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%I #15 Aug 28 2019 10:04:33
%S 1,0,1,-5,0,6,0,-5,0,8,35,0,-112,0,80,0,7,0,-28,0,24,-21,0,126,0,-216,
%T 0,112,0,-15,0,108,0,-216,0,128,165,0,-1584,0,4752,0,-5632,0,2304,0,
%U 55,0,-616,0,2112,0,-2816,0,1280,-143,0,2002,0,-9152,0,18304,0,-16640,0,5632,0,-91,0,1456,0,-7488
%N Coefficient table for polynomials related to the eigenfunctions of a certain Schroedinger problem (Poeschl-Teller I).
%C The eigenenergies are e_n = ((n+1)^2-1)/2, n=1,2,... and the normalized eigenfunctions (modulo phases) are Psi_n(q) which are for odd n: sqrt(1/(Pi*e_n))*((1+n*y^2)*U(n,y)/y - (n+1)*U(n-1,y)) with y:=cos(q) and the Chebyshev polynomials U(n,y) (see A049310). For even n: sqrt(1/(Pi*e_n))*tan(q)*(n*U(n+1,y)+ (n+2)*U(n-1,y))/2, with y:=cos(q).
%C The row polynomials P(n,y):=sum(a(n,m)*y^m,m=0..n-1), n>=1, are related to the eigenfunctions as follows.
%C For odd n: P(n,y)= sqrt(Pi*e_n)*Psi_n(q)/(gcdrow(n)*y^2) with y=cos(q) and gcdrow(n) the greatest common divisor of the row coefficients factored out. It seems that gcdrow(2*k-1)=2*A006519(2*k-1), k=1,2,3,... (twice the highest power of 2 dividing n).
%C For even n: P(n,y)= sqrt(Pi*e_n)*Psi_n(q)/(8*gcdrow(n)*sin(q)*cos(q)) with y=cos(q) and the gcdrow(2*k) sequence is [1,1,2,2,1,1,4,4,1,1,2,2,1,1,8,8,1,1,2,2,1,1,...],k=1,2,... This looks like the doubled sequence A006519 (the first 100 terms have been checked).
%C The one-dimensional, stationary Schroedinger equation with potential V(q) = (2/cos(q)^2-1)/ 2 where |q| < Pi/2 (so-called Poeschl-Teller I potential) can be solved exactly, e.g., via supersymmetry adapted to quantum mechanics (and called D=1+0, N=2 supersymmetry). q:=Pi*x/L, L:=sqrt(m*w/hbar) with the coordinate |x| <= L/2 and m and w are the mass and the frequency of the underlying harmonic Bose oscillator. The reduced Planck constant is hbar = h/(2*Pi). The eigenenergies E_n and the potential are divided by hbar*w to become dimensionless e_n and V(q).
%H W. Lang, <a href="/A130415/a130415.txt">First 10 lines and more.</a>
%F a(n,m) = [y^m] P(n,y), n=1,2,..., m=0,1,2,..., with the polynomials P(n,y) defined above for even and odd n in terms of Chebyshev polynomials and the eigenenergies e_n.
%e Ground state (n=1): e_1=3/2, Psi_1(q)= sqrt(2/(3*Pi))*((1+cos(q)^2)*2 - 2)= sqrt(2/(3*Pi))*2*cos(q)^2; gcdrow(1)=2, P(1,y)=sqrt(Pi*3/2)*(1/(2*y^2)*Psi_1(q=arccos(y)) =1.
%e First excited state (n=2): e_2=4, Psi_2(q)= sqrt(1/(4*Pi))*tan(q)*(2*U(3,cos(q))+4*U(1,cos(q)))/2 = (4/sqrt(Pi))*sin(q)*cos(q)^2; gcdrow(2)=1, P(2,y)= sqrt(Pi*4)*Psi_n(q=arccos(y))/(8*1*sqrt(1-y^2)*y) =y.
%e Triangle begins:
%e [1];
%e [0,1];
%e [-5,0,6];
%e [0,-5,0,8];
%e [35,0,-112,0,80];
%e [0,7,0,-28,0,24];
%e ...
%K sign,tabl,easy
%O 0,4
%A _Wolfdieter Lang_, Jul 13 2007
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