Re: failure to Integrate in orthogonal polynomials
- To: mathgroup at smc.vnet.net
- Subject: [mg67842] Re: failure to Integrate in orthogonal polynomials
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Mon, 10 Jul 2006 06:37:52 -0400 (EDT)
- References: <e8nt4n$k7c$1@smc.vnet.net> <e8qg8f$o63$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Jens-Peer Kuska wrote:
>Hi,
>
>I see no sense using the Legendre polynomials for that problem.
>Anyway the matrix element of the Hamiltonian is
><p[n]|H|p[m]> and that mean you have to compute
>
>Integrate[p[[n]]*(D[p[[m]],{x,2}] + p[[m]]/x), {x, -1, 1}]
>
>instead of
>
>Integrate[p[[n]]*(D[p[[m]],{x,2}] + 1/x), {x, -1, 1}]
>
>or is the wave function in space representation not more
>multiplied by the potential ??
>
>The code
> > If[Integrate[p[[n]]*p[[m]], {x, -1,
> > 1}] - 1 == 0, Integrate[p[[n]]*(D[p[[m]], {x,
> > 2}] + 1/x), {x, -1, 1}]/Integrate[p[[n]]*p[[m]], {x, -1, 1}], 0]
>
>is even more interesting, because first you test that <p[n]|p[m]>==1,
>i.e n==m and it would be quicker to test n==m .. and finally you
>divide by Integrate[p[[n]]*p[[m]], {x, -1, 1}], but you do this only
>in the case where you already know that
>Integrate[p[[n]]*p[[m]], {x, -1, 1}]==1
>
>The whole condition is nonsense because the Hamiltonian must have
>off-diagonal elements. And finally why not use the potential 1/(x+I*eps)
>compute all your integrals and set eps->zero ???
>
>And why you don't start with the LegendreP[0,x] to ensure that your
>basis is complete ???
>
>Regards
> Jens
>
>
>
>
>>
>>
>
>
>
Jens-Peer Kuska,
Your help corrected two errors :
1) the potential
2) not starting at zero
Also I stopped dividing by the <p[n]*p[m]>.
It gives a spectrum for the Legendre particle in a box states that at
least works.
The idea was to set system constants hbar^2/2*m and e^2 to one:
take out Bessel, Fourier and spin partitions in a Schrödinger Hamiltonian
and make it a "pure" sphere-like orthogonal standing wave
in one dimension.
Getting the potential to Integrate involved adding a systematic constant
to the even levels that canceled out the singularity that a 1/x term gave.
The output seem to be resonance frequency / energy levels
at which transitions would take place between Legendre
states in such a simplified system.
Clear[norm, p0, p, Enm, Inm]
p0 = Table[LegendreP[n, x], {n, 0, 5}]
norm = Table[(1/Integrate[p0[[n]]*p0[[n]], {x, -1, 1}])^(1/2), {n, 1, 6}]
p = Table[p0[[n]]*norm[[n]], {n, 1, 6}]
c = Table[CoefficientList[p[[m]], x][[1]], {m, 1, 6}]
v[x_, m_] = c[[m]]/x - p[[m]]/x
Enm = Table[Integrate[p[[n]]*(D[p[[m]], {x, 2}] + v[x, m]), {x, -1, 1}], {
n, 1, 6}, {m, 1, 6}]
MatrixForm[Enm]
Union[Flatten[N[Enm]]]
ListPlot[Union[Flatten[N[Enm]]], PlotJoined -> True, PlotRange -> All]