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Confusion (8) from Weinberg's QFT.(Lippmann-Schwinger eqn)

  1. Nov 3, 2012 #1
    I was reviewing the first few chapters of Weinberg VolI and found a hole in my understanding in page 112, where he tried to show in the asymptotic past [itex]t=-\infty[/itex], the in states coincide with a free state. In particular, he argued the integral [tex]\int d\alpha\frac{e^{-iE_{\alpha}t}g(\alpha)T_{\beta\alpha}^+\Phi_\beta}{E_\alpha-E_\beta+i\epsilon}\ldots(1)[/tex] would vanish, where [itex]d\alpha=d^3\mathbf{p}[/itex](also involves discrete indices like spin, but of no relevance here). In his argument, he used a contour integration in the complex [itex]E_\alpha[/itex] plane, in which the integral of central interest is the integration along real line [tex]\int_{-\infty}^\infty dE_\alpha\frac{e^{-iE_{\alpha}t}g(\alpha)T_{\beta\alpha}^+\Phi_\beta}{E_\alpha-E_\beta+i\epsilon}\ldots(2)[/tex].
    I don't see how to obtain (2) from (1), since the lower bound of energy is the rest mass, in the best case I could get something like [itex]\int_{m}^\infty dE_\alpha\cdots[/itex], but how could one extend this onto the whole real line.
     
    Last edited: Nov 3, 2012
  2. jcsd
  3. Nov 3, 2012 #2

    Bill_K

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    Well I think the answer is that he's not really integrating over dE, he's integrating over dα, which includes an integral over dp. E(p) is treated as just a variable that depends on p.

    The integral over dp goes from 0 to ∞, which can be extended by symmetry to -∞. Then on the large semicircle, since both p and E ≈ pc go to infinity, he can make the argument that eiEt vanishes.
     
  4. Nov 4, 2012 #3
    But in most cases there is no such symmetry because of [itex]g(\alpha)[/itex]
     
  5. Nov 5, 2012 #4
  6. Nov 5, 2012 #5

    Bill_K

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    I take it back. No extension is necessary. The integral is ∫ dα ≡ ∫-∞dpx-∞ dpy-∞dpz. It already extends along the entire real axis in each dimension, and all that needs to be done is to close the contour with a semicircle.

    Repeat: it is not ∫ dEα, as you have it written. E is not the integration variable.
     
  7. Nov 5, 2012 #6
    But then it is hard to see how to apply residue calculus since the denominator is in terms of energy. If we want we can convert the momentum integral into energy integral, since [itex]d^3\mathbf{p}=p^2\sin\theta dpd\theta d\phi[/itex], and [itex]E_\alpha=\sqrt{p^2+m^2}[/itex], since by definition [itex]E_\alpha[/itex] is the energy of a free particle labeled by [itex]\alpha[/itex]
     
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