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I have a Hamiltonian (Z) in matrix form, I solved it for time independent random real numbers, now want to introduce time dependent in such a way at any time the random real numbers change between the range {-Sqrt[3sigma2], Sqrt[3sigma2]}, here is my code

Nmax = 100; (*Number of sites*)

tini = 0; (*initial time*)

tmax = 200; (*maximal time*)

\[Sigma]2 = 0.1; (*Variance*)

n0 = 50; (*initial condition*)

ra = 1; (*coupling range*)

\[Psi]ini = Table[KroneckerDelta[n0 - i], {i, 1, Nmax}];

RR = RandomReal[{-Sqrt[3*\[Sigma]2], Sqrt[3*\[Sigma]2]}, Nmax];

Z = Table[
    Sum[KroneckerDelta[i - j + k], {k, 1, ra}] + 
     Sum[KroneckerDelta[i - j - k], {k, 1, ra}], {i, 1, Nmax}, {j, 1, 
     Nmax}] + DiagonalMatrix[RR];

usol = NDSolveValue[{I D[\[Psi][t], t] == 
     Z.\[Psi][t], \[Psi][0] == \[Psi]ini}, \[Psi], {t, tini, tmax}];

What can I do for introduce this time dependent and solve the differential equation(usol)? I hope my question is clear

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  • $\begingroup$ Can you please clarify, have you included the time independent solution? If so, can you make this explicitly known to the reader? If not, can you please include this? $\endgroup$ Commented Apr 17, 2021 at 8:33
  • $\begingroup$ the line "usol" is the solution of the time independent Hamiltonian. So now, I want to add a time dependent in the part of the random real numbers (the line RR) in which for any time or failing that [Delta]t the random real numbers change always between {3*[Sigma]2], Sqrt[3*[Sigma]2]} and compare the solutions. $\endgroup$ Commented Apr 17, 2021 at 17:25

1 Answer 1

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You discretized the x coordinate. Therefore, you will get a vector of Nmax different interpolation functions of t for a solutions. To combine these Nmax functions into one one single function of x and t you may proceed as (I assume usol is the output from above):

usol = NDSolveValue[{I D[[Psi][t], t] == Z.[Psi][t], [Psi][0] == [Psi]ini}, [Psi], {t, tini, tmax}]

We first calculate the function values on a grid and then interpolate these data:

dat = Flatten[
   Table[{{x, t}, usol[t][[x]]}, {x, 1, Nmax}, {t, 0, tmax}], 1] // Chop;
wf[x_, t_] = Interpolation[dat][x, t];

As your initial condition is a delta function, you have a lot of different momenta, what makes the wave function rather complicated. Here is a slice for x==50 and x==Nmax of Abs[psi]^2:

enter image description here

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  • $\begingroup$ Add these two lines from above code ``` Pusol[t_] = (Re[usol[t]]^2 + Im[usol[t]]^2); ListDensityPlot[Table[Pusol[t][[i]], {t, tini, tmax}, {i, 1, Nmax}], PlotTheme -> "Scientific", FrameLabel -> {"n", "t"}, PlotLabel -> "[Sigma] =", PlotLegends -> Automatic] ``` this is the probability in time in x basis for a time independent Hamiltonian. I would like compare the same picture for time dependent Hamiltonian when the random numbers change at any time or failing that in a [Delta]t. If I understand your answer is like a Fourier Transform, right? $\endgroup$ Commented Apr 17, 2021 at 17:21
  • $\begingroup$ Sorry for the late answer. Abs[wf[t]]^2 and (Re[usol[t]]^2 + Im[usol[t]]^2) are the same thing, namely the position probability density. No, may answer is not a Fourier Transform, it is position, not momenta. If you change the Hamiltonian, usol changes, but the procedure above stays the same. $\endgroup$ Commented Apr 23, 2021 at 7:23

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