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By Heitler W.

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Example text

We now give a few further examples by specifying the nature of the potential energy V. (a) The one-dimensional harmonic oscillator If m is the mass of the oscillator of force constant k, then the Hamiltonian is  2 2 kx2 ^ ¼ À h r þ H 2m 2 ð1:62Þ (b) The atomic one-electron problem (the hydrogen-like system) If r is the distance of the electron of mass m and charge Àe from a nucleus of charge þZe (Z ¼ 1 will give the hydrogen atom), then the Hamiltonian in SI units6 is 2 2 1 Ze2 ^ ¼À h r À H 4p«0 r 2m ð1:63Þ To get rid of all fundamental physical constants in our formulae we shall introduce consistently at this point a system of atomic units7 (au) by posing ð1:64Þ e ¼ h ¼ m ¼ 4p«0 ¼ 1 The basic atomic units of charge, length, energy, and time are expressed in SI units as follows: 8 > charge; e e ¼ 1:602 176 462  10 À 19 C > > > > > h2 > > > length; Bohr a ¼ 4p« ¼ 5:291 772 087  10À11 m 0 0 > 2 > me > < 1 e2 > energy; Hartree Eh ¼ ¼ 4:359 743 802  10À18 J > > 4p« a > 0 0 > > > > h > > > ¼ 2:418 884 331  10À17 s time t¼ > : Eh ð1:65Þ 5 The quantities observable in physical experiments must be real.

28) is divergent at jxj ¼ 1, so that once again the series must be truncated to a polynomial. 29) vanishes, then ðk þ mÞðk þ m þ 1Þ À l ¼ 0 ð3:31Þ giving6 l ¼ ðk þ mÞðk þ m þ 1Þ k; m ¼ 0; 1; 2; . . ð3:32Þ k þ m ¼ ‘ a non-negative integer ð‘ ! 0Þ ð3:33Þ Posing we obtain ‘ ¼ m; m þ 1; m þ 2; . . ‘ ! jmj À‘ m ‘ ð3:34Þ and we recover the well-known relation between angular quantum ^2 numbers ‘ and m. Hence, we obtain for the eigenvalue of L l ¼ ‘ð‘ þ 1Þ ð3:35Þ ‘ ¼ 0; 1; 2; 3; . . ; ðn À 1Þ ð3:36Þ m ¼ 0; Æ1; Æ2; Æ3; .

0Þ ð3:33Þ Posing we obtain ‘ ¼ m; m þ 1; m þ 2; . . ‘ ! jmj À‘ m ‘ ð3:34Þ and we recover the well-known relation between angular quantum ^2 numbers ‘ and m. Hence, we obtain for the eigenvalue of L l ¼ ‘ð‘ þ 1Þ ð3:35Þ ‘ ¼ 0; 1; 2; 3; . . ; ðn À 1Þ ð3:36Þ m ¼ 0; Æ1; Æ2; Æ3; . . 25) 6 Remember that we are using m for |m| ! 0. 40 ATOMIC ORBITALS "ð‘ À mÞ=2 X Q‘m ðxÞ ¼ ð1 À x2 Þm=2 a2k x2k þ ð‘ À X m À 1Þ=2 k¼0 # a2k þ 1 x2k þ 1 ð3:38Þ k¼0 where the first term in brackets is the even polynomial and the second term is the odd polynomial, whose degree is at most kmax ¼ ‘ À mð!

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