# How to expand the existing basis set so that it becomes more complete?

I have a set of anisotropic gaussian basis set which describes the ground state of the system with great accuracy.

The Hamiltonian of the system has the following form: $$H=-\frac{1}{2}\Delta-\frac{1}{r}+\frac{1}{2}\rho^2-1$$ where $$r=(\rho,z,\phi)$$ is a coordinate in the cylindrical system

The anisotropic gaussian basis set which I use to solve this task:$$\psi_j=e^{-b_{0j}\;z^2}e^{-a_{0j}\;\rho^2}$$where $$a_{0j}$$ and $$b_{0j}$$ are parameters

$$a_{0j}$$=0.500074, 0.502676, 0.510451, 0.507502, 0.52768, 0.625164, 0.935248, 1.826161, 3.845598, 8.293859, 18.014295, 39.20736, 85.3836, 185.975252, 405.095196, 882.399269, 1922.095421, 4186.829292, 9120.018288

$$b_{0j}$$=0.026284, 0.057253, 0.124713, 0.184064, 0.271658, 0.429143, 0.799226, 1.740925, 3.792194, 8.2604, 17.993331, 39.194226, 85.375371, 185.970096, 405.091965, 882.397245, 1922.094153, 4186.828497, 9120.01779

The exact value of the ground state energy from the literature is -1.0222124 and this basis set with great accuracy give this value (-1.02221). It's know the excited levels energies from the literature too: -0.174, -0.069, -0.036865, -0.022815, -0.015495. The basis set that I used give the follow values for the excited levels energies: -0.173374, -0.0199577, 0.312572, 1.18162... If the first excited level differs in the third decimal place, then the energies of the subsequent excited levels are completely different. This is because the basis set I am using is not complete and it is necessary to increase the number of basis functions in order to get more accurate values for the excited levels.

Is it possible, having a certain set of basic functions (in my case, it is 19 functions), to supplement it with additional functions so that it becomes more complete? How to do it? Because I would like to get the energies of the excited levels too.

In the code I rename $$\rho\equiv r$$

ClearAll["Global*"]

(*the basis set*)
b00 = {0.026284, 0.057253, 0.124713, 0.184064, 0.271658, 0.429143,
0.799226, 1.740925, 3.792194, 8.2604, 17.993331, 39.194226,
85.375371, 185.970096, 405.091965, 882.397245, 1922.094153,
4186.828497, 9120.01779};
a00 = {0.500074, 0.502676, 0.510451, 0.507502, 0.52768, 0.625164,
0.935248, 1.826161, 3.845598, 8.293859, 18.014295, 39.20736,
85.3836, 185.975252, 405.095196, 882.399269, 1922.095421,
4186.829292, 9120.018288};
b0 [j_] := b00[[j]];
a0 [j_] := a00[[j]];

Psi[r_, z_, j_] := Exp[-b0[j]*z^2]*Exp[-a0[j]*r^2];
nmax = Dimensions[b00][[1]];

Kk[r_, z_, j1_, j2_] :=
FullSimplify[
Psi[r, z, j2]*
Laplacian[Psi[r, z, j1], {r, \[Theta], z}, "Cylindrical"]*r*2*Pi];
Kx[j1_, j2_] := -(1/2)*
NIntegrate[
Kk[r, z, j1, j2], {r, 0, Infinity}, {z, -Infinity, Infinity}];
Kxx = Table[Kx[j1, j2], {j1, 1, nmax}, {j2, 1, nmax}];

Ka[r_, z_, j1_, j2_] :=
FullSimplify[Psi[r, z, j2]*1/2*r^2*Psi[r, z, j1]*r*2*Pi];
KA[j1_, j2_] :=
KA[j1, j2] =
KA[j2, j1] =
NIntegrate[
Ka[r, z, j1, j2], {r, 0, Infinity}, {z, -Infinity, Infinity}];
KAx = Table[KA[j1, j2], {j1, 1, nmax}, {j2, 1, nmax}];

Ks[r_, z_, j1_, j2_] :=
FullSimplify[Psi[r, z, j2]*(-1)*Psi[r, z, j1]*r*2*Pi];
KS[j1_, j2_] :=
KS[j1, j2] =
KS[j2, j1] =
NIntegrate[
Ks[r, z, j1, j2], {r, 0, Infinity}, {z, -Infinity, Infinity}];
KSx = Table[KS[j1, j2], {j1, 1, nmax}, {j2, 1, nmax}];

VP1[r_, z_] := -(1/Sqrt[r^2 + z^2]);
Px1[j1_, j2_] :=
Px1[j1, j2] =
Px1[j2, j1] =
NIntegrate[
Psi[r, z, j2]*VP1[r, z]*Psi[r, z, j1]*r*2*Pi, {r, 0,
Infinity}, {z, -Infinity, Infinity}];
Pxx1 = Table[Px1[j1, j2], {j1, 1, nmax}, {j2, 1, nmax}];

Bb[j1_, j2_] :=
Bb[j1, j2] =
Bb[j2, j1] =
NIntegrate[
Psi[r, z, j2]*Psi[r, z, j1]*r*2*Pi, {r, 0,
Infinity}, {z, -Infinity, Infinity}];
Bxx = Table[Bb[j1, j2], {j1, 1, nmax}, {j2, 1, nmax}];

EEE = Kxx + Pxx1 + KAx + KSx;
Sort[Eigenvalues[{EEE, Bxx}]](*Energies*)

Out[88]= {-1.02221, -0.173374, -0.0199577, 0.312572, 1.18162,
2.72596, 5.63718, 12.0333, 17.9735, 28.2709, 61.4758, 135.447,
298.338, 656.001, 1440.89, 3168.65, 7007.42, 15772.6, 37523.6}

• I think this is a physics/chemistry question, not a Mathematica question. You use the word "basis" a lot, but a basis is a set of linearly independent vectors that span the space they are a basis of. In order to know how to augment a set of vectors that don't span a given space, it can be helpful to consider the action of $H$ on the vectors in your existing set. May 5 at 5:07
• How were the existing $a$'s and $b$'s found? Knowing the answer to that question probably allows you to extend the procedure to include more values, increasing the number of vectors and therefore hopefully span more of the relevant low-energy subspace. May 5 at 5:08
• But again, this is not a Mathematica question on my opinion. May 5 at 5:08
• @Marius Ladegård Meyer, thanks! a's and b's was found from the supplementary materials of this article pubs.aip.org/aip/jcp/article/147/24/244108/195534/… May 5 at 6:16
• Okay, so based on the abstract, you "just" have to follow their approach but don't stop at "so few" values of a and b... They even comment on the existence of a (much) larger basis from 2014. I don't think this is something that Mathematica can do for you, but you can maybe program it to do so. But that is your job ;) May 5 at 10:42

Its a pure mathematical question.

In cylindrical coordinates with an atractive plane 1/r Coulomb potential and and a harmonic oscillator in the x,y plane, in states with plane integer angular momentum M yielding a repulsive potential

 M^2/r^2, M=0,


the natural Hilbert space basis is

{ HermiteH[n, r] Exp[-r^2] e^i k z}


because the system is free in z.

Of course any base in z will do, eg

 HermiteH[m] Exp[-z^2 ]


in order to have a base of normalizable orthogonal functions centered at z=0.

By my knowlege, the plane - r-> 1/r is not an electric potential. The plane central potential of a wire is

q Log (r/r0)


with zero divergence

D[ 1/r D[r Log'[r]] ==0
`

while

2pi r dr * 1/r D[r (1/r)'] =2 pi 1/r dr = 2 pi r dr (1/r^2)

has a cylindrical charge density falling off with 1/r^2 on circles.

• thanks! This explanation is hard to understand. Could you please write it in more detail. May 6 at 15:52