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Potential Barrier

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the transmission and reflection coefficients for:
    [tex]V(x)=\left\{\begin{matrix}
    V_0, & -a\leq x\leq a\\
    0, & elsewhere
    \end{matrix}\right.[/tex]


    2. Relevant equations
    [tex]\frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}\psi+V\psi=E\psi[/tex]


    3. The attempt at a solution
    I have successfully got the solution for:
    [tex]E=V_0[/tex]
    and
    [tex]E>V_0[/tex]
    But I am having trouble with
    [tex]E<V_0[/tex]
    Here's my attempt:
    First, I denoted the left of the barrier as region I, within the barrier as region II, and to the right of the barrier as region III.
    Rearranging the schrodinger Equation I get:
    [tex]\frac{\partial^2 }{\partial x^2}\psi_{I}=\frac{-2m}{\hbar^2}(E)\psi_{I}[/tex]
    [tex]\frac{\partial^2 }{\partial x^2}\psi_{II}=\frac{2m}{\hbar^2}(V_0-E)\psi_{II}[/tex]
    [tex]\frac{\partial^2 }{\partial x^2}\psi_{III}=\frac{-2m}{\hbar^2}(E)\psi_{III}[/tex]
    The solutions are:
    [tex]\psi_{I}=Ae^{ikx}+Be^{-ikx}[/tex]
    [tex]\psi_{II}=Ce^{-lx}+De^{lx}[/tex]
    [tex]\psi_{I}=Fe^{ikx}[/tex]
    where:
    [tex]k=\frac{\sqrt{2mE}}{\hbar}[/tex]
    [tex]l=\frac{\sqrt{2m(V_0-E)}}{\hbar}[/tex]
    The -ikx term omitted since there's assumed to be no incoming wave from +x-direction.
    Applying the boundary conditions:
    [tex]\psi_{I}(-a)=\psi_{II}(-a)\Rightarrow Ae^{-ika}+Be^{ika}=Ce^{la}+De^{-la} \Rightarrow (1)[/tex]

    [tex]\left.\begin{matrix}
    \frac{\partial \psi_{I}}{\partial x}
    \end{matrix}\right|_{x=-a}=\left.\begin{matrix}
    \frac{\partial \psi_{II}}{\partial x}
    \end{matrix}\right|_{x=-a}\Rightarrow ik(Ae^{-ika}-Be^{ika})=l(-Ce^{la}+De^{-la})\Rightarrow (2)[/tex]

    [tex]\psi_{II}(a)=\psi_{III}(a)\Rightarrow Ce^{-la}+De^{la}=Fe^{ika}\Rightarrow (3)[/tex]

    [tex]\left.\begin{matrix}
    \frac{\partial \psi_{II}}{\partial x}
    \end{matrix}\right|_{x=a}=\left.\begin{matrix}
    \frac{\partial \psi_{III}}{\partial x}
    \end{matrix}\right|_{x=a}\Rightarrow l(-Ce^{-la}+De^{la})=ikFe^{ika}\Rightarrow (4)[/tex]

    From (3) we have:

    [tex]C=Fe^{ika}e^la-De{2la}[/tex]

    Putting that into (4) we get:

    [tex]D=\frac{1}{2}(1+\frac{ik}{l})Fe^{ika}e^{-la}[/tex]

    From (3) we also have:

    [tex]D=Fe^{ika}e^-{la}-Ce^{-2la}[/tex]

    Putting that into (4) we now get:

    [tex]C=\frac{1}{2}(1-\frac{ik}{l})Fe^{ika}e^{la}[/tex]

    Rearranging (2) we get:

    [tex] Ae^{-ika}-Be^{ika}=\frac{il}{k}(De^{-la}-Ce^{la})[/tex]

    Summing the equation above with (1) we get::

    [tex]2Ae^{-ika}=Ce^{la}+De^{-la}-\frac{il}{k}(De^{-la}-Ce^{la})=Ce^{la}(1+\frac{il}{k})+De^{-la}(1-\frac{il}{k})[/tex]

    Now we put in the the C and D that we got from playing with (3) and (4) before and get:

    [tex]2Ae^-{ika}=\frac{1}{2}(1-\frac{ik}{l})Fe^{ika}e^{la}e^{la}(1+\frac{il}{k})+\frac{1}{2}(1+\frac{ik}{l})Fe^{ika}e^{-la}e^{-la}(1-\frac{il}{k})[/tex]

    [tex]2Ae^{-ika}=Fe^{ika}\left [\frac{1}{2} (1-\frac{ik}{l})(1+\frac{il}{k}) e^{2la} +\frac{1}{2} (1+\frac{ik}{l})(1-\frac{il}{k}) e^{-2la} \right ][/tex]

    [tex]2Ae^{-ika}=Fe^{ika}\left [\frac{1}{2} (1+\frac{il}{k}-\frac{ik}{l}+1)e^{2la}+\frac{1}{2} (1-\frac{il}{k}+\frac{ik}{l}+1)e^{-2la} \right ][/tex]

    [tex]2Ae^{-ika}=Fe^{ika}\left [\frac{1}{2} (2+ i\frac{l^2-k^2}{kl})e^{2la}+\frac{1}{2} (2-i\frac{l^2-k^2}{kl} )e^{-2la} \right ][/tex]


    [tex]2Ae^{-ika}=Fe^{ika}\left [e^{2la}+ i\frac{l^2-k^2}{kl}\frac{1}{2} e^{2la}+e^{-2la} -\frac{1}{2}i\frac{l^2-k^2}{kl} e^{-2la} \right ][/tex]


    [tex]2Ae^{-ika}=Fe^{ika}\left [(e^{2la}+e^{-2la})+( i\frac{l^2-k^2}{kl})(\frac{1}{2} e^{2la} -\frac{1}{2} e^{-2la} )\right ][/tex]


    [tex]2Ae^{-ika}=Fe^{ika}\left [2cosh(2la)+( i\frac{l^2-k^2}{kl})sinh(2la)\right ][/tex]

    [tex]F=\frac{Ae^{-2ika}}{cosh(2la)+( i\frac{l^2-k^2}{2kl})sinh(2la)}[/tex]

    [tex]T=\frac{\left | F \right |^2}{\left | A \right |^2}=\frac{e^{-2ika}}{cosh(2la)+( i\frac{l^2-k^2}{2kl})sinh(2la)}\frac{e^{2ika}}{cosh(2la)-( i\frac{l^2-k^2}{2kl})sinh(2la)}[/tex]

    [tex]=\frac{1}{cosh^2(2la)+( \frac{l^2-k^2}{2kl})^2sinh^2(2la)}[/tex]

    [tex]=\frac{1}{1+sinh^2(2la)+\frac{1}{4}( \frac{l^2}{k^2}+\frac{k^2}{l^2}-2)sinh^2(2la)}[/tex]


    I am now stuck... I can't get the sinh(2la) that I wanted. Did I do snmething wrong?
     
    Last edited: Oct 3, 2009
  2. jcsd
  3. Oct 3, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Everything look fine to me (although you could simplify [itex]\frac{l^2}{k^2}+\frac{k^2}{l^2}[/itex] significantly)....do you know what your final answer is supposed to look like?
     
    Last edited: Oct 3, 2009
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