Solving for transmission coefficient in the finite square well Consider a finite square well of depth $V_0$ and which extends from $-a$ to $a$. For $|x|>a$, $V=0$. The wavefunction ansatz one can propose for an incoming wave from the left $Ae^{ikx}$ is:
$$ \psi = Ae^{ikx} + Be^{-ikx}, x<a $$
$$ \psi = Ce^{ik_2 x} + Be^{-ik_2 x}, |x|<a $$
$$ \psi = Fe^{ikx}, x>a $$
Where $k=\frac{2mE}{\hbar^2}$ and $k_2=\frac{2mE+V_0}{\hbar^2}$ can be obtained from the Schrodinger Equation. Then, if we define the transmission coefficient to be:
$$ T = \frac{F^2}{A^2}$$
One should be able to find the value of this coefficient if $F$ is written in terms of $A$. We can apply boundary conditions to do so: $\psi$ and $\psi'$ must be continuous at $-a$ and at $a$, so:
$$ Ae^{-ika}+Be^{ika} = Ce^{ik_2 a}+De^{ik_2}a $$
$$ -ik(Ae^{-ika}-Be^{ika}) = -ik_2(Ce^{-ik_2 a}-De^{ik_2a}) $$
$$ Ce^{ik_2a}+De^{-ik_2a} = Fe^{ika} $$
$$ ik_2(Ce^{ik_2a}-De^{-ik_2a}) = ikFe^{ika} $$
Now, I have seen derivations of $T$ where $\psi$ is taken to be $Ccos(k_2x)+Dsin(k_2x)$, but it should be the same here and I am not being able to eliminate the $C$ and $D$ to get equations only in $B$,$F$ and $A$. Am I not seeing a way to do this, are the boundary conditions wrong or is the ansatz wrong (I have been told that either way the value of $T$ should be the same). Thanks!
 A: You have mistakes in the equations for boundary conditions, they would be:
$$\psi\text{ continuous at }x=-a\longrightarrow Ae^{-ika}+B e^{ika}=C e^{-ik_2a}+De^{ik_2 a}$$
$$\psi'\text{ continuous at }x=-a\longrightarrow ik(Ae^{-ika}-B e^{ika})=ik_2(C e^{-ik_2a}-De^{ik_2 a})$$
$$\psi\text{ continuous at }x=a\longrightarrow C e^{ik_2a}+De^{-ik_2 a}=F e^{ika}$$
$$\psi'\text{ continuous at }x=a\longrightarrow ik_2(C e^{ik_2a}-De^{-ik_2 a})=ikFe^{ika}.$$
Then, solving for $D$ using the first equation we get
$$D=Ae^{-i(k+k_2)a}+Be^{i(k-k_2)a}-Ce^{-i2k_2a}.$$
Inserting this in the second one and solving for $C$
$$ik(Ae^{-ika}-B e^{ika})=ik_2(2C e^{-ik_2a}-Ae^{-ika}-Be^{ika})\rightarrow\\C=\frac{A}{2k_2}e^{i(k_2-k)a}(k_2+k)+\frac{B}{2k_2}e^{i(k_2+k)a}(k_2-k),$$
so $D$ now is
$$D=Ae^{-i(k+k_2)a}(1-\frac{1}{2}-\frac{k}{2k_2})+Be^{i(k-k_2)a}(1-\frac{1}{2}+\frac{k}{2k_2}).$$
Similarly, using the third one you can get $F$ in terms of $A $ and $B$, i.e. $F=F(A,B)$, and from the fourth one, $B=B(A)$. Finally, you can use this last relation between the $A$ and the $B$ to get $F=F(A,B(A))=F(A),$ and compute the transmission coefficient
$$\mathbb{T}=\frac{|F|^2}{|A|^2}.$$
