PPairing, Quasi-spin and Seniority
Bijay Kumar Agrawal ∗ and Bhoomika Maheshwari Saha Institute of Nuclear Physics, Kolkata, India. Homi Bhabha National Institute, Mumbai, India. University of Malaya, Kuala Lumpur, Malaysia.June 9, 2020
Abstract
We present our concise notes for the lectures and tutorials on pairing,quasi-spin and seniority delivered at SERB school on Role of Symme-tries in Nuclear Physics, AMITY University, 2019. Starting with somegeneric features of residual nucleon-nucleon interactions, we provide de-tailed derivation of the matrix elements for the δ -interaction which is thebasis for the standard pairing Hamiltonian. The eigen values for stan-dard pairing Hamiltonian are then obtained within the quasi-spin formal-ism. The algebra involving quasi-spin operators is performed explicitlyusing the annihilation and creation operators for single nucleon togetherwith the application of Wick’s theorem. These techniques are expectedto be helpful in deriving the mean-field equations for the Hartree-Fock,BardeenCooperSchrieffer and Hartree-Fock Bogoliubov theories. Role of symmetries is important when no exact solutions to a physical prob-lem are known, such as nuclear force. More often, our problem is analogousto the atomic and molecular structure. However, things are simpler there, asCoulomb forces are well known. In case of the nuclear force, the knowledgeof its mathematical form is still an open question. Fortunately, due to sym-metry properties of the basic interactions in most physical systems, qualitativefeatures of the composite system are not too sensitive to the details of the inter-action itself. Symmetries of physical laws may lead to the laws of conservationof spin, isospin and energy. For example, the orbital, spin and isospin quan-tum operators usually commute with the nuclear interaction Hamiltonian. Thismeans the interaction Hamiltonian commutes with all rotations in orbital, spin ∗ Email: [email protected] (Bijay Agrawal) a r X i v : . [ nu c l - t h ] J un nd isospin spaces. Therefore, the angular momentum coupling and sphericalharmonics play a vital role in understanding various properties of nuclei.One of the most important inputs to the system of interacting nucleonsis the matrix elements for the appropriate nucleon-nucleon interaction. Thesematrix elements can be evaluated by decomposing nucleon-nucleon interactionin to its radial and angular parts. The evaluation of two-body matrix elements,thus, reduces to the calculation of the radial and the angular matrix elements.The procedure for the calculation of the two-body matrix elements for differentinteractions mainly differ in their radial matrix elements. The angular partusually depends on the product of the spherical harmonics corresponding to thetwo nucleons. The angular part of the matrix elements is straightforward butlengthy, since, they are evaluated within the anti-symmetrized coupled statesof two nucleons. Further more, each of the nucleonic wave function correspondto the coupled state of the their orbital angular momentum and intrinsic spin.The derivation is not available in full detail at a single place, which we foundimportant to discuss with students during SERB school.In the present article, we summarize details of our lectures on the evaluationof the direct and exchange terms for the matrix elements of the δ -interaction.Next, we consider the standard pairing Hamiltonian which mimics the short-range nature of the δ -interaction. Its exact eigen values and eigen states areobtained within the quasi-spin formalism. The algebra involving quasi-spin op-erators is performed directly in terms of the annihilation and creation operatorsfor single nucleon. We simplify our algebra using the Wick’s theorem. Thesetechniques are also handy in deriving the mean-field equations corresponding todifferent trial wave-function which leads to the various mean-field approachessuch as Hartree-Fock, Bardeen-Cooper-Schrieffer and Hartree-Fock Bogoliubovapproximations. Aim of the present notes is to provide sufficient details in aself-contained manner. In Appendices, we include various identities involving3j- and 6j- symbols and provide the derivations for the quasi-spin operators andthe reduced matrix elements for the spherical harmonics which were coveredduring tutorials. The binding energy for several hundreds of nuclei are known very precisely. Thevariations of binding energy per nucleon with mass number clearly manifeststhe strong and short range nature of the nuclear force. The Hamiltonian forsuch an strongly interacting system is not exactly solvable. A general nuclearHamiltonian satisfy H Ψ( r , r , r , ...r A ) = E Ψ( r , r , r , ...r A ) (1)2here H comprises the kinetic energy and interaction part, H = A (cid:88) i =1 T ( r i ) + A (cid:88) i 12 ; j || Y K || l (cid:48) 12 ; j (cid:48) )( l 12 ; j || Y K || l (cid:48) 12 ; j (cid:48) ) (16)where C denotes the radial integral as follows: C = 14 π (cid:90) R ∗ n l j ( r ) R ∗ n l j ( r ) δ ( r − r ) r r R n (cid:48) l (cid:48) j (cid:48) ( r ) R n (cid:48) l (cid:48) j (cid:48) ( r ) r r dr dr = 14 π (cid:90) R ∗ n l j ( r ) R ∗ n l j ( r ) R n (cid:48) l (cid:48) j (cid:48) ( r ) R n (cid:48) l (cid:48) j (cid:48) ( r ) r dr (17)The reduced matrix elements appearing in Eq. (16) can be simplified usingEq.(87) as, ( l 12 ; j || Y K || l (cid:48) 12 ; j (cid:48) ) = ( − j − (cid:32) ˆ j (cid:48) ˆ j ˆ K √ π (cid:33) 12 (1 + ( − l (cid:48) + l + K ) × (cid:18) j K j (cid:48) − (cid:19) (18)( l 12 ; j || Y K || l (cid:48) 12 ; j (cid:48) )( l 12 ; j || Y K || l (cid:48) 12 ; j (cid:48) ) = ( − j + j − (cid:32) ˆ j (cid:48) ˆ j ˆ j (cid:48) ˆ j ˆ K π (cid:33) 14 (1 + ( − l (cid:48) + l + K )(1 + ( − l (cid:48) + l + K ) × (cid:18) j K j (cid:48) − (cid:19) (cid:18) j K j (cid:48) − (cid:19) (19)In the Eq.(19) ˆ j i = √ j i + 1, ˆ K = √ K + 1. For simplicity, let us consider l (cid:48) i , j (cid:48) i = l i , j i . So Eq. (19) becomes, 7 l 12 ; j || Y K || l 12 ; j )( l 12 ; j || Y K || l 12 ; j ) = ( − j + j − (cid:32) ˆ j ˆ j ˆ K π (cid:33) × (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) 14 (1 + ( − l + K )(1 + ( − l + K )= ( − j + j − (cid:32) ˆ j ˆ j ˆ K π (cid:33) 12 (1 + ( − K ) × (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) (20)Substituting Eq. (20) in Eq. (16) we get, D = 4 πC (cid:88) K ( − j + J + j (cid:26) j j Jj j K (cid:27) ( − j + j − (cid:32) ˆ j ˆ j ˆ K π (cid:33) 12 (1 + ( − K ) × (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) = C (cid:32) ˆ j ˆ j (cid:33) (cid:88) K ( − J − ˆ K (1 + ( − K ) × (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) = C (cid:32) ˆ j ˆ j (cid:33) (cid:88) K (cid:20) ( − J − ˆ K × (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) +( − J + K − ˆ K × (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19)(cid:21) (21)The sum over K in Eq.(21) can be performed analytically. It can be reduced toproduct of two 3j- symbols using a suitable identity as follows, (cid:88) M (cid:18) j j J m m M (cid:19) (cid:18) j (cid:48) j (cid:48) J m (cid:48) m (cid:48) − M (cid:19) = (cid:88) J (cid:48) M (cid:48) ( − j + j (cid:48) + m + m (cid:48) ˆ J (cid:48) (cid:26) j j J j (cid:48) j (cid:48) J (cid:48) (cid:27)(cid:18) j (cid:48) j J (cid:48) m (cid:48) m M (cid:48) (cid:19) (cid:18) j j (cid:48) J (cid:48) m m (cid:48) − M (cid:48) (cid:19) (22)Substituting J = J , J (cid:48) = K , j (cid:48) = j and j (cid:48) = j in Eq. (22) we get,8 M (cid:18) j j Jm m M (cid:19) (cid:18) j j Jm (cid:48) m (cid:48) − M (cid:19) = (cid:88) KM (cid:48) ( − J + K + m + m (cid:48) ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j j Km (cid:48) m M (cid:48) (cid:19) (cid:18) j j Km m (cid:48) − M (cid:48) (cid:19) = (cid:88) KM (cid:48) ( − J + K + m + m (cid:48) ˆ K (cid:26) j j Jj j K (cid:27) ( − j + j + K (cid:18) j K j m (cid:48) M (cid:48) m (cid:19) ( − j + j + K (cid:18) j K j m − M (cid:48) m (cid:48) (cid:19) = (cid:88) KM (cid:48) ( − J + K + m + m (cid:48) ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j K j m (cid:48) M (cid:48) m (cid:19) (cid:18) j K j m − M (cid:48) m (cid:48) (cid:19) (23)Sums over M and M (cid:48) are constrained by, m + m + M = m (cid:48) + m (cid:48) − M = 0 m (cid:48) + m + M (cid:48) = m + m (cid:48) − M (cid:48) = 0Substituting m = m (cid:48) = − / m = m (cid:48) = +1 / M = M (cid:48) = 0 (cid:18) j j J − + (cid:19) (cid:18) j j J − + (cid:19) = (cid:88) K ( − J + K − ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) (24)9ubstituting m = m = +1 / m (cid:48) = m (cid:48) = − / M = − , M (cid:48) = 0 in Eq.(23) becomes, (cid:18) j j J + − (cid:19) (cid:18) j j J − − (cid:19) = (cid:88) K ( − J + K ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) = (cid:88) K ( − J + K ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j K j − (cid:19) ( − j + K (cid:18) j K j − (cid:19) = (cid:88) K ( − J − ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) (25)Adding Eqs. (24) and (25), the R.H.S. and L.H.S. become, RHS = (cid:88) K ( − J − (1 + ( − K ) ˆ K (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) (26) LHS = (cid:18) j j J + − (cid:19) (cid:18) j j J − − (cid:19) + (cid:18) j j J − + (cid:19) (cid:18) j j J − + (cid:19) = (cid:18) j j J + − (cid:19) + (cid:18) j j J − + (cid:19) (27)Using the recursion relation for 3j- symbol as in Eq. (88), yields, (cid:18) j j J 12 12 − (cid:19) = [ ˆ j + ˆ j ( − j + j + J ] J ( J + 1) (cid:18) j j J − 12 12 (cid:19) LHS = (cid:34) j + ˆ j ( − j + j + J ] J ( J + 1) (cid:35) (cid:18) j j J − 12 12 (cid:19) (28)Using Eqs. (21), (26) and (28), the direct term simply becomes, D = C (cid:32) ˆ j ˆ j (cid:33) × (cid:34) j + ˆ j ( − j + j + J ] J ( J + 1) (cid:35) (cid:18) j j J − 12 12 (cid:19) (29)Let us consider a special case, if j = j = j, and J is even, then, second termin Eq. (29) vanishes, and the direct term becomes, D = C (cid:32) ˆ j (cid:33) (cid:18) j j J − 12 12 (cid:19) (30)10e shall see below, for this special case, odd values of J are not allowed for theidentical nucleons. Since, the contribution from the exchange term in this caseis equal and opposite to that for the direct term. By substituting Eq.(11) in Eq.(13) and separating radial and angular parts ofmatrix elements the exchange term becomes, E = ( − j (cid:48) + j (cid:48) + J + T πC × (cid:88) K (cid:104) α j α j ; JM | Y K (1) · Y K (2) | α (cid:48) j (cid:48) α (cid:48) j (cid:48) ; JM (cid:105) (31)It must be noticed that the radial part of the matrix element C is same asthat for the direct term Eq.(17), because, the δ -interaction acts only when r = r . The angular part can further be decomposed into the products ofmatrix elements for Y K (1) and Y K (2) using Eq. (87). The Eq. (31) simplifiesto, E = ( − j (cid:48) + j (cid:48) + J + T πC (cid:88) K ( − j + J + j (cid:48) (cid:26) j j Jj (cid:48) j (cid:48) K (cid:27) (cid:104) α j || Y K (1) || α (cid:48) j (cid:48) (cid:105)(cid:104) α j || Y K (2) || α (cid:48) j (cid:48) (cid:105)(cid:104) α ; j | | Y K | | α (cid:48) ; j (cid:48) (cid:105) (cid:104) α ; j | | Y K | | α (cid:48) ; j (cid:48) (cid:105) = ( − j + j − (cid:32) ˆ j (cid:48) ˆ j ˆ j (cid:48) ˆ j ˆ K π (cid:33) 14 (1 + ( − l (cid:48) + l + K )(1 + ( − l (cid:48) + l + K ) × (cid:18) j K j (cid:48) − (cid:19) (cid:18) j K j (cid:48) − (cid:19) For α (cid:48) j (cid:48) = α j and α (cid:48) j (cid:48) = α j , (cid:104) α ; j | | Y K | | α ; j (cid:105)(cid:104) α ; j | | Y K | | α ; j (cid:105) = ( − j + j − (cid:32) ˆ j ˆ j ˆ K π (cid:33) 12 (1 + ( − l + l + K ) × (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) (32)11 = ( − T πC (cid:32) ˆ j ˆ j π (cid:33) (cid:88) K ˆ K (1 + ( − l + l + K ) × (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) = ( − T C (cid:32) ˆ j ˆ j (cid:33) (cid:88) K (cid:20) ˆ K × (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) +( − l + l + K ˆ K × (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19)(cid:21) (33)The sum over K can be performed analytically as in the case of direct term.Substituting J = J , J (cid:48) = K , j (cid:48) = j and j (cid:48) = j in Eq. (22), we get, (cid:88) M (cid:18) j j Jm m M (cid:19) (cid:18) j j Jm (cid:48) m (cid:48) − M (cid:19) = (cid:88) KM (cid:48) ( − J + K + m + m (cid:48) ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j j Km (cid:48) m M (cid:48) (cid:19) (cid:18) j j Km m (cid:48) − M (cid:48) (cid:19) = (cid:88) KM (cid:48) ( − J + K + m + m (cid:48) ˆ K (cid:26) j j Jj j K (cid:27) ( − j + j + K (cid:18) j K j m (cid:48) M (cid:48) m (cid:19) (cid:18) j K j m (cid:48) − M (cid:48) m (cid:19) = (cid:88) KM (cid:48) ( − j + j + J + m + m (cid:48) ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j K j m (cid:48) M (cid:48) m (cid:19) (cid:18) j K j m (cid:48) − M (cid:48) m (cid:19) (34)Substituting m = m (cid:48) = − / m = m (cid:48) = +1 / M = M (cid:48) = 0 (cid:18) j j J − + (cid:19) (cid:18) j j J − + (cid:19) = (cid:88) K ( − j + j + J − ˆ K (cid:26) j j Jj j K (cid:27)(cid:18) j K j − (cid:19) (cid:18) j K j + − (cid:19) (35) (cid:18) j j J − + (cid:19) = (cid:88) K ( − J + K − ˆ K (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) (36)12ubstituting m = m = +1 / m (cid:48) = m (cid:48) = − / 2, and the allowed values of M = − , M (cid:48) = 0 in Eq.(34) becomes, (cid:18) j j J + − (cid:19) (cid:18) j j J − − (cid:19) = (cid:88) K ( − j + j + J ˆ K (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19)(cid:18) j j J + − (cid:19) = (cid:88) K ˆ K (cid:26) j j Jj j K (cid:27) (cid:18) j K j − (cid:19) (cid:18) j K j − (cid:19) (37)Substituting Eqs. (36) and (37) in Eq. (33) E = ( − T C (cid:32) ˆ j ˆ j (cid:33) (cid:34)(cid:18) j j J + − (cid:19) + ( − l + l + J − (cid:18) j j J − + (cid:19) (cid:35) = ( − T C (cid:32) ˆ j ˆ j (cid:33) (cid:34)(cid:18) j j J + − (cid:19) − ( − l + l + J (cid:18) j j J − + (cid:19) (cid:35) (38)Using the recursion relation for 3j-symbol, Eq. (88), E = ( − T C (cid:32) ˆ j ˆ j (cid:33) (cid:18) j j J − 12 12 (cid:19) (cid:34) [ ˆ j + ˆ j ( − j + j + J ] J ( J + 1) − ( − l + l + J (cid:35) (39) Total matrix element is the sum of the direct and exchange terms with appro-priate normalization factor. Using Eqs. (13),(29), and (39), the total matrixelement can be written as, (cid:104) α j α j ; JM | δ ( (cid:126)r ) | α j α j ; JM (cid:105) as = N [ D + E ]= N C (cid:32) ˆ j ˆ j (cid:33) (cid:18) j j J − 12 12 (cid:19) × (cid:34) − ( − l + l + J + T + (1 + ( − T ) [ ˆ j + ˆ j ( − j + j + J ] J ( J + 1) (cid:35) (40)For the special case, α i j i = αj and N = , the Eq.(40) becomes, (cid:104) ( αj ) ; JM | δ ( (cid:126)r ) | ( αj ) ; JM (cid:105) as = C (cid:32) ˆ j (cid:33) × (cid:18) j j J − (cid:19) (cid:34) − ( − J + T + (1 + ( − T ) ˆ j (1 − ( − J ) J ( J + 1) (cid:35) . (41)13or the identical nucleons, T = 1, the above matrix element is non-zero only foreven values of J and is given by, (cid:104) ( αj ) ; JM | δ ( (cid:126)r ) | ( αj ) ; JM (cid:105) as = C (cid:32) ˆ j (cid:33) × (cid:18) j j J − (cid:19) = C (cid:18) (2 j + 1) (cid:19) × (cid:18) j j J − (cid:19) Substituting the value of 3j-symbol for J = 0 and 2, (cid:104) ( αj ) ; JM | δ ( (cid:126)r ) | ( αj ) ; JM (cid:105) as = C π (cid:40) (2 j + 1) for J=0(2 j + 1) (cid:16) (2 j − j +3)16 j ( j +1) (cid:17) for J=2The | ( αj ) ; 00 (cid:105) is referred as paired state. The matrix element for the δ -interaction between the paired states is much larger than those between theunpaired sates i.e., (cid:104) ( αj ) ; 00 | δ ( (cid:126)r ) | ( αj ) ; 00 (cid:105) >> (cid:104) ( αj ) ; 2 M | δ ( (cid:126)r ) | ( αj ) ; 2 M (cid:105) (42)The pairing Hamiltonian, as formally defined later in Section 5, mimics the δ -interaction acting on the paired state so, (cid:104) j ; JM | ˆ H pair | j ; JM (cid:105) ∝ (2 j + 1) δ J (43) The creation and annihilation operators for coupled angular momentum statescan be conveniently expressed in terms of these operators for single nucleon.The algebra involving these operators become simple due to the application ofWick’s theorem as described below very briefly. The creation operator ’ a † ’ andannihilation operator ’ a ’ for single-nucleon can be defined as, a † α |−(cid:105) = | α (cid:105) , (44) a β | α (cid:105) = δ αβ |−(cid:105) , (45) a |−(cid:105) = 0 , (46)where |−(cid:105) is the vacuum state. The anti-commutation relations for the creationand annihilation operators for single nucleon ’ a † ’ and ’ a ’ are given by, (cid:110) a † α , a † β (cid:111) = { a α , a β } = 0 (cid:8) a † α , a β (cid:9) = δ αβ (47)14he normal order product is defined as,: a β a † γ : = − a † γ a β (48): a β a † γ a † δ : = a † γ a † δ a β (49): a α a β a † γ a † δ : = a † γ a † δ a α a β (50) (cid:104)−| : a α a β a † γ a † δ : |−(cid:105) = (cid:104)−| a † γ a † δ a α a β |−(cid:105) = 0 (51)The normal ordered product essentially arranges all the creation operators leftto the annihilation operators as should be clear from the above equations. Theover all sign is negative only if number of permutations required to shift all thecreation operators to the left is odd.The contractions of two operators are given by, a β a † γ = (cid:104)−| a β a † γ |−(cid:105) = δ βγ a † γ a β = (cid:104)−| a † γ a β |−(cid:105) = 0 a α a β = a † γ a † δ = 0 (52)A product of set of creation and annihilation operators can be simplified interms of normal order products and contractions using Wick’s theorem. Forexample, the product of a set of four creation and annihilation operators can beexpressed using the Wick’s theorem as, a α a β a † γ a † δ = : a α a β a † γ a † δ : + (cid:88) singles aa † : aa † : + (cid:88) doubles aa † aa † (53)The sums in the above equation run over all the possible single and doublecontractions. The Eq.(53) can easily be extended to a product of any numberof creation and annihilation operators. The coupled state of two nucleons can be expressed in terms of product of thesingle-nucleon states as [2, 3, 4, 5], | j j ; JM (cid:105) = (cid:88) m ,m (cid:104) j m j m | j j JM (cid:105)| j m j m (cid:105) (54)where, j i and m i are the angular momentum and its z -component for the single-nucleon, respectively. The (cid:104) j m j m | j j JM (cid:105) is the Clebsch-Gordan coeffi-cient. The anti-symmetrization of the above equation yields, | j j ; JM (cid:105) as = N (cid:88) m ,m (cid:104) j m j m | j j JM (cid:105) ( | j m j m (cid:105) − | j m j m (cid:105) )= N (cid:88) m ,m (cid:104) j m j m | j j JM (cid:105)| j m j m (cid:105) as (55)15he operators A † ( j , j ; JM ) and A ( j , j ; JM ) which create and annihilate twonucleons in a coupled state can be expressed analogous to Eq. (55) as, A † ( j j ; JM ) = N (cid:88) m ,m (cid:104) j m j m | j j JM (cid:105) a † j m a † j m (56) A ( j j ; JM ) = N (cid:88) m ,m (cid:104) j m j m | j j JM (cid:105) a j m a j m (57) The normalization constant can easily be calculated by applying Wick’s theoremsuch that, (cid:104)| A ( j j ; JM ) A † ( j j ; JM ) |−(cid:105) = 1 . (58)Using Eqs. (56) and (57), (cid:104)−| A ( j j ; JM ) A † ( j j ; JM ) |−(cid:105) = N (cid:88) m m (cid:88) m (cid:48) m (cid:48) (cid:104) j m j m | j j JM (cid:105)(cid:104) j m (cid:48) j m (cid:48) | JM (cid:105)(cid:104)−| a j m (cid:48) a j m (cid:48) a † j m a † j m |−(cid:105) (59)The R.H.S. of the Eq.(59) can be simplified by applying the Wick’s theoremEq.(53). Also, the vacuum expectation value for a normal ordered productvanishes. Only the fully contracted terms will contribute, we have (cid:104)−| a j m (cid:48) a j m (cid:48) a † j m a † j m |−(cid:105) = (cid:104)−| (cid:26) a j m (cid:48) a † j m a j m (cid:48) a † j m − a j m (cid:48) a † j m a j m (cid:48) a † j m (cid:27) |−(cid:105)(cid:104)−| a j m (cid:48) a j m (cid:48) a † j m a † j m |−(cid:105) = δ m m (cid:48) δ m m (cid:48) − δ j j δ m m (cid:48) δ m m (cid:48) . (60)16ubstituting Eq. (60) in Eq. (59), (cid:104)−| A ( j j ; JM ) A † ( j j ; JM ) |−(cid:105) = N (cid:88) m m (cid:88) m (cid:48) m (cid:48) (cid:104) j m j m | j j JM (cid:105)(cid:104) j m (cid:48) j m (cid:48) | j j JM (cid:105) ( δ m m (cid:48) δ m m (cid:48) − δ j j δ m m (cid:48) δ m m (cid:48) )= N (cid:88) m m ( (cid:104) j m j m | j j JM (cid:105)(cid:104) j m j m | j j JM (cid:105)− δ j j (cid:104) j m j m | j j JM (cid:105)(cid:104) j m j m | j j JM (cid:105) )= N (cid:88) m m (cid:16) (cid:104) j m j m | j j JM (cid:105) − δ j j ( − j + j − J (cid:104) j m j m | j j JM (cid:105)(cid:104) j m j m | j j JM (cid:105) (cid:1) = N (cid:88) m m (cid:16) (cid:104) j m j m | j j JM (cid:105) + δ j j ( − − J (cid:104) j m j m | j j JM (cid:105) (cid:17) (61) (cid:104)−| A ( j j ; JM ) A † ( j j ; JM ) |−(cid:105) = N (1 + δ j j ( − − J ) (cid:88) m m (cid:104) j m j m | j j JM (cid:105) (cid:104)−| A ( j j ; JM ) A † ( j j ; JM ) |−(cid:105) = N (1 + δ j j ( − − J ) N = 1 (cid:112) δ j j (62)In Eq.(61) we have substituted j = j in the phase factor due to the presenceof δ j j . It may be easily verified from Eq. (55) that if j = j , only the evenvalue of J are allowed. Two nucleons are said to be in the paired state when, j = j = j, J = M = 0.The creation and annihilation operators for the paired state can be obtainedusing Eqs. (56) and (57) together with (62),i.e., A † ( jj ; 00) = 1 √ (cid:88) m ,m (cid:104) jm jm | jj (cid:105) a † jm a † jm = 1 √ (cid:88) m ( − j − m √ j + 1 a † jm a † j, − m (63) A ( jj : 00) = 1 √ (cid:88) m ( − j − m √ j + 1 a j, − m a jm (64)In the above Eqs. (63) and (64) the operators A † ( jj ; 00) creates a pair while A ( jj ; 00) annihilates it. To facilitate the discussion of pairing Hamiltonian, theoperators A † and A can further be simplified as,17 † ( jj ; 00) = (cid:114) (cid:88) m> ( − j − m a † jm a † j, − m (65) A ( jj : 00) = (cid:114) (cid:88) m> ( − j − m a j, − m a jm (66)where, Ω = (2 j + 1) is the pair degeneracy or the maximum number of pairsin a given j -shell.The pairing Hamiltonian in a single j -shell can be defined as,ˆ H pair = − GS + j S − j (67)where, G is the pairing strength. The operators S + j S − j are the quasi-spin op-erators as they obey the commutation rules analogous to those for the angularmomentum operators (see Appendix A). These operators can be expressed interms of the pair creation and annihilation operators as, S + j = √ Ω A † = (cid:88) m> ( − j − m a † jm a † j, − m (68) S − j = √ Ω A = (cid:88) m> ( − j − m a j, − m a jm (69)Using Eqs. (68) and (69), the pairing Hamiltonian can be written as,ˆ H pair = − G (cid:88) m,m (cid:48) > ( − j − m − m (cid:48) a † jm a † j, − m a j, − m (cid:48) a jm (cid:48) It can be easily verified that the matrix element or the pairing Hamiltonian isnon-zero only between the paired states, i.e., (cid:104) j ; JM | ˆ H pair | j ; JM (cid:105) = − G j + 1) δ J δ M . (70) The eigen values of the pairing Hamiltonian can be obtained once the com-mutation relation between S + j and S − j is known. In what follows, we drop theindex j from the operators S + j and S − j . This commutation relation can be easilyevaluated by applying Wick’s theorem, as briefly described in Appendix A, onegets, (cid:2) S + , S − (cid:3) − = ˆ N − Ω = 2 S (71)where S is a operator analogous to z-component of quasi spin in a way that S + , S − and S follow the SU(2) Lie algebra (Appendix A). Also, the particlenumber operator ( ˆ N ), ˆ N = (cid:88) m> [ a † m a m + a †− m a − m ] . (72)18t can be further shown that,[ S + , S ] − = − S + (73) (cid:2) S − , S (cid:3) − = S − (74)The commutations of S + , S − and S suggest that these operators follow the an-gular momentum algebra and can be called as quasi-spin operators. Therefore, S + S − = S − ( S ) + S = S ( S + 1) − S ( S − S ( S + 1) − (cid:32) ˆ N − Ω2 (cid:33) (cid:32) ˆ N − Ω2 − (cid:33) = S ( S + 1) − (cid:32) Ω − ˆ N (cid:33) (cid:32) Ω − ˆ N (cid:33) (75)where S ( S + 1) is the eigen-value for the operator S with S being an integeror half-integer analogous to the case of angular momentum. In principle, theeigen-state of the pairing Hamiltonian can be characterized by the quasi-spin S and its z -component S . However, it is convenient to express them in terms ofthe total number of nucleons and the seniority quantum number, i.e., numberof the unpaired nucleons.Let us consider a n -nucleon state | j n , ν ; JM (cid:105) with ν number of unpaired nu-cleons or the seniority quantum number. Thus, if we consider a state with all thenucleons unpaired, i.e., n = ν , ˆ N | j ν , ν ; JM (cid:105) = ν | j ν , ν ; JM (cid:105) and S − j | j ν , ν ; JM (cid:105) =0, since, there is no pair to annihilate. By operating L.H.S. and R.H.S. of Eq.(75)on | j ν , ν ; JM (cid:105) , S ( S + 1) − (cid:32) Ω − ˆ N (cid:33) (cid:32) Ω − ˆ N (cid:33) | j ν νJM (cid:105) = S + S − | j ν νJM (cid:105) S ( S + 1) − Ω − ν (cid:18) Ω − ν (cid:19) = 0 S = Ω − ν S + S − = Ω − ν (cid:18) Ω − ν (cid:19) − (cid:32) Ω − ˆ N (cid:33) (cid:32) Ω − ˆ N (cid:33) S + S − = 14 ( ˆ N − ν )(2Ω − ˆ N − ν + 2) (77)19 .2 Eigen values of ˆ H pair The pairing Hamiltonian using Eqs. (67) and (77) becomes,ˆ H pair = − GS + S − = − G N − ν )(2Ω − ˆ N − ν + 2)ˆ H pair | j n , ν ; JM (cid:105) = − G N − ν )(2Ω − ˆ N − ν + 2) | j n , ν ; JM (cid:105) = − G n − ν )(2Ω − n − ν + 2) | j n , ν ; JM (cid:105) = E ( n, ν ) | j n , ν ; JM (cid:105) The pairing eigen value E ( n, ν ) can be obtained as, E ( n, ν ) = − G n − ν )(2Ω − n − ν + 2)For fully paired state, ν = 0, in case of even-even nuclei E ( n, 0) = − G n (2Ω − n + 2)The pairing energy is given as, E ( n, ν ) − E ( n, 0) = G ν (2Ω − ν + 2) (78)It may be noted that the pairing energy is independent of the number of nucleonsin the j -shell. One can understand the energy separation of 0 + and 2 + in even-even nuclei interms of pairing. For instance, in Fig. 2 we display the observed spectra for a fewnuclei around mass number A = 90 with the neutron number N = 50. It maybe noted that the energy separations between the 0 + and 2 + for these nuclei arealmost 1 . − . J = 0 + , ν = 0. First excited-state corresponds to breakingof one proton pair yielding J = 2 + , ν = 2. Therefore, within the pairing model,the energy separation between the ground and the first excited states can beapproximated using Eq.(78) as, E (2 + ) − E (0 + ) ≈ E ( n, − E ( n, E ( n, − E ( n, 0) = G Ω . (79)20 E * ( M e V ) Mo Ru Pd Cd + + + + Figure 2: Energy spectrum of even-even nuclei from mass-90 region.For A = 90 mass region, valence nucleons are in orbit g / , i.e., Ω = 5, typicalvalue of G = A yields, E ( n, − E ( n, ∼ . Consider Ca with three particles in j = 7 / j couple to give final total J values? Theeasiest is to use m-scheme. If we use m-scheme for three particles in j = 7 / J values are 15 / , / , / , / , / , / 2. For the case of J = 7 / J = 0, givinga total J = 7 / / configuration. For the J = 15 / , / , / , / / 2, there are no pairs of particles coupled to J = 0. Since a J = 0 pair isthe lowest configuration for two particles in the same orbital, J = 7 / Ca to Ca, odd-Aisotopes, 7 / − state lies as the ground states. These cases are still simple tounderstand, since as the number of valence nucleons grows, the number of waysof mixing basis states to generate a given J increases rapidly. So, the large-scaleshell model calculations are trending in recent times. However, pairing providesa simple physics understanding of such nuclear features.21 Summary We have presented detailed derivation for the matrix elements of the δ -interactionevaluated between anti-symmetrized coupled states. The multipole expansionmethod used for this purpose enables one to decompose the matrix elementsinto their respective radial and angular parts. The angular part of the matrixelements are evaluated analytically and the similar procedure can be followedfor different interactions. The standard pairing Hamiltonian is then consideredwhose matrix elements between the paired states are similar to those for the δ -interaction. The solution for the pairing Hamiltonian is obtained within thequasi-spin formalism. Finally, simple application of the pairing to explain theenergy separation between the lowest 0 + and 2 + states in the even-even nucleiand the ground state spin for the odd-A nuclei are discussed. Some importantrelations involving matrix elements for the spherical harmonics and the commu-tation relations for the quasi-spin operators are presented in the Appendices inorder to make the derivation self-contained. The quasi-spin algebra is performeddirectly in terms of the creation and annihilation operators for single-nucleontogether with the application of the Wick’s theorem. These techniques are quiteuseful in dealing with the mean-field equations for the more general cases such asthe Hartree-Fock, BardeenCooperSchrieffer and Hartree-Fock Bogoliubov theo-ries. Appendix A Commutation of S + , S − and S op-erators The commutation of S + and S − can be expressed using Eqs. (68) and (69) as, (cid:2) S + , S − (cid:3) − = (cid:88) m,m (cid:48) > ( − j − m − m (cid:48) [ a † m a †− m , a − m (cid:48) a m (cid:48) ] − = (cid:88) m,m (cid:48) > ( − j − m − m (cid:48) ( a † m a †− m a − m (cid:48) a m (cid:48) − a − m (cid:48) a m (cid:48) a † m a †− m ) . (81)By applying Wick’s theorem (Eq. (53)) to the 2nd term in Eq.(81), a − m (cid:48) a m (cid:48) a † m a †− m = a † m a †− m a − m (cid:48) a m (cid:48) + δ − m (cid:48) m a †− m a m (cid:48) − δ − m (cid:48) , − m a † m a m (cid:48) − δ m (cid:48) m a †− m a − m (cid:48) + δ m (cid:48) , − m a † m a − m (cid:48) − δ − m (cid:48) m δ m (cid:48) , − m + δ − m (cid:48) , − m δ m (cid:48) m . (82)The terms involving single and double Kronecker-delta are due to the single anddouble contractions, respectively. Substituting Eq. (82) into Eq. (81) we get, (cid:2) S + , S − (cid:3) − = (cid:88) m,m (cid:48) > ( − j − m − m (cid:48) [ − δ − m (cid:48) m a †− m a m (cid:48) + δ − m (cid:48) , − m a † m a m (cid:48) + δ m (cid:48) m a †− m a − m (cid:48) − δ m (cid:48) , − m a † m a − m (cid:48) + δ − m (cid:48) m δ m (cid:48) , − m − δ − m (cid:48) , − m δ m (cid:48) m ] (83)22ince sum runs over m, m (cid:48) > (cid:2) S + , S − (cid:3) − = (cid:88) m,m (cid:48) > ( − j − m − m (cid:48) [+ δ − m (cid:48) , − m a † m a m (cid:48) + δ m (cid:48) m a †− m a − m (cid:48) − δ − m (cid:48) , − m δ m (cid:48) m ]= (cid:88) m> ( − j − m [ a † m a m + a †− m a − m − (cid:88) m> [ a † m a m + a †− m a − m − (cid:2) S + , S − (cid:3) − = ˆ N − Ω (84) (cid:2) S + , S − (cid:3) − = 2 S (85)where ˆ N = a † m a m + a †− m a − m . Similarly, one can show that,[ S + , S ] − = − S + ; [ S − , S ] − = S − The commutations of S + , S − and S suggest that these operators follow theangular momentum algebra and can be called as quasi-spin operators. Appendix B Important identities/relations One often encounters the matrix elements of the scalar product of two sphericalharmonics to be evaluated between the coupled states. Such matrix elementsare conveniently expressed in terms of the 6j-symbol and the product of thereduced matrix elements of the each of the spherical harmonics. The 6j-symbolcontains the geometrical factor of the angular momentum couplings, whereas,the physics is contained in the reduced matrix elements. Following Eq.(10.27)of Ref. [2],1. Matrix element of Y K · Y K (cid:104) α j α j ; JM | Y K (1) .Y K (2) | α (cid:48) j (cid:48) α (cid:48) j (cid:48) ; JM (cid:105) = ( − j + J + j (cid:48) (cid:26) j j Jj (cid:48) j (cid:48) K (cid:27) ( α j || Y K (1) || α (cid:48) j (cid:48) )( α j || Y K (2) || α (cid:48) j (cid:48) ) (86)( α j || Y K || α (cid:48) j (cid:48) ) ≡ ( 12 l j || Y K || l (cid:48) j (cid:48) )( 12 l j || Y K || l (cid:48) j (cid:48) ) = ( − j − / ˆ j ˆ j (cid:48) ˆ K √ π × 12 [1 + ( − l + l (cid:48) + K ] (cid:18) j K j (cid:48) − (cid:19) (87)where, ˆ x = √ x + 1.2. Recursion relation for 3j-symbols (cid:18) j j J 12 12 − (cid:19) = − (cid:112) J ( J + 1) [ˆ j + ˆ j ( − j + j + J ] (cid:18) j j J − 12 12 (cid:19) (88)23 ppendix C Matrix elements of spherical har-monics We provide some details of the derivation for the matrix elements of the sphericalharmonics (see Eq. (87)). By using Wigner-Eckart’s theorem [1, 2, 3, 4, 5], (cid:104) lm | Y LM | l (cid:48) m (cid:48) (cid:105) = (cid:90) Y lm ∗ Y LM Y l (cid:48) m (cid:48) dτ = ( − l − m (cid:18) l L l (cid:48) − m M m (cid:48) (cid:19) (cid:104) l || Y L || l (cid:48) (cid:105) (89)Also, the scalar product of spherical harmonics, Y l .Y l (cid:48) = (cid:88) m ( − m Y lm ( θ, φ ) Y l (cid:48) , − m (cid:48) ( θ (cid:48) , φ (cid:48) ) (90)is invariant with respect to the rotation of axes. It follows that the scalarproduct must be a function of angle Θ between the directions ( θ, φ ) and ( θ (cid:48) , φ (cid:48) ).This angle Θ is the only quantity independent of the choice of axes. Choosingaxes so that the direction ( θ (cid:48) , φ (cid:48) ) becomes the new z − axis, Y lm ( θ, φ ) → Y lm (0 , 0) = δ m, (91) Y l ( θ (cid:48) , φ (cid:48) ) → ˆ l (cid:114) π P l ( cosθ ) = ˆ l (cid:114) π P l ( cos Θ) (92) Y l .Y l (cid:48) = ˆ l (cid:114) π ˆ l (cid:48) (cid:114) π P l ( cos Θ) (93)where ˆ l = √ l + 1 and ˆ l (cid:48) = √ l (cid:48) + 1. Also, if Y lm ( θ, φ ) and Y l (cid:48) m (cid:48) ( θ (cid:48) , φ (cid:48) ) arespherical harmonics of the same angles ( θ, φ ) then (cid:80) mm (cid:48) (cid:104) LM | lml (cid:48) m (cid:48) (cid:105) ˆ l (cid:113) π ˆ l (cid:48) (cid:113) π Y lm ( θ, φ ) Y l (cid:48) m (cid:48) ( θ, φ )is a tensor of rank L . That is, (cid:80) mm (cid:48) (cid:104) LM | lml (cid:48) m (cid:48) (cid:105) ˆ l (cid:114) π ˆ l (cid:48) (cid:114) π Y lm ( θ, φ ) Y l (cid:48) m (cid:48) ( θ, φ ) = A L ˆ L (cid:114) π Y LM ( θ, φ )= (cid:104) L | l l (cid:48) (cid:105) ˆ L (cid:114) π Y LM ( θ, φ ) (94)where ˆ L = √ L + 1 and the value of A L can be found by putting θ = 0, φ = 0. (cid:88) m,m (cid:48) (cid:104) LM | lml (cid:48) m (cid:48) (cid:105) ˆ l (cid:114) π ˆ l (cid:48) (cid:114) π Y lm (0 , Y l (cid:48) m (cid:48) (0 , 0) = A L ˆ L (cid:114) π Y LM (0 , (cid:88) m,m (cid:48) (cid:104) LM | lml (cid:48) m (cid:48) (cid:105) δ m, δ m (cid:48) , = A L δ M, (cid:104) L | l l (cid:48) (cid:105) = A L | l − l (cid:48) | ≤ L ≤ | l + l (cid:48) | , and M = m + m (cid:48) . So, Y LM ( θ, φ ) = (cid:88) m,m (cid:48) ˆ l ˆ l (cid:48) ˆ L (cid:114) π (cid:104) LM | lml (cid:48) m (cid:48) (cid:105)(cid:104) L | l l (cid:48) (cid:105) Y lm ( θ, φ ) Y l (cid:48) m (cid:48) ( θ, φ ) Y lm ( θ, φ ) Y l (cid:48) m (cid:48) ( θ, φ ) = (cid:88) M (cid:104) lml (cid:48) m (cid:48) | LM (cid:105)(cid:104) l l (cid:48) | L (cid:105) ˆ l ˆ l (cid:48) ˆ L (cid:114) π Y LM ( θ, φ ) Y LM ( θ, φ ) Y l (cid:48) m (cid:48) ( θ, φ ) = (cid:88) m (cid:104) LM l (cid:48) m (cid:48) | lm (cid:105)(cid:104) L l (cid:48) | l (cid:105) ˆ L ˆ l (cid:48) ˆ l (cid:114) π Y lm ( θ, φ )In terms of 3j-symbols, Y LM ( θ, φ ) Y l (cid:48) m (cid:48) ( θ, φ ) = (cid:88) m ( − m (cid:18) l (cid:48) L lm (cid:48) M − m (cid:19) (cid:18) l (cid:48) L l (cid:19) ˆ L ˆ l ˆ l (cid:48) (cid:114) π Y lm ( θ, φ ) (95)This implies that (cid:90) Y lm ∗ Y LM Y l (cid:48) m (cid:48) dτ = (cid:90) Y lm ∗ ( − m (cid:18) l (cid:48) L lm (cid:48) M − m (cid:19)(cid:18) l (cid:48) L l (cid:19) ˆ L ˆ l ˆ l (cid:48) (cid:114) π Y lm dτ (96)= ( − m (cid:18) l (cid:48) L lm (cid:48) M − m (cid:19) (cid:18) l (cid:48) L l (cid:19) ˆ L ˆ l ˆ l (cid:48) (cid:114) π (97)The last step is due to orthogonality of spherical harmonics. Therefore, (cid:104) l || Y L || l (cid:48) (cid:105) = ( − l ˆ L ˆ l ˆ l (cid:48) (cid:114) π (cid:18) l (cid:48) L l (cid:19) (98)In | lj (cid:105) -basis, by using Wigner-Eckart theorem, (cid:104) lj || Y L || lj (cid:48) (cid:105) = (cid:104) lj || Y L || l (cid:48) j (cid:48) (cid:105) = ( − + l (cid:48) + j + L ˆ j ˆ j (cid:48) (cid:26) l j j (cid:48) l (cid:48) L (cid:27) (cid:104) l || Y L || l (cid:48) (cid:105) (99)where (cid:104) l || Y L || l (cid:48) (cid:105) = ( − l ˆ L ˆ l ˆ l (cid:48) (cid:114) π (cid:18) l (cid:48) L l (cid:19) (cid:26) l j j (cid:48) l (cid:48) L (cid:27) (cid:18) l (cid:48) L l (cid:19) : (cid:26) l j j (cid:48) l (cid:48) L (cid:27) (cid:18) l (cid:48) L l (cid:19) = (cid:26) l (cid:48) L lj (cid:48) j (cid:27) (cid:18) l (cid:48) L l (cid:19) = (cid:88) allm (cid:48) s ( − j + j (cid:48) + + m s + m j + m (cid:48) j (cid:18) l (cid:48) L l (cid:19)(cid:18) l (cid:48) j (cid:48) m s − m (cid:48) j (cid:19) (cid:18) j L j (cid:48) − m j m (cid:48) j (cid:19)(cid:18) j lm j − m s (cid:19) (cid:18) l (cid:48) L l (cid:19) (100)Since, (cid:80) m (cid:48) l m L m l (cid:18) l (cid:48) L l (cid:19) (cid:18) l (cid:48) L l (cid:19) = 1, however, m (cid:48) l = m L = m l = 0.Therefore, (cid:26) l j j (cid:48) l (cid:48) L (cid:27) (cid:18) l (cid:48) L l (cid:19) = (cid:88) m s m j m (cid:48) j ( − j + j (cid:48) + + m s + m j + m (cid:48) j (cid:18) l (cid:48) j (cid:48) m s − m (cid:48) j (cid:19)(cid:18) j L j (cid:48) − m j m (cid:48) j (cid:19) (cid:18) j lm j − m s (cid:19) (101)3j-symbol is non-zero only if m j = m (cid:48) j = m s (cid:26) l j j (cid:48) l (cid:48) L (cid:27) (cid:18) l (cid:48) L l (cid:19) = (cid:88) m s ( − j + j (cid:48) + + m s + m s + m s (cid:18) l (cid:48) j (cid:48) m s − m (cid:48) s (cid:19)(cid:18) j L j (cid:48) − m s m (cid:48) s (cid:19) (cid:18) j lm s − m s (cid:19) = ( − j + j (cid:48) + +3 / (cid:18) l (cid:48) j (cid:48) − (cid:19) (cid:18) j L j (cid:48) − (cid:19) (cid:18) j l − (cid:19) +( − j + j (cid:48) + − / (cid:18) l (cid:48) j (cid:48) − 12 12 (cid:19) (cid:18) j L j (cid:48) − (cid:19) (cid:18) j l − 12 12 (cid:19) (102)Since, (cid:18) j j j m m m (cid:19) = ( − j + j + j (cid:18) j j j − m − m − m (cid:19) ,26 l j j (cid:48) l (cid:48) L (cid:27) (cid:18) l (cid:48) L l (cid:19) = [( − j + j (cid:48) + +3 / +( − j + j (cid:48) + − / ( − l (cid:48) + + j (cid:48) + j + L + j (cid:48) + j + + l ] (cid:18) l (cid:48) j (cid:48) − (cid:19) (cid:18) j L j (cid:48) − (cid:19) (cid:18) j l − (cid:19) = [( − j + j (cid:48) + ( − j + j (cid:48) − ( − l + l (cid:48) +1+2 j +2 j (cid:48) + L ] (cid:18) l (cid:48) j (cid:48) − (cid:19) (cid:18) j L j (cid:48) − (cid:19) (cid:18) j l − (cid:19) = ( − j + j (cid:48) [1 + ( − l + l (cid:48) +1+ L ] (cid:18) l (cid:48) j (cid:48) − (cid:19) (cid:18) j L j (cid:48) − (cid:19) (cid:18) j l − (cid:19) = ( − j + j (cid:48) [1 + ( − l + l (cid:48) +1+ L ]( − j (cid:48) − ˆ l (cid:48) √ (cid:18) j L j (cid:48) − (cid:19) ( − j − ˆ l √ (cid:18) l (cid:48) j (cid:48) − (cid:19) = (cid:18) j (cid:48) l (cid:48) − 12 12 (cid:19) = ( − j (cid:48) − ˆ l (cid:48) √ (cid:104) j (cid:48) − 12 12 12 | l (cid:48) (cid:105) (cid:18) j (cid:48) l (cid:48) − 12 12 (cid:19) = ( − j (cid:48) − ˆ l (cid:48) √ √ (cid:18) j l − (cid:19) = ( − j − ˆ l √ √ (cid:26) l j j (cid:48) l (cid:48) L (cid:27) (cid:18) l (cid:48) L l (cid:19) = ( − j +2 j (cid:48) − [1 + ( − l + l (cid:48) + L ] 12ˆ l ˆ l (cid:48) (cid:18) j L j (cid:48) − (cid:19) = − l ˆ l (cid:48) [1 + ( − l + l (cid:48) + L ] (cid:18) j L j (cid:48) − (cid:19) = − l ˆ l (cid:48) [1 + ( − l + l (cid:48) + L ] (cid:18) j (cid:48) L j − (cid:19) (106)Since, (cid:18) j L j (cid:48) − (cid:19) = ( − j + L + j (cid:48) (cid:18) j L j (cid:48) − (cid:19) = ( − j +2 L +2 j (cid:48) (cid:18) j (cid:48) L j − (cid:19) .Therefore, the reduced matrix element of spherical harmonics in | lj > -basis can27e written as (cid:104) lj || Y L || l (cid:48) j (cid:48) (cid:105) = (cid:104) lj || Y L || l (cid:48) j (cid:48) (cid:105) = ( − + l + l (cid:48) + j + L ˆ j ˆ j (cid:48) ˆ l ˆ l (cid:48) ˆ L (cid:114) π × − l ˆ l (cid:48) [1 + ( − l + l (cid:48) + L ] (cid:18) j L j (cid:48) − (cid:19) = − 12 ( − + l + l (cid:48) + j + L [1 + ( − l + l (cid:48) + L ]ˆ j ˆ j (cid:48) ˆ L (cid:114) π (cid:18) j L j (cid:48) − (cid:19) = − 12 ( − j + [1 + ( − l + l (cid:48) + L ]ˆ j ˆ j (cid:48) ˆ L (cid:114) π (cid:18) j L j (cid:48) − (cid:19) = 12 ( − j +3 / [1 + ( − l + l (cid:48) + L ]ˆ j ˆ j (cid:48) ˆ L (cid:114) π (cid:18) j L j (cid:48) − (cid:19) = 12 ( − j +2 − [1 + ( − l + l (cid:48) + L ]ˆ j ˆ j (cid:48) ˆ L (cid:114) π (cid:18) j L j (cid:48) − (cid:19) = 12 ( − j − [1 + ( − l + l (cid:48) + L ]ˆ j ˆ j (cid:48) ˆ L (cid:114) π (cid:18) j L j (cid:48) − (cid:19) (107)If ( − l + l (cid:48) + L is odd then the reduced matrix element results in zero. On theother hand, if ( − l + l (cid:48) + L is even then (cid:104) lj || Y L || l (cid:48) j (cid:48) (cid:105) = ( − j − ˆ j ˆ j (cid:48) ˆ L (cid:114) π (cid:18) j L j (cid:48) − (cid:19) (108)The final result is independent of l and l (cid:48) . References [1] Amos deShalit and Igal Talmi, Nuclear Shell Theory ( Dover PublicationsInc.) (1963).[2] Igal Talmi Simple Model of Complex Nuclei (Harwood Academic Publishers,Switzerland) (1993). 283] M. K. Pal, Theory of Nuclear Structure (Affiliated East West Press Pvt.Ltd. New Delhi-Madras) (1982).[4] Kris L.G. Heyde The Nuclear Shell Model , (Springer-Verlag Berlin Heidel-berg) (1990).[5] R. F. Casten