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Potential Barrier - Special Case!

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A beam of particles, each of mass m and kinetic energy E, is incident on a potential barrier

    [tex] V(x) = V_0 \; \; for \; \; 0 \leq x \leq a [/tex]
    [tex] \; \; \; \; \; \; \; \; \; = 0 \; \; for \; \; x < 0 \; \; and \; \; x > a [/tex]
    [tex] E = V_0 \; \mbox{is the special case}[/tex]

    The part of the problem i'm on is finding the transmission probability


    3. The attempt at a solution

    I've solved the Schrodinger Equation and equated the solutions at the two boundaries which gave me

    [tex] C + D = B [/tex]
    [tex] ik(C - D) = A [/tex]
    [tex] Aa + B = Ge^{ika} [/tex]
    [tex] A = kiGe^{ika} [/tex]

    [tex] A, B, C, D, G \; \mbox{constants}[/tex]
    I feel like im just going round in circles finding the transmission probability, in my notes I have transmission prob as [tex] (\frac{G}{A})^2 [/tex]. However, a hint at the bottom says once the 4 continuity equations have been found, eliminate A and B, which I've tried and doesn't seem to get me any where!

    If anyone has some advice or could push me in the right direction that would be really appreciated. Thanks
     
    Last edited: Apr 23, 2010
  2. jcsd
  3. Apr 23, 2010 #2
  4. Apr 24, 2010 #3
    My constants correspond to those on the wikipedia article. So I have,
    [tex] A = B_2a \\;
    B = B_1 \\;
    C = A_r\\;
    D = A_l\\;
    G = C_r\\; [/tex]

    I understant that the constant I would've had [tex] F [/tex] disappears because there is no particle from the right, but I don't understand why [tex] D \; (or \; A_l) [/tex] becomes 1. By the looks of it's something quite simple that i'm missing!
     
  5. Apr 24, 2010 #4
    They set it to 1 to normalize it. They assume the incident wave is 100%. So the transmission and reflection will be less than 100% and add up to it.
     
  6. Apr 24, 2010 #5
    I realised I mixed up A with C in my first post when I gave the transmission probability. Will it still work if I keep the transmission as a ratio with C as the denominator rather than normalizing it?
     
  7. Apr 24, 2010 #6
    You can keep it as ratios if you like. That works just as well.
     
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