Fourier transform of a wave equation

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.

 

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

 

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

 

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.