Angular momentum algebra J-1J. ANGULAR MOMENTUMSources:- D.C. Harris och M.D. Bertolucci, Symmetry and Spectroscopy, Oxford UniversityPress, 1978.- J. M. Hollas, Modern Spectroscopy, Wiley, Chichester, 1987.- G. Herzberg, Spectra of Diatomic Molecules, Van Nostrand, 1950.- G. Herzberg, Infrared and Raman Spectra, Van Nostrand, 1945.- P.W. Atkins, Molecular Quantum Mechanics, Oxford University Press, 1983.- H.W. Kroto, Molecular Rotation Spectra, Wiley, 1975.- E.B. Wilson, J.C. Decius och P.C. Cross, Molecular Vibrations, McGraw-Hill, 1955.(En ny upplaga: Dover, 1980).- E. U. Condon och G. H. Shortley, Theory of atomic spectra, Cambridge, 1953.- L.A. Woodward, Introduction to the Theory of Molecular Vibrations and VibrationalSpectroscopy, Clarendon Press, 1972.- S. Califano, Vibrational States, Wiley, 1976.- E.F.H. Brittain, W.O. George och C.H.J. Wells, Academic Press, 1970.- M.D. Harmony, Introduction to Molecular Energies and Spectra, Holt & Winston,1972.

J-2 Molecular spectroscopyJ.1. Basic propertiesIn the classical mechanics the angular momentum is de�ned as~L = m~r � ~v = ~r � ~p: (J:1)It is thus a vector quantity with the lengt L. It can be split to three Cartesian components,Lx, Ly and Lz, that are all well de�ned.In quantum mechanics a similar de�nition can be used for the orbital angular momentumbut, e.g., for the spin such a de�nition is not possible. The generalized de�nition ofangular momentum in quantum mechanics is based on the commutation relationsof the components in orthogonal orientations, j1, j2 and j3, of the angular momentumoperator ~j, i.e., [j1; j2] = i�hj3[j2; j3] = i�hj1[j3; j1] = i�hj2: (J:2)The alternative de�nition for the components of the orbital angular momentum followsfrom the classical de�nition in equation (J.1),Lx = �i�h�y @@z � z @@y�Ly = �i�h�z @@x � x @@z�Lz = �i�h�x @@y � y @@x� : (J:3)In atomic units, �h = 1.The equations (J.2) are also equivalent with the de�nition~j � ~j = i~j: (J:4)In other words, any operator that ful�ls this relation it is an angular momentum operator.The commutation relations guarantee that all properties of the angular momentum opera-tors can be obtained mathematically. This branch of quantummechanics is called angularmomentum algebra or Racah algebra.The commutator relations show that the components do not have common eigenfunctions.All components cannot be good quantum numbers at the same time.

Angular momentum algebra J-3The square of the angular momentum is~j2 � ~j � ~j = j21 + j22 + j23 : (J:5)It is a scalar quantity and is therefore denoted by j2. The square commutes with one of thecomponents j1, j2 and j3 but not with all three because the components do not commute.One usually chooses j3 to be the commuting component,[j2; j3] = 0: (J:6)This relation follows directly from the commutation relations above,[j2; j3] = j21 j3 + j22 j3 + j33 � j3j21 � j3j22 � j33= j21 j3 � j1j3j1 + j1j3j1 + j22 j3 � j2j3j2 + j2j3j2 � j3j21 � j3j22= j1(j1j3 � j3j1) + j2(j2j3 � j3j2)� (j3j1 � j1j3)j1 � (j3j2 � j2j3)j2= �ij1j2 + ij2j1 � ij2j1 + ij1j2= 0: (J:7)For the quantum mechanical angular momentum, both the length and the orientation arequantitized and are described by the quantum numbers j and m, respectively. Thelength (or norm of the vector) is related to the scalar product j2 while the orientation isgiven by the component along the z axis, jz. A vector with length jjj cannot have a zcomponent that is larger than the length. Therefore the quantum number m is restrictedto the values +j � m � �j.

J-4 Molecular spectroscopyJ.2. EigenfunctionsThe term angular momentumrefers to circular motion. The orbital angular momentumis best described in spherical coordinates (r; �; �) rather than in Cartesian coordinatesx; y; z. The transformation from the Cartesian to the spherical coordinates is given by theequations r = (x2 + y2 + z2) 12� = arctan �12(x2 + y2) 12 �� = arctan� yx� (J:8)and the inverse transformation by x = r cos(�) sin(�)y = r sin(�) sin(�)z = r cos(�): (J:9)Using these formulae the operators (J.3) and (J.5) can be easily transformed.The operators are in spherical coordinatesjz = �i @@�j2 = � � 1sin2 � @2@�2 + 1sin � @@� �sin � @@��� : (J:10)The common eigenfunction mj (�; �) can be solved from these di�erential equations. It isa product of an exponential and an associated Legendre polynomialmj (�; �) = NjmPmj (�)eim�: (J:11)The function mj (�; �) is called spherical harmonic function. A few of the �rst sphericalharmonic functions are given in the table below.

Angular momentum algebra J-5j m Imaginary wavefunction Alternative real wavefunction0 0 q 14� -1 0 q 34� cos � -1 �1 q 38� sin �e�i� q 34�sin� cos�q 34�sin� sin�2 0 q 516� (3 cos2 � � 1) -�1 q 158� sin � cos �e�i� q 154� sin � cos � cos�q 154� sin � cos � sin��2 q 1532� sin2 �e�2i� q 1516� sin2 � cos 2�q 1516� sin2 � sin 2�3 0 q 6316� ( 53 cos3 � � cos �) -�1 q 2164� sin �(5 cos2 � � 1)e�i� q 2132� sin �(5 cos2 � � 1) cos�q 2132� sin �(5 cos2 � � 1) sin��2 q 10532� sin2 � cos �e�2i� q 10516� sin2 � cos � cos 2�q 10516� sin2 � cos � sin 2��3 q 3564� sin3 �e�3i� q 3532� sin3 � cos 3�q 3532� sin3 � sin 3�

J-6 Molecular spectroscopyJ.3. EigenvaluesTwo new operators are needed in order to calculate the eigenvalues of the eigenvalues of j3and j2. The step-up operator j+ and the step-down operator j� are de�ned throughthe components j1 and j2, j+ = j1 + ij2j� = j1 � ij2: (J:12)A number of mathematical relations can be derived for the operators, such asj+j� = j2 + j3 � j23j�j+ = j2 � j3 � j23[j�; j3] = j�[j+; j3] = �j+: (J:13)The relations are easily proved, e.g.,j+j� = (j1 + ij2)(j1 � ij2)= j21 � ij1j2 + ij2j1 + j22= j21 + j22 + j23 � j23 � i(j1j2 � j2j1)= j21 + j22 + j23 � j23 � i[j1; j2]= j2 + j3 � j23 (J:14)and [j�; j3] = (j1 � ij2)j3 � j3(j1 � ij2)= j1j3 � j3j1 � ij2j3 + ij3j2= [j1; j3]� i[j2; j3]= �ij2 � i � ij1= j1 � ij2= j�: (J:15)Let the eigenvalue of the component j3 be m and the corresponding eigenfunction m,i.e., j3m = mm: (J:16)Operate by the step-down operator j� on both sides of this eigenvalue equation. This willresult in j�j3m = mj�m: (J:17)

Angular momentum algebra J-7Because j�j3 = j3j� + j� (J:18)according to equation (J.13)c one can writej3j�m = (m� 1)j�m: (J:19)This means that the function j�m is an eigenfunction of the component j3 with thecorresponding eigenvalue m � 1. The operator j� transforms a state with the eigenvaluem to a state with eigenvalue m� 1. Hence the name �step-down operator�. Observe thatthe state j�m is not normalized, j�m = Cm�1; (J:20)where C is a normalization factor.Similarly, one can operate on equation (J.16) with j+, which will lead toj+j3m = mj+m: (J:21)As j+j3 = j3j+ � j+ (J:22)according to equation (J.13), one can writej3j+m = (m+ 1)j+m: (J:23)Apart from a normalization factor D, one will obtain the state m+1 by operating withj+, j+m = Dm+1: (J:24)The operator j+ will give a ascending series of states, m, m+1, m+2 : : :. The operatorj� will give descending series m, m�1, m�2 : : :. These two series must terminatebecause m is a common eigenstate of j3 and j2 which means thatj+j�m = (`2 +m�m2)mj�j+m = (`2 �m�m2)m (J:25)where `2 is a real number. FurtherZ�mj+j�m d� = Z(j�m)�(j�m) d�= jCj2 Z�m�1m�1 d�� 0: (J:26)

J-8 Molecular spectroscopyEquality is valid only if j�m = Cm�1 = 0: (J:27)Correspondingly, the latter equation will give the resultZ�mj�j+m d� = Z(j+m)�(j+m) d�= jDj2 Z�m+1m+1 d�� 0: (J:28)Equality is valid only if j+m = Dm+1 = 0: (J:29)The wavefunction m is normalized which implies that`2 +m�m2 � 0;`2 �m�m2 � 0: (J:30)For a �xed value of `2 this is possible only when both the descending and the ascendingseries terminate. In that case one will have a minimum value mmin and a maximum valuemmax of the quantum number m,� `2 +mmin �m2min = 0`2 �mmax �m2max = 0 (J:31)or 8<:mmin = 12 �q14 + `2 mmin�1 = 0mmax = � 12 +q14 + `2 mmax+1 = 0: (J:32)The operators j� alter the quantum number m by �1. Then the di�erence mmax - mminmust be an integer. The number of levels with di�erent m ismmax �mmin = 2r14 + `2 = heltal = 2j + 1; (J:33)where j is the largest allowed value of jmj. As the di�erence must be an integer, j canonly adopt the values j = 0; 12 ; 1; 33 ; 2; : : : : (J:34)By taking the square of each side of the equation one can see that1 + 4`2 = 4j2 + 4j + 1 `2 = j(j + 1): (J:35)The quantum number m varies between �j and j.The quantity `2 was introduced as the eigenvalue of the operator j2. Therefore one canwrite the concurrent eigenvalue equations for the common eigenstate j jm >j2 j jm >= j(j + 1) j jm >; j = 0; 12 ; 1; 32 ; : : :j3 j jm >= m j jm >; m = �j;�j + 1; : : : ; j: (J:36)

Angular momentum algebra J-9J.4. The step operatorsIt is required that all states are normalized. From the equations (J.25), (J.26) and (J.28)it follows that jCj2 < m� 1 j m� 1 >= jCj2 =< j�(jm) j j�(jm) >= `2 +m�m2= j(j + 1)�m(m� 1);jDj2 < m+ 1 j m+ 1 >= jDj2 =< j+(jm) j j+(jm) >= `2 �m�m2= j(j + 1)�m(m+ 1): (J:37)Apart from an arbitrary phase factor, one �nds that� j� j jm >=pj(j + 1)�m(m� 1) j j;m� 1 >j+ j jm >=pj(j + 1)�m(m+ 1) j j;m+ 1 > : (J:38)This expression was derived from the commutation relations and is therefore valid for allquantum mechanical angular momenta.It is also possible to derive from these equations the expectation values of the operators j1and j2, < jm� 1jj1jjm >= 12p(j �m)(j �m+ 1)< jm� 1jj2jjm >= �12 ip(j �m)(j �m+ 1): (J:39)For an orbital angular momentum the eigenvalues of the step-up and step-down operatorscan be derived from the de�nition ~L = ~r � ~p. Only integer quantum numbers are allowedfor ~L.One can switch to a matrix representation where the state vectors corresponding to a givenquantum number j are 2j + 1 dimensional vectorsj j;m = j >= 0BBBB@ 100...0

1CCCCA ; j j;m = j � 1 >= 0BBBB@ 010...01CCCCA ; etc: (J:40)and the angular momentum operators are (2j + 1) � (2j + 1) dimensional matrices. Theonly non-vanishing elements in the matrices corresponding to the operators j� and j+ are(j�)m�1;m =pj(j + 1)�m(m� 1); (j+)m+1;m =pj(j + 1)�m(m+ 1): (J:41)

J-10 Molecular spectroscopyIn the case j = 12 (e.g., the spin of an electron or a proton) one will obtain8>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>:

j 12 ; 12 >= � 10� ; j 12 ;� 12 >= � 01�j� = �+ 12 � 12+ 12 0 0� 12 1 0 �; j+ = � 0 10 0�m = 12 ) (m� 1;m) = (� 12 ; 12 ) ) pj(j + 1)�m(m� 1) =q34 � 12 (� 12 ) = 1m = � 12 ) (m+ 1;m) = ( 12 ;� 12 ) ) pj(j + 1)�m(m+ 1) =q34 + 12 (12 ) = 1j1 = 12(j+ + j�) = 12 � 0 11 0�j2 = � 12 i(j+ � j�) = 12 � 0 �ii 0 �j3 = �i[j1; j2] = � i4 �� 0 11 0�� 0 �ii 0 �� � 0 �ii 0 �� 0 11 0��= � i4 �� i 00 �i�� ��i 00 i ��= 12 � 1 00 �1�j2 = j21 + j22 + j23 = 34 � 1 00 1� : (J:42)In the case j = 1 (e.g., the nuclear spin of a deuteron or 14N , or the total spin of a tripletstate) one will obtain similarly8>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>:j 11 >= 0@ 1001A ; j 10 >= 0@ 0101A ; j 1;�1 >= 0@ 0011Aj� = 0@ 0 0 0p2 0 00 p2 01A ; j+ = 0@ 0 p2 00 0 p20 0 0 1Aj1 = 1p2 0@ 0 1 01 0 10 1 01A ; j2 = 1p2 0@ 0 �i 0i 0 �i0 i 0 1Aj3 = 0@ 1 0 00 0 00 0 �11A ; j2 = 20@ 1 0 00 1 00 0 11A :

(J:43)

Angular momentum algebra J-11J.5. Addition of two angular momentaConsider the two angular momentum vectors ~j1 and ~j2. Both are angular momenta andful�l therefore the condition ~j� � ~j� = i~j� (� = 1; 2): (J:44)One also observes that� j2� j j�;m� >= j�(j� + 1) j j�;m� >; j� = 0; 12 ; 1; : : :j�;3 j j�;m� >= m� j j�;m� >; �j� � m� � j� (J:45)Now consider a physical system whose total angular momentum is the vector sum~j = ~j1 + ~j2: (J:46)Because ~j � ~j = (~j1 + ~j2)� (~j1 + ~j2)= ~j1 � ~j1 + ~j2 � ~j2= i~j1 + i~j2= i~j; (J:47)also ~j is an angular momentum vector. Denote its eigenstates by j jm >, i.e.,� j2 j jm >= j(j + 1) j jm > j = 0; 12 ; 1; : : :j3 j jm >= m j jm > �j � m � j. (J:48)Construct the state j jm > as a linear combination of the states j j1;m1 > and j j2;m2 >,j jm >= j1Xm1=�j1 j2Xm2=�j2 C(j1; j2; j;m1;m2;m) j j1;m1 >j j2;m2 > : (J:49)The coe�cients C are called the Clebsch-Gordan coe�cients or occasionally theWig-ner coe�cients.Every vector j jm > can be expressed by means of the product vectors j j1m1 >j j2m2 >because the product vectors span a vector space with the same dimension as the vectorsj jm >.

J-12 Molecular spectroscopyThis is proven by counting the dimensions. One can assume that j1 � j2. In that case thequantum number j of the sum vector can assume the values j1 + j2, j1 + j2 - 1, : : :, j1 -j2. Each of these possible values gives 2j + 1 linearly independent states j jm > equal to[2(j1 + j2) + 1] + [2(j1 + j2 � 1) + 1] + : : :+ [2(j1 � j2 + 1) + 1] + [2(j1 � j2) + 1]= (2j1 + 1)� number of brackets= (2j1 + 1)(2j2 + 1) QED (J:50)For the component j3 of the angular momentum one can writej3 = j1;3 + j2;3 )m = m1 +m2: (J:51)Therefore the quantities ~j form a vector sum and the quantities m an arithmetic sum. Theequation (J.49) can be written asj jm >=Xm1 C(j1; j2; j;m1;m�m1;m) j j1;m1 >j j2;m2 > (J:52)The phase factor is chosen so that the coe�cients C are real.There are several alternative notations for the Clebsch-Gordan coe�cients,C(j1; j2; j;m1;m2;m) � C(j1; j2; j;m1;m�m1;m)� C(j1; j2; j;m1;m�m1)�< j1;m1; j2;m2 j j;m >�< j1; j2;m1;m2 j j1; j2; j;m > : (J:53)Consider as an example a system that consists of two particles with spin 12 each. Thenumber of states is (2 � 12 + 1)� (2 � 12 + 1) or four. There is only one way of obtaining thequantum number value m = 1,j 1; 1 >=j 12 ; 12 >j 12 ; 12 > )C(12 ; 12 ; 1; 12 ; 12 ; 1) = 1 (J:54)De�ne the step-down operator j� = j1;� + j2;� (J:55)

Angular momentum algebra J-13which is used to generate new states with j = 1. By operating with j� on the state j 1; 1 >one will obtain the normalized statej 1; 0 >= 1p2 �j 12 ;�12 >j 12 ; 12 > + j 12 ; 12 >j 12 ;�12 >�C(12 ; 12 ; 1;�12 ; 12 ; 0) = C(12 ; 12 ; 1; 12 ;�12 ; 0) = 1p2 : (J:56)By operating once more with j� one will note thatj 1;�1 >=j 12 ;�12 >j 12 ;�12 > )C(12 ; 12 ; 1;�12 ;�12 ;�1) = 1: (J:57)The fourth state is obtained by forming an antisymmetric expression with m = 0 so thatit is orthogonal to sthe state j 1; 0 >. Therefore the new state will be a singlet state,j 0; 0 >= 1p2 �j 12 ;�12 >j 12 ; 12 > � j 12 ; 12 >j 12 ;�12 >�C(12 ; 12 ; 1;�12 ; 12 ; 0) = 1p2 C(12 ; 12 ; 1; 12 ;�12 ; 0) = � 1p2 : (J:58)The Clebsch-Gordan coe�cients form the following 4� 4 matrixC = 0BB@m1;m2 = 12 ; 12 �12 ; 12 12 ;� 12 � 12 ;� 12j;m = 1; 1 1 0 0 01; 0 0 1p2 1p2 00; 0 0 1p2 � 1p2 01;�1 0 0 0 1 1CCA: (J:59)

Generally one can show that the Clebsch-Gordan coe�cients C(j1; 12 ; j;m � m2;m2;m)can be expressed as j = j1 + 12 h j1+m+ 122j1+1 i 12 h j1�m+ 122j1+1 i 12j1 � 12 � h j1�m+ 122j1+1 i 12 h j1+m+ 122j1+1 i 12The matrix C is a real and orthogonal matrix because the eigenvectors j jm > must beorthogonal,Xm1 C(j1; j2; j;m1;m�m1;m)C(j1; j2; j0;m1;m0 �m1;m0) = �j;j0�m;m0 : (J:60)

J-14 Molecular spectroscopyThe states are given by the matrix equationj = Cm; (J:61)where m is a column vector.On the other hand, it is known that the transposed matrix of a real orthogonal matrix(actually an adjoint matrix which is the same thing as a transposed for real matrices) isits inverse, Cy = C�1: (J:62)This is seen easily because Xk CikCykj =Xk CikCjk = �ij : (J:63)Also the column vectors of C are orthogonal,Xk CkiCkj =XCyikCkj = �ij : (J:64)The latter equation means thatXj C(j1; j2; j;m1;m�m1;m)C(j1; j2; j;m01;m0 �m01;m0) = �m;m0�m1;m01 : (J:65)Therefore the column vectors of the matrix C give the inverse transformationj j1;m1 >j j2;m2 >=Xj C(j1; j2; j;m1;m2;m1 +m2) j j;m1 +m2 > : (J:66)It is often more practical to de�ne a Wigner 3-j symbol instead of using the Clebsch-Gordan coe�cients,� j1 j2 j3m1 m2 m3� � (�1)j1�j2�m3(2j3 + 1) 12 C(j1; j2; j3;m1;m2;�m3): (J:67)(Observe that also other de�nitions appear in the literature.)The Wigner 3-j symbols obey the following symmetry relations,� j1 j2 j3m1 m2 m3� = � j2 j3 j1m2 m3 m1 � = � j3 j1 j2m3 m1 m2 �= (�1)J � j2 j1 j3m2 m1 m3 � etc: J = j1 + j2 + j3= (�1)J � j1 j2 j3�m1 �m2 �m3 � : (J:68)

Angular momentum algebra J-15The Wigner 3-j symbols vanish except for the cases where�m1 +m2 +m3 = 0 andjj1 � j2j � j3 � j1 + j2 cyclically (J:69)The last relation is often written as �(j1; j2; j3), i.e., the three quantities can form atriangle.Tables of 3-j symbols are given by Rotenberg-Bivins-Metropolis-Wooten in �The 3-j and6-j symbols�, MIT Press, 1959. The values are given as powers of primary numbers,(3� j) = � (2n1 � 3n2 � 5n3 � 7n4 : : :) 12 : (J:70)An asterix indicates that the negative square root should be used. If the exponent ni isnegative it is underlined. A comma is used to separate groups of four primary numbers.As an example, one reads in the table that� 2 1 10 0 0� = �21 � 3�1 � 5�1� 12 ; (J:71)that � 1 1 10 0 0� = 0 (J:72)and that � 1 1 00 0 0� = � 13 12 : (J:73)A few of the �rst 3-j symbols by Rotenberg et al. are reported in the table below.

J-16 Molecular spectroscopyj1 j2 j3 m1 m2 m3 j1 j2 j3 m1 m2 m312 12 0 12 � 12 0 1 2 2 1 1 �1 0 �1111 12 12 0 � 12 12 11 2 2 1 1 0 �1 1011 12 12 1 � 12 � 12 �01 2 2 1 2 �2 0 1111 1 0 0 0 0 �01 2 2 1 2 �1 �1 �0111 1 0 1 �1 0 01 2 2 2 �2 0 2 10111 1 1 �1 0 1 11 2 2 2 �1 �1 2 �01111 1 1 0 �1 1 �11 2 2 2 �1 0 1 10111 1 1 0 0 0 0 2 2 2 0 �2 2 10111 1 1 1 �1 0 11 2 2 2 0 �1 1 10111 1 1 1 0 �1 �11 2 2 2 0 0 0 �101132 1 12 � 12 0 12 �11 2 2 2 1 �2 1 �011132 1 12 12 �1 12 21 2 2 2 1 �1 0 101132 1 12 12 0 � 12 11 2 2 2 1 0 �1 101132 1 12 32 �1 � 12 �2 2 2 2 2 �2 0 101132 32 0 12 � 12 0 �2 2 2 2 2 �1 �1 �011132 32 0 32 � 32 0 2 2 2 2 2 0 �2 101132 32 1 � 12 � 12 1 111 52 32 1 � 12 � 12 1 �20132 32 1 12 � 32 1 �101 52 32 1 12 � 32 1 21132 32 1 12 � 12 0 �211 52 32 1 12 � 12 0 10132 32 1 32 � 32 0 211 52 32 1 32 � 32 0 �01132 32 1 32 � 12 �1 �101 52 32 1 32 � 12 �1 �1012 1 1 �1 0 1 �101 52 32 1 52 � 32 �1 112 1 1 0 �1 1 111 52 2 12 � 12 0 12 1012 1 1 0 0 0 111 52 2 12 12 �1 12 �0112 1 1 1 �1 0 �101 52 2 12 12 0 � 12 �1012 1 1 1 0 �1 �101 52 2 12 32 �2 12 1112 1 1 2 �1 �1 001 52 2 12 32 �1 � 12 1112 32 12 0 � 12 12 �101 52 2 12 52 �2 � 12 �112 32 12 1 � 32 12 201 52 2 32 � 32 0 32 �01112 32 12 1 � 12 � 12 211 52 2 32 � 12 �1 32 22112 32 12 2 � 32 � 12 �001 52 2 32 � 12 0 12 10112 32 32 �1 � 12 32 101 52 2 32 12 �2 32 �00112 32 32 0 � 32 32 �201 52 2 32 12 �1 12 �21112 32 32 0 � 12 12 �201 52 2 32 12 0 � 12 10112 32 32 1 � 32 12 101 52 2 32 32 �2 12 31112 32 32 1 � 12 � 12 0 52 2 32 32 �1 � 12 11112 32 32 2 � 32 � 12 �101 52 2 32 32 0 � 32 �01112 32 32 2 � 12 � 32 101 52 2 32 52 �2 � 12 �11012 2 0 0 0 0 001 52 2 32 52 �1 � 32 10012 2 0 1 �1 0 �001 52 52 0 12 � 12 0 112 2 0 2 �2 0 001 52 52 0 32 � 32 0 �112 2 1 �1 0 1 �101 52 52 0 52 � 52 0 112 2 1 0 �1 1 101 52 52 1 � 12 � 12 1 �01112 2 1 0 0 0 0 52 52 1 12 � 32 1 31112 2 1 1 �2 1 �011 52 52 1 12 � 12 0 1111

Angular momentum algebra J-17It can further be proven that the Clebsch-Gordan coe�cients obey the rulesC(j1; j2; j3;m1;m2;m3) = �m1+m2;m3�(j1; j2; j3) [(2j3 + 1)(j1 +m1)!(j1 �m1)!�(j2 +m2)!(j2 �m2)!(j3 +m3)!(j3 �m3)!] 12 �X� (�1)� [(j1 �m1 � �)!(j3 � j2 �m1 + �)!(j2 +m2 � �)!�(j3 � j1 �m2 + �)!(j1 + j2 � j3 � �)!�!] (J:74)where � adopts all the values for which the faculty is positive and�(abc) � � (a+ b� c)!(a+ c� b)!(b+ c� a)!(a+ b+ c+ 1)! � 12 : (J:75)They are also related to the spherical harmonic functions,Z Y m1j1 (�; �)Ym2j2 (�; �)Ym3j3 (�; �)d� =�(2j1 + 1)(2j2 + 1)(2j3 + 1)4� �� � j1 j2 j30 0 0 �� j1 j2 j3m1 m2 m3 � (J:76)from which it follows that12 Z 1�1 Pj1(�)Pj2(�)Pj3(�)d� = � j1 j2 j30 0 0 �2 : (J:77)In principle one can use as the basis set either the coupled representation j j1; j2; j;m >or the uncoupled representation j j1;m1 >j j2;m2 >. Both representations have fourquantum numbers. In practice the system is closer to one or the other limiting case.Consider as an example an atom with nuclear spin ~I and electron spin ~S. Let the orbitalangular momentum ~L vanish. The two angular momenta are added as~F = ~I + ~S: (J:78)In a weak magnetic �eld the state can be described as almost pure coupled angular mo-menta j I; S; F;MF >. In a strong magnetic �eld the system is almost pure uncoupledstate j I;MI >j S;MS >, i.e., the external �eld will �quantitize both angular momenta�.A more advanced discussion on addition of two angular momenta is found in any textbookin quantum mechanics.

J-18 Molecular spectroscopyJ.6. Rotation matrixA rotation of the coordinate axes through the angle ' around the origin in the space R2is given by the transformation� ab � = � cos' sin'� sin' cos' � �xy � : (J:79)It is customary to de�na a rotation matrix in R2 asSrot = � cos' sin'� sin' cos' � : (J:80)A rotation in R2 is uniquely de�ned because thre is only one rotation axis. In R3, however,there are several possibilities to perform such a transformation. We will not discuss in detailthe various possibilities.1 Here we adopt the following convention.(i) Rotate x and y through the angle ' to x0 and y0; rotation axis z.(ii) Rotate x0 and z through the angle � to x00 and c; rotation axis y0.(iii) Rotate x00 and y0 through the angle � to a and b; rotation axis c.The �rst step in the above process is governed by the equation24x0y0z 35 = 24 cos' sin' 0� sin' cos' 00 0 135 24xyz 35 : (J:81)The second step will transform the new coordinates etc. Thus a complete rotation matrixS = S(3) through the Euler angles '; �; � can be obtained as the productS = 24 cos� sin� 0� sin� cos� 00 0 13524 cos � 0 � sin �0 1 0sin � 0 cos � 3524 cos' sin' 0� sin' cos' 00 0 135= 24 cos� cos � cos'� sin� sin' cos� cos � sin'+ sin� cos' � cos� sin �� sin� cos � cos'� cos� sin' � sin� cos � sin'+ cos� cos' sin� sin �sin � cos� sin � sin� cos � 35 :(J:82)The three matrices in the product are called C, B and A which leads to the de�nitionS = CBA: (J:83)One can show by a simple matrix multiplication thatA�1 = Ay: (J:84)1 For a thorough discussion see, e.g., Wilson, Decius och Cross eller Kroto.

Angular momentum algebra J-19A similar formula applies to the other matrices. It can be easily shown thatS�1 = Sy = ST : (J:85)The last step follows from the fact that S is real.In a rotating molecule the angles '; �; � must be functions of time. The time derivative ofthe rotation matrix will appear in expressions such as S _S�1. A simple calculation showsthat2 S _S�1 = 24 0 �!c !b!c 0 �!a�!b !a 0 35 (J:86)with the angular velocities !a = _� sin�� _' cos� sin �!b = _� cos�+ _' sin� sin �!c = _�+ _' cos �: (J:87)Let the permutation symbol e�� assume the valuesexyz = ezxy = eyzx = 1eyxz = ezyx = exzy = �1e�� = 0 f�or alla andra kombinationer: (J:88)Then the equation (J.86) can be written as�S _S�1�� = !�e�� : (J:89)Here the Einstein summation rule has been utilized. This means that one has implicitlya summation over � in the right hand side of the equation.It can be shown that the scalar product of two arrays does not depend on the coordinatesystem, i.e., ~x � ~y = (S�1 � ~a) � (S�1 �~b) = ~a �~b: (J:90)Proof: ~a �~b =Xi aibi�S � S�1�ij =Xk SikS�1kj = �ij�S�1 � ~a�k =Xi S�1ki ai2 See, e.g., Kroto, Appendix A2.

J-20 Molecular spectroscopy(S�1 � ~a) � (S�1 �~b) =Xk "Xi S�1ki ai# 24Xj S�1kj bj35=Xijk S�1ki S�1kj ai bj S�1=ST= Xijk Sik S�1kj ai bj=Xij ai bjXk Sik S�1kj =Xij ai bj�ij =Xij ai bj= ~a �~b

Angular momentum algebra J-21J.7. Permutation operatorThe permutation symbol e�� operates on a three-dimensional array (x; y; z)T in sucha way that the elements change places. As an exampleexyz 24xyz 35 = 24xyz 35 (J:91)while eyxz 24xyz 35 = 24 yxz 35 : (J:92)If an element appears twice in the result, the operation is not a permutation. The per-mutation has the parity +1 or �1. The parity is de�ned by equation (J.88). From thisde�nition one can see thateijk = [�ix�jy�kz + �iy�jz�kx + �iz�jx�ky � �iz�jy�kx � �ix�jz�ky � �iy�jx�kz] : (J:93)Here ��� stands for the Kronecker symbol��� = � 1 for � = �;0 for � 6= �. (J:94)The expression can be written as a determinant,eijk = ������ �ix �iy �iz�jx �jy �jz�kx �ky �kz ������ : (J:95)Let A and B be quadratic matrices of equal dimension. The product of two determinantsjAj and jBj is obtained by the rules of matrix multiplication and the formulajABj = jAj jBj: (J:96)In addition, one may observe that transposition does not change the value of the determi-nant, jAjT = jAj: (J:97)Thus the product of two permutation operators can easily be calculated aseijkelmn = ������ �il �im �in�jl �jm �jn�kl �km �kn ������ : (J:98)

J-22 Molecular spectroscopyOne important special case is eijkeimn = ���� �jm �jn�km �kn ���� : (J:99)One can further observe that eijkeijn = 2�kn (J:100)and eijkeijk = 6: (J:101)In these two equations the Einstein summation convention is assumed.The vector product ~D = ~A� ~B: (J:102)can be written in the following form by using the permutation symbolsDi =Xjk eijkAj Bk: (J:103)The triple product ~F = ( ~A� ~B)� ~C = ~D � ~C (J:104)is obtained as Fl =Xi;n elinDiCn= Xijkn elineijkAjBkCn=Xjkn (�jn�kl � �jl�kn)AjBkCn=Xj AjBlCj �Xn AlBnCn: (J:105)If the arrays ~A, ~B and ~C commute on obtains the well-known equation~F = ~B � ( ~A � ~C)� ~A � ( ~B � ~C): (J:106)If the arrays do not commute, the commutator [Aj; Bl] will be introduced. This will givethe equation Fl =Xj f[Aj; Bl]Cj +BlAjCjg �Xn AlBnCn: (J:107)The de�nition of tha angular momentum can be written as[ji; jj] = iXk eijk jk: (J:108)

Angular momentum algebra J-23J.8. The laboratory coordinate systemThe positions of the atoms in a molecule can be expressed in a �xed laboratory coordi-nate system ~�n, n = 1, 2, : : :, N. Choose a suitable origin within the molecule (such asthe centre of mass) with the coordinates ~R expressed in the laboraty coordinate system.Then one can use amolecule-based coordinate system where the atoms have constantequilibrium coordinates ~ren in the molecule. The atoms perform small vibrations ~dn neartheir equilibrium positions.The two coordinate systems are related through the expression~�n = ~R+ S�1(~ren + ~dn): (J:109)If the laboratory coordinate system ~R refers to the centre of massNXn=1Mnren = 0: (J:110)The molecular translations are included in the laboratory coordinates ~R and the rota-tions of the whole molecule in the rotation matrix S. The coordinates ~dn stand for thevibrations.The coordinate axes in the laboratory coordinate system are denoted x, y and z. Therotation axes in the molecule-based coordinate system are denoted a, b and c. The angularmomentum vector for a rotating molecule is denoted by ~J . It can be expressed in themolecule-based coordinate system. Then its components are J�. However, the measure-ments are done in the laboratory coordinate system. Therefore the operator ~J must also beexpressed in labiratory coordinates, i.e., with components Ji. The transformation betweenthe two coordinate systems is given by the rotation matrix S asJ� = Xi=x;y;z S�iJi: (J:111)The columns and rows of the matrix S are vector operators of class ~T .3 Such operatorsobey the the commutation relations[Ji; S�i] = iXk eijkS�k[J�; S�i] = �iXk e�� S k: (J:112)3 An operator belongs to class T if [Ji; Tj] = ieijkTk. For a deeper discussion on operators of class T , seeCondon och Shortley, kapitel 83.

J-24 Molecular spectroscopyThese equations are used to prove the theorem[Ji; J�] = 0: (J:113)Proof: [Ji; J�] = JiS�j Jj � S�j Jj Ji= [Ji; S�j]Jj + S�j(JiJj � Jj Ji)= ieijk(S�kJJ + S�j Jk) = 0:The components of the angular momentum in the molecule-based coordinate system, J�,obey commutation relations similar to those for Ji but with the opposite sign,[J�; J� ] = �iX e�� J : (J:114)Proof: [J�; J�] (J:111)= [S�iJi; S�jJj ] = S�iJiS�j Jj � S�j JjS�iJi(J:112)= ieijk(S�iS�kJj + S�iS�j Jk + S�kS�j Ji)= �ie�� J :Even if the angular momentum vector has two representations, ~J = (Ja; Jb; Jc)T och~J = (Jx; Jy; Jz)T , it still is the same physical quantity. Therefore the scalar quantity J2that measures the magnitude of the vector, both with Jc and Jz. This follows directlyfrom equation (J.113). The common eigenstate to J2, Jc and Jz is characterized by threequantum numbers, j, k and m. One can write for such an eigenstate the relationsJ2jjkm >= j(j + 1)jjkm >Jcjjkm >= kjjkm >Jz jjkm >= mjjkm > : (J:115)In addition, the step operators obey the rulesJ�jjkm >=pj(j + 1)� k(k � 1)jjk � 1m >=p(j � k)(j � k + 1)jjk � 1m >J�jjkm >=pj(j + 1)�m(m� 1)jjkm� 1 >=p(j �m)(j �m+ 1)jjkm� 1 > :(J:116)Please note the notations for the step operators.

Angular momentum algebra J-25J.9. Symmetry operationsIt has been noted above that the angular momentum operators J2 and Jz have commoneigenfunctions. They are given in equation (J.11). The angular momentum operator Jcthat applies to the rotating coordinate system commutes with the two operators in thelaboratory coordinate system which implies that the eigenfunctions are also eigenfunctionsof that operator. However, one must include the operatorJc = �i @@� (J:117)in the eigenvalue problem. Therefore one will obtain a modi�ed eigenfunction(JKM) = �JKMeiM�eiK�: (J:118)The quantum theory shows that two commuting operators have common eigenfunctions.The eigenvalues may di�er; states may even be degenerate for one operator but not for theother one.The operators that commute with the Hamilton operator are of great interest. The angularmomentum operators above have this property. There are also other important transfor-mations of coordinate system that commute with Hr. Here the following will be discussed1) R, the operator that rotates the whole molecule, all nuclei and electrons including theirspin functions, by an axis that goes through the centre of mass.2) I, the inversion that changes the signs of all coordinates.3) Pnm, all permutations of both the positions and spins of all identical particles n and m.For the rotation operator in an isotropic space one can write[Hr; R] = 0: (J:119)(This is in fact the de�nition of isotropy). If one performs an in�nitesimal rotation by theangle �' in the function f(~qn) the result will beR�'f(~qn) = f(~qn) +Xi �qni @f(~qn)dqni : (J:120)Let �~' be the rotation vector that lies along the rotation axis. Let ~qn be the spatialcoordinate of point P , the origin being the centre of mass. Then the displacement of pointP is obtained as the vector product �~' � ~qn. If one extracts the operator R�' from theabove equation and inserts the displacement one will see thatR�' = 1 +Xi (�~'� ~qn)irni = 1 + (�~'� ~qn) � ~rn = 1 + �~' � (~qn � ~rn): (J:121)

J-26 Molecular spectroscopyThe notation rni = @=@qni is used here. If one multiplies ~rn with �i�h one will obtain thelinear momentum ~pn. The vector product of the spatial vector and the linear momentumgives the angular momentum ~Jn. Summation over all particles n will give the commutationrelation [Hr; R�'] = [Hr; �~' � ~J ] = 0: (J:122)The rotation vector �~' is a constant vector that commutes with all the relevant operators.The inversion operator I changes the sign of all coordinates,If(~q1; ~q2; : : : ; ~qs) = f(�~q1;�~q2; : : : ;�~qs): (J:123)The kinetic energy only depends on terms that are quadratic in coordinates ~qn. In ahomogeneous space, the potential energy only depends on the distances of the atoms.They will not change during the operation. Therefore the Hamiltonian Hr is invariantunder the operation and [Hr; I] = Hr I � IHr = Hr � Hr = 0: (J:124)Because I2 = I(�) = �2; (J:125)the eigenvalue of the operator must be �1.The permutation operator Pnm swaps the coordinates ~qn and ~qm and the correspondingspin coordinates In and Im,Pnmf(~q1; : : : ; ~qm; : : : ; ~qn; : : : ; ~qs; I1; : : : ; Im; : : : ; In; : : : ; Is) =Pnmf(~q1; : : : ; ~qn; : : : ; ~qm; : : : ; ~qs; I1; : : : ; In; : : : ; Im; : : : ; Is): (J:126)The square of the permutation operator is the identity operator. Therefore the eigenvaluepnm of the permutation operator must be �1 for each swap of identical particles. Forparticles with half-integer spin quantum number the eigenvalue is pnm = �1 and forparticles with integer spin quantum number it is pnm = +1. This is, of course, the Paulirule.The symmetry operations of the most important point groups can now be discussedon the basis of the above results. Each molecule with a lower symmetry than C3v is anasymmetric rotor. However, one does not need to consider all the possible point groups.It su�ces to choose the point group D2. The Hamilton operator for an asymmetric rotorcomprises three terms that are quadratic with respect to the three components of theangular momentum. Therefore the Hamiltonian is not a�ected when two of the threecomponents change signs. There will be a change of signs when one rotates by 180� throughthe third axis. Thus one obtains the symmetry operations C2(a), C2(b) and C2(c) that allcommute with the Hamilton operator. They span together with the identity operator Ethe point group D2.

Angular momentum algebra J-27J.10. Hamiltonian of a semirigid moleculeThe classical kinetic energy isT = 12X�� I��!�!� +Xn X�� Mn!�e�� dn� _dn + 12Xn X� Mn _dn� _dn�: (J:127)It is not a trivial task to transform this into a quantum mechanical operator. The �rststep is to express the Cartesian displacements ~dn in normal coordinates ~Q. The trans-formation is pMn dn� =Xr `(�)nr Qr: (J:128)The matrix elements `(�)nr obey the orthogonality conditionXn `(�)nr `(�)nr0 = �rr0 : (J:129)Thus the kinetic energy can be written in normal coordinates asT = 12X�� I��!�!� +Xn X�� Xrr0 !�e�� `(�)nr `( )nr0Qr _Qr0 + 12Xn _Q2r: (J:130)Introduce now the Coriolis constant ��(�)rr0 =Xn X� e�� `(�)nr `( )nr0 ; (J:131)that measures how much angular momentum by the axis � the vibrational motions r andr0 generate. Using this notation the kinetic energy can be expressed asT = 12X�� I��!�!� +X� Xrr0 !��(�)rr0 Qr _Qr0 + 12Xn _Q2r: (J:132)Then intruduce the classical angular momentum and linear momentumJ� = @T@!� =X� I��!� +Xrr0 �(�)rr0 Qr _Qr0Pr = @T@ _Qr =X� Xr0 !��(�)rr0 Qr0 + _Qr: (J:133)

J-28 Molecular spectroscopyWith these notations the kinetic energy becomesT = 12X� J�!� + 12Xr Pr _Qr= 12X� J�!� + 12Xr P 2r � 12Xr PrX� Xr0 !��(�)rr0 Qr0= 12X� (J� � ��) + 12Xr P 2r : (J:134)The new quantity�� =Xrr0 �(�)rr0 QrPr0=Xrr0 �(�)rr0 QrX� Xr00 !��(�)r00r0Qr00 +Xrr0 �(�)rr0 Qr _Qr0 (J:135)gives the change of the angular momentum caused by the vibrational motions. De�ne suchan I 0�� that J� � �� =X� I 0��!� (J:136)or I 0�� = I�� � Xrr0r00 �(�)rr0 �(�)r00r0Qr00 +Xrr0 �(�)rr0 Qr _Qr0 : (J:137)In matrix form this is written as ( ~J � ~�) = I0~!: (J:138)Because I0 = I0y and � = I0�1 one obtains the expression12~!yI0~! = 12( ~J � ~�)y�( ~J � ~�): (J:139)The classical Hamilton operator isH = 12X�� (J� � ��)���(J� � ��) + 12Xr P 2r + V: (J:140)The corresponding quantum mechanical expression isH = 12X�� ���(J� � ��)(J� � ��) + 12Xr P 2r + V (Qr): (J:141)It is di�cult to express the vibrational contribution to the angular momentum�� =Xrr0 �(�)rr0 Qr _Qr0 (J:142)in quantum chemical form.

Angular momentum algebra J-29J.11. Diatomic semirigid moleculeA simpli�ed derivation for a semirigid diatomic molecule is presented in order to illustratethe e�ects that the centrifugal distortion and the zero-point energy have on the observedrotational constant. Some ideas used in the general derivation can also be indicated.However, the terms that contain the angluar momentum ~� and depend on the Coriolise�ect are excluded from this discussion.A typical rotation frequency is of the order of 109 - 1012 Hz and a typical vibrationalfrequency is approximately 1013 - 1014 Hz. Therefore one must use a vibrational averageof the rotation constant B, < vjBjv >. In a diatomic moleculeB = 12I ; (J:143)where I is the momentaneous moment of intertiaI = �r2: (J:144)The reduced mass is � = M1M2M1 +M2 : (J:145)Then it is seen that < vjBjv >=< vj 12I jv >= 12� < vjr�2jv > : (J:146)Perturbation theoretical derivationStart by considering the vibrational motions. Denote the equilibrium distance of theatoms by re and the small variations due to the vibrations by d similarly to the de�nition(J.109). The linear momentum arising from d is pd = @T=@ _d. The classical Hamiltonianfor a harmonic oscillator is H = T + V = p2d2� � 12fd2: (J:147)By using the angular velocity ! = pf=� one can write the potential energy as � 12!2�d2.Introduce now the normal coordinate Q = dp�. The corresponding linear momentum isP = pd=p�. Using these quantities the Hamilton operator can be written asH = 12(P 2 � !2Q2): (J:148)

J-30 Molecular spectroscopyAnother possible de�nition of the normal coordinate is q = dp�! = Qp!. The linearmomentum corresponding to this coordinate is p = pd=p�! = P=p!. Then one obtainsthe alternative form of the Hamiltonian,H = 12!(p2 � q2): (J:149)This expression for the harmonic oscillator can be extended by a term that describes theanharmonicity as (�3=3!)q3.The moment of inertia of a diatomic molecule isI = mr2 = m(re + d)2 = mr2e + 2mred+md2 = Ie + 2pIeQ+Q2: (J:150)This expression is simpli�ed by introducing two new variables. Firstly, de�ne a = 2pIe.Using this quantity the expression can be written asI = Ie + aQ+ a24IeQ2 = Ie�1 + aQIe + a2Q24I2e � : (J:151)Next, introduce the variable � = aQ=2Ie which results in the expressionI = Ie(1 + �)2: (J:152)The inverse of the moment of inertia can now be written as1I = 1Ie (1� 2� + 3�2 �+ : : :): (J:153)Observe that the Hamilton operator of a rigid rotor is H = BeJ2. Therefore the �rst termrepresents the rotational energies of a rigid rotor and the other terms are corrections dueto the vibrations.THe Hamiltonian of the unperturbed system is taken asH = BeJ2 + 12!(p2 � q2): (J:154)The unperturbed basis functions jJv > = jJ > jv > are obtained as eigenfunctions of thisopertor. The Hamiltonian of the perturbed system isH = Be(1� 2� + 3�2)J2 + 12!(p2 � q2) + �33! q3: (J:155)The perturbation isH 0 = �2Be�J2 + 3Be�2J2 + �33! q3 = �Be aQIe J2 + 34Be�aQIe �2 J2 + �33! q3: (J:156)

Angular momentum algebra J-31By introducing the de�nition = a=Iep! and observing that Qp! = q it is seen thatH 0 = �Be qJ2 + 34Be 2q2J2 + �33! q3: (J:157)The perturbation theoretical energy contributions up to second order are[�Be ]2[�1=2!]J2(J + 1)2[ 34Be 2][v + 12]J(J + 1)2[�Be ][ 16�3][�32(v + 12)=!]J(J + 1): (J:158)The terms that do not depend on the quantum number J have been excluded.The usual form of the rotation energy expression isE(vJ) = BvJ(J + 1)�DJ2(J + 1)2: (J:159)A comparison shows thatD = �4[�Be ]2[�1=2!] = 4B2e � aIep!�2 [1=2!] = 4B3e!2 : (J:160)and that Bv = Be � �(v + 12); (J:161)where the harmonic correction gives the contribution�34Be 2 = �34Be� aIep!�2 = �6B2e! : (J:162)The anharmonic contribution is�2[�Be ][ 16�3][�321=!] = �3Be! � aIep!� 16�3 = �6Be! �36 r 2Be! : (J:163)It is customary to write � = �6B2e! (1 + a1): (J:164)

J-32 Molecular spectroscopyJ.12. Selection rulesThe dipole moment integrals determine the selection rules, i.e., wether a transition ispossible or not. If the factor < nj~�jm > is equal to zero the transition n m is saidto be symmetry forbidden. This is the case for most transitions. If the factor is �nitethe transition is said to be symmetry allowed. The intensity of the spectral band isproportional to the integral.It su�ces to calculate the matrix elements < nj~�jm > for the states jJKM > because theeigenstates of the operator Hr always can be written as linear combinations of these. Thedipole moment transforms as �i =X� S�1i� �� =X� S�i��: (J:165)The components �� of the dipole moment operator in the molecule-based coordinate systemcan be considered as constant parameters. Therefore the problem is reduced to calculationof the matrix elements of the transformation matrix S,< J 0K 0M 0j�ijJ 00K 00M 00 >=X� < J 0K 0M 0jS�ijJ 00K 00M 00 > ��: (J:166)Thus the molecule must have a permanent dipole moment (except for Raman, of course).As the rotation matrix S is a tensor of type T it obeys the rules[Ji; S�j] =Xk ieijkS�k (J:167)and [J�; S�i] =X ie�� S i: (J:168)De�ne now a new matrix S� for the principal axes of the molecule,S�i = Sai � Sbi: (J:169)This matrix ful�ls the commutation relation[Jc; S�] = �S�i : (J:170)Therefore one �nds thatXJ 00K00M 00 n< JKM jJcjJ 00K 00M 00 >< J 00K 00M 00jS�i jJ 0K 0M 0 >� < JKM jS�i jJ 00K 00M 00 >< J 00K 00M 00jJcjJ 0K 0M 0 >o = � < JKM jS�i jJ 0K 0M 0 > :(J:171)

Angular momentum algebra J-33From this follows that (K �K 0 � 1) < JKM jS�i jJ 0K 0M 0 >= 0: (J:172)It is seen that all matrix elements where K 6= K 0 � 1 will vanish. The rotation matrixelements Sai and Sbi are linear combinations of these. Similarly, the commutation rule[Jc; Sci] = 0: (J:173)leads to the expression (K �K 0) < JKM jScijJ 0K 0M 0 >= 0: (J:174)From this follows the tule K = K 0. Combining these results one obtains the selectionrules �K = 0;�1.The steps to derive the selection rules for the quantum number J are basically similar.The starting point is the commutator[ ~J2;S] =Xij Ji[Ji; Sj] + [Ji; Sj ]Ji=Xijk ieijk(2JiTk �Xl ieilkTl)=Xijk �2i(ejikJiSk � iSj)= �2i( ~J � S� iS): (J:175)One �nds in the general theory of vector algebra that[ ~J2; ~J � S] = ~J � [ ~J2;S]: (J:176)Then one can see that the double commutator� ~J2; [ ~J2;S]� = ~J4S� 2 ~J2S ~J2 + S ~J4= �2i��2i ~J � ( ~J � S)� 2 ~J � S� i( ~J2S� S ~J2)�= 2 ~J2S+ 2S ~J2 � 4 ~J( ~J � S): (J:177)Consider now the case �J 6= 0. Here one must calculate the non-diagonal matrix elements.The term ~J( ~J � S) will not contribute because the scalar product of two tensors of type Tcommutes with ~J . Therefore one only needs to evaluate< JKM j ~J4S� 2 ~J2S ~J2 + S ~J4jJ 0K 0M 0 >= 2 < JKM j ~J2S+ S ~J2jJ 0K 0M 0 > : (J:178)

J-34 Molecular spectroscopyThis shows that[J2(J + 1)2 � 2J(J + 1)J 0(J 0 + 1) + J 02(J 0 + 1)2] < JKM jSjJ 0K 0M 0 >= 2]J(J + 1) + J 0(J 0 + 1)] < JKM jSjJ 0K 0M 0 > : (J:179)It is seen that the algebraic factors can be written as[J2(J + 1)2 � 2J(J + 1)J 0(J 0 + 1) + J 02(J 0 + 1)2] = (J � J 0)2(J + J 0 + 1)2 (J:180)and 2]J(J + 1) + J 0(J 0 + 1)] = (J + J 0 + 1)2 + (J � J 0)2 � 1; (J:181)which leads to the relations[(J + J 0 + 1)2 � 1][(J � J 0)2 � 1] < JKM jSjJ 0K 0M 0 >= 0: (J:182)From this follows the selection rule �J = �1.