While trying to solve a stochastic Gross-Piaevskii equation I have found a problem that can be tracked down to something buggy occurring in the simplest Schrodinger equation possible:

$$\partial_t \psi = i \partial^2_x \psi \, .$$

This has the very simple solution of a plane wave:

$$\psi(t) = A e^{i k x - i\omega t}, \quad \text{with } \omega = k^2$$

but in some cases I cannot recover this very simple solution numerically. This is due to the fact that my code uses a very standard fast Fourier transform method to go to $k$-space, where the evolution is trivial, and then does the inverse FFT to go back to real space. This method naturally enforces periodic boundary conditions in your simulation grid since you are expanding your solution as a sum of periodic functions. When the initial condition is a plane wave with a periodicity that does not match the domain size, the algorithm is not able to provide the correct solution.

I show here an example. The problem is clear in $|\psi(t)|$, which should remain constant:

Evolution of the wavefunction amplitude

The effect is less dramatic in the phase but still present. One can see some ripples that come from the bottom edges.

Evolution of the phase

I'm not surprised by the fact that, if I enforce periodic boundary conditions, computing the evolution of a wave which is not periodic in the simulation domain yields problems. However, the method is very standard and use extensively, with packages like XMDS employing it by default. Therefore, I am surprised by the fact that I did not find any mention of this method failing to solve such an extremely simple example. Is there something that I'm missing? Should I just get over it and assume that this is not a good method if I expect a solution of this kind? Does someone know a reference where this is documented?

Edit: I add the code to compute the example reported:

import numpy as np
import matplotlib.pyplot as plt
from numpy import exp, dot, mean, fft, cos,sin,cosh,sqrt,absolute, array, transpose, var, conj

import math
from math import pi
import time

from numpy.fft import fft, ifft,fft2,ifft2

L = 4300
dt = 1000
naxis = 1000
k0 = 0.01
M0 =1

# Define position vector
xvector = np.linspace(-L,L,naxis,retstep=True)
dx = xvector[1]
xvector = xvector[0]

nt = 1000
psi0 =M0*exp(1j*k0*xvector)

klist = np.array([y for y in range(0,naxis/2) + [0] + range(-naxis/2+1,0)])
mu_k = pi * (naxis-1)*klist/(L*naxis)
expmu = exp(-1j*(mu_k**2)*(dt)/2)

psit_list = []
psit = psi0

# Computation 
ti = time.time()
for i in range(1,nt):

    psitft = fft(psit)
    psitft = expmu*psitft
    psit = ifft(psitft)


tf = time.time()
psit_list_abs = map(absolute,psit_list)
psit_list_phase = map(np.angle,psit_list)
print(["Time of computation: ",tf-ti," seconds"])

fig = plt.figure(figsize=(9,8))
ax = fig.add_subplot(1, 1, 1)
gpeplot = plt.imshow((np.absolute(psit_list[:10000])),origin='lower',aspect='auto')
plt.suptitle('Wavefunction amplitude $|\psi|$',fontsize=20)
  • 1
    $\begingroup$ You are basically discovering that many people are doing it wrong and you have doubts about your judgement? It happens that folks are naively using an inappropriate method for years as part of "monkey see, monkey do" type research (especially poorly supervised students without much experience are prone to using tools they don't understand without checking whether they are appropriate) . I would not be surprised that the FFT will not represent the problem correctly. Can you calculate the result another way? $\endgroup$
    – CuriousOne
    Jan 14, 2016 at 20:51
  • 2
    $\begingroup$ Without seeing the code, there's not much one can do to help. The incorrect results may be due to a myriad reasons, e.g. assuming an incorrect memory layout of k space. $\endgroup$
    – alarge
    Jan 14, 2016 at 20:54
  • 3
    $\begingroup$ @CuriousOne Most of the time when one looks at something a lot of people are doing and thinks they're all wrong, it's actually the opposite. The discrete Fourier transform (DFT) fails in this case because OP is trying to use it in a system where the implicit periodic boundary condition of the DFT doesn't make sense and matters because the wave functions go to the edges of the simulation boundary. If we were to study e.g. a particle in bound states of a potential well then you could just go far enough out in position such that the boundary conditions don't matter. $\endgroup$
    – DanielSank
    Jan 14, 2016 at 20:54
  • 1
    $\begingroup$ @DanielSank: I was reading the OP the other way around. He says "Therefore, I am surprised by the fact that I did not find any mention of this method failing to solve such an extremely simple example.". Exactly because inappropriate numerical methods tend to fail on even simple counterexamples he is right that this should be documented. Now, I haven't read the documentation and maybe it's in there or maybe you don't think that numerical tools should be doing the handholding. That's fair, I accept that. $\endgroup$
    – CuriousOne
    Jan 14, 2016 at 21:39
  • 1
    $\begingroup$ Would Computational Science be a better home for this question? $\endgroup$
    – Qmechanic
    Jan 14, 2016 at 22:17

1 Answer 1


This is more of a comment than an answer, but I can't fit this into the amount of characters; Writing a quick bit of code, it looks to me like there's not much wrong with the method: The numerical and the analytical solution go on top of one another.

N = 256
T = 256*128
L = 1.
dt = 0.000001

x = linspace(0., N-1, N)*L/N
psix = exp(1j*2*pi*x)

psik = 1j*ones((T, N))*0
psik[0, :] = fft.fft(psix)

kSQ = (hstack((linspace(0, N/2, N/2+1), linspace(-N/2+1, -1, N/2-1)))*2*pi/L)**2

def rkstep(y):
    k1 = -1j*kSQ*y
    k2 = -1j*kSQ*(y+k1*dt/2)
    k3 = -1j*kSQ*(y+k2*dt/2)
    k4 = -1j*kSQ*(y+k3*dt)
    return y + dt/6*(k1 + 2*k2 + 2*k3 + k4)

for t in range(T)[1:]:
    psik[t, :] = rkstep(psik[t-1, :])

psix = fft.ifft(psik)

plot(x, angle(psix[T-1, :]))
plot(x, angle(exp(1j*2*pi*x - 1j*(2*pi)**2*T*dt)))

This I ran in ipython -pylab, using whatever libraries (likely NymPy, SciPy, matplotlib) it loads on my computer to whichever namespace (global). Your mileage may vary unless you use proper imports.

  • $\begingroup$ Thanks for taking the effort of checking :) The only difference between your code and mine is that you use Runge-Kutta to compute the evolution in k-space, which is not necessary since the solution is analytic there. In any case, I checked your code with my parameters and still get the same behaviour that I reported (as expected). If you are interested, the parameters are: N = 1000 T = 1000 L = 2*4300 dt = 10 k = 0.01/pi x = linspace(0., N-1, N)*L/N psix = exp(1j*k*pi*x) $\endgroup$
    – Carlos_San
    Jan 15, 2016 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.