{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter6_32

# Chapter6_32 - h iEt e x t x − = Ψ ψ Substituting Ψ(x,t...

This preview shows page 1. Sign up to view the full content.

PHY3063 R. D. Field Department of Physics Chapter6_32.doc University of Florida Barrier Problems – Finite Step Potential x x 0 E Potential Barrier V(x) V 0 KE Classically Allowed Classically Forbidden x x 0 = 0 E Step Potential Classically Forbidden Region V 0 Consider a particle with total energy E entering from the left and encountering the potential V(x) . The particle’s kinetic energy is ) ( 2 1 2 2 2 x V E mv m p KE x x = = = . At the point where V(x) = E the particle has zero velocity and it turns around and heads back out in the direction it came in. The point x 0 is called the “classical turning point” and the region with V(x) > E is called the “classically forbidden region” . Time-Independent Schrödinger Equation: Look for solutions of the time- dependent Schrödinger equation of the form
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: h / ) ( ) , ( iEt e x t x − = Ψ ψ . Substituting Ψ (x,t) into the time dependent equation yields ) ( )) ( ( ) ( 2 2 2 2 x x V E dx x d m Φ − = − h . Step Potential: Consider a potential V(x) such that V(x) = 0 for x < 0 and V(x) = V for x ≥ . Left Region (x < 0): In this region ) ( ) ( 2 2 2 x k dx x d L L − = with 2 2 h mE k = and the most general solution is ikx L ikx L L e B e A x − + + = ) ( Right Region (x ≥ 0): In this region ) ( ) ( 2 2 2 x dx x d R R κ = with 1 ) ( 2 2 − = − = E V k E V m h and the most general solution is x R x R R e B e A x + − + = ) ( . The second term is unphysical since ∞ → ∞ → x x e and hence B R = 0 ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online