Given the following:

enter image description here

The potential is:

$V(x) = 0$ , $x<0$

$V(x) = V_o$ , $0\leq x\leq a$

$V(x) = 0$ , $x > a$

In the three regions the solutions to Schrondinger Eq. are:

$\psi_1 = Ae^{ik_1x} + Be^{-ik_1x}$

$\psi_2 = Ce^{ik_2x} + De^{-ik_2x}$

$\psi_3 = Fe^{ik_3x} + Ge^{-k_3x}$


$k_1 = k_3 = \sqrt{\frac{2mE}{\hbar^2}}$

$k_2 = \sqrt{\frac{2m(E-V_o)}{\hbar^2}}$

The goal is to show that

$T = \frac{j_{transmitted}}{j_{incident}}= \frac{|F|^2}{|A|^2}$

reduces to

$T = \frac{1}{1 + \frac{1}{4}\frac{V_o^2}{E(E-V_o)}sin^2k_2a}$ (Eq. 2.37 in the textbook)

I am unable to make any progress after the below steps

$A + B = C + D = F + G$

$A + B = C + D = F + 0$

$k_1(A - B) = k_2(C - D) = k_3F$

$F = A - B = \frac{k_2(C - D)}{k_3}$

$A = \frac{k_2(C - D)}{k_3} + B = F + B$

$T = \frac{(\frac{\sqrt{\frac{2m(E-V_o)}{\hbar^2}}(C-D)}{\sqrt{\frac{2mE}{\hbar^2}}})^2}{(\frac{\sqrt{\frac{2m(E-V_o)}{\hbar^2}}(C-D)}{\sqrt{\frac{2mE}{\hbar^2}}}+B)^2}$

How can I get the above into the desired form of Eq. 2.37?


1 Answer 1


Remember that for the wavefunction, we have two sorts of boundary conditions: continuity, AND continuity of its derivatives (where the latter applies if we do not have an infinite potential). So when you write the first equation comes from continuity of $\psi$:

$$\psi_1(x=0) = \psi_2(x=0) \implies A + B = C + D$$

But we can acquire even more boundary conditions if we consider the continuity of the derivatives:

$$\psi_1'(x=0) = \psi_2'(x=0)$$

and there will also be similar equations for $\psi_2$ and $\psi_3$. (And since this is a homework question, you have the fun of working it out!)

  • $\begingroup$ Give me a bit to work on it and I'll return with my progress $\endgroup$
    – snowg
    Aug 20, 2020 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.