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Number of states with given spin I and isospin T for n fermions in a j orbit

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X. Yin and Y. M. Zhao. Number of states with given spin I and isospin T for n fermions in a j orbit[J]. Chinese Physics C. doi: 10.1088/1674-1137/ac7eb3
X. Yin and Y. M. Zhao. Number of states with given spin I and isospin T for n fermions in a j orbit[J]. Chinese Physics C.  doi: 10.1088/1674-1137/ac7eb3 shu
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Received: 2022-06-11
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Number of states with given spin I and isospin T for n fermions in a j orbit

    Corresponding author: Y. M. Zhao, ymzhao@sjtu.edu.cn
  • 1. Shanghai Key Laboratory of Particle Physics and Cosmology, School of Physics and Astronomy, Shanghai Jiao Tong University, Shanghai 200240, China
  • 2. Collaborative Innovation Center of IFSA (CICIFSA), Shanghai Jiao Tong University, Shanghai 200240, China

Abstract: In this study, we investigate formulas of the number of states with a given total spin I and isospin T for n nucleons in a single-j shell denoted by DIT(j,n). Talmi's recursion formulas for the number of states with a given z-axis projection of total spin are generalized by further considering the isospin couplings and are applied to derive explicit formulas of DIT(j,n).

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    I.   INTRODUCTION
    • Determining the number of states with a given spin I for identical nucleons in a single-j shell [denoted by DI(j,n)] is a common practice in nuclear structure theory and atomic physics. In the nuclear shell model, DI(j,n) is usually tabulated for relatively small j and n. In cases with larger j or n, DI(j,n) might be obtained by subtracting the number of states with total angular momentum projection M=I+1 from that with M=I [1], where M equals m1+m2++mn, and mi [i=1,2,,n] is the projection of j on the z-axis for the i-th nucleon. Other recipes to extract DI(j,n) include Racha's formulas in terms of the seniority scheme [2] and generating function method studied extensively by Katriel et al. [3] and Sunko and collaborators [46].

      The first explicit and compact formula of DI(j,n) was obtained specifically for I=0 and n=4 by Ginocchio and Haxton while studying the fractional quantum Hall effect [7]. In Ref. [8], DI(j,n) was constructed empirically for n=3 and 4, and for a few I values with n=5. The formulas for n=3 was soon proved by Talmi, who introduced his recursive formulas [9]. These formulas were later derived based on the reduction rule from SU(4) to SO(3) group [10]; the formula for n=4 was derived by the reduction rule from SU(5) to the SO(3) group [11], with a demonstration that DI(j,n) can be actually derived based on the reduction rule from SU(n+1) to SO(3) group. The Ginocchio-Haxton formula of D0(j,4) was also revisited by Zamick and Escuderos [12]. An explicit recursion formula from n1 particles to n particles was obtained in Ref. [13] by introducing "pseudo" particles which allow fermions take integer spins. In Refs. [14, 15], the number of states, denoted by DI(j,n), was applied to derive sum rules of six-j and nine-j symbols, some of which were also revisited in Ref. [ 16]. In the last decade, Pain and collaborators extensively studied the odd-even staggering of DI(j,n) [17, 18], and compact formulas of DI(j,n) for n=3,4, and 5 [19]. The enumeration of the number of states with given spin was also extended to boson systems in Ref. [20]. The study of DI(j,n) motivated a number of studies related to a single-j shell; here, we mention Refs. [2134] without providing further details for completeness.

      Given that the nuclear shell model Hamiltonian respects the isospin symmetry, it is natural to generalize the enumeration of DI(j,n) to the number of states with given spin I and isospin T, denoted by DIT(j,n), for nucleons in a single-j shell. In Ref. [35], Zamick and Escuderos found a few simple relations between DIT(j,4) with T=0 and DIT(j,4) with T=2. In Ref. [36], compact and explicit formulas of DIT(j,n) for n=3 and 4 were derived in terms of sum rules of six-j and nine-j symbols involving the expression of DI(j,n) given in Ref. [11].

      Similar to the enumeration of DI(j,n), DIT(j,n) can be obtained in terms of the number of states with given spin projection M and isospin projection MT. This number is denoted by NMMT(j,n). For n fermions in a single-j shell,

      DIT(j,n)=[NM=I,MT=T(j,n)NM=I+1,MT=T(j,n)][NM=I,MT=T+1(j,n)NM=I+1,MT=T+1(j,n)]=NM=I,MT=T(j,n)+NM=I+1,MT=T+1(j,n)NM=I+1,MT=T(j,n)NM=I,MT=T+1(j,n).

      (1)

      In this study, we investigate DIT(j,n) in terms of NMMT(j,n), which is obtained by generalizing Talmi's recursion formula of NM(j,n), in which isospin is not considered. Compact formulas of DIT(j,n) are derived for n=3 as an exemplification. The rest of this paper is organized as follows. In Sec. II, we generalize Talmi's formulas for NM(j,n) by further considering the isospin symmetry, and present recursion formulas of NMMT. In Sec. III, we apply our generalized recursion formulas of NMMT and extract NMMT(n,j) for n=3. In Sec. IV, DIT(j,3) is derived by using Eq. (1). In Sec. V, we summarize our study. In Appendix A, we provide an alternative mathematical proof of our generalized Talmi's recursion formula by using the generating function method. In Appendix B, we tabulate DIT(j,n) with n=3 and 4, j=5/2,7/2, and 9/2.

    II.   NUMBER OF STATES WITH GIVEN z-COMPONENT PROJECTIONS OF SPIN AND ISOSPIN
    • We use the convention that the z-axis projection of the isospin for a neutron and a proton equals 1/2 and 1/2, respectively. For n nucleons in a single-j shell, we denote the projections of total spin I and total isospin T, respectively, as M and MT. With this convention, the neutron number equals n/2+MT, and the proton number equals n/2MT. Let us denote the z-axis spin projections of neutrons and protons by using (mν1,mν2,...,mν2j+1) and (mπ1,mπ2,...,mπ2j+1), respectively, with the convention that j=mν1>mν2>...>mν2j+1=j for neutrons and j=mπ1>mπ2>...>mπ2j+1=j for protons. We define nνi=1 (or nπi=1) for i=1,...,2j+1 if the orbit of mνi (or mπi) is filled; otherwise, it equals zero.

      According to the above conventions, we have

      M=nν1mν1+nν2mν2+...+nν2j+1mν2j+1+nπ1mπ1+nπ2mπ2+...+nπ2j+1mπ2j+1,

      and the maximum of I equals

      Imax=Mmax=(j+(j1)+...+(jn/2MT+1))+(j+(j1)+...+(jn/2+MT+1))=n(2j+1)/2n2/4M2T.

      (2)

      We denote the number of states with given M and MT for n nucleons in a single-j shell by using NMMT(j,n). Clearly, we can divide the values of (nν1,nν2j+1,nπ1,nπ2j+1) into sixteen cases, as tabulated in Table 1. Here, we exemplify them by using ( nν1=1,nν2j+1=1,nπ1=1,nπ2j+1=1). In this case, there are (n4) nucleons distributed in orbits for which the absolute values of mi values equal or are below (j1). Consequently, the number of states with given M and MT, i.e., NMMT(j,n), equals the number of states of the same M and MT, but with (n4) nucleons in a (j1) shell, namely, NMMT(j1,n4). According to this classification, we are able to obtain NMMT(j,n) by using the values of NMMT(j,n), where either j is smaller than j, or n is smaller than n, or both (j,n) are smaller than (j,n). In other words, from Table 1 we have

      nν1nν2j+1nπ1nπ2j+1Number of states
      1111NM,MT(j1,n4)
      1110NMj,MT1/2(j1,n3)
      1101NM+j,MT1/2(j1,n3)
      1100NM,MT1(j1,n2)
      1011NMj,MT+1/2(j1,n3)
      1010NM2j,MT(j1,n2)
      1001NM,MT(j1,n2)
      1000NMj,MT1/2(j1,n1)
      0111NM+j,MT+1/2(j1,n3)
      0111NM,MT(j1,n2)
      0101NM+2j,MT(j1,n2)
      0100NM+j,MT1/2(j1,n1)
      0011NM,MT+1(j1,n2)
      0010NMj,MT+1/2(j1,n1)
      0001NM+j,MT+1/2(j1,n1)
      0000NM,MT(j1,n)

      Table 1.  Number of states corresponding to the available values of nν1,nν2j+1,nπ1,nπ2j+1.

      NMMT(j,n)={nν1,nν2j+1,nπ1,nπ2j+1}NMj(nν1nν2j+1+nπ1nπ2j+1),MT12(nν1+nν2j+1nπ1nπ2j+1)(j1,nnν1nν2j+1nπ1nπ2j+1).

      (3)

      Here, the summation over {nν1,nν2j+1,nπ1,nπ2j+1} refers to the sixteen cases listed in Table 1. Note that the above formula holds also for n=1,2,3,4, with the convention that NMMT(j,n)=0 if n<0 and NMMT(j,0)=δM,0δMT,0. Note also that Eq. (3) can be proved alternatively based on the generating-function technique described in Ref. [3], as presented in Appendix A.

    III.   RECURSION FORMULA OF \boldsymbol{N_{M M_T}(j, 3)}
    • The value of N_{M M_T}(j, n) with n=1 or 2 is trivially determined as follows. For M_T = \pm {1}/{2} and |M| \leq j , N_{M, M_T = \pm \frac{1}{2}}(j, 1) = 1 ; otherwise, it equals zero. For n=2 with M_T = \pm 1 , the highest value of M equals M_{\rm max} = 2j - 1 and for M_T = 0 , M_{\rm max} = 2j . For M_T = \pm 1 and M \leq 2j - 1 ,

      \begin{eqnarray} N_{M, M_T = \pm 1}(j, 2) = \left \lfloor \frac{2j + 1 - |M|}{2} \right \rfloor\; , \end{eqnarray}

      (4)

      where " \left \lfloor "\; {\rm and} \; \; "\right \rfloor" means taking the largest integer without exceeding the value inside; for M_T = 0 and M \leq 2j ,

      \begin{eqnarray} N_{M, M_T = 0}(j, 2) = 2j + 1 - |M|\; . \end{eqnarray}

      (5)

      Otherwise, N_{M M_T}(j, 2) = 0 . With these results, we can construct explicit formulas of N_{M M_T}(j, 3) by using Eq. (3).

    • A.   M_T = 1/2 and j < M \leq M_{\max}

    • We first discuss N_{M M_T}(j, 3) with the requirement that M_T = \frac{1}{2} and j < M \leq M_{\rm max} . We recursively apply Eq. (3) to this term and obtain the value of N_{M M_T}(j, n) :

      \begin{aligned}[b] N_{M M_T}(j, n) =& N_{M M_T}(j-1, n) + \sideset{}{^{\prime}}\sum_{\left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}} \right\}} N_{M - j \left(n_{\nu_1} - n_{\nu_{2j + 1}} + n_{\pi_1} - n_{\pi_{2j + 1}}\right), M_T - \frac{1}{2} \left(n_{\nu_1} + n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\right)} \Big(j - 1, n - n_{\nu_1} - n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\Big) \\ =& N_{M M_T}(j - l, n) + \sum_{i = 0}^{l - 1} \sideset{}{^{\prime}}\sum_{\left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}} \right\}} \\& N_{M - (j - i) \left(n_{\nu_1} - n_{\nu_{2j + 1}} + n_{\pi_1} - n_{\pi_{2j + 1}}\right), M_T - \frac{1}{2} \left(n_{\nu_1} + n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\right)} \Big(j - i - 1, n - n_{\nu_1} - n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\Big) \; , \end{aligned}

      (6)

      where \sideset{}{^{\prime}}\sum denotes summations that exclude the case \left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}}\right\} = \left\{0, 0, 0, 0\right\} and decouple the term N_{M, M_T}(j - 1, n) [the last term in the summation of the sixteen terms of Table 1]; l is fixed with the requirement that N_{M M_T}(j - l, n) = 0 while N_{M M_T} (j - l + 1, n) \neq 0 , which means that

      \begin{aligned}[b] & M > \frac{n(2j + 1)}{2} - \frac{n^2}{4} - M_T^2 - l n, \\ & M \leq \frac{n(2j + 1)}{2} - \frac{n^2}{4} - M_T^2 - (l - 1) n, \end{aligned}

      or equivalently,

      \begin{eqnarray} l = \left \lfloor \frac{2j + 1}{2} - \frac{n}{4} - \frac{M_T^2 + M}{n} \right \rfloor + 1. \end{eqnarray}

      (7)

      For n=3 and M_T= 1/2 , the value of l becomes

      \begin{eqnarray} l = \left \lfloor j - \frac{M+1}{3} \right \rfloor + 1 \; . \end{eqnarray}

      (8)

      According to Eq. (6), with the conditions N_{M M_T}(j - l, n) = 0 and N_{M M_T} (j - l + 1, n) \neq 0 , we have

      \begin{aligned}[b] N_{M, M_T = \frac{1}{2}} (j, 3) = \sum_{i = 0}^{l - 1} \sideset{}{^{\prime}}\sum_{\left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}} \right\}} N_{M - (j - i) \left(n_{\nu_1} - n_{\nu_{2j + 1}} + n_{\pi_1} - n_{\pi_{2j + 1}}\right), \frac{1}{2} - \frac{1}{2} \left(n_{\nu_1} + n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\right)} \Big(j - i - 1, 3 - n_{\nu_1} - n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\Big). \end{aligned}

      (9)

      On the right hand side of the above formula, there are many terms that break the requirement of Eq. (2) for n \leq 2 . Let us exemplify this by using two cases: \left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}}\right\} = \left\{0, 0, 0, 1\right\} and \left\{1, 1, 1, 0\right\} . For the first case, \left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}}\right\} = \left\{0, 0, 0, 1\right\} , the corresponding contribution to N_{M, M_T = \frac{1}{2}} (j, 3) on the right hand of Eq. (10) is N_{M + (j - i), M_T = 1}(j - i - 1, 2) . Given that M > j here, the value of M + (j - i) is larger than the maximum M_{\rm max} of two nucleons in a single j - i - 1 shell, i.e., M_{\rm max} = 2j -2i -2 . Similarly, for \left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}}\right\} = \left\{1, 1, 1, 0\right\} , the corresponding contribution on the right hand side is N_{M - (j - i), M_T = 0} (j - i - 1, 0) . Given that M - (j - i) > 0 , N_{M - (j - i), M_T = 0} (j - i - 1, 0) = \delta_{M - (j - i), 0} = 0 . As a result, we obtain

      \begin{aligned}[b] N_{M, M_T = \frac{1}{2}}(j, 3) =& \sum_{i = 0}^{l - 1} \Big(N_{M - (j - i), M_T = 0}(j - i - 1, 2) + N_{M - (j - i), M_T = 1}(j - i - 1, 2) + N_{M - 2(j - i), M_T = \frac{1}{2}}(j - i - 1, 1) \Big) \\ =& \sum_{i = 0}^{l - 1} \Bigg(3j - 3i - 1 - M + \left \lfloor \frac{3j - 3i - 1 - M}{2} \right \rfloor + 1 \Bigg) = (3j - 1 - M)l - \frac{3l(l - 1)}{2} + \sum_{i = 0}^{l - 1} \left \lfloor \frac{3j - 3i - 1 - M}{2} \right \rfloor + l \; . \end{aligned}

      (10)

      The summation of \left \lfloor \dfrac{3j - 3i - 1 - M}{2} \right \rfloor is tedious but straightforward. One has

      \begin{eqnarray} l - 1 = \left\{ \begin{aligned} &2k & & \mathrm{if} \; {\rm when}\; 3j - M = 6k + 1, 6k + 2, 6k + 3, \nonumber \\ &2k + 1& & \mathrm{if} \; {\rm when}\; 3j - M = 6k + 4, 6k + 5, 6k + 6. \end{aligned} \right. \end{eqnarray}

      When 3j - M = 6k + 1 , the summation becomes

      \begin{aligned}[b] \sum_{i = 0}^{l - 1} \left \lfloor \frac{3j - 3i - 1 - M}{2} \right \rfloor =&\sum_{i = 0}^{2k} \left( 3k - 2i + \left \lfloor \frac{i}{2} \right \rfloor \right) \\ =&\sum_{i = 0}^{2k} \left( 3k - 2i \right) + 2 \sum_{i = 0}^{k - 1} i + k =3k^2 + k . \end{aligned}

      This summation for the case 3j - M = 6k + 2, 3, 4, 5, 6 can be obtained similarly, as follows:

      \begin{aligned}[b] \sum_{i = 0}^{l - 1} \left \lfloor \frac{3j - 3i - 1 - M}{2} \right \rfloor = \left\{ \begin{aligned} &3k^2 + k & & \mathrm{if}\; 3j - M = 6k + 1, \\ &3k^2 + 2k & & \mathrm{if}\; 3j - M = 6k + 2, \\ &3k^2 + 3k + 1& & \mathrm{if}\; 3j - M = 6k + 3, \\ &3k^2 + 4k + 1& & \mathrm{if}\; 3j - M = 6k + 4, \\ &3k^2 + 5k + 2& & \mathrm{if}\; 3j - M = 6k + 5, \\ &3k^2 + 6k + 3& & \mathrm{if}\; 3j - M = 6k + 6. \end{aligned} \right. \end{aligned}

      (11)

      Substituting these results into Eq. (10), we obtain

      \begin{aligned}[b] \\[-4pt] N_{M, M_T = \frac{1}{2}}(j, 3) = \left\{ \begin{aligned} &9k^2 + 6k + 1 & & \mathrm{if}\; 3j - M = 6k + 1, \\ &9k^2 + 9k + 2 & & \mathrm{if}\; 3j - M = 6k + 2, \\ &9k^2 + 12k + 4 & & \mathrm{if}\; 3j - M = 6k + 3, \\ &9k^2 + 15k + 6 & & \mathrm{if}\; 3j - M = 6k + 4, \\ &9k^2 + 18k + 9 & & \mathrm{if}\; 3j - M = 6k + 5, \\ &9k^2 + 21k + 12 & & \mathrm{if}\; 3j - M = 6k + 6 \end{aligned} \right. = \dfrac{1}{4}(3j - M)^2 + \dfrac{1}{2}(3j - M) + \eta(3j - M), \end{aligned}

      (12)

      where \eta(3j - M) = \left(0, \dfrac{1}{4}\right) if \mathrm{mod} (3j - M, 2) = (0, 1) , respectively.

    • B.   T_z = 1/2 and 0 < M \leq j

    • Let us now address N_{M, M_T = \frac{1}{2}}(j, 3) with T_z = 1/2 and 0 < M \leq j . In this case, Eq. (3) is applied recursively for j - M + 1 times. As a result, we have

      \begin{aligned}[b] N_{M M_T}(j, n) =& N_{M M_T}(j-1, n) + \sideset{}{^{\prime}}\sum_{\left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}} \right\}} \\ &N_{M - j \left(n_{\nu_1} - n_{\nu_{2j + 1}} + n_{\pi_1} - n_{\pi_{2j + 1}}\right), M_T - \tfrac{1}{2} \left(n_{\nu_1} + n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\right)} \Big(j - 1, n - n_{\nu_1} - n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\Big) \\ =& N_{M M_T}(M - 1, n) + \sum_{i = 0}^{j - M} \sideset{}{^{\prime}}\sum_{\left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}} \right\}} \\ &N_{M - (j - i) \left(n_{\nu_1} - n_{\nu_{2j + 1}} + n_{\pi_1} - n_{\pi_{2j + 1}}\right), M_T - \tfrac{1}{2} \left(n_{\nu_1} + n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\right)} \Big(j - i - 1, n - n_{\nu_1} - n_{\nu_{2j + 1}} - n_{\pi_1} - n_{\pi_{2j + 1}}\Big) \; , \end{aligned}

      (13)

      where, again, \sideset{}{^{\prime}}\sum denotes summation among the sixteen sets of \left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}}\right\} excluding \left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}}\right\} = \left\{0, 0, 0, 0\right\} . For M_T = \frac{1}{2} and n = 3 , the above formula is reduced to

      \begin{aligned}[b] N_{M , M_T = \frac{1}{2}}(j, 3) =& 1 + N_{M, M_T = \frac{1}{2}}(M - 1, 3) + \sum_{i = 0}^{j - M} \Big( N_{M - (j - i), M_T = 0}(j - i - 1, 2) + N_{M + (j - i), M_T = 0}(j - i - 1, 2) \\ &+ N_{M - (j - i), M_T = 1}(j - i - 1, 2) + N_{M + (j - i), M_T = 1}(j - i - 1, 2) \\ & + N_{M, M_T = -\frac{1}{2}}(j - i - 1, 1) + N_{M - 2(j - i), M_T = \frac{1}{2}}(j - i - 1, 1) + N_{M, M_T = \frac{1}{2}}(j - i - 1, 1) \\ &+ N_{M, M_T = \frac{1}{2}}(j - i - 1, 1) + N_{M + 2(j - i), M_T = \frac{1}{2}}(j - i - 1, 1) + N_{M, M_T = \frac{3}{2}}(j - i - 1, 1) \Big)\\=& 1 + N_{M, M_T = \frac{1}{2}}(M - 1, 3) + \sum_{i = 0}^{j - M} \left(j - i + M - 1\right) + \sum_{i = 0}^{j - M - 2} \left(j - i - M - 1\right)\end{aligned}

      \begin{aligned}[b] \quad\quad + \sum_{i = 0}^{j - M}\left \lfloor \frac{j - i + M - 1}{2} \right \rfloor + \sum_{i = 0}^{j - M - 3}\left \lfloor \frac{j - i - M - 1}{2} \right \rfloor + 3 \sum_{i = 0}^{j - M - 1} 1, \end{aligned}

      (14)

      where the first term " 1 " is given by N_{M = 0, M_T = 0}(j - i - 1, 0)= 1 with \left\{n_{\nu_1}, n_{\nu_{2j + 1}}, n_{\pi_1}, n_{\pi_{2j + 1}}, i\right\} = \left\{1, 1, 1, 0, j - M\right\} ; the second term N_{M, M_T = \frac{1}{2}}(M - 1, 3) is given in Eq. (12) with j = M - 1 , i.e.,

      \begin{eqnarray} N_{M, M_T = \frac{1}{2}}(M - 1, 3) = M^2 - 2 M + \dfrac{3}{4}. \end{eqnarray}

      (15)

      The summations in Eq. (14) equal zero when the upper limit of the given summation index is larger than its lower limit. On the right hand side, the summations of \left \lfloor \frac{j - i + M - 1}{2} \right \rfloor and \left \lfloor \frac{j - i - M - 1}{2} \right \rfloor are classified into two cases, j - M = 2k and 2k + 1 . For j - M = 2k , we obtain

      \begin{aligned}[b] \sum_{i = 0}^{j - M} \left \lfloor \frac{j - i + M - 1}{2} \right \rfloor =&\sum_{i = 0}^{2k} \left \lfloor \frac{2k - i + 2M - 1}{2} \right \rfloor \\ =&\sum_{i = 0}^{2k} \left(k - i + \frac{2M - 1}{2} + \left \lfloor \frac{i}{2} \right \rfloor \right) \\ =&\sum_{i = 0}^{2k} \left(k - i + \frac{2M - 1}{2} \right) + 2\sum_{i = 0}^{k - 1} i + k \\ =&(j - M + 1)\left(M - \dfrac{1}{2}\right) + \left(\frac{j - M}{2}\right)^2 . \end{aligned}

      The summation with j-M = 2k+1 is obtained similarly. The results of these two cases are unified as follows:

      \begin{aligned}[b] \sum_{i = 0}^{j - M} \left \lfloor \frac{j - i + M - 1}{2} \right \rfloor =&\left(j - M + \dfrac{1}{4}\right)\left(M - \dfrac{1}{2}\right) + \left(\frac{j - M}{2}\right)^2 \\ & +\left\{ \begin{aligned} & 0& & \mathrm{if\,mod} (j - M, 2) = 0 , \\ & - \dfrac{1}{4}& & \mathrm{if\,mod} (j - M, 2) = 1 . \end{aligned} \right. \end{aligned}

      (16)

      Following a similar procedure, we obtain

      \begin{aligned}[b] \sum_{i = 0}^{j - M - 3} \left \lfloor \frac{j - i - M - 1}{2} \right \rfloor =&\left(\frac{j - M}{2}\right)^2 - \frac{j - M}{2} \\ &+ \left\{ \begin{aligned} &0 & & \mathrm{if\,mod} (j - M, 2) = 0 , \\ &+ \dfrac{1}{4}& & \mathrm{if\,mod} (j - M, 2) = 1 . \end{aligned} \right. \end{aligned}

      (17)

      Substituting Eqs. (15)–(17) into Eq. (14), we obtain N_{M, M_T = \frac{1}{2}}(j, 3) with 0 < M \leq j in a compact form:

      \begin{eqnarray} N_{M, M_T = \frac{1}{2}}(j, 3) = \dfrac{3}{2} j^2 + j + \dfrac{1}{4} - \frac{M^2}{2} . \end{eqnarray}

      (18)
    • C.   M_T = 3/2 and 0 < M

    • The results for N_{M M_T}(j, n) are reduced to N_{M} (j, n) for identical particles, and Eq. (3) is reduced to Talmi's recursion formula. For this case, N_{M, M_T = \frac{3}{2}}(j, 3) was obtained in Ref. [19], which we cite for completeness. For j < M \leq M_{\mathrm{max}} , according to Eqs. (2.11a)–(2.11b) in Ref. [19],

      \begin{eqnarray} N_{M, M_T = \frac{3}{2}}(j, 3) = \frac{1}{12} \left(3j - M\right)^2 + \alpha(3j - M), \end{eqnarray}

      (19)

      where

      \begin{eqnarray} \alpha(3j - M) &=& 0, -\frac{1}{12}, -\frac{1}{3}, \frac{1}{4}, -\frac{1}{3}, -\frac{1}{12}, \end{eqnarray}

      for \mathrm{mod}(3j - M, 6) = 0, 1, 2, 3, 4, 5 , respectively. For 0< M \leq j , according to Eqs. (2.19a)-(2.19b) in Ref. [19],

      \begin{eqnarray} N_{M, M_T = \frac{3}{2}}(j, 3) = \frac{1}{2} j^2 - \frac{1}{6} M^2 + \beta \left(M - \frac{1}{2}\right), \end{eqnarray}

      (20)

      where

      \begin{eqnarray} \beta(3j - M) &=& -\frac{1}{12}, \frac{1}{4}, -\frac{1}{12}, \end{eqnarray}

      for \mathrm{mod}(M - \dfrac{1}{2}, 3) = 0, 1, 2 , respectively.

    IV.   NUMBER OF STATES WITH GIVEN I AND T
    • Provided the results of N_{M M_T}(j, 3) , i.e., Eqs. (12), (18), (19), and (20), we can readily substitute them into Eq. (1) to obtain the number of states with given I and T, i.e., D_{I T}(j, 3) .

      We first consider the case T = {1}/{2}. When I > j and T = {1}/{2}, we obtain

      \begin{aligned}[b] D_{I, T = \frac{1}{2}}(j, 3) =& N_{I, \frac{1}{2}} (j, 3) + N_{I + 1, \frac{3}{2}} (j, 3) \\ &- N_{I + 1, \frac{1}{2}} (j, 3) - N_{I, \frac{3}{2}} (j, 3). \end{aligned}

      (21)

      According to Eqs. (12) and (19), the above formula is reduced to

      \begin{aligned}[b] D_{I, T = \frac{1}{2}}(j, 3) =&-\frac{1}{3} I + j + \frac{1}{3} + \eta(3j - I) - \eta(3j - I - 1) \\ &+ \alpha(3j - I - 1) - \alpha(3j - I) \\ =&-\frac{1}{3} I + j + \left\{ \begin{aligned} &0 & &\mathrm{if}\,\mathrm{mod}(3j - I, 3)=0 , \\ & \frac{2}{3} & &\mathrm{if}\,\mathrm{mod}(3j - I, 3)=1 , \\ & \frac{1}{3} & &\mathrm{if}\,\mathrm{mod}(3j - I, 3)=2 \\ \end{aligned} \right. \\ =& \left \lfloor \frac{3j - I + 2}{3} \right \rfloor . \end{aligned}

      (22)

      For I = j and T = {1}/{2} ,

      \begin{aligned}[b] D_{I, T = \frac{1}{2}}(j, 3) =& N_{I, \frac{1}{2}} (j, 3) + N_{I + 1, \frac{3}{2}} (j, 3) \\ &- N_{I + 1, \frac{1}{2}} (j, 3) - N_{I, \frac{3}{2}} (j, 3) . \end{aligned}

      (23)

      By using Eqs. (12), (18), (19), and (20), we have

      \begin{aligned}[b] D_{I, T = \frac{1}{2}}(j, 3) =&\frac{2}{3} j + \frac{7}{12} + \alpha(2j - 1) - \eta(2j - 1) \\ & - \beta\left(j - \frac{1}{2}\right) \\ =&\frac{2}{3} j + \left\{ \begin{aligned} & \frac{2}{3} & & \mathrm{if}\, \mathrm{mod}(2j, 3) = 1 , \\ &0 & & \mathrm{if}\, \mathrm{mod}(2j, 3) = 3 , \\ & \frac{1}{3} & & \mathrm{if}\, \mathrm{mod}(2j, 3) = 5 \\ \end{aligned} \right. \\ =& \left \lfloor \frac{2j + 2}{3} \right \rfloor. \end{aligned}

      (24)

      Similarly, for 0< I < j and T = {1}/{2},

      \begin{aligned}[b]D_{I, T = \frac{1}{2}}(j, 3) =& N_{I, \frac{1}{2}} (j, 3) + N_{I + 1, \frac{3}{2}} (j, 3) \\ &- N_{I + 1, \frac{1}{2}} (j, 3) - N_{I, \frac{3}{2}} (j, 3) . \end{aligned}

      (25)

      By using Eqs. (18) and (20), the above formula is reduced to

      \begin{aligned}[b] D_{I, T = \frac{1}{2}}(j, 3) =& \frac{2}{3} J + \frac{1}{3} + \beta\left(I + \frac{1}{2}\right) - \beta\left(I - \frac{1}{2}\right) \\ =& \left\{ \begin{aligned} &\frac{2}{3} I + \frac{2}{3} & & \mathrm{if}\, \mathrm{mod}(2I, 3) = 1 , \\ &\frac{2}{3} I & & \mathrm{if}\, \mathrm{mod}(2I, 3) = 3 , \\ &\frac{2}{3} I + \frac{1}{3} & & \mathrm{if}\, \mathrm{mod}(2I, 3) = 5 \\ \end{aligned} \right. \\ =& \left \lfloor \frac{2I + 2}{3} \right \rfloor. \end{aligned}

      (26)

      Eqs. (24)–(26) can be rewritten as follows. For I \geq j and T = {1}/{2},

      \begin{eqnarray} D_{I, T = \frac{1}{2}}(j, 3) = \left \lfloor \frac{3j - I + 2}{3} \right \rfloor\; , \end{eqnarray}

      (27)

      and for 0 < I \leq j and T = {1}/{2},

      \begin{eqnarray} D_{I, T = \frac{1}{2}}(j, 3) = \left \lfloor \frac{2I + 2}{3} \right \rfloor. \end{eqnarray}

      (28)

      These results [Eqs. (27)–(28) in this paper] are consistent with Eqs. (25)–(26) in Ref. [ 36], which were obtained in terms of sum rules of six-j symbols, except that the formulas derived in this paper are expressed in a more transparent and understandable form.

      The T = {3}/{2} case is much simpler than the T=1/2 case. One uses Talmi's recursion formulas for N_{M=I, \frac{3}{2}} (j, n) and obtains D_{I, T = \frac{3}{2}}(j, 3) straightforwardly. The resulting formulas are consistent with those given in Ref. [ 8] and Ref. [19]. We show them next for completeness. According to Eq. (1) in Ref. [ 8], for 0< I \leq j and T = {3}/{2},

      \begin{eqnarray} D_{I, T = \frac{3}{2}}(j, 3) = \left \lfloor \frac{2I + 3}{6} \right \rfloor\; ; \end{eqnarray}

      (29)

      and according to Eq. (2) in Ref. [ 8], for I \geq j and T = {3}/{2},

      \begin{eqnarray} D_{I, T = \frac{3}{2}}(j, 3) = \left \lfloor \frac{3j - 3 - I}{6} \right \rfloor + \delta_I. \end{eqnarray}

      (30)

      where

      \begin{eqnarray} \delta_I = \left\{ \begin{aligned} 0& & \mathrm{if}\ \mathrm{mod}(3j - 3 - I, 6)=1,\nonumber \\ 1& & \mathrm{otherwise}. \nonumber \\ \end{aligned} \right. \end{eqnarray}

      In principle, one can follow the above procedure as explained in Eqs. (1)–(3) for larger n values. However, this procedure becomes formidably tedious with lengthy formulas and tables for larger n values. Instead of those formulas, we calculate D_{IT}(j, n) for n=3 and 4 and j=5/2 , 7/2, and 9/2 and tabulate them in Table B1 and Table B2 in Appendix B.

    V.   SUMMARY
    • To summarize, in this paper we generalize Talmi's recursion formula of the number of states with given spin I by further consideration of the isospin symmetry. This generalization is also proved alternatively based on the generating function method. We also exemplify the generalized Talmi's recursion formulas obtained in this paper by applying our generalized Talmi's recursion formulas in constructing the number of states with given spin and isospin of three nucleons in a single-j shell.

    APPENDIX A: GENERATING-FUNCTION TECHNIQUE
    • In this Appendix, we present an alternative proof of the recursion formula presented in Eq. (3) based on the generating-function method.

      A generating function is a polynomial function for which the expansion coefficients are related to the number of states with a given requirement. We define the generating function corresponding to N_{M M_T}(j, n) , similar to that in Ref. [17],

      \begin{aligned}[b] f_{j} (x, y, z) = \sum_{n = 0}^{\infty} \sum_{M = -\infty}^{\infty} \sum_{M_T = -\infty}^{\infty} z^n x^M y^{M_T} N_{M M_T}(j, n) , \end{aligned}\tag{A1}

      which yields

      \begin{aligned}[b] f_{j} (x, y, z) =& \sum_{n = 0}^{\infty} \sum_{M = -\infty}^{\infty} \sum_{M_T = -\infty}^{\infty} z^n x^M y^{M_T} \\ &\sum_{\left\{ n_{\nu_1}, ... , n_{\nu_{2j+1}}, n_{\pi_1}, ..., n_{\pi_{2j+1}}\right\}} \\ &\Big( \delta_{n, n_{\nu_1} + ... + n_{\pi_{2j+1}}} \delta_{M_T, \frac{1}{2} \left( n_{\nu_1} + ... - n_{\pi_{2j + 1}}\right)} \\ & \delta_{M, n_{\nu_1}m_{\nu_1} + ... + n_{\pi_{2j+1}}m_{\pi_{2j+1}}} \Big) \\ =& \prod_{i = 1}^{2j + 1} \left(1 + z x^{m_{\nu_i}} y^{\frac{1}{2}}\right)\left(1 + z x^{m_{\pi_i}} y^{-\frac{1}{2}}\right) \\ =& f_{j-1}(x, y, z) g_{i}(x, y, z) \end{aligned} \tag{A2}

      with

      \begin{aligned}[b] f_{j}(x, y, z) =& \prod_{i = 2}^{2j} \left(1 + z x^{m_{\nu_i}} y^{\frac{1}{2}}\right) \left(1 + z x^{m_{\pi_i}} y^{-\frac{1}{2}}\right) \\ g_{j}(x, y, z) =& \left(1 + z x^{-j} y^{\frac{1}{2}}\right) \left(1 + z x^{j} y^{\frac{1}{2}}\right) \\ &\times \left(1 + z x^{-j} y^{-\frac{1}{2}}\right) \left(1 + z x^{j} y^{-\frac{1}{2}}\right) . \end{aligned}\tag{A3}

      The generating function is expanded in powers of z as follows:

      f_{j} (x, y, z) = \sum_{n = 0}^{\infty} z^{n} f_{j, n} (x, y) \tag{A3}

      with

      \begin{aligned}[b] f_{j, n} (x, y) =& \frac{1}{n!} \frac{\partial^n}{\partial z^n} f_{j} (x, y, z) \Big |_{z = 0} \\ =& \sum_{M = -\infty}^{\infty} \sum_{M_T = -\infty}^{\infty} N_{M M_T}(j, n) x^M y^{M_T} . \end{aligned}\tag{A5}

      Using the Leibniz formula, one obtains

      \begin{aligned}[b] f_{j,n}(x, y) =& \frac{1}{n!} \sum_{k = 0}^{n} \binom{n}{k} \frac{\partial^{n-k}}{\partial z^{n-k}} f_{j-1}(x, y, z) \\ &\times \frac{\partial^k}{\partial z^k} g_{j}(x, y, z) \Big |_{z=0} . \end{aligned}\tag{A6}

      By using Eqs. (A1), (A5), the above relation yields

      \begin{aligned}[b] &\sum_{M = -\infty}^{\infty} \sum_{M_T = -\infty}^{\infty} N_{M M_T} (j, n) x^M y^{M_T} \\ =&\frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} \sum_{M = -\infty}^{\infty} \sum_{M_T = -\infty}^{\infty} N_{M M_T} (j-1, n-k) \\ & x^M y^{M_T}\times \frac{\partial^k}{\partial z^k} g_{i}(x, y, z) \Big |_{z=0} . \end{aligned}\tag{A7}

      Given that g_{j}(x, y, z) is a fourth-order polynomial, the only derivatives that are nonzero correspond to

      \begin{aligned}[b] k = 0: \frac{\partial^0}{\partial z^0} g_{j}(x, y, z) \Big |_{z=0} =& 1 , \\ k = 1: \frac{\partial^1}{\partial z^1} g_{j}(x, y, z) \Big |_{z=0} =& x^{-j}y^{\frac{1}{2}} +x^{j}y^{\frac{1}{2}} \\ &+ x^{-j}y^{-\frac{1}{2}} + x^{j}y^{-\frac{1}{2}} , \\ k = 2: \frac{\partial^2}{\partial z^2} g_{j}(x, y, z) \Big |_{z=0} =& y + y^{-1} + x^{-2j} + x^{2j} \\ & + 1 + 1 , \\ k = 3: \frac{\partial^3}{\partial z^3} g_{j}(x, y, z) \Big |_{z=0} =& x^{-j}y^{\frac{1}{2}} + x^{j}y^{\frac{1}{2}} \\ & + x^{-j}y^{-\frac{1}{2}} + x^{j}y^{-\frac{1}{2}} , \end{aligned}

      \begin{aligned}[b] k = 4: \frac{\partial^4}{\partial z^4} g_{j}(x, y, z) \Big |_{z=0} = 1 . \end{aligned}\tag{A8}

      Eq. (A7) holds for any independent variables x and y; therefore, the coefficients of homogeneous terms on both sides are equal, which immediately leads to Eq. (3).

    APPENDIX B: \boldsymbol{D_{IT} (J, N)} WITH \boldsymbol{J=3/2-9/2} AND \boldsymbol{N=3-4}
    • In this Appendix, we tabulate D_{IT} (j, n) with j=3/2 – 9/2 and n=3–4, for convenience.

      jT I^D
      \frac{5}{2} \frac{1}{2} (\frac{1}{2})^{1}, (\frac{3}{2})^{1}, (\frac{5}{2})^{2}, (\frac{7}{2})^{2}, (\frac{9}{2})^{1}, (\frac{11}{2})^{1}, (\frac{13}{2})^{1}
      \frac{5}{2} \frac{3}{2} (\frac{3}{2})^{1}, (\frac{5}{2})^{2}, (\frac{9}{2})^{1}
      \frac{7}{2} \frac{1}{2} (\frac{1}{2})^{1}, (\frac{3}{2})^{1}, (\frac{5}{2})^{2}, (\frac{7}{2})^{3}, (\frac{9}{2})^{2},
      (\frac{11}{2})^{2}, (\frac{13}{2})^{2}, (\frac{15}{2})^{1},(\frac{17}{2})^{1}, (\frac{19}{2})^{1}
      \frac{7}{2} \frac{3}{2} (\frac{3}{2})^{1}, (\frac{5}{2})^{2}, (\frac{7}{2})^{3}, (\frac{9}{2})^{1}, (\frac{11}{2})^{1}, (\frac{15}{2})^{1}
      \frac{9}{2} \frac{1}{2} (\frac{1}{2})^{1}, (\frac{3}{2})^{1}, (\frac{5}{2})^{2}, (\frac{7}{2})^{3}, (\frac{9}{2})^{3}, (\frac{11}{2})^{3},
      (\frac{13}{2})^{3}, (\frac{15}{2})^{2}, (\frac{17}{2})^{2}, (\frac{19}{2})^{2}, (\frac{21}{2})^{1}, (\frac{23}{2})^{1}, (\frac{25}{2})^{1}
      \frac{9}{2} \frac{3}{2} (\frac{3}{2})^{1}, (\frac{5}{2})^{1}, (\frac{7}{2})^{1}, (\frac{9}{2})^{2}, (\frac{11}{2})^{1}, (\frac{13}{2})^{1}, (\frac{15}{2})^{1}, (\frac{17}{2})^{1}, (\frac{21}{2})^{1}

      Table B1.  D_{IT}(j,\,n) for j=5/2-9/2, with n=3. ID in this Table represents that there are D states with spin I for the jn configuration.

      jT I^D
      \frac{5}{2} 0 0^{2}, 2^{3}, 3^{1}, 4^{3}, 5^{1}, 6^{2}, 8^{1}
      \frac{5}{2} 1 1^{2}, 2^{2}, 3^{3}, 4^{2}, 5^{2}, 6^{1}, 7^{1}
      \frac{5}{2} 2 0^{1}, 2^{1}, 4^{1}
      \frac{7}{2} 0 0^{3}, 2^{4}, 3^{2}, 4^{5}, 5^{2}, 6^{5}, 7^{2}, 8^{3}, 9^{1}, 10^{2}, 12^{1}
      \frac{7}{2} 1 1^{3}, 2^{3}, 3^{5}, 4^{4}, 5^{5}, 6^{4}, 7^{4}, 8^{2}, 9^{2}, 10^{1}, 11^{1}
      \frac{7}{2} 2 0^{1}, 2^{2}, 4^{2}, 5^{1}, 6^{1}, 8^{1}
      \frac{9}{2} 0 0^{3}, 2^{6}, 3^{2}, 4^{7}, 5^{4}, 6^{7}, 7^{4},
      8^{7}, 9^{3}, 10^{5}, 11^{2}, 12^{3}, 13^{1}, 14^{2}, 16^{1}
      \frac{9}{2} 1 1^{4}, 2^{4}, 3^{7}, 4^{6}, 5^{8}, 6^{7}, 7^{8},
      8^{6}, 9^{6}, 10^{4}, 11^{4}, 12^{2}, 13^{2}, 14^{1}, 15^{1}
      \frac{9}{2} 2 0^{2}, 2^{2}, 3^{1}, 4^{3}, 5^{1}, 6^{3}, 7^{1}, 8^{2}, 9^{1}, 10^{1}, 12^{1}

      Table B2.  The same as Table B1 except for n=4 .

Reference (36)

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