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Quantum Physics 2005 Notes-6 Solving the Time Independent Schrodinger Equation (A Numerical Approach) Notes 6 Quantum Physics F2005 1 The Schrodinger Equation Hˆ ! = Eˆ! h2 " 2 "! # ! + V ( x, t )! = ih 2 "t 2m "x 2 " ! ( x) 2m = # 2 ( E # V ( x))! ( x) 2 "x h Notes 6 Quantum Physics F2005 2 A dimensionless form for calculations #The coefficient 2m 38 has a value of ~ 2 $ 10 for an electron 2 h in MKS units . #Big and small numbers are problems in repetitive numerical calculations, so we will scale it out by redefining making the original length and energy variables dimensionless. #The choice of scale is arbitrary, so we can make it to fit the problem. Notes 6 Quantum Physics F2005 3 A specific example #For a finite square well of width L, we expect the energies to be h2$ 2 of order: E0 = and useful lengths to be of order L. 2 2mL #As a concrete example, let's take a finite square well: L = 10 nm and m = 9 x10-31 kg 1.242 Thus, E0 = ~ 0.00375 eV 4.09 •100 We will also take V0 = 4 E0 to finish setting up the example. Notes 6 Quantum Physics F2005 4 Letting z = x / L and scaling % =E / E0 and W = V / E0 we have: " 2! ( z ) 2 (V ( x ) # E ) 2 = ( $ ) ! ( ) = ( $ ) z (W # % )! . 2 "z E0 We are now ready to solve this TISE numerically. Convert the TISE to a finite difference equation. a) z ⇒ z j = j &z (step size chosen for accuracy and time) b) ! ( z ) ⇒ ! ( z j ) ' ! j c) W ( z ) ⇒ W ( z j ) ' W j ! ' j+1/2 " j ! = Notes 6 ! ( ' j +1 #! j ) &z ' ' ! j+1/2 #! j-1/2 &z = ! j +1 # 2! j +! j #1 &z Quantum Physics F2005 5 Making a difference equation " 2! ( z ) 2 Plugging the differences into = ( $ ) (W # % )! . 2 "z We get: ! j+1 = 2 # &z 2 ($ ) 2 ( % # W j ) ! j #! j #1 If we know the values of ! j and ! j #1 near some point, we can solve for ! j+1. We can usually get ! j and ! j #1 from the continuity or symmetry conditions at a point. The only parameter with which to achieve agreement of the wavefunction with expected behavior is % . Notes 6 Quantum Physics F2005 6 The starting point • We can choose the starting point based on symmetry. • For the finite well, even solutions have zero slope at the center. (Let’s take !0= !1=1.) • Expected asymptotic behavior: For z growing to large values (outside the well) we expect the wavefunction to go to zero. Notes 6 Quantum Physics F2005 7 A spreadsheet example Position !j Wj 0 1 0 1 1 0 2 2 # &z 2 ($ ) 2 ( % # W j ) ! j #! j #1 0 3 .. 0 .. .. 4 n Vary E to get the right behavior out here. 4 file: C:\Documents and Settings\Peter Persans\My Documents\classes\quantumphysics\lectures\numerical_integration_TISE_even_square_well.xls Notes 6 Quantum Physics F2005 8