Question

A spin-orbit interaction between a spin and an orbital motion has the form HS-o = Bs. (1) Assume that S= } for the spin and 1 = 1 for the orbital angular momentum. A. Write down the basis states without the spin-orbit interaction in terms of a direct product of spin eigenstates and orbital angular momentum eigenstates, such as 1,1 > > B. Using angular momentum summation rules, write down all possible values of j and the corresponding eigenvalues m; for the jy. Using total angular momentum - 9 = åt - (2) C.Rewrite the interaction Hs-o in terms of 2, 7 and 52 D. Write down all eigen energies E, of Hs-o in terms of j,l and s. E. Derive (2j+1)E; (3) Where the summation is on all the values of j.

Part D and E

Answer 1

Page-/ in traction Solution :-(A): SPin - Droit H5-0 = B Ŝ. :: SERL=2 لها با Sme =3 ž[ -)5,ms >= |2 :->3372.4333) md 2-1, 0, 1 =) 14, mw>12,-13,11,0>,[1,1> .: Basis states in terms of divect Products : of Il mes o 15, ms) = 11,-1)/3;32:11,13131>;1-1713,); 11.-23132); 11.0713:3>;/2,0712; ): 1 2,0>12. ): 11.0717.1); 12,1913; }); 12,1312;=>; 11,2012,4>;/1,1713,33 Toted States are = (25+1) (24+1) = (2xź +1) (2x1+1) = (2x2 +1) (2x2) - (axş+1) (4) = 12 Scanned with CamScanner

Page 2 B) : 1=1, 8 = 3/4 13.5'+ =) 15-11zis Istdl in step of 1 =) 9 = 2 3 2 1 5 / : m; = -5 to +5 .: atļ: m3 = { [ % = 3 : m; = 3 2 1 1 2 3 4 j. şim , ŷ = ô t =)). 3 = (Stî). (Ŝ têj = 3 2 = § 2+2 ² + 25 . î =) Ŝ î = ² (2î² - Ŝ²) Hs-o = ß § HS-D Scanned with CamScanner Hs-o = (3-2-549)