A268759 - OEIS

%I #36 Nov 08 2016 21:02:47

%S 0,0,0,1,0,0,3,3,0,0,6,9,6,0,0,10,18,18,10,0,0,15,30,36,30,15,0,0,21,

%T 45,60,60,45,21,0,0,28,63,90,100,90,63,28,0,0,36,84,126,150,150,126,

%U 84,36,0,0,45,108,168,210,225,210,168,108,45,0,0,55,135,216,280,315

%N Triangle T(n,k) read by rows: T(n,k) = (1/4)*(1 + k)*(2 + k)*(k - n)*(1 + k - n).

%C Off-diagonal elements of angular momentum matrices J_1^2 and J_2^2.

%C Construct the infinite-dimensional matrix representation of angular momentum operators (J_1,J_2,J_3) in the block-diagonal, Jordan-Schwinger form (cf. Harter, Klee, Schwinger). The triangle terms T(n,k) satisfy:(1/2)T(n,k)^(1/2) = <n(n+1)/2+k+1|J_1^2|n(n+1)/2+k+3> = <n(n+1)/2+k+3|J_1^2|n(n+1)/2+k+1> = - <n(n+1)/2+k+1|J_2^2|n(n+1)/2+k+3> = - <n(n+1)/2+k+3|J_2^2|n(n+1)/2+k+1>. In the Dirac notation, we write elements m_{ij} of matrix M as <i|M|j>=m_{ij}. Matrices for J_1^2 and J_2^2 are sparse. These equalities and the central-diagonal equalities of A141387 determine the only nonzero entries.

%C Notice that a(n) = T(n,k) is always a multiple of the triangular numbers, up to an offset. Conjecture: the triangle tabulating matrix elements <n(n+1)/2+k+1|J_1^p|n(n+1)/2+k+p+1> is determined entirely by the coefficients: binomial(n,p) (cf. A094053). Various sequences along the diagonals of matrix J_1^p lead to other numbers with geometric interpretations (Cf. A000567, A100165).

%H W. Harter, <a href="http://www.uark.edu/ua/modphys/markup/PSDS_Info.html/">Principles of Symmetry, Dynamics, Spectroscopy</a>, Wiley, 1993, Ch. 5, page 345-346.

%H B. Klee, <a href="http://demonstrations.wolfram.com/QuantumAngularMomentumMatrices/">Quantum Angular Momentum Matrices</a>, Wolfram Demonstrations Project, 2016.

%H J. Schwinger, <a href="http://www.ifi.unicamp.br/~cabrera/teaching/paper_schwinger.pdf"> On Angular Momentum </a>, Cambridge: Harvard University, Nuclear Development Associates, Inc., 1952.

%F T(n,k) = (1/4)*(1 + k)*(2 + k)*(k - n)*(1 + k - n).

%F G.f.: x^2/((1-x)^3(1-x*y)^3)

%e 0;

%e 0,  0;

%e 1,  0,  0;

%e 3,  3,  0,  0;

%e 6,  9,  6,  0,  0;

%e 10, 18, 18, 10, 0,  0;

%e 15, 30, 36, 30, 15, 0, 0;

%e ...

%t Flatten[Table[(1/4) (1 + k) (2 + k) (k - n) (1 + k - n), {n, 0, 10, 1}, {k, 0, n, 1}]]

%Y Cf. A114327, A094053, A141387.

%K nonn,tabl

%O 0,7

%A _Bradley Klee_, Feb 20 2016