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Dirac delta potential

  1. Nov 25, 2009 #1
    why in the problem of dirac delta potential, the integral

    [tex]\int^{\epsilon}_{-\epsilon}\phi''(x)dx[/tex] is equal to [tex]\phi'(\epsilon)-\phi'(-\epsilon)[/tex]???

    but [tex]\int^{\epsilon}_{-\epsilon}\phi(x)dx[/tex] is equal to 0

    if, for example[tex]\phi(x)=e^x[/tex]

    then [tex]\phi(x)''=\phi(x)[/tex]

    but, the firts integral is [tex]e^{\epsilon}-e^{-\epsilon}[/tex]
    and the second integral would be to zero

    i dont understand
     
    Last edited by a moderator: Nov 25, 2009
  2. jcsd
  3. Nov 25, 2009 #2

    George Jones

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    Do you know why this is required?
    If [itex]\phi[/itex] is continuous, this is true.

    In this situation, you want to work with wavefunctions that satisfy the above two conditions.
    If [itex]\phi \left( x \right)=e^x[/itex], then [itex]\phi \left( x \right)[/itex] does not satisfy the above two conditions, and thus [itex]\phi \left( x \right)=e^x[/itex] is not a possible wavefunction.
     
  4. Nov 25, 2009 #3
    then :
    ¿ only if: [tex]\phi(x)[/tex] is continuous then [tex]\int^{\epsilon}_{-\epsilon}\phi(x)dx=0???[/tex]

    if [tex]\phi(x)[/tex] is discontinuous then the integral is not zero???
     
    Last edited: Nov 25, 2009
  5. Nov 26, 2009 #4

    George Jones

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    There is a subtle difference between this and what I wrote. I did not write "only if", I wrote "if".
    I am not sure what you mean.

    Consider a function somewhat related to the problem, [itex]\phi \left( x \right) = \left| x \right|[/itex]. What are: [itex]\phi' \left( x \right)[/itex]; [itex]\phi'' \left( x \right)[/itex];

    [tex]\lim_{\epsilon \rightarrow 0} \int^\epsilon_{-\epsilon} \phi'' \left( x \right) dx?[/tex]
     
  6. Nov 26, 2009 #5
    I think:

    [tex]\int^{\epsilon}_{-\epsilon} \phi''(x) =\phi'(x)|^{\epsilon}_{-\epsilon}[/tex]

    [tex]=(\theta(x)-\theta(-x))|^{\epsilon}_{-\epsilon}[/tex]

    Heaviside function:
    [tex]\theta(x)=1; x>0[/tex] [tex]\theta(x)=0; x<0[/tex]

    then
    [tex]\int^{\epsilon}_{-\epsilon} \phi''(x) =2\theta(\epsilon)-2\theta(-\epsilon)=2-0=2[/tex]

    I'm not sure.............
     
  7. Nov 26, 2009 #6
    [tex]\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0[/tex]

    [tex]V(x)=-\alpha\delta(x),\alpha>0[/tex]

    You will have two region; region 1 for [tex]x<0[/tex] and region 2 for [tex]x\geq 0[/tex]. In the boundary of this two regions wave function must me continuous. Integrate equation

    [tex]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)-\alpha\delta(x)\psi(x)=E\psi(x)[/tex] from [tex]-\epsilon[/tex] to [tex]\epsilon[/tex] you will get

    [tex]\psi'(0^+)-\psi'(0^-)=-\frac{2m\alpha}{\hbar^2}\psi(0)[/tex]
     
  8. Nov 26, 2009 #7
    thanks , but my main question is about the properties of integrals:

    ¿¿when

    [tex]\lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon} \phi(x)'' dx,[/tex][tex]\lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon} \phi(x)' dx,[/tex][tex]\lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon} \phi(x) dx,[/tex]

    are zero or non zero???
     
  9. Nov 27, 2009 #8

    DrDu

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    That is a question of definition. The delta function alone is not a well defined operator.
    However, the kinetic energy operator with a delta function can be shown to be one.
    To define an operator, you have to specify the domain of functions on which it is defined. In one dimension you can either choose psi to be continous with discontinous first derivative, which then is called a delta function type potential, or to be discontinous, in the latter case, one speaks of a delta' interaction.
     
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