Tunneling and transmission Lets say we have a tunelling problem in the picture, where $W_p$ is a finite potential step: 

If particle is comming from the left a general solutions to the Schrödinger equations for sepparate intervals I, II and II are: 
\begin{align}
\text{I:}& & \psi_1 &= \overbrace{A e^{i\mathcal L x}}^{\psi_{in}} + \overbrace{Be^{-i \mathcal L x}}^{\psi_{re}}& \mathcal L &= \sqrt{\tfrac{2mW}{\hbar^2}}\\
\text{II:}& & \psi_2 &= C e^{\mathcal K x} + De^{-\mathcal K x}& \mathcal K &= \sqrt{-\tfrac{2m(W-W_p)}{\hbar^2}}\\
\text{III:}& & \psi_3 &= \underbrace{E e^{i \mathcal L x}}_{\psi_{tr}}& &\\
\end{align}
Where $\psi_{in}$ is an incomming wave, $\psi_{re}$ is a reflected wave and $\psi_{tr}$ is transmitted wave. I used the boundary conditions and got a system of 4 equations: 
\begin{align}
{\tiny\text{boundary}}&{\tiny\text{conditions at x=0:}} & {\tiny\text{boundary conditions}}&{\tiny\text{at x=d:}}\\
A + B &= C + D & Ce^{\mathcal K d} + De^{-\mathcal K d} &= E e^{i \mathcal L d}\\
i \mathcal L A - i \mathcal L B &= \mathcal KC - \mathcal K D & \mathcal K C e^{\mathcal K d} - \mathcal K D e^{-\mathcal K d}&= i \mathcal L E e^{i \mathcal L d}
\end{align}
So now i decided to calculate coefficient of transmission $T$: 
\begin{align}
T &= \dfrac{|j_{tr}|}{|j_{in}|} \!=\! \Bigg|\dfrac{\dfrac{\hbar }{2mi}\!  \left( \dfrac{d\overline{\psi}_{tr}}{dx}\, \psi_{tr} - \dfrac{d \psi_{tr}}{dx}\, \overline{\psi}_{tr} \right)}{\dfrac{\hbar}{2mi}  \!\left( \dfrac{d\overline{\psi}_{in}}{dx}\, \psi_{in} - \dfrac{d\psi_{in}}{dx}\, \overline{\psi}_{in} \right) }\Bigg| \!=\! \Bigg|\dfrac{\frac{d}{dx}\big(\overbrace{Ee^{-i\mathcal L x}}^{\text{konjug.}}\big) Ee^{i\mathcal L x} - \frac{d}{dx} \left( Ee^{i\mathcal L x}\right)\! \overbrace{Ee^{-i\mathcal L x}}^{\text{konjug.}}}{ \frac{d}{dx}\big(\underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}\big) Ae^{i\mathcal L x} - \frac{d}{dx} \left( Ae^{i\mathcal L x}\right)\! \underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}}\Bigg|\! = \nonumber\\
&=\Bigg|\dfrac{-i\mathcal L Ee^{-i\mathcal L x} E e^{i \mathcal L x} - i\mathcal L E e^{i \mathcal L  x} Ee^{-i \mathcal L x}}{-i \mathcal L A e^{-i\mathcal L x} Ae^{i \mathcal L x} - i \mathcal L A e^{i \mathcal L x}Ae^{-i \mathcal L x} }\Bigg|=\Bigg|\dfrac{-i\mathcal L E^2 - i\mathcal L E^2}{-i \mathcal L A^2 - i \mathcal L A^2}\Bigg|=\Bigg|\dfrac{-2 i \mathcal L E^2}{-2i\mathcal L  A^2}\Bigg| = \frac{|E|^2}{|A|^2}
\end{align}
It accured to me that if out of 4 system equations i can get amplitude ratio $E/A$, i can calculate $T$ quite easy. Could anyone show me how do i get this ratio?
 A: Strictly speaking, you have 4 equations and 5 unknowns. However, given that the coefficient A is applied to the incoming wave-function, you could arbitrarily set it equal to 1 (because it represents 100% of the wave) and solve the system of equations for E.  Then $T=E$. This is how the problem is handled in most cases.  Alternatively, if you absolutely cannot set $A=1$, then try assuming A is a given and solve the 4 equations for B, C, D, and E in terms of A. Then, again, perform $T=E/A$.
In theory, the ratio for any A will be the same as for A=1.
(I checked, it is, the A divides out in the end).
EDIT
You can easily solve for B,C,D, and E using matrices, where your four system equations are:
$$\begin{pmatrix}-1 & 1 & 1 & 0 \\ i \mathcal L & \mathcal K & -\mathcal K & 0 \\ 0 & e^{\mathcal Kd} & e^{-\mathcal Kd} & -e^{i\mathcal Ld} \\ 0 & \mathcal Ke^{\mathcal Kd} & -\mathcal Ke^{-\mathcal Kd} & -i\mathcal Le^{i\mathcal Ld}\end{pmatrix} \begin{pmatrix}B \\ C \\ D \\ E\end{pmatrix}=\begin{pmatrix}A \\ i\mathcal LA \\ 0 \\ 0\end{pmatrix} $$
Optionally, $A=1$. But if you invert the matrix and solve for E, you should get:
$$E={4iA\mathcal K\mathcal L\over\mathcal K^2 e^{i\mathcal Ld-\mathcal Kd}-\mathcal K^2 e^{d\mathcal K+id\mathcal L}+2i\mathcal L\mathcal Ke^{id\mathcal L-d\mathcal K}+2i\mathcal L\mathcal Ke^{id\mathcal L+d\mathcal K}-\mathcal L^2 e^{id\mathcal L-d\mathcal K}+\mathcal L^2 e^{id\mathcal L+d\mathcal K}}$$
And, of course, A=1
