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08square - Finite Square Well Potential For V=finite...

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P460 - square well 1 Finite Square Well Potential For V=finite “outside” the well. Solutions to S.E. inside the well the same. Have different outside. The boundary conditions (wavefunction and its derivative continuous) give quantization for E<V 0 longer wavelength, lower Energy. Finite number of energy levels Outside: h h h h ) ( 2 2 ) ( 2 0 2 ) ( 0 0 2 2 2 0 2 2 2 ) ( cos , sin ) ( E V m x k V E m m k dx d m k e V E k V E and k x k V E V E ± = < = = > = ψ ψ h mE k 2 1 =
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P460 - square well 2 Boundary Condition Want wavefunction and its derivative to be continuous often a symmetry such that solution at +a also gives one at -a Often can do the ratio (see book) and that can simplify the algebra x boundary x boundary x boundary x boundary boundary boundary = = = + + + + ) ( 1 ) ( 1 ) ( ) ( ) ( ) ( ψ ψ ψ ψ ψ ψ ψ ψ
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P460 - square well 3 Finite Square Well Potential 0 , 0 ) ( : ) ( : cos sin ) ( : 2 2 2 2 2 2 1 1 = ±∞ + = > + = < + = F D x as as Ge Fe x x De Ce x x x k B x k A x inside x k x k a x k x k a ψ ψ ψ ψ Equate wave function at boundaries 2 / 2 2 2 / 2 2 2 1 1 2 1 1 cos sin cos sin a k a k a k a k a k a k Ce B A Ge B A = + = + And derivative 2 / 2 2 1 2 1 2 / 2 2 1 2 1 2 1 1 2 1 1 sin cos sin cos a k a k a k a k a k a k e Gk Bk Ak e Ck Bk Ak = = +
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P460 - square well 4 Finite Square Well Potential on quantizati k k x e e B x well in x k B x II on quantizati k k x e e A x well in x k A x I a k a x k a k a k a k a x k a k a k 2 2 1 2 2 / 2 1 2 2 1 2 2 / 2 1 1 2 2 1 1 2 2 1 tan ) cos( ) ( cos ) ( : cot ) sin( ) ( sin ) ( : = > = = = > = = ψ ψ ψ ψ E+R does algebra. 2 classes. Solve numerically k1 and k2 both depend on E. Quantization sets allowed energy levels 0 2 2 2 2 2 V E ma n n π h
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P460 - square well 5 Finite Square Well Potential Number of bound states is finite. Calculate assuming “infinite” well energies. Get n. Add 1 Electron V=100 eV width=0.2 nm 2 2 0 2 2 2 2 2 2 2 0 2 π π h h V ma ma n n n V E ) ( 4 7 . 10 2 2 2 ) 14 . 3 ( ) 197 ( 100 ) 2 (. 51 . 2 2 levels of number N n nm ev eV nm MeV = = Deuteron p-n bound state. Binding energy 2.2 MeV radius = 2.1 F (really need 3D S.E………) state bound only N n F MeV MeV F MeV 1 1 1 . 0
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