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Particle in a Central Potential: Calculate Energies

  1. Feb 2, 2014 #1
    Hello,

    This latest homework I have been doing has been very confusing to me and I have spent hours trying to complete it. Here is a problem that I really don't know where to start. If anyone could just point me in the right direction, or let me know if my ideas are correct or not, it would be so great!

    1. The problem statement, all variables and given/known data
    Consider a particle in a central potential of the form..

    [tex]V(r)=-U\frac{a}{r}e^{-\frac{r}{a}}[/tex]

    Where U>0 and a>0.

    (i) Show that, if ##U >> \frac{\hbar^2}{ma^2}##, in the zeroth order approximation the eigenkets and eigenenergies of the low-lying levels have the same form as in the Coulomb potential. Determine the eigenenergies to zeroth order.

    (ii) Compute all corrections to those eigenenergies in first order perturbation theory. Hint: Estimate the order of magnitude of second order corrections and compute all first order corrections that are greater.



    2. Relevant equations

    The problem says the following equations might be useful.

    [tex]<r^k>=\int_0^{∞}dr r^{2+k}[R_{nl}(r)]^2[/tex]
    [tex]<r>=\left( \frac{a_0}{2Z} \right)[3n^2-l(l+1)][/tex]
    [tex]<r^2>=\left( \frac{a_0^2 n^2}{2Z^2} \right)[5n^2+1-3(l+1)][/tex]
    [tex]<1/r>=\frac{Z}{n^2 a_0}[/tex]
    [tex]<1/r^2> = \frac{Z^2}{n^3a_0^2\left( l+\frac{1}{2} \right)}[/tex]


    3. The attempt at a solution

    For some reason, "zeroth order" is confusing me. We talked about calculating things to first and second order, and I am having some sort of "disconnect" as to what I am supposed to do. The only thing I can think of is the following.

    Expand the exponential term in the central potential function.

    [tex]V(r) \approx -U\frac{a}{r}\left[ 1-\frac{r}{a}+\frac{r^2}{2!a^2}+ \dots \right][/tex]

    The zeroth order would be only the first term. Meaning the potential function would become..

    [tex]V(r) \approx -U\frac{a}{r}[/tex]

    Which is of the same form as the Coulomb potential. I can then just make a substitution in the final results of the Coulomb potential and obtain the eigenenergies and eigenvectors. But I don't know where the ##U >> \frac{\hbar^2}{ma^2}## comes into play. Also, I don't understand the "Hint" in part (ii).

    If anyone could steer me in the right direction it would be great!!
     
  2. jcsd
  3. Feb 4, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I think "zeroth order" means the powers of r here, so the zeroth order of the potential includes the first order of the exponential function. Then U>>,,, should be a reasonable thing to look at.

    If you know the wavefunction (from (i)), you can let the second order of the potential (the term with r^2) act on this wavefunction. Numerical prefactors do not matter here I think, just find the correct order of whatever appears in the result (probably powers of ##\frac{\hbar^2}{ma^2U} \ll 1##), and then calculate everything that has a smaller power than that.
     
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