# Fourier Transformation in 4D space

In mathematical physics course, I see Fourier transformation of function f(t) as

$$\bar{f}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)\ e^{-i\omega t} dt.$$

I wanted to know the significance of the factor $$\frac{1}{\sqrt{2\pi}}$$ in front of the integral.

also,

is the following correct for the Fourier transformation of function $$f(x^{\mu})$$ in Minkowski space ?

$$\bar{f}(k^\mu) = {(some\ constant)}\int f(x^\mu)\ e^{-ik_\mu \cdot x^\mu} d^4x$$

what must be the constant in front of integral in this case?

• In the convention you're using, the coefficient is $(2\pi)^{-2}$ because the integral is four-dimensional.
– J.G.
Commented Oct 11, 2022 at 13:20
• Would Mathematics be a better home for this question? Commented Oct 11, 2022 at 13:29
• I just want to advocate for the convention $$\bar{f}(k) = \int\mathrm{d}^{d}x \exp(-2\pi i\; k\cdot x) f(x),$$ which is symmetric, i.e. $$f(x) = \int\mathrm{d}^d k \exp(2\pi i\; k\cdot x) \bar{f}(k).$$ Commented Oct 11, 2022 at 13:45
• @ɪdɪətstrəʊlə. That's fine sometimes, but at least in quantum mechanics, the exponential being of the form $e^{ikx}$ or $e^{ipx/hbar}$ is somewhat desirable so that we can directly interpret the quantity in the exponent as either $k$ (the wave vector) or $p$ (the momentum). Otherwise, we'd have to carry around factors of $2\pi$ on the physical quantities (e.g., eigenvalues of $\hat{p}$), which is annoying. Commented Oct 11, 2022 at 16:35

The basic mathematical inconvenience is that (for functions where the Fourier and Inverse Fourier Transforms exist) $$$$\int_{-\infty}^\infty d\omega e^{-i \omega T} \int_{-\infty}^\infty dt e^{i \omega t} f(t) = 2\pi f(T)$$$$ So you need to put that $$2\pi$$ somewhere in your definition of a Fourier transform, if you want the Inverse Fourier Transform of the Fourier Transform to give you back the original function.

There are two common conventions in physics for dealing with this annoying factor of $$2\pi$$.

(1) Define the FT and IFT "symmetrically" $$\begin{eqnarray} \tilde{F}(\omega) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty d t e^{i \omega t}f(t) \\ f(t) &=& \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty d \omega e^{-i\omega t} \tilde{F}(\omega) \end{eqnarray}$$

(2) Always choose to associate factors of $$2\pi$$ with the "momentum space" measure, $$d\omega$$ or $$dk$$ $$\begin{eqnarray} \tilde{F}(\omega) &=& \int_{-\infty}^\infty d t e^{i \omega t}f(t) \\ f(t) &=& \int_{-\infty}^\infty \frac{d \omega}{2\pi} e^{-i\omega t} \tilde{F}(\omega) \end{eqnarray}$$

In my experience in quantum field theory and cosmology (others may differ), option (1) is common in teaching quantum mechanics, but at a research level option (2) is almost always used.

• Indeed, the first is desirable in QM so that we can interpret the Fourier transform as a momentum space wave function, which need be normalized so that its square is a probability density function. Since we're not generally taking Fourier transforms of wave functions in QFT (instead, we have Fourier transforms of amplitudes and field operators and such), it's not quite so important to have that normalization condition. Commented Oct 11, 2022 at 16:38

The usual convention in physics is that

(some constant)=1

which requires that in the definition of the inverse FT one has to put a constant different from 1:

$$f(x^\mu) = \int \bar{f}(k_\mu)\ e^{ik_\mu \cdot x^\mu} \frac{d^4k}{(2\pi)^4}$$.