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Triple delta potential

  1. Apr 17, 2014 #1
    1. The problem statement, all variables and given/known data

    I'm looking for the bound energy of a triple delta potential:

    [tex]V(x) = -w \left [ \delta(x-a) + \delta(x) + \delta(x+a) \right ][/tex]

    What is the correct transcendental equation for kappa?

    2. Relevant equations

    My wave function is [tex]\psi_1(x) = A e^{\kappa x}[/tex] for x < -a, [tex]\psi_2(x) = B \cosh(\kappa(x+a/2)[/tex] for -a < x < 0, [tex]\psi_3(x) = C \cosh(\kappa(x-a/2))[/tex] for 0< x < a, [tex]\psi_4(x) = D e^{-\kappa x}[/tex] forx > a.

    We use this continuity formula [tex]\psi'( z+\epsilon) - \psi'(z-\epsilon) = -\frac{2mw}{\hbar^2} \psi(z)[/tex]

    3. The attempt at a solution

    Calculating the continuity formula at x = 0 gives [tex]\kappa \tanh(\frac{\kappa a}{2}) = \frac{m w}{\hbar^2}[/tex]

    This means, you get the same bound energy as for one delta potential if a is very large, and something weird for small a?
     
    Last edited: Apr 17, 2014
  2. jcsd
  3. Apr 17, 2014 #2

    mfb

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    Very large a just mean your particle is in the middle and the other parts are not important.

    Shouldn't there be a second equation from the second potential? And I think your approach for the inner wavefunctions is not general enough to find all solutions.
     
  4. Apr 18, 2014 #3
    Is my equation for x = 0 correct? For x = a or x = -a, you get the same transcendental equation as for the double-delta potential?

    [tex]1 + \tanh(\frac{\kappa a}{2}) =\frac{2 m w}{\hbar^2 \kappa}[/tex]
     
    Last edited: Apr 18, 2014
  5. Apr 18, 2014 #4

    mfb

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    For your chosen wavefunctions, it looks fine.

    Up to factors of two maybe, but you have to check your ansatz for the wavefunctions.
     
  6. Apr 18, 2014 #5
    So after all, you get two different bound energy values. One for the delta at x = 0 and one for two deltas at x = -a and x = a.
     
    Last edited: Apr 18, 2014
  7. Apr 18, 2014 #6

    mfb

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    That just shows your wavefunction is not general enough. The bound energy states have to satisfy both equations at the same time.
     
  8. Apr 18, 2014 #7
    Now I see where I did wrong.

    The correct wave function is:
    [tex]\psi_1(x) = A e^{\kappa x}[/tex]
    for x < -a,

    [tex]\psi_2(x) = B e^{\kappa x} + C e^{-\kappa x}[/tex]
    for -a < x < 0,

    [tex]\psi_3(x) = D e^{\kappa x} + E e^{-\kappa x}[/tex]

    for 0< x < a,

    [tex]\psi_4(x) = F e^{-\kappa x}[/tex]
    for x > a.
     
  9. Apr 18, 2014 #8

    mfb

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    That is better, right.

    I would use ##\psi_1(x) = A e^{\kappa (x+a)}## and similar expressions to make the coefficients easier to evaluate, but that is a detail.
     
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