Quantum Mechanics - II Angular Momentum - II : Addition of Angular Momentum - Clebsch-Gordan Coefficients Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai 400098 September 1, 2019 1 Introduction Consider a spin-half particle. The state space of the particle is spanned by position kets fj ~xig and two dimensional spin space spanned by j "i and j #i. If the spin-orbit coupling is weak, the Hilbert space is the product of the position space and the spin space, and is spanned by j ~x; ±i =j ~xi⊗ j "; #i The rotation operator is still given by exp −iJ~ · nθ=^ ~ , where J~ is given by J~ = L~ ⊗ I ⊕ I ⊗ S~ which is usually abbreviated as J~ = L~ + S~. The rotation operator in the product space is given by the product of the operators for the orbital and spin parts : ~ ~ UR(^n; θ) = exp −iL · nθ=^ ~ exp −iS · nθ=^ ~ 1 The total wavefunction is a a product of the space part and a spin part α(~x) = (~x)χα where α =" or #. The wavefunction is a two component tensor "(~x) #(~x) We have seen that the orbital angular momentum is described by the operator L2 and 2 Lz, while the spin operators by S and Sz. The set of operators required to describe the 2 2 composite system is fL ;Lz;S ;Szg. Since the state space is a product of independent states, we can write j l; s; ml; msi =j l; mmi⊗ j s; msi where 2 2 L j l; mli = l(l + 1)~ j l; mli Lz j l; mli = ml~ j l; mli 2 2 S j s; msi = s(s + 1)~ j s; msi Sz j s; msi = ms~ j s; msi Instead of considering spin and orbital angular momenta of a single particle, we could con- sider more complex system consisting of two or more particles. We could, for instance, talk about orbital momenta of two spinless particles or angular momenta of more com- plex systems with many particles with different angular momenta. However, the basic formulation remains the same as for adding two angular momenta. Let us consider two angular momenta J1 and J2. We define the total angular momenta of the system by ~ ~ ~ J = J1 ⊗ I ⊕ I ⊗ J2 abbreviated as ^ ^ ^ J = J1 + J2 [The hat or the vector signs (which is more commonly used) indicating the operator nature of angular momentum will be frequently omitted. The commutation relations satisfied by 2 J1 and J2 are [J1i;J1j] = i~ijkJ1k [J2i;J2j] = i~ijkJ2k [J1i;J2j] = 0 (1) Since the components of J1 commute with those of J2, they can have common eigenstates. 2 2 We can choose, J1 ;J1z;J2 and J2z to be a set of operators which have common eigenstates. Let us denote the common eigenstates by j j1; j2; m1; m2i ≡| j1; m1i j j2; m2i [ Note : This notation is not standard, several books will denote this as j j1; m1; j2; m2i. We follow Sakurai here.]. We then have 2 2 J1 j j1; j2; m1; m2i = j1(j1 + 1)~ j j1; j2; m1; m2i z J1 j j1; j2; m1; m2i = m1~ j j1; j2; m1; m2i 2 2 J2 j j1; j2; m1; m2i = j2(j2 + 1)~ j j1; j2; m1; m2i z J2 j j1; j2; m1; m2i = m2~ j j1; j2; m1; m2i The dimension of the space to which J1 and J2 belong is (2j1 + 1)(2j2 + 1). The set of states j j1; j2; m1; m2i form a complete and orthonormal set +j1 +j2 X X j j1; j2; m1; m2ihj1; j2; m1; m2 j= 1 (2) m1=−j1 m2=−j2 0 0 0 0 hj ; j ; m ; m j j ; j ; m ; m i = δ 0 δ 0 δ 0 δ 0 (3) 1 2 1 2 1 2 1 2 j1;j1 j2;j2 m1;m1 m2;m2 As mentioned for the case of the orbital and spin angular momenta, the rotation operator in this case is given by ~ ! ~ ! J1 · nθ^ J2 · nθ^ U1R(^n; θ) ⊗ U2R(^n; θ) = exp −i exp −i ~ ~ As a consequence of (1), the components of the total angular momentum J^ satisfy [Ji;Jj] = i~ijkJk (4) which shows that J^ is an angular momentum operator. It is easy to check that J 2 com- 2 2 mutes with Jz;J1 and J2 . However, Jz does not commute with either J1z or J2z. As a re- 2 2 2 sult, it is possible to consider an alternate set of commuting operators, viz., fJ ;z ;J1 ;J2 g 3 2 2 which would describe the same space as did our old set fJ1 ;J2 ;J1z;J2zg. We denote the simultaneous eigenstates represented by this new set by j j1; j2; j; mi. In terms of this new set, we have 2 2 J j j1; j2; j; mi = j(j + 1)~ j j1; j2; j; mi (5) Jz j j1; j2; j; mi = m~ j j1; j2; j; mi (6) 2 2 These kets are also eigenkets of J1 and J2 , though not of J1z and J2z. They satisfy the completeness relation j X j j1; j2; j; mihj1; j2; j; m j= 1 (7) m=−j They can be taken to be orthonormalized 0 0 0 0 hj1; j2; ; j; m j j1; j2; j ; m i = δm;m0 δj;j0 (8) It is tacitly assumed that j1 and j2 in a given problem are given and fixed . We have not yet found out what values j can take. However, the sum over j in (7) must be over all values of j consistent with given values of j1 and j2. Using the completeness property(2) of the old set fj j1; j2; m1; m2ig, we may express a member of the new set fj j1; j2; j; mig as j1 j2 X X j j1; j2; j; mi = j j1; j2; m1; m2ihj1; j2; m1; m2 j j1; j2; j; mi (9) m1=−j1 m2=−j2 The coefficients hj1; j2; m1; m2 j j1; j2; j; mi defined in (9) are called Clebsch-Gordan (C-G)Coefficients.. The basis defined here as represented by the kets on the left hand side of (9) will be also referred to as the Coupled Representation. 2 Properties of C-G Coefficients 1. The C-G coefficients hj1; j2; m1; m2 j j1; j2; j; mi are zero unless m = m1 + m2. To show this, note that Jz − J1z − J2z is a null operator, hj1; j2; m1; m2 j Jz − J1z − J2z j j1; j2; j; mi = 0 4 because Jz acts to the ket on its right giving m times the ket, while J1z − J2z acts to the bra on the left giving m1 − m2 times the bra, (m − m1 − m2)hj1; j2; m1; m2 j j1; j2; j; mi = 0 which shows that of m 6= m1 + m2, the C-G coefficient would be zero. 2. By convention, the C-G coefficients are taken to be real, so that hj1; j2; m1; m2 j j1; j2; j; mi = hj1; j2; jm j j1; j2; m1; m2i 3. Orthonormality of C-G Coefficients : In the expression (8) given above, if we insert a complete set given by (2) of old states, we get 0 0 0 0 hj1; j2; ; j; m j j1; j2; j ; m i = δm;m0 δj;j0 +j1 +j2 X X 0 0 0 0 hj1; j2; j; m j j1; j2; m1; m2ihj1; j2; m1; m2 j j1; j2; j ; m i = δm;m0 δj;j0 m1=−j1 m2=−j2 (10) Substituting j = j0 and m = m0 in the above, we get X 2 j hj1; j2; m1; m2 j j1; j2; j; mi j = 1 (11) m1;m2 In a similar way may start with the orthonormality condition on the old basis set (3) and insert the completeness condition (7) of the new basis set to obtain +j X X 2 j hj1; j2; m1; m2 j j1; j2; j; mi j = 1 (12) j m=−j 4. The C-G coefficients vanish unless j j1 − j2 |≤ j ≤ j1 + j2 (13) max max The inequality on the right is obvious because m1 = j1 and m2 = j2. Hence max m = j1 + j2. Since the maximum value of m is j, the right inequality follows 5 and jmax = j1 + j2. The inequality on the left requires a bit of working out. We know that there are (2j1 + 1)(2j2 + 1) number of states. For each value of j, there are 2j + 1 states. Thus we must have jmax X (2j + 1) = (2j1 = 1)(2j2 + 1) j=jmin Left hand side of above is an arithmetic series of finite number of terms. The sum works out as follows : j − j + 1 l:h:s: = max min (2j + 1 + 2j + 1) 2 max min = (jmax − jmin + 1)(jmax + Jmin + 1) 2 2 = (jmax + 1) − jmin 2 2 = (j1 + j2 + 1) − jmin 2 2 Equating the above to (2j1 + 1)(2j2 + 1), we get jmin = (j1 − j2) , which gives jmin =j j1 − j2 j Thus the values of j satisfies j j1 − j2 |≤ j ≤ j1 + j2 (14) 5. For a given j, the possible m values are −j ≤ m ≤ +j. A notation, primarily used by the nuclear physicists is known as Wigner's 3-j symbol and is related to C-G coefficients as follows. While the C-G coefficients are for adding two angular moments, the 3-j symbols are for addition of three angular momenta such that their sum gives zero angular momentum state.