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1 WKB Wavefunctions for Simple Harmonics Masatsugu WKB wavefunctions for simple harmonics Masatsugu Sei Suzuki Department of Physics, SUNY at Binmghamton (Date: November 19, 2011) _________________________________________________________________________ Gregor Wentzel (February 17, 1898, in Düsseldorf, Germany – August 12, 1978, in Ascona, Switzerland) was a German physicist known for development of quantum mechanics. Wentzel, Hendrik Kramers, and Léon Brillouin developed the Wentzel– Kramers–Brillouin approximation in 1926. In his early years, he contributed to X-ray spectroscopy, but then broadened out to make contributions to quantum mechanics, quantum electrodynamics, and meson theory. http://en.wikipedia.org/wiki/Gregor_Wentzel _________________________________________________________________________ Hendrik Anthony "Hans" Kramers (Rotterdam, February 2, 1894 – Oegstgeest, April 24, 1952) was a Dutch physicist. http://en.wikipedia.org/wiki/Hendrik_Anthony_Kramers _________________________________________________________________________ Léon Nicolas Brillouin (August 7, 1889 – October 4, 1969) was a French physicist. He made contributions to quantum mechanics, radio wave propagation in the atmosphere, solid state physics, and information theory. http://en.wikipedia.org/wiki/L%C3%A9on_Brillouin _______________________________________________________________________ 1. Determination of wave functions using the WKB Approximation 1 In order to determine the eave function of the simple harmonics, we use the connection formula of the WKB approximation. V x E III II I x b O a The potential energy is expressed by 1 V (x) m 2 x 2 . 2 0 The x-coordinates a and b (the classical turning points) are obtained as 2 2 a 2 , b 2 , m0 m0 from the equation 1 V (x) m 2 x2 , 2 0 or 1 1 m 2a 2 m 2b2 , 2 0 2 0 where is the constant total energy. Here we apply the connection formula (I, upward) at x = a. 2 Connection formula-I (upward): 2A a B a cos( k(x)dx ) sin( k(x)dx )] II k(x) x 4 k(x) x 4 A x B x exp( (x)dx) exp( (x)dx)] I (x) a (x) a In I , B should be equal to zero: B = 0. Then we have A x exp( (x)dx) , I (x) a for x>b. The wave function II is obtained as 2A a cos( k(x)dx ) II k(x) x 4 2A b a cos( k(x)dx k(x)dx ) k(x) x b 4 2A x a cos( k(x)dx k(x)dx ) k(x) b b 2 4 2A a x sin{ k(x)dx [ k(x)dx ]} k(x) b b 4 2A a x a x [sin( k(x)dx)cos( k(x)dx ) cos( k(x)dx)sin( k(x)dx )] k(x) b b 4 b b 4 Next we use the connection formula (II, downward) at x = b. Connection formula-II 3 C b D b exp( (x)dx) exp( (x)dx)] III 2 (x) x 2 (x) x C x D x cos( k(x)dx ) sin( k(x)dx )] II k(x) b 4 2 k(x) b 4 The comparison of this equation with 2A a x 2A a x sin( k(x)dx)cos( k(x)dx ) cos( k(x)dx)sin( k(x)dx )] , II k(x) b b 4 k(x) b b 4 yields the relation between C, D, and A, 2A a a C sin( k(x)dx) , and D 4Acos( k(x)dx) , k(x) b b In III , D should be equal to zero: D = 0. This means that a a D 4Acos( k(x)dx) 0 , cos( k(x)dx) 0 , b b or a 1 k(x)dx (n ) , b 2 where n is a positive integer. This integral can be calculated as 4 a 2m a 1 k(x)dx m 2 (a 2 x 2 )dx 0 b a 2 m a 0 a 2 x 2 dx a m a 2 1 0 (n ) 2 2 which means that 1 2 2 1 2 2 1 1 m0 a m0 (n ) 0 (n ) 2 2 m0 2 2 In other words, the energy is quantized. This is amazing. Noting that a 1 sin[ k(x)dx] sin[(n ) ] cos(n ) (1)n , b 2 we have the final forms of the wave functions A a b sin( k(x)dx)exp( (x)dx) III k(x) b x A(1)n b exp((x)dx) k(x) x for x<b and 2A a x sin( k(x)dx)cos( k(x)dx ) II k(x) 4 b b 2A(1)n x cos( k(x)dx ) k(x) b 4 for b<x<a. 5 2. WKB wave functions for simple harmonics We introduce a new variable (dimensionless) as, x , with m 0 . The parameters a and b are rewritten as 1 2(n )0 2 2 (2n 1) 2n 1 a 2 2 , m0 m0 m0 and 2n 1 b . We also note that (x) and k(x) are expressed by 2m 1 (x) V (x) 0 (n ) 2 2m 1 2 2 1 m0 x 0 (n ) 2 2 2 (2n 1) 6 2m 1 k(x) 0 (n ) V (x) 2 2m 1 1 2 2 0 (n ) m0 x 2 2 (2n 1) 2 Using these parameters, we have x (x)dx s 2 (2n 1)ds , a 2n1 x k(x)dx s 2 (2n 1)ds . b 2n1 The wavefunction in the region II; 2A(1)n x cos( k(x)dx ) II k(x) b 4 A 2(1)n cos[ s 2 (2n 1)ds ] 2 1/ 4 [(2n 1) ] 2n1 4 ____________________________________________________________________ The wavefunction in the region I A x exp[ (x)dx] I (x) a A 1 exp[ s 2 (2n 1)ds] 2 1/ 4 [ (2n 1)] 2n1 The wavefunction in the region III 7 A(1)n b exp[ (x)dx] III (x) x A (1)n 2n1 exp[ s 2 (2n 1)ds] 2 1/ 4 [ (2n 1)] We assume that the parameter A/ is determined from the normalization condition of the wave function. 3. Result from Mathematica: n () 6 5 4yn x 3 n=0 2 1 x -4 -2 2 4 Fig. n = 0 (ground state). The blue lines are classical turning points. y4 n x n=1 2 x -6 -4 -2 2 4 6 -2 -4 8 Fig. n = 1. The blue lines are classical turning points. y4 n x 3 n=2 2 1 x -6 -4 -2 2 4 6 -1 Fig. n = 2. The blue lines are classical turning points. y4 n x n=3 2 x -6 -4 -2 2 4 6 -2 -4 Fig. n = 3. The blue lines are classical turning points. 9 y4 n x 3 n=4 2 1 x -6 -4 -2 2 4 6 -1 Fig. n = 4. The blue lines are classical turning points. y4 n x n=5 2 x -6 -4 -2 2 4 6 -2 -4 Fig. n = 5. The blue lines are classical turning points. 10 y4 n x 3 n=6 2 1 x -6 -4 -2 2 4 6 -1 Fig. n = 6. The blue lines are classical turning points. In summary; we see that the WKB solution agrees well with the solutions from the quantum mechanics 2 4. Result from Mathematica: n ( ) 7 6 5 4 2 3yn x n=0 2 1 x -6 -4 -2 2 4 6 11 4 2 3yn x n=3 2 1 x -6 -4 -2 2 4 6 4 2 3yn x n=4 2 1 x -6 -4 -2 2 4 6 2.0 2 1.5yn x n1.0=15 0.5 x -10 -5 5 10 12 2.0 2 1.5yn x n1.0=20 0.5 x -10 -5 5 10 _________________________________________________________________________ REFERENCES 1. David J. Griffiths, Introduction to Quantum Mechanics (Prentice Hall, Englewood Cliff, NJ, 1995). 2. David. Bohm, Quantum Theory (Dover Publication, Inc, New York, 1979). 3. Eugen Merzbacher, Quantum Mechanics, 3rd edition (John Wiley & Sons, New York, 1998). 4. Leonard Schiff, Quantum Mechanics (McGraw-Hill Book Company, Inc, New York, 1955). 5. Richard L. Liboff, Introductory Quantum Mechanics, 4th edition (Addison Wesley, 2002). _________________________________________________________________________ APPENDIX-1: Mathematica program 13 We calculate the following integrals separately, since it takes a quite long time to calculate. Here we use the results. g3 Integrate s2 2n1 1 , s, 1, 2n1 1 Simplify , n1 0, 1 2n1 1 &; g1 Integrate s2 2n1 1 , s, 2n1 1, 1 Simplify, n1 0, 1 2n1 1 &; g2 Integrate 2n11 s2 , s, 2n1 1, 1 Simplify, n1 0, 2n1 1 1 2n1 1 &; 1 g11 2 1 1 2 n1 12 1 2 n1 Log 1 2 n1 2 Log 1 1 2 n1 12 ; 4 1 1 g21 2 n1 2 1 1 2 n1 12 2 4 n1 ArcTan ; 4 1 2 n1 12 g31 1 1 1 2 n1 12 1 2 n1 Log 1 2 n1 2 1 2 n1 Log 1 1 2 n1 12 ; 1 1 Exp g11 ; 12 2 n1 1 1 n1 3 Exp g31 ; 12 2 n1 1 2 1 n1 2 Cos g21 ; 4 12 2 n1 1 14 n_, _ : 3 UnitStep 1 2 n1 1 2 UnitStep 1 2 n1 1 UnitStep 1 2 n1 1 1 UnitStep 1 2 n1 1 . n1 n, 1 Simplify; P n_, s_ : Module h1, h2, h3, n1 , n1 n; h1 Plot n1, , , s, s , PlotStyle Hue 0.1 n1 , Thick , Background LightGray ; h2 Graphics Blue, Thick, Line 2 n1 1, 0 , 2 n1 1, 1 n1 4 , Line 2 n1 1, 0 , 2 n1 1, 4 , Text Style"", Black, 15 , s 0.5, 0 , Text Style " ", Black, 15 , 0.3, 4 , n Text Style "n" ToString n1 , Black, 15 , 3, 3 ; h3 Show h1, h2 , PlotRange All ; Q n_, s_ : Module h1, h2, h3,n1 , n1 n; h1 Plot n1, 2, , s, s, PlotStyle Hue 0.1 n1 , Thick , Background LightGray ; h2 Graphics Blue, Thick, Line 2 n1 1, 0, 2 n1 1, 4 , Line 2 n1 1, 0 , 2 n1 1, 4 , Text Style"", Black, 15 , s 0.5, 0 , 2 TextStyle " n ", Black, 15 , 0.5, 3 , Text Style "n" ToString n1 , Black, 15 , 3, 3 ; h3 Show h1, h2 , PlotRange All ; _________________________________________________________________________ APPENDIX-2: Connection formula 15 16 .
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