- #1
T-7
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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!
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!
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