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Integrals in the derivation of the Optical Theorem

  1. Feb 12, 2009 #1
    Hi everyone,

    I'm trying to work out all the intricate details of the limits involved in computing the integrals leading to the Optical Theorem in nonrelativistic Quantum Mechanics. In doing so, I have arrived at the following integrals which are to be evaluated in the [itex]r \rightarrow \infty[/itex] regime:

    [tex]
    r^{2}\int_{0}^{2\pi}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta)d\phi = I_{1}
    [/tex]
    [tex]
    r^{2}\int_{0}^{2\pi}\int_{-1}^{+1}f^{*}(\theta)e^{-ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)+\frac{1}{r^2}\right)d(\cos\theta)d\phi = I_{2}
    [/tex]

    Modulo the [itex]\phi[/itex] integral, the expression for [itex]I_{1}[/itex] is

    [tex]r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta) = r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)\right)d(\cos\theta) - \int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}d(\cos\theta)[/tex]

    I am stuck at these integrals. I know what the "final" term in the optical theorem is, but without using the asymptotic solution, I don't know the right direction to proceed in. Specifically, if I consider [itex]f(\theta)(1+\cos\theta)[/itex] as the "first" function in the first integral, then integrating by parts I get

    [tex]\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}(1+\cos\theta)d(\cos\theta) = \left[f(\theta)(1+\cos\theta)\frac{e^{ikr(1-\cos\theta)}}{-ikr}\right]_{-1}^{+1} - \int_{-1}^{+1}\frac{e^{ikr(1-\cos\theta)}}{-ikr}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)} + f(\theta)\right)d(\cos\theta)[/tex]

    Therefore,

    [tex]r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)\right)d(\cos\theta) = ikr\left[\mbox{b.d.}\right] + \int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)} + f(\theta)\right)d(\cos\theta)[/tex]

    where b.d. denotes the boundary term in the second last equation. Put together, these equations give

    [tex]r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta) = ikr\left[\mbox{b.d.}\right] + \int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)}\right)d(\cos\theta)[/tex]

    Clearly the boundary term tends to zero as [itex]r \rightarrow \infty[/itex], so my original question reduces to the following question:

    How does one evaluate the following integrals

    [tex]\int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)}\right)d(\cos\theta)[/tex]
    [tex]\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}d(\cos\theta)[/tex]

    in the [itex]r \rightarrow \infty[/itex] regime? I want to get some more insight into the physics as [itex]r \rightarrow \infty[/itex] so I'm hoping that an explicit evaluation of all these terms will be helpful.

    Thanks.
     
    Last edited: Feb 12, 2009
  2. jcsd
  3. Feb 13, 2009 #2
    Someone please help...
     
  4. Feb 14, 2009 #3
    Doesn't the saddle point method work? If r is very large, then the contribution to the integral will come from a narrow range of theta near the point where the aregument of the exponential is extremal.

    So, you expand i k r (1-cos(theta)) around its extremal values and then, for large r, you end up with Gaussian integrals.
     
  5. Feb 15, 2009 #4
    That seems intuitive. But I don't know how to rigorously justify it. Is this the saddle point method?

    (I can plug in the asymptotic form of the spherical bessel functions in

    [tex]e^{ikr\cos\theta} = \sum_{l=0}^{\infty}(2l+1)i^{l}j_{l}(kr)P_{l}(\cos\theta)[/tex]

    and get the limit.)
     
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