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Here we are solving a set of non-linear equations. And the parameters are

SetDirectory@NotebookDirectory[];
hbar = 1 ;
h = 2 Pi hbar;
num = 8 10^6;
{TFradiix, TFradiiy, TFradiiz} = {230, 110, 2.4} 10^-6;
n0 = 1;
q = 47 h;
c2n0 = 39 h;

{\[Omega]x, \[Omega]y, \[Omega]z} = 2 Pi {4.3, 8.8, 420};
rcx = 1;
rcy = 1/Sqrt[M \[Omega]y/hbar];
x0 = 20 rcx;
halfLx = x0;
halfLy = x0 TFradiiy/TFradiix;
M = 1;
c2 = c2n0/n0/(hbar \[Omega]x);
c0 = 27.6 c2;
numDiscrete = 200.;
\[Delta]x = 2 halfLx/(numDiscrete);
\[Delta]y = 2 halfLy/(numDiscrete);

datax = Table[x, {x, -halfLx, halfLx, \[Delta]x}];
datay = Table[y, {y, -halfLy, halfLy, \[Delta]y}];

And the initial state for \[Psi]0 is

datapsi0 = ToExpression@Import["datapsi0.csv"];

nx = Length@datapsi0;
ny = Length@First@datapsi0;
data\[Psi]Interpolation = 
  Catenate[
   ParallelTable[{{datax[[ix]], datay[[iy]]}, 
     datapsi0[[ix, iy]] // Chop}, {ix, 1, nx}, {iy, 1, ny}]];
initial\[Psi]1[x_, y_] := Interpolation[data\[Psi]Interpolation][x, y]

where the file "datapsi0.csv" can be downloaded here.

The code for solving equations is

Sx = 1/Sqrt[2] ( {
     {0, 1, 0},
     {1, 0, 1},
     {0, 1, 0}
    } );
Sy = I/Sqrt[2] ( {
     {0, -1, 0},
     {1, 0, -1},
     {0, 1, 0}
    } );
Sz = ( {
    {1, 0, 0},
    {0, 0, 0},
    {0, 0, -1}
   } );
vecF = {Sx, Sy, Sz};
vec\[Psi][t_, x_, 
   y_] := {\[Psi]1[t, x, y], \[Psi]0[t, x, y], \[Psi]minus1[t, x, 
    y]};
vecFbar[t_, x_, 
   y_] := {(Conjugate[vec\[Psi][t, x, y]] . Sx . 
     vec\[Psi][t, x, y]), (Conjugate[vec\[Psi][t, x, y]] . Sy . 
     vec\[Psi][t, x, y]), (Conjugate[vec\[Psi][t, x, y]] . Sz . 
     vec\[Psi][t, x, y])};

Vtrap[x_, y_] := 
  1/2 M (x^2 + (\[Omega]y/\[Omega]x)^2 y^2); (*normlized potential*)
\[CapitalOmega] = 2 Pi 150;(*Rabi frequency*)
\[Omega]0 = 2 Pi 291 10^3;(*Larmor frequency*)
\[Omega] = \[Omega]0;(*Frequency of RF *)
\[CapitalDelta] = \[Omega] - \[Omega]0;
\[Delta]0 = 1000 2 Pi;
f = 60;
\[Delta][t_] := \[Delta]0 Sin[2 Pi f t/\[Omega]x];
Hs[t_] := (hbar (\[CapitalDelta] - \[Delta][t]) Sz + q Sz . Sz - 
     hbar \[CapitalOmega] Sx)/(hbar \[Omega]x); (*the factor 1/(hbar \
\[Omega]x) is for dimensionless*)
eqs = {I hbar D[\[Psi]1[t, x, y], 
      t] == -(hbar^2/(2 M)) Laplacian[\[Psi]1[t, x, y], {x, y}] + 
     Vtrap[x, y] \[Psi]1[t, x, y] + ((Hs[t] . vec\[Psi][t, x, y])[[
       1]]) + c0 (Abs[\[Psi]1[t, x, y]]^2 + Abs[\[Psi]0[t, x, y]]^2 + 
        Abs[\[Psi]minus1[t, x, y]]^2) \[Psi]1[t, x, y] + 
     c2 ((vecFbar[t, x, y] . vecF) . vec\[Psi][t, x, y])[[1]], 
   I hbar D[\[Psi]0[t, x, y], 
      t] == -(hbar^2/(2 M)) Laplacian[\[Psi]0[t, x, y], {x, y}] + 
     Vtrap[x, y] \[Psi]0[t, x, y] + ((Hs[t] . vec\[Psi][t, x, y])[[
       2]]) + c0 (Abs[\[Psi]1[t, x, y]]^2 + Abs[\[Psi]0[t, x, y]]^2 + 
        Abs[\[Psi]minus1[t, x, y]]^2) \[Psi]0[t, x, y] + 
     c2 ((vecFbar[t, x, y] . vecF) . vec\[Psi][t, x, y])[[2]], 
   I hbar D[\[Psi]minus1[t, x, y], 
      t] == -(hbar^2/(2 M)) Laplacian[\[Psi]minus1[t, x, y], {x, y}] +
      Vtrap[x, y] \[Psi]minus1[t, x, 
       y] + ((Hs[t] . vec\[Psi][t, x, y])[[3]]) + 
     c0 (Abs[\[Psi]1[t, x, y]]^2 + Abs[\[Psi]0[t, x, y]]^2 + 
        Abs[\[Psi]minus1[t, x, y]]^2) \[Psi]minus1[t, x, y] + 
     c2 ((vecFbar[t, x, y] . vecF) . vec\[Psi][t, x, y])[[3]]};
implicitRegion = Rectangle[{-halfLx, -halfLy}, {halfLx, halfLy}];
bc = {\[Psi]1[0, x, y] == 0, \[Psi]0[0, x, y] == 
    initial\[Psi]1[x, y], \[Psi]minus1[0, x, y] == 0, 
   DirichletCondition[{\[Psi]1[t, x, y] == 0, \[Psi]0[t, x, y] == 
      0, \[Psi]minus1[t, x, y] == 0}, True]};

solv = NDSolve[{eqs, bc}, {\[Psi]1, \[Psi]0, \[Psi]minus1}, {t, 0, 
    2}, {x, y} \[Element] implicitRegion];

The results show

Table[Plot3D[
  Evaluate[ Abs[\[Psi]0[t, x, y] /. solv[[1]]]^2], {x, -halfLx, 
   halfLx}, {y, -halfLy, halfLy}, PlotLabel -> t, PlotRange -> All, 
  PlotPoints -> 100], {t, 0, 2, .5}]

enter image description here

My question is

The boundary condition DirichletCondition[{\[Psi]1[t, x, y] == 0, \[Psi]0[t, x, y] == 0, \[Psi]minus1[t, x, y] == 0}, True] requires \[Psi]0==0 at the edge, why does the wave function \[Psi]0 grows in the four corners? This phenomenon is unreasonable physcally. And what should we solve this problem?

Note: More mathematical details can be found in this paper.

Added:

Decreasing x0 from 20 rcx to 10 rcx and run the code again, we find the fearure of growing in the corners disappers

Table[Plot3D[
  Evaluate[ Abs[\[Psi]0[t, x, y] /. solv[[1]]]^2], {x, -halfLx, 
   halfLx}, {y, -halfLy, halfLy}, PlotLabel -> t, PlotRange -> All, 
  PlotPoints -> 100], {t, 0, 10, 2}]

enter image description here

And what is the reason here?

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2 Answers 2

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@Roland Franzius, I use the PeriodicBoundaryCondition

bc = {\[Psi]1[0, x, y] == 0, \[Psi]0[0, x, y] == 
    initial\[Psi]1[x, y], \[Psi]minus1[0, x, y] == 
    0, \[Psi]1[t, -halfLx, y] == \[Psi]1[t, halfLx, y], \[Psi]1[t, 
     x, -halfLy] == \[Psi]1[t, x, halfLy], \[Psi]0[t, -halfLx, 
     y] == \[Psi]0[t, halfLx, y], \[Psi]0[t, x, -halfLy] == \[Psi]0[t,
      x, halfLy], \[Psi]minus1[t, -halfLx, y] == \[Psi]minus1[t, 
     halfLx, y], \[Psi]minus1[t, x, -halfLy] == \[Psi]minus1[t, x, 
     halfLy](*,DirichletCondition[{\[Psi]1[t,x,y]==0,\[Psi]0[t,x,y]==
   0,\[Psi]minus1[t,x,y]==0},True]*)};

The results show

solv = NDSolve[{eqs, bc}, {\[Psi]1, \[Psi]0, \[Psi]minus1}, {t, 0, 
    8}, {x, -halfLx, halfLx}, {y, -halfLy, halfLy}(*,{x,
   y}\[Element]implicitRegion*)];
Table[DensityPlot[
  Evaluate[ Abs[\[Psi]0[t, x, y] /. solv[[1]]]^2], {x, -halfLx, 
   halfLx}, {y, -halfLy, halfLy}, PlotLabel -> t, PlotRange -> All, 
  PlotPoints -> 100], {t, 0, 8, 2}]

enter image description here

Although the wave function \[Psi]0 doesn‘t grow in the corners, but the obvious expanding feature is not desired.

So where is the problem?

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    $\begingroup$ Perhaps, this should be edited into your question or asked as a separate question. $\endgroup$
    – bbgodfrey
    Commented Mar 26, 2023 at 18:42
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You model a system of three fields, each obeying Schödinger equation with a osillator potential and interacting via the third order interaction term (as far as I see)

i d_t psi_k == - Lap psi_k + V psi_k + WW(psi_1,psi_2,psi_3) 

The one-particle linear potential term Vtrap*psi is a confining for all energies, so the psi_k move in the pit insofar as the total energy conserved.

But even with energy conservation for all fields, I dont't see on inspection, if the scattering is producing higher frequency components, that naturally blur out a spatial concentrated initial field.

This fact is behind the spatial wave-packet spreading of a Gaussin wave packet, that is a Fourier composition of wave components of all energies in (0, infinity)

Comment on code: If a function like Hs depends on variable t,x you should define it as Hs[t_,x_] in order to enable the NDSolve algorithm to read and transfers a transparent map to its code. Otherwise the first argument of NDSolve should be enclosed in Evaluate[] in order to make dependenecies transparent.

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    $\begingroup$ Your answer probably would be more appropriate as a comment. $\endgroup$
    – bbgodfrey
    Commented Mar 27, 2023 at 1:39

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