I'm trying to solve the boundary conditions of a 1D potential barrier for a free particle. The boundary conditions on the continuity of the wave function and its first derivatives leads to four equations:


$ik(A-B) = q(F-G)$

$Fe^{qa}+Ge^{-qa} = Ce^{ika}$

$q(Fe^{qa} - Ge^{-qa}) = ikCe^{ika}$

the book I'm reading doesn't solve explicitly this system of equations, so I decided to give it a try. The main objective is to eliminate B, F and G because I want to calculate the transmission rate through a one-dimensional potential barrier:

$T = \frac{|C|^2}{|A|^2}$

So the first thing I did was to use the last 2 equations to solve F and G in relation to C, the result is:

$F = \frac{Ce^{ika}}{2}(e^{-qa}+\frac{ik}{q}e^{-qa})$

$G = \frac{Ce^{ika}}{2}(e^{qa}-\frac{ik}{q}e^{qa})$

After it, I used the first 2 equations to write A in relation to F and G and used the above 2 equations to write A in relation to C, the result is:

$A = \frac{Ce^{ika}}{2}(2cosh(qa) + sinh(qa)\frac{k^2+q^2}{qk}i)$

Replacing A and C in the transmission equation I get this result:

$T = (cosh^2(qa) + sinh^2(qa)(\frac{k^2+q^2}{2qk})^2)^{-1}$

I'm really distressed by this result because the result from the book is

$T = (1 + sinh^2(qa)(\frac{k^2+q^2}{2qk})^2)^{-1}$

I already revised all my calculations 5 times, I can't find any error and I don't see how I can use some hyperbolic identity to get rid of the $cosh^2(qa)$ and make it 1. I searched the web for it and didn't find any website showing the steps for this calculation.I know this is not a proper physics question, but could you please help me?

  • $\begingroup$ There is a solution here, although the boundary for the potential is different. $\endgroup$ – Hanting Zhang Feb 11 at 18:44

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