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Regarding double well potential

  1. Dec 14, 2014 #1
    A particle is in the following double-well potential with E<0:
    V(x)=0 for x<-a, x>a; -V0 for -a<x<-b, b<x<a; 0 for -b<x<b
    I am asked to show that the eigenvalues conditions may be written in the form:
    tan(q(a-b))=qα(1+tanh(αb)/(q22tanh(αb))$$
    and
    tan(q(a-b))=qα(1+coth(αb)/(q22coth(αb))$$
    for the even and odd solutions, where -E=ħ2α2/2m and E+V02q2/2m.

    I first tried to define the wave function in the various regions, focusing on the positive x axis only and demanding odd solutions:
    Ψ(x)=Ae-αx for x>a; Bsin(q(x-b)) for b<x<a; Ce-αx + Deαx for 0<x<b

    Is that correct thus far?
     
  2. jcsd
  3. Dec 14, 2014 #2

    mfb

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    Looks good so far.
     
  4. Dec 14, 2014 #3
    Alright, but as I impose continuity of wave function and its derivatives I obtain the four following equations, from which I am unable to derive the desired expression:
    (1) Ae-αa = Bsin(q(a-b))
    (2) -αAe-αa = Bqcos(q(a-b))
    (3) 0 = Ce-αb + Deαb
    (4) Bq = -αCe-αb + Dαeαb

    What am I doing wrong?
     
  5. Dec 14, 2014 #4

    mfb

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    You could explain what you did. Where does equation 3 come from?
    Where is the odd/even condition?
     
  6. Dec 14, 2014 #5
    Well, equation 3 stems from the demand for continuity at x=b. I have decided to focus on the odd solutions first, hence the use of sine. Is that wrong?
     
  7. Dec 14, 2014 #6

    mfb

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    Ah, I see the issue.
    The way you defined the wave function in the well, you get a mixture of odd and even functions (the sine would have some non-zero value at x=0). If you want odd functions, you need something like sin(qx).
     
  8. Dec 14, 2014 #7
    So which function ought I to choose for Ψ(x) in the well? Ought it to be a combination of sine and cosine at once (like harmonic oscillator)?
     
  9. Dec 14, 2014 #8
    In other words, should I then simply choose sin(qx) for the odd and cos(qx) for the even, within the well?
     
  10. Dec 14, 2014 #9

    mfb

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    You can do that, if you choose them without that offset it is easier to split them into odd and even contributions.
    That is the easierst approach I think.
     
  11. Dec 15, 2014 #10
    Yet please note the expressions I am expected to arrive at. It seems I am indeed expected to use forms as sin(q(x-b)). Wouldn't you agree?
     
  12. Dec 15, 2014 #11

    mfb

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    I don't see how you get that impression. Yes there is an q(a-b), but that can appear in other ways, too.
     
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