-
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 projectionM=I+1 from that withM=I [1], where M equalsm1+m2+⋯+mn , andmi [i=1,2,⋯,n ] is the projection of j on the z-axis for the i-th nucleon. Other recipes to extractDI(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 [4–6].The first explicit and compact formula of
DI(j,n) was obtained specifically forI=0 andn=4 by Ginocchio and Haxton while studying the fractional quantum Hall effect [7]. In Ref. [8],DI(j,n) was constructed empirically forn=3 and 4, and for a few I values withn=5 . The formulas forn=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 forn=4 was derived by the reduction rule from SU(5) to the SO(3) group [11], with a demonstration thatDI(j,n) can be actually derived based on the reduction rule from SU(n+1) to SO(3) group. The Ginocchio-Haxton formula ofD0(j,4) was also revisited by Zamick and Escuderos [12]. An explicit recursion formula fromn−1 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 byDI(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 ofDI(j,n) [17, 18], and compact formulas ofDI(j,n) forn=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 ofDI(j,n) motivated a number of studies related to a single-j shell; here, we mention Refs. [21–34] 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 byDIT(j,n) , for nucleons in a single-j shell. In Ref. [35], Zamick and Escuderos found a few simple relations betweenDIT(j,4) withT=0 andDIT(j,4) withT=2 . In Ref. [36], compact and explicit formulas ofDIT(j,n) forn=3 and 4 were derived in terms of sum rules of six-j and nine-j symbols involving the expression ofDI(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 projectionMT . This number is denoted byNMMT(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 ofNMMT(j,n) , which is obtained by generalizing Talmi's recursion formula ofNM(j,n) , in which isospin is not considered. Compact formulas ofDIT(j,n) are derived forn=3 as an exemplification. The rest of this paper is organized as follows. In Sec. II, we generalize Talmi's formulas forNM(j,n) by further considering the isospin symmetry, and present recursion formulas ofNMMT . In Sec. III, we apply our generalized recursion formulas ofNMMT and extractNMMT(n,j) forn=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 tabulateDIT(j,n) withn=3 and 4,j=5/2,7/2 , and 9/2. -
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 andMT . With this convention, the neutron number equalsn/2+MT , and the proton number equalsn/2−MT . 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 thatj=mν1>mν2>...>mν2j+1=−j for neutrons andj=mπ1>mπ2>...>mπ2j+1=−j for protons. We definenνi=1 (ornπi=1 ) fori=1,...,2j+1 if the orbit ofmνi (ormπi ) 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+(j−1)+...+(j−n/2−MT+1))+(j+(j−1)+...+(j−n/2+MT+1))=n(2j+1)/2−n2/4−M2T.
(2) We denote the number of states with given M and
MT for n nucleons in a single-j shell by usingNMMT(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 (n−4 ) nucleons distributed in orbits for which the absolute values ofmi values equal or are below (j−1 ). Consequently, the number of states with given M andMT , i.e.,NMMT(j,n) , equals the number of states of the same M andMT , but with (n−4 ) nucleons in a (j−1 ) shell, namely,NMMT(j−1,n−4) . According to this classification, we are able to obtainNMMT(j,n) by using the values ofNMMT(j′,n′) , where eitherj′ is smaller than j, orn′ is smaller than n, or both(j′,n′) are smaller than(j,n) . In other words, from Table 1 we havenν1 nν2j+1 nπ1 nπ2j+1 Number of states 1 1 1 1 NM,MT(j−1,n−4) 1 1 1 0 NM−j,MT−1/2(j−1,n−3) 1 1 0 1 NM+j,MT−1/2(j−1,n−3) 1 1 0 0 NM,MT−1(j−1,n−2) 1 0 1 1 NM−j,MT+1/2(j−1,n−3) 1 0 1 0 NM−2j,MT(j−1,n−2) 1 0 0 1 NM,MT(j−1,n−2) 1 0 0 0 NM−j,MT−1/2(j−1,n−1) 0 1 1 1 NM+j,MT+1/2(j−1,n−3) 0 1 1 1 NM,MT(j−1,n−2) 0 1 0 1 NM+2j,MT(j−1,n−2) 0 1 0 0 NM+j,MT−1/2(j−1,n−1) 0 0 1 1 NM,MT+1(j−1,n−2) 0 0 1 0 NM−j,MT+1/2(j−1,n−1) 0 0 0 1 NM+j,MT+1/2(j−1,n−1) 0 0 0 0 NM,MT(j−1,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}NM−j(nν1−nν2j+1+nπ1−nπ2j+1),MT−12(nν1+nν2j+1−nπ1−nπ2j+1)(j−1,n−nν1−nν2j+1−nπ1−nπ2j+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 forn=1,2,3,4 , with the convention thatNMMT(j,n)=0 ifn<0 andNMMT(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. -
The value of
N_{M M_T}(j, n) withn=1 or 2 is trivially determined as follows. ForM_T = \pm {1}/{2} and|M| \leq j ,N_{M, M_T = \pm \frac{1}{2}}(j, 1) = 1 ; otherwise, it equals zero. Forn=2 withM_T = \pm 1 , the highest value of M equalsM_{\rm max} = 2j - 1 and forM_T = 0 ,M_{\rm max} = 2j . ForM_T = \pm 1 andM \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; forM_T = 0 andM \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 ofN_{M M_T}(j, 3) by using Eq. (3). -
We first discuss
N_{M M_T}(j, 3) with the requirement thatM_T = \frac{1}{2} andj < M \leq M_{\rm max} . We recursively apply Eq. (3) to this term and obtain the value ofN_{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 termN_{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 thatN_{M M_T}(j - l, n) = 0 whileN_{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 andM_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 andN_{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 toN_{M, M_T = \frac{1}{2}} (j, 3) on the right hand of Eq. (10) isN_{M + (j - i), M_T = 1}(j - i - 1, 2) . Given thatM > j here, the value ofM + (j - i) is larger than the maximumM_{\rm max} of two nucleons in a singlej - 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 isN_{M - (j - i), M_T = 0} (j - i - 1, 0) . Given thatM - (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. -
Let us now address
N_{M, M_T = \frac{1}{2}}(j, 3) withT_z = 1/2 and0 < M \leq j . In this case, Eq. (3) is applied recursively forj - 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\} . ForM_T = \frac{1}{2} andn = 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 byN_{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 termN_{M, M_T = \frac{1}{2}}(M - 1, 3) is given in Eq. (12) withj = 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 and2k + 1 . Forj - 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) with0 < 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) -
The results for
N_{M M_T}(j, n) are reduced toN_{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. Forj < 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. For0< 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. -
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} . WhenI > j andT = {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 andT = {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 andT = {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 andT = {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 andT = {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 theT=1/2 case. One uses Talmi's recursion formulas forN_{M=I, \frac{3}{2}} (j, n) and obtainsD_{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], for0< I \leq j andT = {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 andT = {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) forn=3 and 4 andj=5/2 , 7/2, and 9/2 and tabulate them in Table B1 and Table B2 in Appendix B. -
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.
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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).
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In this Appendix, we tabulate
D_{IT} (j, n) withj=3/2 – 9/2 and n=3–4, for convenience.j T 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.j T 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 .
Number of states with given spin I and isospin T for n fermions in a j orbit
- Received Date: 2022-06-11
- Available Online: 2022-11-15
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