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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 ....
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This note was uploaded on 05/29/2011 for the course PHY 4064 taught by Professor Fields during the Spring '07 term at University of Florida.
 Spring '07
 FIELDS
 mechanics, Energy

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