Dirac Delta Function

Consider what happens to the previously mentioned real-space Gaussian, and its Fourier transform, in the limit $\sigma_x\rightarrow \infty$, or, equivalently, $\sigma_k\rightarrow 0$. There is no difficulty in seeing, from Equation (8.15), that

$$F(x)\rightarrow 1.$$ (8.21)

In other words, the real-space Gaussian morphs into a function that takes the constant value unity everywhere. The Fourier transform is more problematic. In the limit $\sigma_k\rightarrow 0$, Equation (8.16) yields a $k$-space function that is zero everywhere apart from $k=0$ (because the function is negligible for $\vert k\vert\gtrsim \sigma_k$), where it is infinite [because the function takes the value $(2\pi\,\sigma_k)^{-1/2}$ at $k=0$]. Moreover, according to Equation (8.20), the integral of the function over all $k$ remains unity. Thus, the Fourier transform of the uniform function $F(x)=1$ is a sort of integrable “spike” located at $k=0$. This unusual function is known as the Dirac delta function, and is denoted $\delta (k)$. Thus, one definition of a delta function is

$$\delta (k)= \lim_{\sigma_k\rightarrow 0}\frac{1}{(2\pi\,\sigma_k^{\,2})^{1/2}}\,\exp\left(-\frac{k^{\,2}}{2\,\sigma_k^{\,2}}\right).$$ (8.22)

As has already been mentioned, $\delta(k)=0$ for $k\neq 0$, and $\delta(0)=\infty$. Moreover,

$$\int_{-\infty}^\infty\delta (k)\,dk = 1.$$ (8.23)

Consider the integral

$$\int_{-\infty}^\infty F(k)\,\delta(k)\,dk,$$ (8.24)

where $F(k)$ is an arbitrary function. Because of the peculiar properties of the delta function, the only contribution to the previous integral comes from the region in $k$-space in the immediate vicinity of $k=0$. Furthermore, provided $F(k)$ is well behaved in this region, we can write

$$\int_{-\infty}^{\infty} F(k)\,\delta(k)\,dk = \int_{-\infty}^\infty F(0)\,\delta (k)\,dk=F(0)\,\int_{-\infty}^{\infty} \delta(k)\,dk = F(0),$$ (8.25)

where use has been made of Equation (8.23).

A change of variables allows us to define $\delta(k-k')$, which is a “spike” function centered on $k=k'$. The previous result can be generalized to give

$$\int_{-\infty}^{\infty} F(k)\,\delta(k-k')\,dk = F(k'),$$ (8.26)

for all $F(k)$ that are well behaved at $k=k'$. Indeed, this expression can be thought of as an alternative definition of a delta function. Incidentally, a Dirac delta function is sometimes referred to as a generalized function. This nomenclature arises because the actual value of the function is ill defined (for instance, it would be impossible to graph the function), whereas its integral is perfectly well defined (as evidenced by the previous equation).

We have seen that the delta function $\delta (k)$ is the cosine Fourier transform of the uniform function $F(x)=1$. It, thus, follows from Equation (8.11) that

$$\delta (k) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \cos(k\,x)\,dx.$$ (8.27)

This result represents yet another definition of the delta function. By symmetry, we also have

$$0 = \frac{1}{2\pi}\int_{-\infty}^{\infty} \sin(k\,x)\,dx.$$ (8.28)

It follows that

$$\frac{1}{2\pi}\int_{-\infty}^\infty \cos(k\,x)\,\cos(k'\,x)\,dx =... ...}^\infty\left\{\cos\left[(k-k')\,x\right]+\cos\left[(k+k')\,x\right]\right\}dx,$$ (8.29)

which yields

$$\frac{1}{2\pi}\int_{-\infty}^\infty \cos(k\,x)\,\cos(k'\,x)\,dx= \frac{1}{2}\left[\delta(k-k') + \delta(k+k')\right],$$ (8.30)

where use has been made of Equation (8.27), and a standard trigonometric identity. (See Appendix B.) Likewise,

$$\frac{1}{2\pi}\int_{-\infty}^\infty \sin(k\,x)\,\sin(k'\,x)\,dx = \frac{1}{2}\left[\delta(k-k') - \delta(k+k')\right],$$ (8.31)

$$\frac{1}{2\pi}\int_{-\infty}^\infty \cos(k\,x)\,\sin(k'\,x)\,dx =0.$$ (8.32)

(See Exercise 4.) Incidentally, Equations (8.30)–(8.32) can be used to derive Equations (8.11) and (8.12) directly from Equation (8.10). (See Exercise 5.)