Why does each delta function come with a 2π and each 'd' in integration come with a 1/2π?

Question

I have been taking for granted in the past when encountering that. But I just found I cannot seriously justify that.

Answer 1

The position of $2\pi$ comes from conventions associated with the Fourier transform. The presence of an overall factor of $2\pi$ after applying a Fourier transform and its inverse is a basic property of the Fourier transform itself.

Typically in (particle) physics, the Fourier transform and its inverse are defined as follows:

$$f(x)=\frac{1}{2\pi}\int dk\; \tilde{f}(k)e^{-ikx}$$ $$\tilde{f}(k)=\int dx\; f(x)e^{ikx}$$

This leads to the following definition of the position-space and momentum-space delta functions:

$$\delta(x)=\frac{1}{2\pi}\int dk\; e^{-ikx}$$

$$\delta(k)=\int dx\; e^{ikx}$$

In other words, $\int dk\; e^{-ikx}=2\pi\delta(x)$, and $\int dx\; e^{ikx}=\delta(k)$. This asymmetric convention works because physicists typically know whether a function takes arguments in position space or in momentum space. There is also a "symmetric" Fourier transform convention:

$$f(x)=\frac{1}{\sqrt{2\pi}}\int dk\; \tilde{f}(k)e^{-ikx}$$

$$\tilde{f}(k)=\frac{1}{\sqrt{2\pi}}\int dx\; f(x)e^{ikx}$$

which leads to the following: $\int dk\;e^{-ikx}=\sqrt{2\pi}\delta(x)$ and $\int dx\; e^{ikx}=\sqrt{2\pi}\delta(k)$. These are two of the more common conventions in physics, but there are many others. The physics doesn't change if you use a different convention, but you will make mistakes if your conventions aren't consistent, so usually it's wisest to stick to the asymmetric convention above, since that's very likely what your textbooks and problem sets will be assuming.

Answer 2

As has been mentioned, the crucial point is $$\int e^{ikx}dk=2\pi\delta(x).$$I'll sketch a proof. Define $$f(x):=\int_0^\infty\frac{\sin y}{y}e^{-xy}dy$$so $f(\infty)=0$ and $$f'(x)=-\int_0^\infty \sin y\,e^{-xy}dy=-\frac{1}{1+x^2}.$$ Hence $f(x)=\frac{\pi}{2}-\arctan x$ and $\int_{-\infty}^\infty\frac{\sin y}{y}dy=2f(0)=\pi$. For any $\epsilon > 0$, $$\int_{-\infty}^\infty\frac{\sin (y/\epsilon)}{\pi y}dy=1.$$Taking the $\epsilon \to 0^+$ limit (rigorously, this should use measures) sends the integrand to $\infty$ for $y=0$ or $0$ for $y\ne 0$, so we have a nascent delta function. Hence $$\delta(x)=\lim_{N\to\infty}\frac{\sin Nx}{\pi x}=\lim_{N\to\infty}\frac{1}{2\pi}\int_{-N}^N e^{ikx}dk=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}dk.$$

Answer 3

I'll add another perspective to the other answers by employing the sifting property:

$$\int_{-\infty}^\infty\mathrm{d}t\,f(t)\,\delta(t-\tau) = f(\tau)$$

For the case that $\tau = 0$, we have

$$\int_{-\infty}^\infty\mathrm{d}t\,f(t)\,\delta(t) = f(0)$$

Now, consider the following double integral (and assume $f(t)$ is 'nice' enough)

$$\int_{-\infty}^\infty\mathrm{d}t\,f(t)\int_{-\infty}^\infty\mathrm{d}\omega\,e^{-i\omega t} = \int_{-\infty}^\infty\mathrm{d}\omega\,\int_{-\infty}^\infty\mathrm{d}t\,f(t)e^{-i\omega t} = \int_{-\infty}^\infty\mathrm{d}\omega\,F(\omega)$$

where $F(\omega)$ is the Fourier transform of $f(t)$. Next, recognize that

$$\int_{-\infty}^\infty\mathrm{d}\omega\,F(\omega) = 2\pi \left(\frac{1}{2\pi}\int_{-\infty}^\infty\mathrm{d}\omega\,F(\omega)e^{i\omega t}\right),\quad t = 0$$

But the term in parenthesis is just the inverse Fourier transform of $F(\omega)$ which, when evaluated at time $t=0$, is just $f(0)$, and it follows that

$$\int_{-\infty}^\infty\mathrm{d}t\,f(t)\int_{-\infty}^\infty\mathrm{d}\omega\,e^{-i\omega t} = 2\pi\,f(0) = 2\pi\,\int_{-\infty}^\infty\mathrm{d}t\,f(t)\,\delta(t)$$

And so conclude that

$$ \Rightarrow \int_{-\infty}^\infty\mathrm{d}\omega\,e^{-i\omega t} = 2\pi\,\delta(t) = \int_{-\infty}^\infty\mathrm{d}\omega\,e^{i\omega t}$$

Now, change variables using $\omega = 2\pi\,\nu$ and see that

$$\int_{-\infty}^\infty\mathrm{d}\omega\,e^{i\omega t} = \int_{-\infty}^\infty 2\pi\,\mathrm{d}\nu\,e^{i2\pi\nu t} = 2\pi\,\delta(t)$$

or

$$\int_{-\infty}^\infty\mathrm{d}\nu\,e^{i2\pi\nu t} = \delta(t)$$

so the factors of $2\pi$ in the Fourier transforms arise from a scaling in the frequency domain from ordinary frequency $\nu$ in cycles per second to angular frequency $\omega$ in radians per second (or, in the spatial case, the scaling from inverse wavelength $\frac{1}{\lambda}$ in cycles per meter to (angular) wavenumber $k$ in radians per meter)