# ch14 - 1 CHAPTER 14 1 The spin-part of the wave function is...

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CHAPTER 14 1. The spin-part of the wave function is the triplet m s = 1 χ + (1) χ + (2) m s = 0 1 2 ( χ + (1) χ - (2) + χ - (1) χ + (2) ) m s =- 1 χ - (1) χ - (2) This implies that the spatial part of the wave function must be antisymetric under the interchange of the coordinates of the two particles. For the lowest energy state, one of the electrons will be in an n = 1, l = 0 state. The other will be in an n = 2, l = 1, or l = 0 state. The possible states are 1 2 u 100 ( r 1 ) u 21 m ( r 2 ) - u 100 ( r 2 ) u 21 m ( r 1 ) ( 29 m = 1 ,0 , - 1 1 2 u 100 ( r 1 ) u 200 ( r 2 ) - u 100 ( r 2 ) u 200 ( r 1 ) ( 29 Thus the total number of states with energy E 2 + E 1 is 3 x 4 = 12 2. For the triplet state, the first order perturbation energy shifts are given by E 21 m = d 3 r 1 d 3 r 2 | 1 2 u 100 ( r 1 ) u 21 m ( r 2 ) - u 100 ( r 2 ) u 21 m ( r 1 ) ( 29 | 2 e 2 4 πε 0 | r 1 - r 2 | E 200 = d 3 r 1 d 3 r 2 | 1 2 u 100 ( r 1 ) u 200 ( r 2 ) - u 100 ( r 2 ) u 200 ( r 1 ) ( 29 | 2 e 2 4 πε 0 | r 1 - r 2 | The l = 1 energy shift uses tw-electron wave functions that have an orbital angular momentum 1. There is no preferred direction in the problem, so that there cannot be any dependence on the eigenvalue of L z . Thus all three m values have the same energy. The l = 0 energy shift uses different wave functions, and thus the degeneracy will be split. Instead of a 12-fold degeneracy we will have a splitting into 9 + 3 states. The simplification of the energy shift integrals reduces to the simplification of the integrals in the second part of Eq. (14-29). The working out of this is messy, and we only work out the l = 1 part. The integrals d 3 r 1 d 3 r 2 r 1 2 d r 1 0 r 2 2 d r 2 0 d 1 d 2 and the angular parts only come through the u 210 wave function and through the 1/ r 12 term. We use Eqs. (14-26) – (14-29) to get, for the direct integral 1

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e 2 4 πε 0 r 1 2 dr 1 0 r 2 2 dr 2 0 R 10 ( r 1 ) 2 R 21 ( r 2 ) 2 d 1 d 2 1 4 π 2 3 4 π cos θ 2 2 P L (cos θ 12 ) r < L r L + 1 L where θ 12 is the angle between r 1 and r 2 . We make use of an addition theorem which reads P L (cos θ 12 ) = P L (cos θ 1 ) P L (cos θ 2 ) + 2 ( L - m )! ( L + m )! m = 1 P L m (cos θ 1 ) P L m (cos θ 2 )cos m φ 2 r < L r L + 1 Since the sum is over m = 1,2,3,…the integration over φ 2 eliminates the sum, and for all practical purposes we have P L L (co s θ 12 ) r < L r L + 1 = P L L (co s θ 1 ) P L (co s θ 2 ) r < L r L + 1 The integration over d 1 yields 4 π δ L0 and in our integral we are left with d 2 (cos θ 2 ) 2 = 4 π /3 . The net effect is to replace the sum by 1/ r to be inserted into the radial integral. (b) For the exchange integral has the following changes have to be made: In the radial integral, R 10 ( r 1 ) 2 R 21 ( r 2 ) 2 R 10 ( r 1 ) R 21 ( r 1 ) R 10 ( r 2 ) R 21 ( r 2 ) In the angular integral 1 4 π 3 4 π (co s θ 2 ) 2 3 (4 π ) 2 co s θ 1 co s θ 2 In the azimuthal integration again the m ≠ 0 terms disappear, and in the rest there is a product of two integrals of the form d 3 4 π cos θ P L (cos θ ) = 4 π 3 δ L 1 The net effect is that the sum is replaced by 1 3 r < r 2 inserted into the radial integral.
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• Spring '05
• mokhtari
• wave function, ground state

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