# 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

$$\psi = Ce^{ik_2 x} + Be^{-ik_2 x}, |x|

$$\psi = Fe^{ikx}, x>a$$ 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!

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

$$\mathbb{T}=\frac{|F|^2}{|A|^2}.$$

• Thanks for pointing that out. I already corrected that. However this does not answer my question, as I still don’t know how to proceed from here. What strategy should be used to solve that system of equations to get B and F in terms of A? Commented Apr 7, 2021 at 22:52
• I have edited my answer to make it more explicit.
– AFG
Commented Apr 8, 2021 at 7:39
• Understood. Thanks! The problem I had was that I expected to be able to divide equations by one another, but it turns out that it is necessary to do the whole process subsituting one by one Commented Apr 8, 2021 at 12:45