Lets say we have a finite square well symetric around ##y## axis (picture below).(adsbygoogle = window.adsbygoogle || []).push({});

I know how and why general solutions to the second order ODE (stationary Schrödinger equation) are as follows for the regions I, II and III.

\begin{align}

\text{I:}& & \psi_{\text{I}}&= Ae^{\kappa x} \\

\text{III:}& & \psi_{\text{III}}&= Be^{-\kappa x} \\

\text{II:}& & \psi_{\text{II}}&= C \cos(k x) + D\sin(kx)

\end{align}

But now i got to a point where i have to start applying a boundary conditions to get a speciffic solution. So i start with the 1st boundary condition which is ##\psi_{\text{I}}\left(-\frac{d}{2}\right)=\psi_{\text{II}}\left(-\frac{d}{2}\right)## for the left potential shift and ##\psi_{\text{II}}\left(\frac{d}{2}\right)=\psi_{\text{III}}\left(\frac{d}{2}\right)## for the right potential shift. These leave me with a system of 2 equations (one for left and one for right potential shift):

\begin{align}

{\scriptsize\text{left potential shift:}}& & Ae^{-\kappa \frac{d}{2}} &= C \cos\left(k\tfrac{d}{2}\right) - D\sin\left(k \tfrac{d}{2}\right)\\

{\scriptsize \text{right potential shift:}}& & Be^{-\kappa \frac{d}{2}} &= C \cos\left(k\tfrac{d}{2}\right) + D\sin\left(k \tfrac{d}{2}\right)\\

\end{align}

Question 1:

From here on authors of most books don't seem to explain much. Most of them only say that we must use ##\boxed{D\!=\!0}## to solve foreven solutionsand ##\boxed{C\!=\!0}## to solve for *odd solutions*. What is this argument based on?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Finite potential well

**Physics Forums | Science Articles, Homework Help, Discussion**