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

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