This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: CHEM 356, Lecture 14, Fall 2009 1 Effect of a Step Potential. Idealized Step Potential There are two cases to be considered for a simple step potential of the type illustrated in the figure: either the total energy for a particle moving towards the potential step is less than or equal to the step-height, ? ≤ , or it is greater than the step–height, ? > . It is natural to place the origin for the –axis at the potential step, thereby splitting the –axis into two regions, which we shall label as regions I (with ≤ 0), and II (with > 0). As indicated, we shall consider motion from region I towards region II. i) For ? < , we have ? = 2 / (2 ) + ( ) < ( ), > 0. Classically: particle encounters the barrier, and is reﬂected back. Quantum Mechanically: more complicated. ≤ 0: − ℏ 2 2 2 I 2 = ? I ( ) I ( ) = ? e 1 + ? e − 1 ℏ 1 = √ 2 ? > 0: − 2 II 2 = 2 ( ? − ) ℏ 2 II ( ) II ( ) = ? e 2 + ? e − 2 ℏ 2 = √ 2 ( − ? ) We shall, of course, require that ( ) be well–behaved, which implies that ( ) < ∞ ∀ , and this implies that ? = 0. CHEM 356, Lecture 14, Fall 2009 2 Moreover, we have proposed to solve the Schr¨ odinger equation separately in each of the two regions that we have labelled I and II and, in order that the solution to the Schr¨ odinger equation be smooth and well–behaved over the entire domain (in this case, for all values of between −∞ and + ∞ ), we must join the two solutions at their common boundary (or boundaries, in more complicated cases in which we find more than two solutions separately). This means that we need to introduce appropriate matching conditions , which are: that ( ) be finite and continuous, and that ′ ( ) be finite and continuous across a boundary ....
View Full Document
This note was uploaded on 02/28/2011 for the course CHEM 356 taught by Professor Prof.iaskjd during the Fall '09 term at Waterloo.
- Fall '09