I have a problem with the argument of a finite square well.

enter image description here enter image description here

The stuff I read has mentioned that the Curvature " second derivative " is opposite sign of the wave function only when the E larger than V where E = energy of the particle V = potential.This is how we obtained oscillatory solution (when E is larger than V) ( I understood this part )

And For E smaller than V ,it is true that we obtain exponential solutions, but can we use the same argument I used above? Is it true that as u = ( exp(-ikx ) decays the curvature has the same sign, so the negative gradient starts getting to be more positive but less and less positive until it tails down to infinity? since now second derivative u depends on the value of u and the difference between E and V.

Is it a correct way of interpreting this phenomena?

Appreciate any idea or help from every one.


I think when $E-V<0$ then the solution is $u = Ce^{-kx}$ (for constant $V$ in that region), not $e^{-ikx}$ and the second derivative necessarily has the same sign as the solution..

$u=Ce^{kx}$ is also a solution in the $E-V<0$ region but this is unphysical as it represents unlimited growth of the wave-function for increasing $x$.

  • $\begingroup$ Sorry I should have mentioned my k here is imaginary... It's a k = i gamma. My question is If anyone agree with my second derivative argument? for the exponential decay $\endgroup$ – el psy Congroo Oct 14 '13 at 9:32

In the right part of the graph, $x \ge x_0$, you have $E < V$ and then : $\frac{d^2u}{dx^2} = \Lambda u$, with $\Lambda >0$.

You may note, that for $x \ge x_0$, $u >0$ , and then $\frac{d^2u}{dx^2} >0$.

So, assuming $u>0$ the gradient $\frac{du}{dx}$ becomes less and less negative, or more and more positive, or, a succession of these 2 cases.

With the graph you provide, it is clear that $u>0$ and that the gradient is zero for $x \to +\infty$, so necessarily the gradient has to be negative for $x=x_0^+$, and this gradient becomes less and less negative, when $x$ is increasing, and finally going to zero for $x \to +\infty$


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.