Spin states, many-electron atoms and valence bond theory - Cobalt

spin functions1fÝ1,2Þ = 1sÝ1Þ1sÝ2Þ2Interchanging the two electrons, we getfÝ2,1Þ = 1sÝ2Þ1sÝ1Þ12= 1sÝ1Þ1sÝ2Þ12= ?1sÝ1Þ1sÝ2Þ12ÝJÝ1ÞKÝ2Þ ? KÝ1ÞJÝ2ÞÞ.ÝJÝ2ÞKÝ1Þ ? KÝ2ÞJÝ1ÞÞ,ÝKÝ1ÞJÝ2Þ ? JÝ1ÞKÝ2ÞÞ,ÝJÝ1ÞKÝ2Þ ? KÝ1ÞJÝ2ÞÞ,= ?fÝ1,2Þ.The wave function is antisymmetric with respect to an interchange of two electrons.)(c) Show that the presence of spin makes no difference to the expression within the one-electronmodel of the ground state.(ANS. This follows directly from the fact that the 1-electron Hamiltonians, HÝjÞ, areindependent of spin. Therefore the spin function will simply divide out.)2. The total spin angular momentum, S, for a two-electron system is defined byS = å S xæ e 1 + å S yæ e 2 + å S zæ e 3 = S 1 + S 2where æ e 1,æ e 2, and æ e 3 are the three mutually orthogonal unit vectors oriented along the x-, y-, andz-axes, respectively, and å S x , å S y , and å S z are the three components of total spin angularmomentum. The spin angular momentum for each electron isS j = å S xjæ e 1 + å S yjæ e 2 + å S zjæ e 3 , j = 1,2.(a) Show that(ANS.å S x = å S x1 + å S x2 ,å S y = å S y1 + å S y2 ,å S z = å S z1 + å S z2 .S = å S xæ e 1 + å S yæ e 2 + å S zæ e 3 = S 1 + S 2 = å S x1 + å S x2æ e 1 + å S y1 + å S y2æ e 2 + å S z1 + å S z2æ e 3 .Identifying the coefficients of the unit vectors gives us the result that we want.)(b) Prove that the square of the total spin angular momentum iså S2=å S 12+å S 22+ 2S 1 6 S 2(ANS.å S2= S 6 S,= S 1 + S 2 6 S 1 + S 2 ,= S 1 6 S 1 + S 2 6 S 2 + S 1 6 S 2 + S 2 6 S 1 ,= å S 12+å S 22+ 2S 1 6 S 2 .We have S 1 6 S 2 = S 2 6 S 1 since the spin operators for electron 1 commute with the spinoperators for electron 2, i.e., ßS x1 ,S x2 à = ßS x1 ,S y2 à = ßS x1 ,S z2 à = ` = 0.)(c) Show thatwhere S j± = S xj ± iS yj , with j = 1,2, and henceS 1 6 S 2 = 1 2 Ýå S 1?å S 2+ + å S 1+å S 2? Þ + å S z1å S z2 ,

which implies that M S = ?1. Next, we get åSz f 3 = 0,å S z f 1 = å S z1 + å S z2 JÝ1ÞJÝ2Þ,= å S z1 JÝ1Þ JÝ2Þ + JÝ1Þ å S z2 JÝ2Þ ,= + 1 2 ¥JÝ1ÞJÝ2Þ + 1 2 ¥JÝ1ÞJÝ2Þ,= +¥JÝ1ÞJÝ2Þ,= +¥f 1 ,= M S ¥f 1 .Therefore, we have M S = +1. Similarly, we getå S z f 2 = ?¥f 2 ,and so M S = 0. Finally, we can show thatåS 2f4 = 0,which means that S = 0 and 2S + 1 = 1. Therefore, f 4 corresponds to a singlet state. Also, wehaveå S z f 4 = 0,and hence, M S = 0. These eigenstates of å S 2 and å S z are the states of two electrons whose spinsare coupled. This is an example of the addition of angular momenta for two electrons.)