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Homework Help: How do I simplify this?

  1. Mar 1, 2005 #1
    Hi, I've been doing a QM problem to do with a potential barrier going from x = 0 to x = a. I need to work out an expression for the transmission coefficient T and have a horrible looking expression:

    I can't use LaTeX properly, so:

    A = (F/4k_{1}k_{2}).e^ik_{1}a[e^-k_{2}a (k_{2} + ik_{1})(k_{1} - ik_{2}) + e^k_{2}a (k_{2} - ik_{1})(k_{1} + ik_{2})]

    where k_{1} = (1/hbar)(2mE)^1/2, k_{2} = (1/hbar)[2m(V_{0} - E)]^1/2

    How on Earth do I simplify this further? Thanks.
     
    Last edited: Mar 1, 2005
  2. jcsd
  3. Mar 1, 2005 #2

    dextercioby

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    Is this what u meant...?
    [tex] A = (F/4k_{1}k_{2})\cdot e^{ik_{1}a}[e^{-k_{2}a} (k_{2} + ik_{1})(k_{1} - ik_{2}) + e^{k_{2}a} (k_{2} - ik_{1})(k_{1} + ik_{2})] [/tex]

    ,where [tex]k_{1} = (1/\hbar)(2mE)^{1/2}, k_{2} = (1/\hbar)[2m(V_{0} - E)]^{1/2} [/tex]

    Daniel.

    P.S.Won't u be needing the square modulus of this expression?
     
    Last edited: Mar 1, 2005
  4. Mar 1, 2005 #3

    Galileo

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    Yikes. I've done this problem before. I know it's not pretty.

    Anyway, if you work out the brackets:
    [tex]e^{-k_2a}(k_{2} + ik_{1})(k_{1} - ik_{2})[/tex]
    and
    [tex]e^{k_2a}(k_{2} - ik_{1})(k_{1} + ik_{2})][/tex]
    you can rewrite the sum as:

    [tex]2k_1k_2\left(e^{k_2a}+e^{-k_2a}\right)+i(k_2^2-k_1^2)(e^{k_2a}-e^{k_2a})[/tex]
    The exponential combinations form hyperbolic sinusoids. The expression is equal to:
    [tex]4k_1k_2\cosh(k_2a)+2i(k_2^2-k_1^2)\sinh(k_2a)[/tex]

    So the original expression becomes (divide both sides by F, since you're interested in T=|F/A|^2, right?):

    [tex]\frac{A}{F}=e^{ik_1a}\left(\cosh(k_2a)-\frac{i}{2}\left(\frac{k_2^2-k_1^2}{k_1k_2}\right)\sinh(k_2a)\right)[/tex]

    Take the modulus squared on both sides (the term in brackets is in the form a+bi with a and b real, so it's simply a^2+b^2):

    [tex]T^{-1}=\frac{|A|^2}{|F|^2}=\cosh^2(k_2a)+\frac{1}{4}\left(\frac{k_2^2-k_1^2}{k_1k_2}\right)^2\sinh^2(k_2a)[/tex]

    Now use the expressions for [itex]k_1, k_2[/itex] to show that:

    [tex]\left(\frac{k_2^2-k_1^2}{k_1k_2}\right)^2=-4+\frac{V_0^2}{E(V_0-E)}[/tex]

    Use the identity [itex]cosh^2(x)-sinh^2(x)=1[/itex] and write out [itex]k_2[/itex] in the argument to get the final answer.
    ..whew
     
    Last edited: Mar 1, 2005
  5. Mar 1, 2005 #4
    Ahh, cheers both of you.
     
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