0
$\begingroup$

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:

$A+B=F+G$

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

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

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

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?

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.