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Quantum mechanics, potential barrier

  1. Oct 22, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A particle of mass m goes toward the unidimensional barrier potential of the form [itex]V(x)=0[/itex] for [itex]x\leq 0[/itex] and [itex]a\leq x[/itex] and [itex]V(x)=V_0[/itex] for [itex]0<x<a[/itex].
    1)Write the corresponding Schrödinger's equation.
    2)Calculate the transmission coefficient for the cases [itex]0<E<V_0[/itex] and [itex]E>V_0[/itex]. Hint: Check out quantum mechanics books.


    2. Relevant equations
    Schrödinger's equation.


    3. The attempt at a solution
    I solved the Schrödinger's equation for Psi.
    The result I have is
    [itex]\Psi _I (x)=Ae^{ik_1 x}+Be^{-i _k1x}[/itex]
    [itex]\Psi _{II}(x)=De^{-k_2x}[/itex]
    [itex]\Psi _{III}(x)=Fe^{ik_1x}[/itex].
    The continuity conditions give:
    (1) [itex]A+B=D[/itex].
    (2) [itex]i k_1 A-i k_1 B=-k_2 D[/itex].
    (3) [itex]De^{-k_2a}=Fe^{ik_1a}[/itex]
    (4) [itex]-k_2 De^{-ak_2}=ik_1 Fe^{ik_1a}[/itex].

    I isolated B in function of A. I reached [itex]B=A \cdot \frac{\left (1 +\frac{ik_1}{k_2} \right ) }{ \left ( \frac{ik_1}{k_2} -1\right )}[/itex].
    Since [itex]D=A+B[/itex], I got D in function of A only.
    And since [itex]F=De^{-a(k_2+ik_1)}[/itex], I reached F in function of A only.
    Now the coefficient of transmission is [itex]\frac{F}{A}[/itex]. This gives me [itex]\left [ 1+ \frac{\left ( 1+ \frac{ik_1}{k_2} \right ) }{\left ( \frac{ik_1}{k_2}-1 \right ) } \right ] e^{-a(k_2+i_k1)}[/itex].
    I checked out in Cohen-Tanoudji's book about this and... he says that F/A represent the transmission coeffient just like me but then in the algebra, he does [itex]T= \big | \frac{F}{A} \big | ^2[/itex]. He reaches [itex]T=\frac{4k_1^2k_2^2}{4k_1^2k_2^2+(k_1^2-k_2^2)^2 \sin k_2 a}[/itex].
    So I don't understand why he says F/A but does the modulus of it to the second power. And he seems to get a totally different answer from mine.
    By the way my [itex]k_1=\sqrt {\frac{-2mE}{\hbar ^2}}[/itex] and my [itex]k_2=\sqrt {\frac{2m(V_0-E)}{\hbar ^2}}[/itex].
    What am I doing wrong?!
     
    Last edited: Oct 22, 2011
  2. jcsd
  3. Oct 23, 2011 #2

    vela

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    It's possible to get reflections at x=a, so the region II wave function needs to be modified.

    The transmission coefficient is |F/A|2.
     
  4. Oct 23, 2011 #3

    vela

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    Where did he say this? I can never find anything in that encyclopedia of a textbook.

    The closest thing I saw was that he said you could get the transmission coefficient from the ratio, but he never actually said the ratio itself was the transmission coefficient.
     
  5. Oct 23, 2011 #4

    fluidistic

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    Ah whoops, I totally missed this.
    Again oops once again. You're right, he says that the ratio allows us to determine the coefficients.
    Ok I understand now why it's the modulus to the second power. I'll rework on that.
     
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