Answer Happy

The operator for the square of the total spin of two electrons is Stotal= (S1 + S2)² = S² + S² + 2(Ŝ1æŜ2x + Ŝ1yŜ2y + Ŝ1zŜ2z). Consider that ħ iħ ħ Šta Sya = Śza 2 2 2 ħ Saß= Syß = SzB=B = B ħ 2a B iħ 2 α α Part A Complete the sequence of steps to show that [a(1)ß(2) + ß(1)a(2)] /√2 and [a(1)ß(2) — ß(1)a(2)] /√√2 are eigenfunctions of Stot total Solve this problem by acting on a (1)B(2) and B(1) a (2) separately, and combining the results. Match the items in the left column to the appropriate blanks in the expressions on the right. Reset Help a(1) ħ² |Stotap(1)B(2) = Ŝ₁a(1)ß(2) + Ŝ₂a(1)B(2) + 2 [Ŝ1xŜ2xa(1)ß(2) + Ŝ1yŜ2ya(1)B(2) + Ŝ1zŜ2zα(1)B(2)] |s₁| + S²₂2 Ŝ2x B(2) + Sly a(1)B(2) + a(1)B(2)] + 2 [S1 S2y + S1z Ŝ2z B(1) a (2) a(1)B(2) +2 | a(1)B(2) + B(1)a(2) a(2) i²B(1)a(2) - = 2ħ² ħ² |Stotaß(1)a(2) = Ŝ₁ß(1)a(2) + Ŝ₂ß(1)a(2) + 2 [Ŝ1xŜ2xß(1¹)a(2) + Ŝ1yŜ2yß(1)a(2) + Ŝ¹1z2z³(1)a(2)] 4 $²₁ + $²2 Ŝ2x ħ² + Ŝly B(1)a(2) + B(1) a (2)] S2y + Ŝ1z B(1)a(2) +2 +2[$ur Ŝ2x a(1)B(2) 2 B(1) 7ħ² i²a(1)B(2) - = a(1)B(2) + 4 Stotal [a(1)B(2) + ß(1)a(2)] : = × [a(1)ß(2) + ß(1)a(2)] 2 Stotal [a(1)ß(2) — ß(1)a(2)] ×[a(1)B(2) — B(1)a(2)] |||2 ħ 3ħ² 4 0 B(1)a(2)