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Unformatted text preview: CHAPTER 14 1. The spinpart of the wave function is the triplet m s = 1 χ + (1) χ + (2) m s = 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  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  r 1 r 2  The l = 1 energy shift uses twelectron 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 12fold 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. (1429). 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 dr 1 ∞ ∫ r 2 2 dr 2 ∞ ∫ d Ω 1 d Ω 2 ∫ ∫ and the angular parts only come through the u 210 wave function and through the 1/ r 12 term. We use Eqs. (1426) – (1429) to get, for the direct integral 1 e 2 4 πε r 1 2 dr 1 ∞ ∫ r 2 2 dr 2 ∞ ∫ R 10 ( r 1 ) 2 R 21 ( r 2 ) 2 d Ω 1 ∫ d Ω 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 ∑ (cos θ 12 ) r < L r L + 1 = P L L ∑ (cos θ 1 ) P L (cos θ 2 ) r < L r L + 1 The integration over d Ω 1 yields 4 π δ L0 and in our integral we are left with d Ω 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....
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This note was uploaded on 03/09/2008 for the course PHYS 401 taught by Professor Mokhtari during the Spring '05 term at UCLA.
 Spring '05
 mokhtari

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