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Copy from here https://mathematica.stackexchange.com/questions/284809/why-does-the-minimum-eigenvalue-change-dramatically-when-one-basis-function-is-a

I have a basis set which describes with high accuracy the ground state of this system:$$H=-\frac{1}{2}\Delta-\frac{1}{r}+\frac{25}{8}\rho^2-5/2$$where $r=(\rho,z,\phi)$ is a coordinate in the cylindrical system.

Ground state from the literature $E_{min}=-1.3803975$

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

$a_{0j}$={1.25, 1.25003, 1.25006, 1.251393, 1.252726, 1.2587955, 1.264865, 1.260499, 1.303186, 1.546232, 2.337185, 4.507307, 9.426298, 20.270036, 43.977048, 95.675215, 208.32642, 453.736619, 988.32236, 2152.803005, 4689.357171, 10214.647096}

$b_{0j}$={0.026284, 0.0417685, 0.057253, 0.09098300000000001, 0.124713, 0.19818550000000001, 0.271658,0.417292, 0.61588, 1.008925, 1.949878, 4.24735, 9.251849, 20.15297, 43.898488,95.622497, 208.291042, 453.712878`, 988.306428, 2152.792314,4689.349997, 10214.642282}

This basis gives the energy of the ground state $E_{min}=-1.3803971$, which agrees perfectly with the value from the literature.

Now if we add only one element to the list $a_{0j}$ and list $b_{0j}$ (1.2626819999999999 and 0.344475 respectively), then the absolute value of minimum energy becomes huge ($E_{min}=-7.1824042$).
Why is this happening?

It is known from theory that an increase in the number of basis functions should either increase the accuracy or not change the values of the eigenvalues.

The code had written in Wolfram Mathematica. In the code I have renamed $\rho ≡ r$

ClearAll["Global`*"]
b00 = {0.026284`, 0.0417685`, 0.057253`, 0.09098300000000001`, 
   0.124713`, 0.19818550000000001`, 0.271658`,(*0.344475`,*)0.417292`,
    0.61588`, 1.008925`, 1.949878`, 4.24735`, 9.251849`, 20.15297`, 
   43.898488`, 95.622497`, 208.291042`, 453.712878`, 988.306428`, 
   2152.792314`, 4689.349997`, 10214.642282`};
a00 = {1.25`, 1.25003`, 1.25006`, 1.251393`, 1.252726`, 1.2587955`, 
   1.264865`,(*1.2626819999999999`,*)1.260499`, 1.303186`, 1.546232`, 
   2.337185`, 4.507307`, 9.426298`, 20.270036`, 43.977048`, 
   95.675215`, 208.32642`, 453.736619`, 988.32236`, 2152.803005`, 
   4689.357171`, 10214.647096`};

nmax = Dimensions[b00][[1]];

a0 [j_] := a00[[j]];
b0 [j_] := b00[[j]];

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


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]*25/8*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]*(-5/2)*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}];

EE1 = Eigenvalues[{Kxx + Pxx1 + KAx + KSx, Bxx}];
Min[EE1]

Out[48]= -1.3804
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  • $\begingroup$ Wouldn’t it be helpful to readers to mention someplace that you are using, I think, the Ritz method? $\endgroup$
    – Ghoster
    Commented May 6, 2023 at 23:30

1 Answer 1

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It may be that the additional basis function is numerically very close to a linear combination of the existing basis functions.

This would cause your matrix $B$ to become nearly singular, and most numerical methods break down when that happens.

I'm not a Mathematica expert, but this page seems to describe a similar problem, which was fixed by increasing the numerical precision.

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