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Reflectionless Potential Eigentates

  1. Jul 2, 2013 #1
    I know that the bound eigenstates for the reflectionless potential (Poschl-Teller potential) is



    [itex]P^{\lambda}_{\mu}[/itex] are the associated Legendre polynomials and [itex]\lambda[/itex] is a positive integer while [itex]\mu[/itex] is an integer able to take on values from [itex]\lambda, \lambda-1, ... , 1[/itex]

    Is this the same equation for unbound states. For example, if [itex]\lambda=1.1[/itex], would I be able to use the top equation or is a different equation be necessary?

    I'm attempting to compare the eigenstates for [itex]\lambda=1[/itex] to [itex]\lambda=1.1[/itex] as seen on:


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  3. Jul 2, 2013 #2


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    Have you tried reading some source ? Problems nr. 38 and 39 of Fl├╝gge should give the whole solution.
  4. Jul 2, 2013 #3
    I've looked at Flugge, but, unfortunately Google Books is missing a few crucial pages and my local university library does not carry it. I've requested it via interlibrary loan, but i do not know how long it will take to send to the University.

    I've looked at other references, but many of them are too advanced for me.

    I'm pretty sure the best way to compare [itex]\lambda=1[/itex] to [itex]\lambda=1.1[/itex] would be to solve the Schrodinger equation. What would be the best way to solve that equation (scaling [itex]2m=\hbar=1[/itex])? When I solve it in Mathematica or Maple, I get an answer that seems to make sense, but how do I evaluate the constants? According to Reflectionless eigenstates of the sech(x)2 potential by Lekner, he gives the constants for the first three bound states, but does not explain where he got them from (equations 18-20).


    ([itex]\lambda=1[/itex] is shown)

    Attached Files:

    Last edited: Jul 2, 2013
  5. Jul 2, 2013 #4


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    Lekner's paper: http://www.victoria.ac.nz/scps/about/staff/pdf/Reflectionless_eigenstates.pdf

    Lekner explains what to do just after his eq.(51): in eq.(3), replace k by iq, where E=-q^2. E is now quantized (and hence so is q) by eq.(51). There is a solution that goes to zero for both [itex]x\to-\infty[/itex] and [itex]x\to+\infty[/itex] only for these values of E.

    Eqs.(18-20) are scattering states with positive energy, not bound states with negative energy.

    Also, your mathematica result has an error: the first argument of LegendreP and LegendreQ should be j, not 1. Lekner's solution can be expressed in this form for some particular values of C[1] and C[2].

    EDIT: I've realized that I'm not sure what you're asking for. Changing [itex]\lambda[/itex] changes the potential, and there are both scattering and bound states for all values of [itex]\lambda[/itex].
    Last edited: Jul 2, 2013
  6. Jul 3, 2013 #5
    I'll try to explain my question better.

    I understand how Lekner calculates the odd and even eigenstates for [itex]\nu=0,1,2[/itex] from equation 1. Lekner then states that reflectionless energy eigenstates can be formed from the superposition of the odd and even eigenstates. I'm not sure how he managed to determine the three constants, [itex]ika[/itex], [itex]\frac{i[1+(ka)^{2}]}{ka}[/itex], and [itex]\frac{ika[4+(ka)^{2}]}{ka}[/itex].

  7. Jul 3, 2013 #6


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    In eqs.(8-17), Lekner uses properties of hypergeometric functions to express the even and odd positive-energy eigenfunctions in terms of elementary functions for [itex]\nu=0,1,2[/itex]. Then he finds the linear combinations that approach [itex]e^{ikx}[/itex] as [itex]x\to+\infty[/itex]; this requirement is how he determines those three constants. Because the potential is reflectionless for integer [itex]\nu[/itex], this linear combination also approaches [itex]e^{ikx}[/itex] as [itex]x\to-\infty[/itex].

    For non-integer [itex]\nu[/itex], the potential is not reflectionless. In that case, the best you can do is find a linear combination of even and odd eigenfunctions that approaches [itex]e^{ikx}[/itex] as [itex]x\to+\infty[/itex], and approaches [itex]Ae^{ikx}+Be^{-ikx}[/itex] as [itex]x\to-\infty[/itex], for some constants [itex]A[/itex] and [itex]B[/itex]. Conservation of probability flux can be used to show that [itex]|A|^2-|B|^2=1[/itex].

    For non-integer [itex]\nu[/itex], you will need to find the appropriate asymptotic forms of the hypergeometric functions in order to do this analysis.
  8. Jul 3, 2013 #7
    Thank you so much! That really helps me.
  9. Jul 23, 2013 #8
    I've recently obtained Flugge's "Practical Quantum Mechanics." I've been going through his solution for Problem 39, but I'm having trouble getting equation (39.14) to match up with equations (29.10a-b) in the same graph. I've attached my Maple worksheet. It's interesting that for v=1, the plots match up exactly, but begins to mismatch by v=2.

    Any suggestions? Am I missing something important?


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