%I #11 Sep 10 2013 02:51:12
%S 2,3,3,3,3,5,11,17,23,7,-1,-9,-17,-25,11,21,31,41,51,61,13,1,-11,-23,
%T -35,-47,-59,17,31,45,59,73,87,101,115,19,35,51,67,83,99,115,131,147,
%U 23,-13,-49,-85,-121,-157,-193,-229,-265,-301,29,69,109,149,189,229,269
%N An integer differential Schroedinger Hamiltonian for primes as a triangular sequence: g(n)]= K=integer<=n: Ep(n,K) = (K n Prime[n] - 2 K n Prime[1 + n] + K n Prime[2 + n] + Prime[n] g(n))/(n*Prime[n]).
%C The equation derivation goes like this: If you have a function of integers n : f[n] then the derivative equivalents are: f[n] f[n+1]-f[n] f[n+2]-2*f[n+1]+f[n] f[n+3]-3*f[n+2]+2*f[n+1]-f[n] Something like: d[n_,k_]=Sum[(-1)^m*Binomial[n,m]*f[n+m],{m,0,k}] So the second order differential equation equivalent would look like: a*(f[n+2]-2*f[n+1]+f[n])+b*(f[n+1]-f[n])+c*f[n]=0.
%C I ran across this in an old problems book by Ulam where they were using the "new" computers ( late 50's / early 60's) to simulate different differential equations. So quantum mechanic is based on Schroedinger's equation: Hamiltonian : Kinetic energy term: K =( hbar2/(2*m))* d2/dx2. Potential energy term: V=g[x]/x. The total Hamiltonian energy ( a constant with quantum levels of n): E K*phi(x)+V*phi(x)=E*phi(x) Putting that into f[n] terms: ( hbar2/(2*m))*(f[n+2]-2*f[n+1]+f[n])+g[n]*f[n]/n=E[n]*f[n].
%C Making ( hbar2/(2*m)) a constant K Clear[f, E0, K, n, g] Solve[K*(f[n + 2] - 2*f[n + 1] + f[n]) + g[n ]*f[n]/n - E0*f[n] == 0, E0] E0=(K n f[n] - 2 K n f[1 + n] + K n f[2 + n] + f[n] g[n])/(n*f[n]) So a quantum prime second order differential Hamiltonian might be: Ep[n_] = (K n Prime[n] - 2 K n Prime[1 + n] + K n Prime[2 + n] + Prime[n] g[n])/(n*Prime[n]).
%F g(n)]= K=integer<=n: Ep(n,K) = (K n Prime[n] - 2 K n Prime[1 + n] + K n Prime[2 + n] + Prime[n] \ g(n))/(n*Prime[n]); t(n,m)=n*Prime[n]*Ep(n,m).
%e {2, 3},
%e {3, 3, 3},
%e {5, 11, 17, 23},
%e {7, -1, -9, -17, -25},
%e {11, 21, 31, 41, 51, 61},
%e {13, 1, -11, -23, -35, -47, -59},
%e {17, 31, 45, 59, 73, 87, 101, 115},
%e {19, 35, 51, 67, 83, 99, 115,131, 147},
%e {23, -13, -49, -85, -121, -157, -193, -229, -265, -301},
%e {29, 69, 109, 149, 189, 229, 269, 309, 349, 389, 429},
%e ...
%t Clear[K, g, Ep]; g[n_] = 1; Ep[n_, K_] = (K n Prime[n] - 2 K n Prime[1 + n] + K n Prime[2 + n] + Prime[n] g[n])/(n*Prime[n]); Table[n*Prime[n]*Ep[n, m], {n, 1, 10}, {m, 0, n}]; Flatten[%]
%K sign,tabf,less
%O 1,1
%A _Roger L. Bagula_, Nov 13 2008
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