Fourier transform of a wave equation

  • Thread starter mvillagra
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  • #1
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Hello, I have a question about the following problem:

Given a wave equation [tex]\Psi(n,t)[/tex] where t is the time, and n is an integer. What is the fourier transform?

I'm trying to reproduce this paper: One-dimensional Quantum Walks by Ambainis et al. (http://citeseer.ist.psu.edu/old/514019.html"), and here says that the spatial fourier transform of such a wave is [tex]\widetilde{\Psi}(k,t)=\sum_n \Psi(n,t)e^{ikn}[/tex], without a "-" minus sign on the exponential. Why is that?

The fourier transform is defined with a minus there!! Or am I missing some property?

I am no expert in fourier analysis but can we interchange the use of the signs between the transform and the inverse transform?
 
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  • #2
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Notice that n is in Z (integers) not N, so assuming that the wavefunction is even, it makes no difference whether we put the minus sign there or not. I don't claim to know what the author is getting at with this line of reasoning, though.
The author refers to [11] when that transform is put forward. I daresay if you look at that reference you would probably get a definitive answer to your question.
 
  • #3
Hans de Vries
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The fourier transform is defined with a minus there!! Or am I missing some property?

I am no expert in fourier analysis but can we interchange the use of the signs between the transform and the inverse transform?
The plus sign is because it's the time component of the wavefunction.

The 4D Quantum mechanical Fourier transform is defined as:

[tex]
\mbox{\Large F} (p^\mu) \ \stackrel{\mathrm{def}}{=}\ ~~\int\limits_{-\infty}^{+\infty}~dx^4~ \mbox{\Large f}(x^\mu) ~~ e^{+i\, p_\mu\, x^\mu}
[/tex]

and its inverse is defined as:

[tex]
~\mbox{\Large f}(x^\mu) ~~\stackrel{\mathrm{def}}{=} ~~ \int\limits_{-\infty}^{+\infty} \frac{dp^4}{(2 \pi)^4}~\mbox{\Large F}(p^\mu) \ e^{-i\, p_\mu\,x^\mu}
[/tex]

With the standard definition:

[tex]p_\mu\, x^\mu ~~=~~ Et-p_x x-p_y y-p_z z[/tex]


So for the spatial coordinates you do have the normal minus sign.

Regards, Hans
 
  • #4
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The plus sign is because it's the time component of the wavefunction.

So for the spatial coordinates you do have the normal minus sign.
I understand these definitions, but its not clear why we have two different transformations, one for time and one for space. How do we know when to use one or the other?

Is there any reference on this? I don't have the reference given in the paper in my library. I tried looking in Griffiths book and Ballentine's book but didn't find anything.
 
  • #5
Hans de Vries
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I understand these definitions, but its not clear why we have two different transformations, one for time and one for space. How do we know when to use one or the other?

Is there any reference on this? I don't have the reference given in the paper in my library. I tried looking in Griffiths book and Ballentine's book but didn't find anything.
This is because in QM we interprete the 4D Fourier domain as the Energy-momentum
domain and the de Broglie matter waves are defined as:


[tex]
\psi ~~=~~ \exp\Big\{\,\tfrac{1}{\hbar}\left(-Et+p_x x+p_y y+p_z z\right)\,\Big\}
[/tex]

Regards, Hans
 
  • #6
f95toli
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Maybe it is worth pointing out that it doesn't really matter if we put the minus sign in the forward transform or the inverse as long as you are consistent; the final result of the calculations will be the same. The fact that you are free to choose is just due to way the Fourier transform is defined; it has nothing to do with physics.
So the answer to the OP is yes; we are free to interchange the signs.

This can be quite confusing since different authors use different conventions but it is rarely a problem as long as you are aware of it.
E.g. electrical engineers and physicists generally use different conventions for historical reasons.
 
  • #7
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aaah, I see now, crystal clear!!

As a computer scientist I'm not very use to these technical details in QM.

Thank you very much guys for all your answers, now I can continue with this :-)
 

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