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Infinite well with delta well in the middle

  1. Sep 6, 2015 #1
    1. The problem statement, all variables and given/known data
    I Have tried to solve a problem about infinite potential well with a delta well in the middle, but I haven't the results and so I can't check if the proceeding is wrong. I post the steps that I have followed hoping someone can help me to understand.

    We have a particle in 1D that can moves only on ##[-a.a]## because of the potential ##V(x)=\begin{cases}-\lambda \delta (x), x\in(-a,a)\\ \infty, otherwise\end{cases}##

    (##\lambda>0)##

    2. Relevant equations
    The Schroedinger's Equation is:
    ##\psi''(x)=\frac{2m}{\hbar^2} (V(x)-E) \psi (x)##

    3. The attempt at a solution
    So we have:##\psi''(x)=-\frac{2m}{\hbar^2} E \psi (x)##
    I have assumed E>0

    Then I have translated the axis origin, and the segment [-a, a] now is [0,2a]

    We have two wavefunctions:

    ##\psi_-=A\sin kx +B\cos kx## for ##0<x<a##

    ##\psi_+=C\sin kx +D\cos kx## for ##a<x<2a##

    Conditions:

    ##\psi_-(0)=0 \rightarrow B=0##

    ##\psi_+(a)=\psi_- (a) \rightarrow D=0## (so I have obtained ##\psi(x)=A \sin k(x)##)

    ##\psi(2a)=0\rightarrow k=\frac{n\pi}{2a}##

    So I have obtained: ##\displaystyle k=\sqrt \frac{2mE}{\hbar} \rightarrow E={\frac{n^2 \pi^2 \hbar^2}{8ma^2}}##

    The energy spectrum is limited on the lower side, but non in the upper one.

    For hight energy, we have a continuous spectrum.

    Anyway, ##\psi_n(x)=A \sin \frac{n \pi}{2a}##, where n=1, 2, ..

    Many thanks for your help!
     
  2. jcsd
  3. Sep 6, 2015 #2

    TSny

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    OK
    D does not have to be zero.

    Also, note that nowhere in your analysis have you taken into account that there is a delta function potential in the middle of the well.

    It's OK to shift the origin to the left end of the well. It should work out. But I find it easier to keep the origin at the center of the well.
     
  4. Sep 6, 2015 #3
    If it isn't zero, I obtain 2 equations, one for tan(ka) and the other for tan(2ka).. but then I don't know how to proceed...
    for the condition on the derivative?
    I don't know how to set it, because if D isn't zero, I can't write the wavefunction in x=a..

    What should I do?
     
  5. Sep 6, 2015 #4

    TSny

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    Can you show more detail for how you arrive at 2 equations and state the 2 equations you obtain?
     
  6. Sep 6, 2015 #5
    ##\psi_+(2a)=Csin2ka+Dcos 2ka=0 \rightarrow tan (2ka)=-D/C##
    ##\psi_-(a)=\psi_+(a); Asin ka =Csinka+D cos ka \rightarrow tan ka= D/(A-C)##
     
  7. Sep 6, 2015 #6
    I have thought to do the axis translation in order to avoid splitting the problem in odd and even solutions...
     
  8. Sep 6, 2015 #7

    TSny

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    Even if you shift the origin to the left side of the well, you will still get two sets of solutions: one set will be even about the center of the well and the other set will be odd about the center. You can see that you are now getting trig functions of both ka and 2ka. So, you will probably need to invoke some "double angle" trig identities.

    Anyway, if you want to proceed with your choice of origin, your 2 equations for A, C, and D look correct to me. One of these constants, A say, can be used to normalize the wavefunction at the end. So, you can use your two equations to determine C and D in terms of A.

    However, there is another condition due to the existence of the delta potential. This will give you another equation that must be satisfied.
     
  9. Sep 6, 2015 #8
    I'm sorry, but my textbook, dealing with the infinite well, explicitly says "In order to avoid different steps for odd and even autofunctions, we choose the coordinates origin in one of the extreme points of the segment" and so does in the next steps. Is it wrong?
     
  10. Sep 6, 2015 #9

    TSny

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    No, the book is correct. However, for the problem with the delta function I believe you will see that the solutions break up into two sets.

    To see how this comes about, you will need to set up the condition that the wavefunction must satisfy at the location of the delta function.
     
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