2
$\begingroup$

My question is about understanding the different solutions of the potential square well.

Imagine a square well defined this way:

$$ V(x) = \begin{cases} ∞&\,{\rm if} x<0 \\ 0&\,{\rm if}\,x\in\left(0,L\right) \\ ∞&\,{\rm if} x>L. \end{cases} $$

When applying the time-independent Schrodinger Equation we get:

$$ -\frac{\hbar^2}{2m} \frac{∂^2\psi}{∂x^2} =E\psi,$$ and then:

$$ \frac{∂^2\psi}{∂x^2} =-\frac{2mE}{\hbar^2}\psi.$$

If we define $k_1^2=-\frac{2mE}{\hbar^2}$ we get the solution $$\psi_1=Ae^{k_1x}+Be^{-k_1x}.$$

But if we define $k_2^2=\frac{2mE}{\hbar^2}$ we get the solution $$\psi_2=Ae^{ik_2x}+Be^{-ik_2x}.$$

The way I see it one solution describes the wave-function using real-valued exponential functions, while the other describes it using complex-valued exponentials (or sines and cosines, using Euler's formula).

With this "mathematical difference", can anyone help me understand if there is any "physical difference" between both solutions? Do they describe different wave-functions? Is there something simple I'm missing?

PS: This question is not homework. Is about me trying to understand the solutions of the infinite potential square well.

$\endgroup$
3
$\begingroup$

As $E$ is always positive, your $k_1$ is imaginary. It incorporates the $i$ that is visible in the second solution. They are really the same solution, with $k_1=ik_2$. If your square well is finite, outside the well we have $E \lt V$ and the $E$ in your solutions becomes $E-V$. Then the first solution has real $k_1$ and represents the tunneling into the walls.

$\endgroup$
  • $\begingroup$ Oh my god. Thank you so much! I was confusing this for so long! Thanks. You sure opened my eyes. $\endgroup$ – Dayman75 Jan 9 '14 at 19:01

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.