Re: failure to Integrate in orthogonal polynomials
- To: mathgroup at smc.vnet.net
- Subject: [mg67837] Re: failure to Integrate in orthogonal polynomials
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Mon, 10 Jul 2006 06:37:39 -0400 (EDT)
- References: <e8nt4n$k7c$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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
Roger Bagula wrote:
> I thought that I might simulate a
> one dimensional quantum Legendre system ( particle in a box) in
> Mathematica.
> It works for odd levels but fails for even levels on the interval/
> domain {-1,1}:
>
> p0 = Table[LegendreP[n, x], {n, 1, 5}]
> norm = Table[(1/Integrate[p0[[n]]*p0[[n]], {x, -1, 1}])^(1/2), {n, 1, 5}]
> p = Table[p0[[n]]*norm[[n]], {n, 1, 5}]
> Enm = Table[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], {n,
> 1, 5}, {m, 1, 5}]
> MatrixForm[Enm]
> Inm = Table[N[Integrate[p[[n]]*p[[m]], {x, -1, 1}]], {n, 1, 5}, {m, 1, 5}]
> MatrixForm[Inm]
> Union[Flatten[N[Enm]]]
>
> If anybody can get an esimate of the even levels it would be nice.
> the odd levels are very near +/-Sqrt[6]. I'm using an 1/x potential
> function and a second derivative operator.
> Roger Bagula
>