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Finding the Boundary Conditions of a Potential Well

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the conditions for a bound state, Eqn1 and Eqn2, can be obtained by requiring the vanquishing of the denominators in Eqn3 at k=i[tex]\kappa[/tex]. Can you give the argument for why this is not an accident?


    2. Relevant equations

    Eqn1: [tex]\alpha[/tex]=q*tan(qa)
    Eqn2: [tex]\alpha[/tex]=-q*cot(qa)

    Eqn3:

    R= i*e[tex]^{-2ika}[/tex][tex]\frac{(q^{2}-k^{2})*sin(2qa)}{2kq*cos(2qa)-i(q^{2}+k^{2})*sin2qa}[/tex]
    T=e[tex]^{-2ika}[/tex][tex]\frac{2kq}{2kq*cos(2qa)-i(q^{2}+k^{2})*sin2qa}[/tex]

    The problem concerns a potential well of V[tex]_{0}[/tex] from -a to a.
    I believe that k[tex]^{2}[/tex]=[tex]\frac{2mE}{h(bar)^{2}}[/tex] and that [tex]\kappa[/tex][tex]^{2}[/tex]=-[tex]\frac{2m(E-V_{0})}{h(bar)^{2}}[/tex]

    I also know that in order to reach these boundaries through another method in the book, they let -[tex]\alpha[/tex][tex]^{2}[/tex]=[tex]\frac{2mE}{h(bar)^{2}}[/tex]...


    3. The attempt at a solution
    I removed the demoninators from Eqn3 and tried to substitute k=[tex]\sqrt{\frac{2mE}{h(bar)^{2}}}[/tex] into Eqn3... but not only was this result messy but I couldn't get the boundary conditions, Eqn1 and Eqn2 to come out. Help?
     
    Last edited: Oct 7, 2009
  2. jcsd
  3. Oct 9, 2009 #2

    gabbagabbahey

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    I'm not 100% clear on your problem statement....First, what is [itex]q[/itex]? Second are these the equations you meant to write:

    [tex]R= ie^{-2ika}\frac{(q^{2}-k^{2})\sin(2qa)}{2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)}[/tex]

    [tex]T=e^{-2ika}\frac{2kq}{2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)}[/tex]

    [tex]k=\left(\frac{2mE}{\hbar^2}\right)^{1/2}[/tex]

    [tex]\kappa=\left(\frac{2m(V_0-E)}{\hbar^2}\right)^{1/2}[/tex]

    [tex]\alpha=\left(-\frac{2mE}{\hbar^2}\right)^{1/2}[/tex]

    ?
     
  4. Oct 9, 2009 #3
    [tex]
    q=\left(\frac{2m(E+V_0)}{\hbar^2}\right)^{1/2}
    [/tex]
     
  5. Oct 9, 2009 #4

    gabbagabbahey

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    Okay, let's look at your eqn 3..."the vanquishing of the denominators" is just a fancy way of saying that [itex]2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)=0[/itex]...so [itex]\tan(2qa)=[/itex]___?
     
  6. Oct 10, 2009 #5
    [tex]=\left(\frac{sin(2qa)}{cos(2qa)}\right)?[/tex]
     
  7. Oct 10, 2009 #6

    gabbagabbahey

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    Well of course, but what is that equal to when [itex]2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)=0[/itex]?
     
  8. Oct 10, 2009 #7
    Alright, I'll take a closer look at this at a later time to make sure that I understand it but I think the missing link is that I initially overlooked q and [tex]\alpha[/tex]. I'm sure that the solution will be clear once I make it through the alegbra, thanks
     
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