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Transmission over a finite square well

  1. Aug 17, 2007 #1

    T-7

    User Avatar

    Hi,

    My Quantum textbook loves to skip the algebra in its derivations. It claims that the solution for 'T', the transmission coefficient of the wave function (E>0, ie. unbound) is

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

    Its prior step is to offer two equations (which match my own derivation) for T in terms of R (the reflection coefficient):

    [tex]
    e^{-ika}+R^{ika} = \frac{T}{2}((1+\frac{k}{q})e^{ika-2iqa}+(1-\frac{k}{q})e^{ika+2iqa})
    [/tex]

    [tex]
    e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1-\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})
    [/tex]

    My own attempt to fill in the gap between the two proceeds as follows, but I didn't quite arrive at that final expression...

    [tex]
    2e^{-ika} = T/2.e^{ika}[(1+k/q).e^{-2iqa} + (1-k/q).e^{2iqa} + (q/k-1).e^{-2iqa} - (q/k-1).e^{2iqa}]
    [/tex]

    [tex]
    4e^{-2ika} = T.[(1+k/q+q/k-1).e^{-2iqa} + (1-k/q-q/k+1).e^{2iqa}]
    [/tex]

    [tex]
    4qke^{-2ika} = T.[(k^{2}+q^{2}).e^{-2iqa} + (2kq-k^{2}-q^{2}).e^{2iqa}]
    [/tex]

    [tex]
    4qke^{-2ika} = T.[(k^{2}+q^{2}).(e^{-2iqa}- + e^{2iqa}) + 2kq.e^{2iqa}]
    [/tex]

    [tex]
    4qke^{-2ika} = T.[(k^{2}+q^{2}).(-2isin(2qa)) + 2kq.(cos 2qa + isin sqa)]
    [/tex]

    [tex]
    T = \frac{4qke^{-2ika}}{2kq.(cos 2qa + isin 2qa)-2i(k^{2}+q^{2})sin(2qa)}
    [/tex]


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

    (Hopefully I've copied all that down correctly!).

    Can anyone spot a false move, or what it is I need to do to make my derivation match the textbook formula.

    Cheers!
     
    Last edited: Aug 17, 2007
  2. jcsd
  3. Aug 18, 2007 #2

    StatusX

    User Avatar
    Homework Helper

    It seems like your answer would match theirs if:

    [tex]e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1-\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})[/tex]

    was changed to:

    [tex]e^{-ika}-R^{ika} = \frac{q}{k}.\frac{T}{2}((1+\frac{k}{q})e^{ika-2iqa}-(1-\frac{k}{q})e^{ika+2iqa})[/tex]

    which also matches the other line better. Are you sure you derived/copied this down right?
     
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