Gaussian wave packet in position space

[KFC]

Hi all,
The Gaussian wave packet is widely discussed in the text. I got the following expression for wave packet in momentum space

$\psi(p, 0) = A \exp\left[-(p-p_0)^2/ (2\sigma_p^2)\right]$
with $A=\sqrt{2\sigma_p/\sqrt{2\pi}}$

As my understanding, the corresponding wave packet in position space should be inverse Fourier transformation of $\psi(p, 0)$. I plot the $\psi(p, 0)$ in matlab, I saw the Gaussian profile. But when I take the inverse Fourier transformation ifftshift(ifft(psi)), I saw something of the order $10^{-6}$. I have no idea why is it.

 

[blue_leaf77]

I saw something of the order 10−610^{-6}

So long as it's normalized, what's the problem? It depends on how you set the parameter values anyway.

 

[KFC]

The problem is after the numerical transformation, all samples for the packet is about $10^{-6}$ instead of ranged from 0 to $10^{-6}$ that means it is almost zero everywhere. Let me restate my question with some code and data. I think I made an mistake in the first post, here is the Guassian in momentum

$\psi(p) = \sqrt{\dfrac{1}{\sqrt{2\pi}\sigma_p}}\exp\left[-\dfrac{(p-p_0)^2}{4\sigma_p^2}\right]$

If you integrate this in mathematica, $\int_{-\infty}^{+\infty} |\psi(p)|^2dp = 1$. I also tried to find the wave packet in position by inverse Fourier transformation, I got

$\phi(x) = \sqrt{\dfrac{2\sigma_p}{\sqrt{2\pi}}} \exp\left[-\sigma_p x^2\right]\exp(ip_0x)$

and $\int_{-\infty}^{+\infty} |\phi(x)|^2dx = 1$. Plotting those functions with $p_0=0, \sigma_p=1$, we observe that $|\psi(p)|^2$ with minimum and maximum as 0 and about 0.9; the minimum and maximum for $|\phi(x)|^2$ is about 0 to 1.2

Now doing numerical evaluation by matlab, I try

p0 = 0;
sigma_beta = 1;
psi = sqrt(1/sqrt(2*pi)/sigma_beta);*exp(-((p-p0).^2)/(4*sigma_beta^2));
phi = ifftshift(ifft(psi));

plot(abs(phi).^2);

The range for the plot is about $10^{-5}$ and is we sum over the $|\phi|^2$, I've got 5.9921e-04 while the sum of $|\psi|^2$ gives 1. See the attachment (don't know why it upload twice and seems I cannot remove one of them)

 

[blue_leaf77]

I also tried to find the wave packet in position by inverse Fourier transformation, I got

$\phi(x) = \sqrt{\dfrac{2\sigma_p}{\sqrt{2\pi}}} \exp\left[-\sigma_p x^2\right]\exp(ip_0x)$

I haven't checked your calculation, but I just want to confirm, have you considered that $p=\hbar k$? Forgetting this may also affect the scaling of the wavepacket in position space, and hence the numerical integration.

 

[KFC]

Thanks for your reply. I rescale the system with $\hbar=1$. Moreover, in both analytical calculation (mathematica) and numerical one (matlab), I have the same scale ($\hbar=1$). so the result should be identical. But they are not. My question is why the numerical one doesn't give the same one from analytical expression.

 

[Avodyne]

Your analytic formulas seem to be correct, provided that you set $\hbar=1$. So you have a MATLAB problem, not a physics problem.

 

[DrClaude]

I've got 5.9921e-04 while the sum of $|\psi|^2$ gives 1. See the attachment (don't know why it upload twice and seems I cannot remove one of them)

Do you simply sum up the components, or do you also multiply by $dx$ or $dp$?

While ifft(fft(psi)) should recover the original psi, I don't know what normalization of the single action of fft or ifft should be. From what I can gather from the manual, there is a $1/N$ factor in ifft, with nothing for the forward fft, so that might be the origin of the scaling you see.

 

[KFC]

Thanks for reply. Yes, I sum up all components and multiply it with dx or dp to test normalization.
For fft part, yes, it has the 1/N in front of series, I've already taking that factor in consideration, but still no help.

 

[DrClaude]

it has the 1/N in front of series, I've already taking that factor in consideration, but still no help.

How?

p0 = 0;
sigma_beta = 1;
psi = sqrt(1/sqrt(2*pi)/sigma_beta);*exp(-((p-p0).^2)/(4*sigma_beta^2));
phi = ifftshift(ifft(psi));

plot(abs(phi).^2);

I don't see anything in there that involves N.