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Double delta function potential

  1. May 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the double delts-function potential
    [tex]
    V(x)=-\alpha[\delta(x+a)+\delta(x-a)]
    [/tex]
    How many bound states does this possess? Find the allowed energies for
    [tex]\alpha=\frac{\hbar^{2}}{ma^{2}}[/tex]and[tex]\alpha=\frac{\hbar^{2}}{4ma^{2}}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I divided the region into three parts x<-a(Region 1) ; -a<x<+a(Region 2) ; x>+a(Region 3)
    Since we consider bound states, E<0 and solving the SE yields
    [tex]Ae^{kx} (x<-a)[/tex]
    [tex]Be^{kx}+Ce^{-kx} (-a<x<a)[/tex]
    [tex]De^{-kx}(x>a)[/tex]
    where [tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex]
    Continuity at x=-a and x=+a respectively give
    [tex]A-B=Ce^{2ka}..............(1)[/tex]
    [tex]D-C=Be^{2ka}..............(2)[/tex]

    For this infinite potentials at points x=-a and x=+a,
    [tex]\Delta(\frac{d\psi}{dx})=-\frac{2m\alpha}{\hbar^{2}}\psi(\underline{+}a)[/tex]
    So these give two more BC

    [tex]A(1-\frac{2m\alpha}{k\hbar^{2}})=B-Ce^{2ka}................(3)[/tex]
    [tex]D(1-\frac{2m\alpha}{k\hbar^{2}})=C-De^{2ka}................(4)[/tex]
    So I tried to solve these (eqns 1 to 4)and what I got was A=D and B=C
    and taking A/B ratios from 1 and 3, i get [tex]ke^{4ka}=\frac{2m\alpha}{\hbar^{2}}[/tex]

    And Im not able to solve this equation for k..i'd be grateful for any help :)
     
    Last edited: May 22, 2008
  2. jcsd
  3. May 22, 2008 #2
    A breakthrough!!
    I tried graphical solution and its clear that there is only one bound state. But how do I calculate energy corresponding to that state without analytically solving this?
     
  4. May 22, 2008 #3

    malawi_glenn

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    so you obtained [tex] k [/tex]?

    [tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex] ?
     
  5. May 22, 2008 #4
    No thats where Im stuck . How do I solve this equation for k?

    [tex]
    ke^{4ka}=\frac{2m\alpha}{\hbar^{2}}
    [/tex]

    By taking exp(4ka) to the other side, we can graphically find that only one solution exists. But how can we calculate that solution without using computers and numerical methods?
     
    Last edited: May 22, 2008
  6. May 22, 2008 #5

    malawi_glenn

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    ah just solve it nummerical. Newton-Rhapson method is good.

    It is quite often that in physics we encounter equations that are not analytically solvable, so we have to solve them by nummerical methods. Same holds for integrals and differential equations.
     
    Last edited: May 22, 2008
  7. May 22, 2008 #6
    ok..thanks a lot!!

    actually i have not encountered ant problem in griffiths book SO FAR which requires numerical solution..so i was thinking if there is any way to solve it analytically. havent studued Newton-Raphson method yet..I'll use matlab may be for now..
     
  8. May 22, 2008 #7

    malawi_glenn

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    I am not 100% sure that a solution to x*exp(x) = konstant, exists analytically.

    Newton Raphson is something you do if you only have a pocket calculator. Matlab is easier if you just study on your own.

    Good luck!
     
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