Chapter 4 Spin-Orbit Coupling 4.1 Addition of Angular Momenta 4.1.1 Alternate set of angular momentum operators We consider two independent systems that each carries angular momenta (each could be either orbital angular momentum or spin). Before the two systems are coupled, each system is described by the angular momentum operators Jˆ1 and Jˆ2 and their eigenstates j1,m1 and j ,m satisfying the conditions | i | 2 2i ˆ2 2 Ji ji,mi = ~ ji(ji +1)ji,mi , ˆ | i | i Jiz ji,mi = ~mi ji,mi , (4.1) ˆ | i | i Ji ji,mi = ~ ji(ji +1) mi(mi 1) ji,mi 1 , ±| i − ± | ± i for i =1, 2. When we consider thep combination of these two systems, the basis of the combined system is created by j ,j ; m ,m = j ,m j ,m , (4.2) | 1 2 1 2i | 1 1i⌦| 2 2i where indicates the tensor product of two states that describe the two independent systems. ⌦ When the total angular momentum of the two systems is considered, it is described by the sum of the two angular momentum operators Jˆ = Jˆ1 + Jˆ2.Becausethetwosystemsare independent, the operators Jˆ1 and Jˆ2 commute with each other, and the components of each system satisfies the commutation relations of the angular momentum operator. It is easy to verify that the components of the total angular momentum operator Jˆ also satisfies the commutation relationship ˆ ˆ ˆ [Ji, Jj]=i~"ijkJk. (4.3) We now look for four independent operators that commute with each other that includes 2 the operators Jˆ and Jˆz,toprovideanalternativedescriptionofthecombinedsystemas compared to the four commuting operators Jˆ2, Jˆ , Jˆ2, Jˆ . { 1 1z 2 2z} 31 32 CHAPTER 4. SPIN-ORBIT COUPLING ˆ ˆ ˆ2 ˆ2 ˆ Since J1 and J2 each commute with J1 and J2 ,sodoesJ.Morespecifically,wenotethat ˆ2 ˆ ˆ2 ˆ2 J and Jz both commute with the operators J1 and J2 ; [Jˆ2, Jˆ2]=[Jˆ2, Jˆ2]=0, 1 2 (4.4) ˆ ˆ2 ˆ ˆ2 [Jz, J1 ]=[Jz, J2 ]=0. The operators Jˆ1z and Jˆ2z commute with the operator Jˆz = Jˆ1z + Jˆ2z. However, since ˆ2 ˆ2 ˆ2 ˆ ˆ ˆ2 ˆ2 ˆ ˆ ˆ ˆ ˆ ˆ J = J1 + J2 +2J1 J2 = J1 + J2 +2J1zJ2z + J1+J2 + J1 J2+, (4.5) · − − ˆ ˆ ˆ2 ˆ2 ˆ2 ˆ2 ˆ the operators J1z and J2z do not commute with J .So,thefouroperators J1 , J2 , J , Jz commute with each other, and forms a natural basis with which to describe the{ total angular} momentum of the combined system. 4.1.2 Addition rule for the angular momentum states Classically, angular momentum is a vector, so when two angular momenta L~ 1 and L~ 2 are added together, the magnitude of the resulting angular momentum can be between L~ L~ | 1 − 2| when the two vectors are aligned in opposite directions, and L~ 1 + L~ 2 when the two vectors are aligned in the same direction. Quantum mechanically, the| “magnitude”| of the angular momentum operator Jˆ is roughly represented by its eigenvalue j,sowhenthetwoangular momentum operators Jˆ1 with eigenvalue j1 and Jˆ2 with eigenvalue j2 are added (without loss of generality, we assume j j ), one can speculate that the resulting operator Jˆ will 1 ≥ 2 have its eigenvalue j between the values j1 j2 and j1 +j2.Itturnsoutthatthisspeculation is exactly true, and the addition rule for angular− momentum can be stated as follows: When two systems each with angular momentum of eigenvalues j and j are added, • 1 2 the angular momentum eigenvalue of the added system runs from j1 j2 to j1 + j2 in increments of 1, and | − | For each resulting j value, mj runs from j to j in increments of 1, i.e., mj = • j, j +1,...,j 1,j. − − − − Although a formal proof is quite complicated, we provide some examples in the next section to illustrate the results. We note that the total number of states in the two systems is (2j1 +1)(2j2 +1),sincethedegeneracy(totalnumberofmj states) are 2j1 +1forthefirst system and 2j2 +1forthesecondsystem,respectively.Inthedescriptionofthecombined system, the total number of mj states is 2j +1 foreach j value, so the total number of states is given by j1+j2 j1+j2 j1 j2 1 N = (2i +1)= (2i) − − (2i)+(j + j ) (j j )+1 i=j1 j2 i=0 0 1 2 1 2 − − − − =(j + j +1)(j + j ) (j j )(j j 1) + 2j +1 (4.6) P1 2 1 2P 1 2 P1 2 2 =(j + j )2 (j j )2−+2j −+2j +1=(2− − j +1)(2j +1), 1 2 − 1 − 2 1 2 1 2 matching the total number of states in the description of individual systems. 4.1. ADDITION OF ANGULAR MOMENTA 33 4.1.3 Addition of two spins In this section, we consider addition of angular momentum for two spin 1/2particles.In ˆ2 ˆ ˆ2 ˆ the uncoupled representation, the four quantum numbers are S1 , S1z, S2 , S2z , and the four possible states are { } 1 1 1 1 ⇠ = , , , . (4.7) | i |2 2 ±2 ±2i These corresponds to the four basis spin states , , , . {| ""i |"#i |#"i |##i} In the coupled representation, the states are simultaneous eigenstates of the four operators Sˆ2, Sˆ , Sˆ2, Sˆ2 ,where { z 1 2 } Sˆ2 =(Sˆ + Sˆ )2 = Sˆ2 + Sˆ2 +2Sˆ Sˆ , 1 2 1 2 1 · 2 (4.8) Sˆz = Sˆ1z + Sˆ2z. First, we note that the values of s,eigenvalueoftheSˆ2 operator, greater than 1 are not possible. This is because that would require ms (eigenvalue of the operator Sˆz)ofvalue larger than 1, but m = m + m and both m and m can only take values of 1/2. s 1s 2s 1s 2s ± Next, we note that s =1ispossible,sincems =1ispossible.Sincems =1canonlyhappen when m1s = m2s =+1/2, this state is not degenerate. From this argument, we note that s =1,m =1 = . | s i |""i Applying the total spin lowering operator Sˆ = Sˆ1 + Sˆ2 yields, using Eq. 4.1, − − − ˆ S s =1,ms =1 = ~p2 s =1,ms =0 −| i | i ˆ ˆ 1 3 1 1 =(S1z + S2z) = ~ + ( + )= + . |""i 2 2 2 2 |"#i |#"i |"#i |#"i q (4.9) So, we conclude that s =1,m =0 =( + )/p2. | s i |"#i |#"i Applying the total spin lowering operator once again on this state gives ˆ S s =1,ms =0 = ~p2 s =1,ms = 1 −| i | − i = 1 (Sˆ + Sˆ )( + ) p2 1z 2z |"#i |#"i (4.10) = ~ 1 3 + 1 1 ( + )=~p2 , p2 2 2 2 2 |##i |##i |##i q and therefore s =1,ms = 1 = ,asexpected.Thesethreestatesformthespin triplet states| − i |##i s =1,m =1 = | s i |""i s =1,ms =0 =( + )/p2 (4.11) |s =1,m = i1 = |"#i |#"i | s − i |##i The next possible s value is s =0,accompaniedbyms =0astheonlypossiblems value. This fourth state is formed by a linear superposition of the four basis spin states , , , ,andisorthogonaltoallthreetripletstatesinEq.4.11.Weseethatthisfourth{| ""i |"# statei |#"i cannot|##i} contain nor states, and therefore has to be a linear superposition of only and |""i.Inorderforsuchasuperpositionstatetobeorthogonaltothe|##i s =1,m|"#i=0 state,|#"i we derive the final state, known as the singlet state,as | s i s =0,m =0 =( )/p2. (4.12) | s i |"#i − |#"i 34 CHAPTER 4. SPIN-ORBIT COUPLING 4.1.4 Clebsch-Gordan coefficients The basis state j, m of the coupled system can be expressed in a linear superposition of the | ji uncoupled basis states j1,j2; m1,m2 , as by definition, basis states span the entire Hilbert space for the quantum| system. By notingi that the identity operator is given by the sum of the outer product of all basis states j1 j2 Iˆ = j ,j ; m ,m j ,j ; m ,m , (4.13) | 1 2 1 2ih 1 2 1 2| m1= j1 m2= j2 X− X− we can express the basis states of the coupled system as j1 j2 j, m = j ,j ; m ,m j ,j ; m ,m j, m . (4.14) | ji | 1 2 1 2ih 1 2 1 2| ji m1= j1 m2= j2 X− X− Inner product of two quantum states is a complex number, and the complex numbers j ,j ; m ,m j, m are called Clebsch-Gordan coefficients. h 1 2 1 2| ji 4.2 Spin-Orbit Coupling 4.2.1 Orbiting spins and the term notation For the case of a hydrogen atom, an electron with spin 1/2orbitsaroundthenucleus(proton), itself with spin 1/2. For now, we ignore the nuclear spin, and focus on the interaction between the electron spin and its orbit around the nucleus. This leads to fine structures in the atomic spectrum. The interaction between the resulting energy levels with the nuclear spin is called the hyperfine interaction: this interaction is typically several orders of magnitude smaller, and we will consider that in later chapters. The total angular momentum Jˆ under these conditions arise from two sources of angular momenta, the spin Sˆ and the orbital angular momentum Lˆ: Jˆ = Lˆ + Sˆ.TheoperatorsLˆ and Sˆ commute, as they are completely independent degrees of freedom. The eigenstates we 2 2 2 consider are simultaneous eigenstates of the operators Jˆ , Jˆz, Lˆ , Sˆ ,whichallcommute with each other. The eigenvalue equations are given by{ } ˆ2 2 J j, mj,l,s = ~ j(j +1)j, mj,l,s | i | i Jˆ j, m ,l,s = m j, m ,l,s z j ~ j j (4.15) ˆ2| i 2 | i L j, mj,l,s = ~ l(l +1)j, mj,l,s ˆ2| i 2 | i S j, mj,l,s = ~ s(s +1)j, mj,l,s , | i | i where j runs from l s to l + s,andforeachj, m runs from j to j.Wenotethat | − | j − ˆ ˆ 1 ˆ2 ˆ2 ˆ2 L S j, mj,l,s = 2 (J L S ) j, mj,l,s · | i 2 − − | i (4.16) = ~ [j(j +1) l(l +1) s(s +1)] j, m ,l,s .