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Scattering problem process

  1. Aug 4, 2012 #1
    Hi everyone, I'm trying to get my head around scattering still. The books I've been looking at derive it, but then kind of skip some steps when they actually do problems with it. This is long but I'd really love it if someone could help me understand!

    From what I can tell, this seems to be the general formalism Shankar uses:
    1. Expand f (the scattering amplitude) in terms of Legendre polynomials:
    [tex]f(\theta,k) = \sum _{l = 0} ^\infty (2l + 1)a_l(k)P_l(cos\theta)[/tex]
    (l is angular momentum, k is wave number, a is coefficient)

    2. We expand the incoming wave like this too:
    [tex]e^{ikz} = e^{ikr cos(\theta)} = \sum _{l = 0} ^\infty i^l (2l + 1)j_l(kr)P_l (cos\theta)[/tex]

    3. We take the limit of the incoming wave as r → ∞ (using limit of j_l):
    [tex]\frac{1}{2ik}\sum _{l = 0}^{\infty} (2l + 1)(\frac{e^{ikr}}{r} - \frac{e^{-(ikr - l\pi)}}{r})P_l(cos(\theta))[/tex]

    4. We know that as r → ∞, the radial wave function has to become the same as the one for a free particle, so we take that limit but note that there can be a "phase shift" in it:
    [tex]R_l(r) = \frac{A_l sin[kr - l\pi/2 + \delta_l(k)]}{r}[/tex]

    5. Then, we basically combine these two, compare some coefficients, and get the form that the whole wave function should have as r → ∞:
    [tex]\psi_k(\vec{r}) = e^{ikz} + [\sum _{l = 0} ^\infty (2l + 1) (\frac{e^{2i \delta_l} - 1}{2ik}) P_l(cos\theta)]\frac{e^{ikr}}{r}[/tex]

    So this much makes sense to me, minus a few sketchy little things. But I can buy it. This is the formalism. Now, here's as much as I've gleamed about actually SOLVING a real problem:

    1. We're given the potential the electron is going to scatter off of.
    2. We solve the Schrodinger equation for that potential.
    3. We look at boundary conditions to get more info about some coefficients
    4. We take the limit as r → ∞ of this solution, and put the solution in the form of [itex]f(\theta,k)[/itex] like above.
    5. This gives us [itex]\delta_l[/itex].
    6. Now we have f, and we can solve for stuff like the cross section.

    But in the problems I've done, they don't explain it around step 4 very well, which is where it connects to the formalism. Here's a simple example from Shankar. It's a hard sphere such that [itex]V(r) = \infty[/itex] if r<r_0, and V = 0 otherwise (step 1).

    So, outside the potential, the particle is basically a free particle and has the form:

    [tex]R_l (r) = A_l j_l(kr) + B_l n_l(kr)[/tex]
    (step 2)

    But we need R = 0 at r = r_0, so that boundary condition demands that:

    [tex]\frac{B_l}{A_l} = -\frac{j_l(kr_0)}{n_l(kr_0)}[/tex]
    (step 3)

    As r → ∞, R becomes

    [tex]R_l(r) = \frac{1}{kr}[A_l sin(kr - l\pi/2) - B_l cos(kr - l\pi/2)] = \frac{\sqrt{A_l^2 + B_l^2}}{kr}[sin(kr - l\pi/2 + \delta_l)][/tex]
    (step 4)

    With [itex]\delta_l = tan^{-1}(-B_l/A_l)[/itex] (step 5?)

    From here, I'm pretty confused. I see how he did all this, but I'm not exactly sure what the equivalence between R_l and f is, which seems to be the point of all the formalism he did. He never explicitly says it. Also, in his final expression for f (basically taken from the r → ∞ expression above), he has

    [tex]f(\theta) = (1/k)\sum_{l = 0}^{\infty} (2l + 1)e^{i\delta_l} sin(\delta_l) P_l(cos\theta)[/tex]

    So it seems like each R_l is one of these terms...but then where are the (2l + 1) and [itex]e^{i\delta_l}[/itex] terms? I guess maybe we don't care about the exponent because you can always shift a function by an arbitrary (complex) phase shift?

    Can anyone help me out??

    Thanks!
     
  2. jcsd
  3. Aug 5, 2012 #2
    hello my friend. i am going to answer very specific questions, because you seem to be very lost and this is a very advanced area. whether you understand it or not depends on whether you have done a lot of previous work or not...
    The solutions that we dreamed up at the start for f, are only valid for a spherically symmetric potential. they are already the well known Legendre polynomials and do not need to be calculated. in his example, he takes the potential to be hard sphere. that is to say, you are throwing your particle at a ball, and your particle can absolutely not penetrate the barrier of the ball. outside the ball, theres no potential -> free particle solutions are valid -> hence the R.
    R_l is the solutions to the radial potential case. f is the scattering amplitude. by definition the solutions to the schrodinger equation (squared) are probability density functions, and since f is defined in θ space, it tells you the probability of the wave function evaluated at the potential barrier to be in the interval θ + dθ. if you take the probablity of the incident particle to be 1, along θ=0, then the radial solution gives the scattering amplitude of the outgoing particle. i.e. if the particle interacts with the potential, the scattered solution defines the scattering behaviour of the particle with 100% certainty
     
  4. Aug 5, 2012 #3
    Hi, thanks for the response!

    I guess my question is how they're sure what they get at the end actually determines the phase shift. In all the problems I get, they basically get R_l in the form of [itex](some stuff)\times sin(kr + (some extra term))[/itex], and then say "that extra term in the sin must be the phase shift".
     
  5. Aug 5, 2012 #4
    setting the LHS of 5 to 3 will give you the final solutions on your last step of workings, provided we put in R that we have deduced using boundary conditions in the LHS. unfortunately the notation in your book, for some reason does not include R in the wavefunction formulae. essentially the whole wavefunction is the given wavefunctions times R (the radial wavefunction). the phase is evaluated by requiring R_l to be linear combination of spherical Bessel functions, j and n. if you let A be cos(delta) and B sin(delta), then you arrive at the final phase equation.
    the radial wavefunction has a phase part which will become apparent as you rewrite the sine functions in exponential form. YOU MAY NOT shift the wavefunctions by an arbitrary phase in this case. the final scattering is sinosoidally dependant on phase of the system, which gives the wavelike behaviour of the outgoing particle (much like diffraction in optics). may I remind you that the wavefunction here is describing both the scatterer and the scattered particle combined, albeit the scatterer is essentially not affected by the incoming particle.
    I must say my initial answer was somewhat confused. Essentially we take the free particle solution, we then assume the same solution but phase shifted and add to it a spherical solution, and we solve for the phase and scattering amplitude.
    I recommend a more complete description from textbooks like Quantum Mechanics by L. Ballentine (chapter 16)
     
    Last edited: Aug 5, 2012
  6. Aug 7, 2012 #5
    Hrm, ok...I've gotta say, I'm still a little confused. I think I can go through and do the process, but the theory is still sketchy for me.
     
  7. Aug 8, 2012 #6
    If you want to know in detail how to solve these problems, you need to read about partial differential equations, which is a whole subject in maths. else you're going to have to believe in a lot of hocus pocus that the books throw at you. Memorising the key bits is the best way to pass the tests, or write a program to solve these things.
     
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