6Fourier transforms

IB Methods

6.5 Fourier transformation of distributions

We have

F[δ(x)] =

Z

∞

−∞

e

ikx

δ(x) dx = 1.

Hence we have

F

−1

[1] =

1

2π

Z

∞

−∞

e

ikx

dk = δ(x).

Of course, it is extremely hard to make sense of this integral. It quite obviously

diverges as a normal integral, but we can just have faith and believe this makes

sense as long as we are talking about distributions.

Similarly, from our rules of translations, we get

F[δ(x − a)] = e

−ika

and

F[e

−i`x

] = 2πδ(k − `),

Hence we get

F[cos(`x)] = F

1

2

(e

i`x

+ e

−i`x

)

=

1

2

F[e

i`x

]+

1

2

F[e

−i`x

] = π[δ(k−`)+δ(k+`)].

We see that highly localized functions in

x

-space have very spread-out behaviour

in k-space, and vice versa.