0
$\begingroup$

Barrier Potential

Where $k_1=\frac{\sqrt{2mE}}{\hbar}$ and $\alpha=\frac{\sqrt{2m(V_0-E)}}{\hbar}$

I'm quite confused as to why the exponentials in regions I and III are complex functions while in region II the exponentials are real valued. I'm pretty sure in region II it's because $\alpha>0$ although this doesn't appear to be a good explanation since $k_1$ is also positive.

$\endgroup$
1
$\begingroup$

In region (1) since $V=0$ the SE becomes

$$\frac{h^2}{2m}\frac{\partial^2 \psi}{\partial x^2}=-E\psi$$

since $E>0$ we always need $k$ to be real so we take $k=\frac{\sqrt{2mE}}{\hbar}$

This will give solutions of the form $\psi=Ae^{ikx}+Be^{-ikx}$

In region(2) SE becomes $$\frac{h^2}{2m}\frac{\partial^2 \psi}{\partial x^2}=-(E-V_0)\psi$$

but observe that $E<V_0$ so we cannot define $k=\frac{\sqrt{2m(E-V_0)}}{\hbar}$ since k will become an imaginary number, so we have to define

$k=\frac{\sqrt{2m(V_0-E)}}{\hbar}$

Which will give solutions of the kind $\psi=Ae^{kx}+Be^{-kx}$

$\endgroup$

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.