Double Delta Potential

A particle of mass $m$ and energy $E>0$ is incident from the left on the double delta potential 

$V(x)=V_0delta(x+a)-V_0delta(x-a)$

Find the wave function describing the stationary states of the particle as well as the transmission and the reflection coefficients for this potential.

$psi=left{begin{matrix}Ae^{ikx}+Be^{-ikx}&x<a\Ce^{ikx}+De^{-ikx}&-a<x<a\Fe^{ikx}&x>aend{matrix}right.$

where $k=sqrt{2mE}/hbar$

Continuity in $psi$ and $(dpsi/dx)$ at $a$ and$-a$ so one has the following 4 system of equations that need to be solved:

$left{begin{matrix}Ae^{-ika}+Be^{ika}-Ce^{- ika}De^{ika}=0\0+0+Ce^{ika}+De^{-ik a}=Fe^{ika}\(2mV/ℏ^2 +ik)Ae^{-ika}+(2mV/ℏ^2 -ik)Be^{ika}-ikCe^{-ika}+ikDe^{ika}=0\0+0-ikCe^{ika}+ikDe^{-ika}=-(2mV/ℏ^2 +ik)Fe^{ika}end{matrix}right.$

Notations

$e^{-ika}=x$,      $e^{ika}=y$,    $(2mV/ℏ^2 +ik)=z$,     $(2mV/ℏ^2 -ik)=u$
Take F as variable. Solve the system:

$A=Det(A)/Det$

$Det=2ikx^2y^2(u-z)$  and

$Det(A)=F(x^2(-k^2y^2+z(-uy^2+iky^2)-ikuy^2)+k^2y^4+z(uy^4+iky^4)-ikuy^4)$

Since $|x|=|y|=|i|=1$  one has

$|A|=|F|frac{-k^2+z(k-u)-ku+k^2+z(k+u)-ku)}{k(u-z)(C+D)}=frac{|F|*2k(z-u)}{-2k(u-z)}|F|$

$Transmission= frac{|F|^2}{|A|^2} =1$

$B=Det(B)/Det$

$Det(B) =F(k^2+z^2)(x^3y-xy^3)$

Since $|x|=|y|=|i|=1$  one has

$|B|=frac{(|F|(k^2+z^2))}{(-2k(u-z))}$

$frac{|B|^2}{|F^2|} =frac{(k^2+z^2 )^2}{(4k^2 (u-z)^2)}$

Reflection

$R=frac{|B|^2}{|A|^2} =frac{|B|^2}{|F|^2} *frac{|F|^2}{|A|^2} =frac{(k^2+z^2 )^2}{(4k^2 (u-z)^2)}$