% Latex file % On the algebra of coupled SO(3) tensors % K. Srinivasa Rao, J. Van der Jeugt and G. Vanden Berghe % J. Math. Phys. 33 (1992), 15-18. % \documentstyle[12pt]{article} \textheight=22,5cm \textwidth=16cm \headsep=0cm \topmargin=0cm \oddsidemargin=0cm \parindent=30pt \parskip=\medskipamount \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bea{\begin{eqnarray}} \def\eea{\end{eqnarray}} \def\beas{\begin{eqnarray*}} \def\eeas{\end{eqnarray*}} \def\para{\parallel} \def\nn{\nonumber} \def\p{\prime} \hyphenation{Rijks-uni-ver-si-teit} \hyphenation{pa-ra-metri-za-tion} \begin{document} \addtolength{\baselineskip}{0.6\baselineskip} \addtolength{\topsep}{-.3\topsep} % % \begin{center} {\LARGE On the algebra of coupled SO(3) tensors}\\[2cm] K. Srinivasa Rao$^{a)}$\,,\ J. Van der Jeugt$^{b)}$ and G. Vanden Berghe \\[.5cm] {\em Laboratorium voor Numerieke Wiskunde en Informatica,\\ Rijksuniversiteit Gent, Krijgslaan 281--S9, B-9000 Gent, Belgium}\\ \end{center} \vspace{3cm} \noindent \underline{Abstract} \begin{minipage}{14cm} The commutation relation for coupled SO(3) tensor operators is constructed. This gives rise to a Lie algebra for which $9j$ symbols appear in the structure constants. In an example it is shown how an exceptional Lie subalgebra can be realised. The closure requirement for the exceptional Lie algebra does not seem to be related to non--trivial $9j$ symbols but rather to vanishings of relations involving $6j$ symbols. \end{minipage} \vfill \noindent-----------------------------------\\ $^{a)}$ {\footnotesize Permanent address~: The Institute of Mathematical Sciences, Madras-600\,113, India}\\ $^{b)}$ {\footnotesize Research Associate of N.F.W.O. (National Fund for Scientific Research of Belgium)} \newpage Let $T^k_q(j_1 j_2)$ $(q=-k,\,-k+1,\,\ldots,\,+k)$ denote a set of SO(3) tensor operators of rank $k$ mapping the $(2j_2+1)$-dimensional vector space with angular momentum basis $|j_2\,,\,m_2\rangle $ into the $(2j_1+1)$-dimensional vector space with basis vectors $|j_1\,,\,m_1\rangle $. Such tensor operators were originally introduced by Elliott $^1$ in his description of collective motion in the nuclear shell model. These tensor operators are completely defined by means of their reduced matrix element \be \langle \, j_b\parallel T^k(j_1\,j_2)\parallel j_a\,\rangle = [k]^{1/2} \delta_{j_1j_b}\delta_{j_2j_a} \ee where $[k]$ stands for $(2k+1)$. Judd $^2$ calculated the commutation relations between the SO(3) tensor operators $T^k_q(jj)$ (i.e. the case $j_1=j_2=j$), which was later extended to the general case by Vanden Berghe and De Meyer $^3$ to the following expression~: \bea \lefteqn{ \left[ T^k_q(j_1j_2),T^{k'}_{q'}(j_3j_4)\right] = \sum_{k'',q''}[k\,k'\,k'']^{1/2} \left( \begin{array}{ccc} k & k' & k''\\ q & q' & -q'' \end{array} \right) (-1)^{2j_4 + j_3-j_2 -q''} }\nn\\ && \times \left( \delta_{j_2j_3}(-1)^{k+k'+k''+j_1+j_2+j_3+j_4} \left\{ \begin{array}{ccc} k & k' & k''\\ j_4 & j_1 & j_2 \end{array} \right\} T^{k''}_{q''}(j_1j_4)\right.\nn\\ &&\qquad\left. - \delta_{j_1j_4} \left\{ \begin{array}{ccc} k & k' & k''\\ j_3 & j_2 & j_1 \end{array} \right\} T^{k''}_{q''}(j_3j_2) \right)\,. \eea In (2), the standard notation for SO(3) $3j$ and $6j$ symbols $^4$ is used. The commutation relation (2) has been used to provide realisations of certain Lie algebras. By identifying some exceptional Lie subalgebras, a relation between so-called non-trivial zeros $^5$ of $6j$ symbols appearing in (2) and embeddings of exceptional Lie algebras has been established $^{6,7,8}$. It is the purpose of the present note to extend the commutation relation (2) to the case of coupled tensor operators. The $9j$ symbol seems to play a key role in this situation. The Lie algebraic structure of coupled tensor operators is discussed, and some possible applications are commented on. % %3 % Coupled SO(3) tensor operators are of the form \be \Bigl( T^{k_1}(j_1j_2) \otimes T^{k'_1}(j'_1j'_2)\Bigr)^k_q = \sum_{q_1,q'_1} \langle k_1\;q_1\;k'_1\;q'_1|k\;q\rangle T^{k_1}_{q_1}(j_1j_2)\otimes T^{k'_1}_{q'_1}(j'_1 j'_2)\,, \ee where $\langle k_1\;q_1\;k'_1\;q'_1|k\;q\rangle $ is an SO(3) coupling coefficient $^4$, and the SO(3) tensor operators act on independent angular momentum vectors~: \bea \lefteqn{ T^{k_1}_{q_1}(j_1j_2) \otimes T^{k'_1}_{q'_1}(j'_1 j'_2) (|j_a\,m_a\rangle \otimes |j'_a\, m'_a \rangle )= }\nn\\ && T^{k_1}_{q_1}(j_1j_2)|j_a\,m_a \rangle \otimes T^{k'_1}_{q'_1}(j'_1j'_2)|j'_a\,m'_a \rangle \,. \eea The action of a tensor operator on an angular momentum state follows from (1) and the Wigner-Eckart theorem~: \be T^k_q(j_1j_2)|j_a\,m_a\rangle = \sum_{j_bm_b} (-1)^{j_b-m_b} \left( \begin{array}{ccc} j_b & k & j_a\\ -m_b & q & m_a \end{array}\right) [k]^{1/2}\delta_{j_1j_b}\, \delta_{j_2 j_a}|j_bm_b\rangle \,. \ee In order to find an expression for the commutator of two coupled SO(3) tensor operators, one calculates the action of their product on a vector $|j_a\,m_a\rangle \otimes |j'_a\,m'_a\rangle $\,, i.e. \be \Bigl( T^{k_1}(j_1j_2) \otimes T^{k'_1}(j'_1j'_2) \Bigr)^k_q \Bigl( T^{k_2}(j_3j_4) \otimes T^{k'_2}(j'_3j'_4) \Bigr)^{k'}_{q'} (|j_a\,m_a\rangle \otimes |j'_a\,m'_a\rangle )\,. \ee To determine (6), the couplings $(k_1\,k'_1\,k)$ and $(k_2\,k'_2\,k')$ are first decoupled, giving rise to two $3j$ symbols $\left( \begin{array}{ccc}k_1 & k'_1 & k \\ q_1 & q'_1 & -q \end{array} \right)$ and $\left( \begin{array}{ccc}k_2 & k'_2 & k' \\ q_2 & q'_2 & -q' \end{array} \right)$. Then, the following expression for the product of two tensor operators, deduced from (5), is used~: \bea \lefteqn{ T^{k_1}_{q_1}(j_1j_2)T^{k_2}_{q_2}(j_3j_4)|j_a\,m_a\rangle = \sum_{K,Q} \delta_{j_2j_3}(-1)^{2j_2+j_4-j_1-Q+k_1+k_2+K} }\nn\\ && [k_1\;k_2\;K]^{1/2} \left( \begin{array}{ccc} k_1 & k_2 & K\\ q_1 & q_2 & -Q \end{array} \right) \left\{ \begin{array}{ccc} k_1 & k_2 & K\\ j_4 & j_1 & j_2 \end{array} \right\} T^K_Q(j_1j_4)|j_a\,m_a\rangle \,. \eea % %4 % Applying (7) both for the $|j_a\,m_a\rangle $ state and the $|j'_a\,m'_a\rangle $ state yields the two $3j$ symbols $\left( \begin{array}{ccc} k_1 & k_2 & K\\ q_1 & q_2 & -Q \end{array} \right)$ and $\left( \begin{array}{ccc} k'_1 & k'_2 & K'\\ q'_1 & q'_2 & -Q' \end{array} \right)$. Finally, recoupling $T^K_Q(j_1j_4)\otimes T^{K'}_{Q'}(j'_1\,j'_4)$ to a coupled tensor of rank $k''$ gives rise to the $3j$ symbol $\left( \begin{array}{ccc} K & K' & k''\\ -Q & -Q'& q'' \end{array} \right)$. With these five $3j$ symbols, the following summation can be performed $^9$~: \bea && {\small \addtolength{\arraycolsep}{-.5\arraycolsep} \sum_{\scriptstyle q_1 , q'_1, q_2 , q'_2 , Q , Q'} \left( \begin{array}{ccc} k_1 & k'_1 & k\\ q_1 & q'_1 & -q \end{array} \right) \left( \begin{array}{ccc} k_2 & k'_2 & k'\\ q_2 & q'_2 & -q' \end{array} \right) \left( \begin{array}{ccc} k_1 & k_2 & K\\ q_1 & q_2 & -Q \end{array} \right) \left( \begin{array}{ccc} k'_1 & k'_2 & K'\\ q'_1 & q'_2 & -Q' \end{array} \right) \left( \begin{array}{ccc} K & K' & k''\\ -Q & -Q' & q'' \end{array} \right) } \nn\\ && \qquad\qquad\qquad = \left( \begin{array}{ccc} k & k' & k''\\ -q & -q' & q'' \end{array} \right) \left\{ \begin{array}{ccc} k_1 & k'_1 & k\\ k_2 & k'_2 & k'\\ K & K' & k'' \end{array} \right\}\,. \eea Hence, keeping track of all the phase factors, one finds~: \bea \lefteqn{ \Bigl( T^{k_1}(j_1j_2) \otimes T^{k'_1}(j'_1j'_2)\Bigr)^k_q \Bigl( T^{k_2}(j_3j_4) \otimes T^{k'_2}(j'_3j'_4)\Bigr)^{k'}_{q'} |j_a\,m_a\rangle \otimes |j'_a\,m'_a\rangle = }\nn\\ && \sum_{K,K',k'',q''} (-1)^{2j_2-j_1+j_4+2j'_2-j'_1+j'_4+K+K'+k''+q''} [k_1\,k_2\,K\,k'_1\,k'_2\,K'\,k\,k'\,k'']^{1/2}\delta_{j_2j_3}\delta_{j'_2j'_3}\nn\\ && \times \left( \begin{array}{ccc} k & k' & k''\\ -q & -q' & q'' \end{array} \right) \left\{ \begin{array}{ccc} k_1 & k_2 & K\\ j_4 & j_1 & j_2 \end{array} \right\} \left\{ \begin{array}{ccc} k'_1 & k'_2 & K'\\ j'_4 & j'_1 & j'_2 \end{array} \right\} \left\{ \begin{array}{ccc} k_1 & k'_1 & k\\ k_2 & k'_2 & k'\\ K & K' & k'' \end{array} \right\}\nn\\ && \times \Bigl( T^K(j_1j_4) \otimes T^{K'}(j'_1j'_4)\Bigr)^{k''}_{q''} |j_a\,m_a\rangle \otimes |j'_a\,m'_a\rangle \,. \eea Using the abbreviation \be J=j_1+j_2+j_3+j_4\,, \quad J'=j'_1+j'_2+j'_3+j'_4\,, \ee the expression for the commutator of two coupled SO(3) tensor operators can be written as follows~: % %5 % \bea \lefteqn{ \Bigl[ \Bigl( T^{k_1}(j_1j_2) \otimes T^{k'_1}(j'_1j'_2)\Bigr)^k_q\,,\, \Bigl( T^{k_2}(j_3j_4) \otimes T^{k'_2}(j'_3j'_4)\Bigr)^{k'}_{q'}\Bigr]= }\nn\\ && \sum_{K,K',k'',q''} (-1)^{K+K'+k''+q''+J+J'} [k_1\,k_2\,K\,k'_1\,k'_2\,K'\,k\,k'\,k'']^{1/2} {\small \addtolength{\arraycolsep}{-.5\arraycolsep} \left( \begin{array}{ccc} k & k' & k''\\ -q & -q' & q'' \end{array} \right) \left\{ \begin{array}{ccc} k_1 & k'_1 & k\\ k_2 & k'_2 & k'\\ K & K' & k'' \end{array} \right\} } \nn\\ && \times \Biggl( (-1)^{2j_1+2j'_1}\delta_{j_2j_3}\delta_{j'_2j'_3} \left\{ \begin{array}{ccc} k_1 & k_2 & K\\ j_4 & j_1 & j_2 \end{array} \right\} \left\{ \begin{array}{ccc} k'_1 & k'_2 & K'\\ j'_4 & j'_1 & j'_2 \end{array} \right\} \Bigl( T^K(j_1j_4)\otimes T^{K'}(j'_1j'_4)\Bigr)^{k''}_{q''} \nn\\ && -(-1)^{k_1+k'_1+k_2+k'_2+K+K'}(-1)^{2j_3+2j'_3}\delta_{j_1j_4}\delta_{j'_1j'_4} \left\{ \begin{array}{ccc} k_1 & k_2 & K\\ j_3 & j_2 & j_1 \end{array} \right\} \left\{ \begin{array}{ccc} k'_1 & k'_2 & K'\\ j'_3 & j'_2 & j'_1 \end{array} \right\}\nn\\ && \times \Bigl( T^K(j_3j_2)\otimes T^{K'}(j'_3j'_2)\Bigr)^{k''}_{q''}\Biggr)\,. \eea A case which is of special interest is that for which $j_1=j_2=j_3=j_4=j$ and $j'_1=j'_2=j'_3=j'_4=j'$. Then, the $j$- or $j'$-dependence of the tensor operators can be suppressed in the notation. The commutation rule becomes~: \bea && \Bigl[ \Bigl( T^{k_1}\otimes T^{k'_1} \Bigr)^k_q\,,\, \Bigl( T^{k_2}\otimes T^{k'_2} \Bigr)^{k'}_{q'} \Bigr] = \nn\\ && \sum_{K,K',k'',q''} (-1)^{2j+2j'+K+K'+k''+q''} [k_1\,k_2\,K\,k'_1\,k'_2\,K'\,k\,k'\,k'']^{1/2} {\small \addtolength{\arraycolsep}{-.5\arraycolsep} \left( \begin{array}{ccc} k & k' & k''\\ -q & -q' & q'' \end{array} \right) \left\{ \begin{array}{ccc} k_1 & k'_1 & k\\ k_2 & k'_2 & k'\\ K & K' & k'' \end{array} \right\} } \nn\\ &&\times \left\{ \begin{array}{ccc} k_1 & k_2 & K\\ j & j & j \end{array} \right\} \left\{ \begin{array}{ccc} k'_1 & k'_2 & K'\\ j' & j' & j' \end{array} \right\} \Bigl[ 1-(-1)^{k_1+k'_1+k_2+k'_2+K+K'}\Bigr] \Bigl( T^K \otimes T^{K'}\Bigr)^{k''}_{q''}\,. \eea It can be verified that for $k'_1=k'_2=0$, (11) reduces to (2). Equation (11) can also be seen as the coupled version of formula (2.2) of Reference~7. From (12) it follows that the set of operators \bea \Bigl( T^{k_1} \otimes T^{k'_1}\Bigr)^k_q & , & k_1=0,1,\ldots,2j,\quad k'_1=0,1,\ldots,2j'\,,\nn\\ && k=|k_1-k'_1|,\ldots,k_1+k'_1\,,\quad q=-k,\ldots,+k \eea close under commutation. In fact, they are the generators of $U(N)$, % %6 % where $N={{(2j+1)}\times(2j'+1)}$. The operator $(T^0\otimes T^0)^0_0$ commutes with all other operators, and deleting it from the set (13) leaves the generators of $SU(N)$. The operators $( T^{k_1}\otimes T^0 )^{k_1}_{q_1}$ and $(T^0\otimes T^{k'_1})^{k'_1}_{q'_1}$ clearly commute, and are the generators of $U(2j+1)$ and $U(2j'+1)$ respectively. Thus (12) describes the chain \be U\Bigl( (2j+1)(2j'+1)\Bigr) \supset U(2j+1) \times U(2j'+1) \supset SO(3) \times SO(3) \supset SO(3). \ee It can also be seen from (12) that the subset of operators in (13) with $k_1+k'_1$ odd also closes under commutation. They form the generators of $SO(N)$ if $j$ and $j'$ are both integers or both half-integers, and of $Sp(N)$ otherwise. We believe that (11) or (12) is the first expression yielding a $9j$ symbol in the structure constants of a Lie algebra. Trying to extend the ideas of Refs. 6, 7 and 8, where the identification of exceptional Lie algebras lead to an explanation of non-trivial zeros of $6j$ symbols appearing in (2), one would at first sight expect to be in a position to relate non-trivial zeros of $9j$ symbols to exceptional Lie algebras realised as subalgebras of (11) or (12). However, this happens not to be the case since the internal $j$-values ($j$ and $j'$) do not appear in the $9j$ symbol. An example can clarify the situation. From Ref.7 one can deduce that the generators of the exceptional Lie algebra $F_4$ can be written as follows~: \bea L^1_q &=& \bigl( T^1(1,1)\otimes T^0(3,3)\bigr)^1_q + \sqrt{5\over7}\bigl( T^1(2,2)\otimes T^0(0,0)\bigr)^1_q\,,\nn\\[.3cm] K^1_q &=& \bigl( T^0(1,1)\otimes T^1(3,3) \bigr)^1_q\,,\nn\\[.3cm] P^5_q &=& \bigl( T^0(1,1) \otimes T^5(3,3)\bigr)^5_q\,,\nn\\[.3cm] G^k_q &=& \bigl( T^2(1,1)\otimes T^3(3,3)\bigr)^k_q + {1\over \sqrt{2}}\Bigl( \bigl( T^2(1,2)\otimes T^3(3,0)\bigr)^k_q + \bigl( T^2(2,1)\otimes T^3(0,3)\bigr)^k_q\Bigr)\,,\nn\\ && \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad k=1,2,3,4,5. \eea % %7 % Using (11), one can calculate the commutator $\Bigl[ G^k_q\,,\,G^{k'}_{q'}\Bigl]$, and require that it should close as a linear combination of the operators (15). In particular, $\Bigl[ G^1_q\,,\,G^3_q\Bigr]$ yields a term in $\bigl( T^2(1,1) \otimes T^5(3,3)\bigr)^3_{q''}$\,, with a factor proportional to \be \left\{ \begin{array}{ccc} 2 & 3 & 1\\ 2 & 3 & 3\\ 2 & 5 & 3 \end{array} \right\} \left( \left\{ \begin{array}{ccc} 2 & 2 & 2\\ 1 & 1 & 1 \end{array} \right\} \left\{ \begin{array}{ccc} 3 & 3 & 5\\ 3 & 3 & 3 \end{array} \right\} + {1\over 2} \left\{ \begin{array}{ccc} 2 & 2 & 2\\ 1 & 1 & 2 \end{array} \right\} \left\{ \begin{array}{ccc} 3 & 3 & 5\\ 3 & 3 & 0 \end{array} \right\} \right)\,. \ee As this term should vanish, (16) must be zero. At first sight one would deduce \be \left\{ \begin{array}{ccc} 2 & 3 & 1\\ 2 & 3 & 3\\ 2 & 5 & 3 \end{array} \right\} =0 \,, \ee which is indeed valid. However, it so happens that the expression in terms of $6j$ symbols is also equal to zero\,! Since this expression appears also in other terms, it would be wrong to deduce the vanishing of a $9j$ symbol from (16). In fact, all one can say from (16) and similar expressions is that the relation \be \left\{ \begin{array}{ccc} 2 & 2 & K\\ 1 & 1 & 1 \end{array} \right\} \left\{ \begin{array}{ccc} 3 & 3 & K'\\ 3 & 3 & 3 \end{array} \right\} + {1\over 2} \left\{ \begin{array}{ccc} 2 & 2 & K\\ 1 & 1 & 2 \end{array} \right\} \left\{ \begin{array}{ccc} 3 & 3 & K'\\ 3 & 3 & 0 \end{array} \right\}=0 \ee holds for $(K,K')\in \{(0,3),(1,2),(1,4),(1,6),(2,1),(2,5)\}$ (and one can also deduce the vanishing of the $6j$ symbol discussed in Ref.7). So the closure of the exceptional Lie algebra involves the vanishing of those symbols (or combination of symbols) which have internal $j$-values as arguments. \vskip 5mm \noindent {\bf Acknowledgements} One of us (K.S.R.) wishes to thank the N.F.W.O.\ Belgium for financial support and the Rijksuniversiteit Gent for its hospitality. The authors thank Dr.\ H.~De Meyer for helpful discussions. \newpage % %8 % \noindent {\bf References}. \begin{enumerate} \item J.P. Elliott, {\em Proc. R. Soc. London, Ser. A} {\bf 245}, 128 (1958). \item B.R. Judd, {\em Operator Techniques in Atomic Spectroscopy} (McGraw-Hill, New York, 1963), p. 102. \item G. Vanden Berghe and H. De Meyer, {\em J. Math. Phys.} {\bf 25}, 772 (1984). \item M. Rotenberg, R. Bivins, N. Metropolis and J.K. Wooten Jr., {\em The $3j$ and $6j$ symbols} (The Technology Press, MIT, Cambridge, 1959). \item L.C. Biedenharn and J.D. Louck, The Racah-Wigner Algebra in Quantum Theory, {\em Encyclopedia of Mathematics} {\bf 9} (Addison-Wesley, Massachusetts, 1981), chapter~5, Topic 10. \item J. Van der Jeugt, G. Vanden Berghe and H. De Meyer, {\em J. Phys. A~: Math. Gen.} {\bf 16}, 1377 (1983). \item H. De Meyer, G. Vanden Berghe and J. Van der Jeugt, {\em J. Math. Phys.} {\bf 25}, 751 (1984). \item G. Vanden Berghe, H. De Meyer and J. Van der Jeugt, {\em J. Math. Phys.} {\bf 25}, 2585 (1984). \item A. de--Shalit and I. Talmi, {\em Nuclear Shell Theory} (Academic Press, N.Y., 1963), p.~517. \end{enumerate} \end{document}