Chapter2_20 - Also notice that for → ε we have 2 2 2 2 2...

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PHY4604 R. D. Field Department of Physics Chapter2_20.doc University of Florida Delta-Function Potential – Bound States Consider a potential of the form ) ( ) ( x x V αδ = . We look for solutions of the time-independent Schrödinger equation ( i.e. stationary state solutions): ) ( ) ( ) ( ) ( 2 2 2 2 x E x x dx x d m ψ ψ αδ ψ = h with h / ) ( ) , ( iEt e x t x = Ψ ψ . This equation yields both bound states (E < 0) and scattered states (E > 0). Bound State Solutions: For E < 0 in the region x < 0 we have ) ( ) ( 2 ) ( 2 2 2 2 x x mE dx x d ψ κ ψ ψ = = h with 2 2 h mE = κ and m E 2 2 2 κ h = The most general solution is x x L Be Ae x κ κ ψ + + = ) ( , but −∞ x x e κ and hence B = 0 . Similarly for x > 0 , x x R De Ce x κ κ ψ + + = ) ( , but +∞ + x x e κ and hence C = 0 . Boundary Conditions: Require ψ (x) be continuous and d ψ (x)/dx to be continuous ( except where V is infinite ). Thus, at x = 0
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Unformatted text preview: . Also notice that for → ε we have ) ( 2 ) ( ) ) ( ( 2 ) ( ) ) ( ( 2 ) ( 2 2 2 2 2 α h h h m dx x E x m dx x E x V m dx dx x d − = − − = − = ∫ ∫ ∫ + − + − + − Thus, ) ( 2 ) ( 2 ) ( ) ( 2 2 L R x L x R m m dx x d dx x d h h − = − = − − = + = which implies that D m A D 2 2 h − = − − but A = D and hence 2 h m = . Thus there is only one allowed energy ( i.e. one bound state) given by 2 2 2 h m E − = and h h / | | ) ( x m e m x − = , where I have normalized ψ (x) so that < ψ | ψ > = 1. x V -αδ (x)...
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