Fourier transform using exponential with positive argument as basis

Question

Fourier transform is normally performed with $e^{ikx}$ as basis. However, in a textbook which I am using, it used $e^{-ikx}$ as basis. Is this something valid to do?

Answer 1

It makes no difference, providing consistency is maintained (rules like those for the derivative and shift have signs changed): since all $k$ are covered, and $k \mapsto -k$ is a bijection, you can replace $k$ with $-k$ without affecting the content of the theory, although as mentioned, lots of the rules you are used to have to be modified to be consistent in terms of the actual numbers that appear in formulae.

Indeed, there are other conventions that scale $k$, normally using $e^{-2\pi i k}$, or change the constants in front of the transform. One has a general Fourier transform dependent on two parameters, $$ \mathcal{F}_{a,b}(f)(k) = \sqrt{\frac{\lvert b \rvert}{(2\pi)^{1-a}}} \int_{-\infty}^{\infty} f(x) e^{ibkx} \, dx, $$ with inverse $$ \mathcal{F}_{a,b}^{-1}(g)(k) = \sqrt{\frac{\lvert b \rvert}{(2\pi)^{1+a}}} \int_{-\infty}^{\infty} g(k) e^{-ibkx} \, dk, $$ where $(a,b)$ has different values depending on the convention in the field: Mathematica's documentation lists $(0,1)$ (modern physics), $(1,-1)$ (pure mathematics; systems engineering), $(-1,1)$ (classical physics), and $(0,-2\pi)$ (signal processing), for example. I like $(0,-2\pi)$ myself: keeps the $2\pi$s to a minimum in things like Parseval. But there is no universal convention, so it is normally defined when introduced, even in papers.