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I am trying to solve stochastic Schrodinger equation (Schrodinger equation in the presence of Ornstein Uhlenbeck Process) $$i\frac{d}{dt}\begin{pmatrix}c_1(t)\\ c_2(t)\end{pmatrix}=H(t)\begin{pmatrix}c_1(t)\\ c_2(t)\end{pmatrix},$$

with

$H(t) = \begin{bmatrix} h_1+v t+f(t)- \cos(k) & -i \sin(k) \\ i \sin(k) & -h_1-v t-f(t)+\cos(k) \end{bmatrix}$

where $f(t)$ is Ornstein Uhlenbeck Process with noise correlation $\langle f(t)f(t')\rangle=\frac{\xi^2}{2\tau_n}e^{-|t-t'|/\tau_n}$ in which $\xi$ is noise intensity and $\tau_n$ is noise characteristic time.

The parameters $\xi=0.1$, $\tau_n=0.01$, $v=0.1$, $h_1=-50$, $h_2=0.5$, the initial time $t_1=0$ and the final time is $t_2=(h_2-h_1)/v$. The initial values of $c_1(t_1)=1$ and $c_2(t_1)=0$.

I have solved the differential equation using NDSolveValue, and I want to obtain $|c_2(t2)|^2$. My code (please see blow) works. But the results I have obtained is very strange.

Although I have calculated $|c_2(t)|^2$ at $t_2$ for all $k$ and $k$ changes very slowly ($k$ step is very small $dk/\pi=0.001$) but $|c_2(t)|^2$ shows very fast oscillation and very large difference between two nearby $k$ (please see black dots in figure) while the noise is the same for all $k$.

Could you please help me to understand why $|c_2(t)|^2$ shows very fast oscillation. Is my result correct or my code has a problem?

enter image description here

$PreRead = (# /. 
     s_String /; 
       StringMatchQ[s, NumberString] && 
        Precision@ToExpression@s == MachinePrecision :> s <> "`50." &);
Clear["Global`*"];
L = 200;
v = 0.1;
h1 = -50;
h2 = 0.5;
xi = 0.1;
tn = 0.01;

t1 = 0.;

t2 = (h2 - h1)/v;



f = Interpolation[
   Normal[RandomFunction[
      OrnsteinUhlenbeckProcess[0, xi/tn, 1/tn, 0], {t1, t2, 0.01}]][[
    1]]];


For[m = 1, m <= L/2, m = m + 1,
  
  
  
  k = ((2 m - 1)*\[Pi])/L;
  
  
  
  {x, y} = 
   NDSolveValue[{I*Derivative[1][c1][t] == 
      2*(h1 + v*t - Cos[k] + f[t])*c1[t] + 2*(-I*Sin[k])*c2[t], 
     I*Derivative[1][c2][t] == 
      2*(I*Sin[k])*c1[t] + 2*(-h1 - v*t + Cos[k] - f[t])*c2[t], 
     c1[t1] == 1, c2[t1] == 0}, {c1, c2}, {t, t1, t2}];
  
  
  Pk22 = Norm[y[t2]]^2;
  
  Print[k,"       ",Pk22]
  
  
  ];

@Daniel Huber @Parsifal Dear Daniel and dear parsifal, I have used your comments Stochastic process: Understanding Ornstein Uhlenbeck Process , Continuous noise representation to construct the continuous noise function. I think you are expert in this field. I was wondering if you wold be able to let me know you idea and feedback.

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  • $\begingroup$ I feel like sharing that I tend to avoid posts that include $PreRead =... (unless it's a question about $PreRead, perhaps) and ClearAll["Global`*"] or other things that will affect my setup. My eyesight's not that great. It's irritating when I don't notice something like that. I spend time figuring out why things aren't working. Then it turns out because it's because something trashed the state of my session. And the time was spent on something pointless. I don't know how to sandbox ClearAll["Global`*"] and the like. $\endgroup$
    – Michael E2
    Commented Jul 26, 2023 at 16:16

1 Answer 1

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The amplitude of random function f you use is too higher compare to 1. With less amplitude we have

Clear["Global`*"]; SeedRandom[1234];
L = 200;
v = 0.1;
h1 = -50;
h2 = 0.5;
xi = 0.1;
tn = 0.01;
t1 = 0.;
t2 = (h2 - h1)/v;
f = Interpolation[
   Normal[RandomFunction[
      OrnsteinUhlenbeckProcess[0, xi/tn, 1/tn, 0], {t1, t2, 
       0.01}]][[1]]];
eps=1/10;

Numerical solution

Do[k = ((2 m - 1)*\[Pi])/L;
   {x[m], y[m]} = 
    NDSolveValue[{I*
        c1'[t] == (h1 + v*t - Cos[k] + eps f[t])*c1[t] + (-I*Sin[k])*
         c2[t], I*
        c2'[t] == (I*Sin[k])*c1[t] + (-h1 - v*t + Cos[k] - eps f[t])*
         c2[t], c1[t1] == 1, c2[t1] == 0}, {c1, c2}, {t, t1, 
      t2}];, {m, 1, L/2}
   ]; 

Visualization

ListLinePlot[
 Table[{((2 m - 1)*\[Pi])/L, Evaluate[Abs[y[m][t2]]^2]}, {m, 1, L/2}],
  PlotRange -> All, PlotStyle -> Red, Mesh -> All, 
 AxesLabel -> {"k", 
   "|\!\(\*SubscriptBox[\(c\), \(2\)]\)\!\(\*SuperscriptBox[\(|\), \
\(2\)]\)"}]

Figure 1

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  • $\begingroup$ Thank you so much for your comment. Could you please let me know why we should consider the amplitude of noise less than 1? As far as I know in the colored noise the amplitude of noise is not restricted and I have considered the amplitude of the noise 1, not very larger than 1. In theoretical calculation the noise intensity $\xi$ and the noise characteristic time $\tau_n$ (noise correlation) are important and the amplitude of noise does not appear in theoretical calculation. $\endgroup$
    – Radmehr
    Commented Jun 21, 2023 at 14:06
  • $\begingroup$ We have a deal with numerical computation, and not with theoretical model. In numerical computation the amplitude of noise should be restricted. Anyway, if you like you can increase eps and check what happened with numerical solution. $\endgroup$ Commented Jun 21, 2023 at 15:10
  • $\begingroup$ Dear Alex, thank yo so much for your instructive and valuable comments. $\endgroup$
    – Radmehr
    Commented Jun 21, 2023 at 15:12
  • $\begingroup$ @Radmehr You are welcome! $\endgroup$ Commented Jun 21, 2023 at 15:14

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