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Double Delata Function Potential

  1. Feb 13, 2006 #1
    Off the top of their head, does anyone know how many bound states the potential v(x)=-c[d(x+a)+d(x-a)] might have?

    I went through the problem as follows: got a growing exponential to the left of -a, got a growing plus a decaying exponential inbetween -a and a, and a decaying exponential to the right of a.

    I set them equal to eachother at the delta barriors (aka the one left of -a equal to the one inside, and the one right of a equal to the one inside).

    I then solved the schrodinger equation for both of the barriors, and then normalized the three components.

    With this set of equations i did 2pgs of alegbra lol, and finally got a E equal to -[(16ma^3)/(2h^2)+4a+(h^2)/(2m))

    i dont knwo how to interpret this energy for the number of bound states, because its always negative (i think). So either its all wrong :( or that is the one and only state, or what? i really dont know if my strategy or my excecution is right, Please help!!!
     
  2. jcsd
  3. Feb 14, 2006 #2

    Gokul43201

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    I don't recall getting analytical solutions to delta-potentials.

    The forms of your wavefunctions are correct. So, you should have 4 arbitrary coefficients (correct?). You also have the parameter 'c' for the strength of the potentials. And you have 4 equations from the BCs. If you solved the equations correctly, you'd get the even and odd solutions (ie : you would end up with something like [itex]A^2 = B^2 \implies A = \pm B[/itex], where A,B are the coefficients of the wavefunction between the two spikes) as expected from the symmetry of the problem. Plugging these two cases back in will give you two transcendental equations. You can solve these graphically, find the solutions, and check if they give rise to normalizable wave functions.

    PS : According to what you've written above, there's only one bound state energy. This may very well be troe, if one of the two solutions gives something unphysical (and needs to be thrown away).
     
    Last edited: Feb 14, 2006
  4. Feb 14, 2006 #3
    Sorry my E= -[(16mc^3)/(2h^2)+4c+(h^2)/(2m)) where c is actually labled alpha in the problem (i dont know how to do that fancy stuff lol so i just choose a random constant and forgot what i labled it)

    Yes i have four coefficients, which ended up pairing off. I got the result that the two tied up in the space -a<x<a were equal, and the two to the right and left of a, and negative a (respectivly) were equal.

    Im not sure what you mean by transcendental equations, and i managed to solve directly not graphically, which is nice if i did it right :)

    My solutions to the scrodinger equation looked like this: Bke^(-ka)-Cke^(ka)-Ake^-ka)=(-2mc)/h^2*(wavefn(-a)+wavefn(a)), (this was at -a). After i solved for the constants, i just chose one of my region specific functions, and plugged in -a and a respectivly for the values of wavefun(-a) and wavefun(a). and i also plugged in for the constants B,C,and A you see up there. This is what led to my energy value, because k=sqrt(-2mE)/h
     
    Last edited: Feb 14, 2006
  5. Feb 14, 2006 #4
    And can someone tell me how to do the graphic equation writing. (lol no clue how to describe it)

    The A^2=B^2 stuff in Gokul's post
     
  6. Feb 15, 2006 #5
    Due thursday :)
     
  7. Feb 15, 2006 #6

    Gokul43201

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    Haven't read your earlier post yet...I'm still at work and have little time.

    For LaTex formatting see : https://www.physicsforums.com/showthread.php?t=8997

    Also, to see the code for what I've written, just click the text (in my post) and a pop-up will display the code used to generate that text.
     
  8. Feb 15, 2006 #7

    Gokul43201

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    Are you sure equality is the only possibility you arrived at ? Like I said before, you should have gotten that their squares are equal. This gives rise to two possible solutions.
     
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