Why do different Fourier transform conventions not make a diﬀerence in physical applications? In physics, we are used to at least two Fourier transform conventions. These are
$$\tilde{f}(k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{ikx}dx,\\
f(x)=\int_{-\infty}^{\infty}\tilde{f}(k)e^{-ikx}dk$$ or,
$$\tilde{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{ikx}dx,\\
f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde{f}(k)e^{-ikx}dk,$$
and we are often told that it is just a convention and it does not make a difference. But why is this so? The final result for $\tilde{f}(k)$ using them will differ by a factor of $\sqrt{2\pi}$. How can we then say that FT "conventions" don't make a difference? Please help.
 A: I think about the Fourier transform a lot, and I like to use a convention that’s different than both the conventions you mentioned. The way I think of the Fourier transform is that it should be essentially an isometry of $L^2$ (possibly up to a constant), and it should have other nice properties like turning convolution into multiplication, and intertwining differentiation and multiplication by polynomials—which makes it useful for analyzing PDE.
I think the reason the constants and different conventions don’t cause us much fuss is because any convention we pick is going to be very useful in applications via those essential properties I mentioned, and translating between conventions is very simple. To switch between conventions we only need to do easy changes of variable in the integrals that define the Fourier transform.
It’s almost like using different bases to do linear algebra—there is no canonical choice of basis for a vector space, but different choices may be more useful in different contexts, and ultimately the choice does no real harm.
