain function, the result is the value of this function at zero plus i*pi. The calculation of lahlh is not correct because the integration along the semicircle is not zero, in spite of the fact that the radius of the circle goes to infinity. If this were the case, integration along a complete circle must be zero, and it is not the case due to the pole.
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