Lecture 15 Review. Fractal dimension of Sierpinski s gasket. Affine transformations and fractal examples. ection algorithm.

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1 Lecture 15 Review Fractal dimension of Sierpinski s gasket. Affine transformations and fractal examples. Root finding: N-R and bi-se ection algorithm. TE Coan/SMU 1

2 HW aside: Newton-Raphson Beware Software ain t magic... Potential pathologies using N-R root finding technique df(x)/dx = 0 S -shaped curves nasty NR solution potentially sensitive to initial guess. Beware. (Graph, too.) TE Coan/SMU 2

3 Particle in a Quantum Box -a V a x -V 0 h2 2m d 2 ψ(x) dx +V(x)ψ(x)= =Eψ(x) 2 a/3 P= ψ(x) 2 dx dx ψ(x) 2 =1 V(x)= { b/3 V0 = 83MeV, for x a=2fm 0, for x >a=2fm d 2 ψ(x) dx 2 + 2m (E+V h 2 0 )ψ(x)=0 for x a d 2 ψ(x) dx 2 + 2m h 2 Eψ(x)=0 for x > a 2m h 2 = 2mc2 ( hc) 2 = 2 940MeV (197Mev fm) 2 2 =0.0483MeV 1 fm 2 TE Coan/SMU 3

4 ψ(x)= Particle in a Quantum Box Ce βx, for <x< aa α= 2m(E+V 0 ) = (E+83) h Bcosαx, for a<x<a 2 Ce βx, fora<x<+ β= = E 2mE h 2 Bcosαa=Ce βa ψ continuity αbsinαa= βce βa ψ continuity αatanαa βa=0 V 0 2m(E+V0 )tan 2m(E+V ) a 2mE=0 h 2 ξtanξ η=0 with ξ=αa η=βa ξ 2 +η 2 = 2mV 0a 2 h 2 =16.08 f(e)=ξtanξ η=0 Even parity solutions only. TE Coan/SMU 4

5 ψ(x)= Particle in a Quantum Box (2) Ce βx, for <x< a α= 2m(E+V 0 ) = (E+83) h Bsinαx, for a<x<a 2 Ce βx, for a<x<+ β= = E 2mE h 2 Bsinαa=Ce βa ψ continuity αbsinαa= βce βa ψ continuity αacotαa+βa=0 2m(E+V0 )cot 2m(E+ +V 0 ) h 2 2 a+ 2mE=0 ξcotξ+η=0 with ξ=αa η=βa ξ 2 +η 2 = 2mV 0a 2 h 2 =16.08 f(e)=ξcotξ+η=0 Odd parity solutions only. TE Coan/SMU 5

6 Numerov Method for ODE Solution Runge-Kutta (4 th order) is our preferred solver for ODEs. Why not use it to solve Schrödinger equation? For special case of dy/dx missing (e.g., Schrödinger eqn) Use Numerov s method: More accurate than RK-4 th order AND w/ fewer steps. When applied to potential wells, Numerov s method is an example of solving ODE w/ boundary conditions (as opposed to ICs). TE Coan/SMU 6

7 Numerov Method for ODE Solution (2) h2 d 2 ψ(x) 2m dx 2 +V(x)ψ(x) = Eψ(x) d 2 ψ(x) dx 2 +k 2 (x)ψ(x)=0 k 2 (x) ( 2m h 2) { E E V 0 x <a E x >a Recall that E < 0 for bound states ( 2m )= Mev 1 fm 2 h 2 -a V a x -V 0 TE Coan/SMU 7

8 Numerov Method for ODE Solution (3) ψ(x+h)=ψ(x)+hψ + h 2 ψ(x h)=ψ(x) hψ + h2 2 2 ψ h3 3! ψ + h4 4! ψiv +... ψ(x+h)+ψ(x h) 2ψ(x)+h 2 ψ + h4 12 ψiv +O(h 6 ) ψ (x) ψ(x+h) ψ(x h) 2ψ(x) h 2 h 2 2 ψ + h3 3! ψ + h4 4! ψiv +... ) h2 12 ψiv (x)+o(h 6 ) Trick: Apply 1+ h2 12 d dx 2 d 2 to TISE ( 1+ h2 12 )( ) d 2 d 2 ψ dx 2 dx 2 +k 2 (x)ψ =0 TE Coan/SMU 8

9 courage TE Coan/SMU 9

10 Numerov Method for ODE Solution (5) ψ(x+h) 2[ h2 k 2 (x x)]ψ(x) [1+ h2 12 k2 (x h)]ψ(x h) 1+ h2 12 k2 (x+h) ψ 2( i+1 12 h2 k k 2 i )ψ i (1+ h2 1+ h2 12 k2 i+1 k2 )ψ 12 i 1 i 1 See numerov.cc TE Coan/SMU 10

11 numerov.ccc discussion ψ i+1 2( h2 k 2 i )ψ i (1+ h2 1+ h2 12 k2 i+1 12 k2 i 1 )ψ i 11 Basic algorithm. Start w/ ψ 0 and ψ 1 specified. Symmetric QM wells have ψ s of definite parity. Recall, ψ(-x) = ψ(x), even parity ψ(-x) = -ψ(x), odd parity -a a x ψ(x) 2 and ψ(-x) 2 have physical significance. -V 0 ψ and ψ must be continuous. numerov.cc lets you pick parity. (By selecting sign of ψ 1.) numerov.cc as coded does not check for continuity of ψ. TE Coan/SMU 11

12 Numerov Technique Lab Exercise Modify numerov.cc Limit number of possible iterations to, say, Modification should tell you if you reached this limit. Alter potential from a square well to a V-shaped well. -a V a -V 0 x Find energy eigenvalues and eigenstates (i.e., find permissible energy levels & corresponding wavefunctions.) TE Coan/SMU 12

13 Summary Application of root finding to particle in a quantum box Numerov technique for ODE solution. Numerical solution of square well potential. Don t suffer in silence. Scream for help!!! TE Coan/SMU 13

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