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Scattering theory - Regular solution

  1. Nov 23, 2011 #1
    Hello physicsforum, this is my first post

    I am reading up on scattering theory and I'm having difficulty rederiving some of the results.
    In 'Inverse problems in Quantum Scattering theory' (2nd Ed.) by Chadan and Sabatier they state that the solution for the s-wave regular solution, which is defined by the boundary condition [itex]{\phi(k,0) = 0}[/itex] and [itex]{\phi'(k,0) = 1}[/itex], is

    [itex]\phi(k,r) = \frac{sin(kr)}{r} + \int^r_0 \frac{sink(r-r')}{k} V(r')\phi(k,r')dr'[/itex].

    It is clear to me that this is a solution of the reduced radial Schrodinger equation for l=0, and I also follow the argument in the book that establishes convergence. But they state that this solution follows from 'the variation of the constants of Lagrange'. As far as I know this is simply the variation of parameters used to solve nonhomogeneous linear differential equations. I have tried to solve the radial equation (for l=0) using this approach but it doesn't work. 'Scattering theory of waves and particles' by Newton gives the same result using the more sophisticated approach of Green's functions, with which I am not very comfortable yet.

    I could sketch my (incorrect) attempt using variation of parameters but I think a few comments should make it clear why it fails.

    Generally, for a 2nd order DE of the type

    [itex]\phi'' + p(r)\phi' + q(r)\phi = f(r)[/itex]

    we look for solutions of the form [itex]\phi_p = u_1(r)\phi_1(r) + u_2(r)\phi_2[/itex]

    where [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are solutions of the homogeneous equation

    [itex]\phi'' + p(r)\phi' + q(r)\phi = 0[/itex].

    The problem is... the reduced Schrodinger is a homogeneous equation. So obviously the method can't work. I managed to get something like the correct solution by pretending that [itex]f(r) = V(r)\phi(r)[/itex] (in which case the homogeneous equation is the harmonic oscillator) but the procedure is mathematically flawed. As a reminder, the reduced radial Schrodinger equation for s-wave is

    [itex]\phi'' + k^2\phi=V(r)\phi[/itex].

    The only restriction on the potential is that [itex]\int_b^∞ |V(r)|rdr<\infty[/itex] where [itex]b≥0[/itex]

    I would appreciate suggestions as to the proper method to approach this problem. Any ideas? (This is NOT a homework problem)
     
  2. jcsd
  3. Nov 23, 2011 #2

    Bill_K

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    Science Advisor

    personperson, You seem to have all the right ideas. My advice would be to find an easier book, especially if you're new to Green's functions. The two books you mention are quite sophisticated, useful as references but not IMO as introductions to the subject. Almost any book on Quantum will have a chapter on scattering.
     
  4. Nov 24, 2011 #3
    Thank you for your reply. I am already somewhat familiar with scattering theory. The stated book seems to suggest that there is a way to solve the equation using variation of parameters rather than Green's functions, but it seems to me at the moment that since the equation is similar to the Helmholtz equation and because of the problem I mentioned earlier I will have to stick with Green's functions. The question arose because I want to generalise the result to two coupled fields with a scattering equation (in one spatial dimension) like

    [itex] (\partial_x^2 \mathbb{I}_2 +
    \left( \begin{array}{cc}
    k^2 & 0 \\
    0 & \tilde{k}^2 \\
    \end{array} \right) )

    \left( \begin{array}{c}
    \psi \\
    \phi \\
    \end{array} \right )

    = V\left( \begin{array}{c}
    \psi \\
    \phi \\
    \end{array} \right )
    [/itex]

    with a solution like


    [itex]\left( \begin{array}{c}
    \psi(x) \\
    \phi(x) \\
    \end{array} \right )

    = \left( \begin{array}{cc}
    sinkx & 0 \\
    0 & sin\tilde{k}x \\
    \end{array} \right)

    +\left( \begin{array}{cc}
    1/k & 0 \\
    0 & 1/\tilde{k}^2 \\
    \end{array} \right)
    \int_0^x dx'
    \left( \begin{array}{cc}
    sink(x-x') & 0 \\
    0 & sin\tilde{k}(x-x') \\
    \end{array} \right)
    V(x')\left( \begin{array}{c}
    \psi(x') \\
    \phi(x') \\
    \end{array} \right )
    [/itex]


    You can sort of already see that this is probably a solution. I wanted to make sure that I understand any subtleties that arise from the simpler case before I continued.
     
  5. Nov 24, 2011 #4
    I don't think it's really a solution, it's something you get after converting schrodinger differential equation into an integral equation, since both sides contain [itex]\phi[/itex]. However variation of parameters, if I remember correctly, would give you an explicit solution to the ODE, so it'll surprise me if variation of parameters works.
     
  6. Nov 24, 2011 #5
    It is a solution in the sense that

    a) if you substitute it on the left hand side of the reduced radial Schrodinger equation I wrote in the first post you get the right hand side. It's exact form depends on the potential V, but that doesn't not make it a solution

    b) it is called the regular solution in the reference ('Inverse scattering...')

    And yes, as I wrote, variation of parameters doesn't seem to work unless you cheat, so my question is why the reference says you can use it to get the result when you apparently can't i.e. what do they mean? In the mean time I have found a different definition of the 'variation of the constants' that involves resolvents but I think that is better posted in the math forums.
     
  7. Nov 24, 2011 #6

    Bill_K

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    Science Advisor

    You are using the term "solution" in a way with which I am previously unfamiliar. :smile: "Solution" in the normal sense means I give you V(x) and you tell me φ(x). The fact that you can substitute it into the radial equation and get an identity does not mean you have solved anything, it simply means it is an equivalent reformulation.

    As Newton's book point out, this is an integral equation, a particular instance of the Lippman-Schwinger equation.
     
  8. Nov 24, 2011 #7
    I see your point, I'm not sure about the answer, and I'm pondering if letting [tex]f(r)=V(r)\phi[/tex] is really "mathematically flawed".
     
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