6Fourier transforms

IB Methods

6.2 The Fourier inversion theorem

We need to be able to express our original function

f

(

x

) in terms of its Fourier

transform

˜

f

(

k

). Recall that in the periodic case where

f

(

x

) =

f

(

x

+

L

), we have

f(x) =

X

n∈Z

ˆ

f

n

e

2inxπ/L

.

We can try to obtain a similar expression for a non-periodic function

f

:

R → C

by taking the limit L → ∞. Recall that our coefficients are defined by

ˆ

f

n

=

1

L

Z

L/2

−L/2

e

−2inπx/L

f(x) dx.

Now define

∆k =

2π

L

.

So we have

f(x) =

X

n∈Z

e

inx∆k

∆k

2π

Z

L/2

−L/2

e

−inx∆k

f(u) du

As we take the limit L → ∞, we can approximate

Z

L/2

−L/2

e

−ixn∆k

f(x) dx =

Z

∞

−∞

e

−ix(n∆k)

f(x) dx =

˜

f(n∆k).

Then for a non-periodic function, we have

f(x) = lim

∆k→∞

X

n∈Z

∆k

2π

e

in∆kx

˜

f(n∆k) =

1

2π

Z

∞

−∞

e

ikx

˜

f(k) dk.

So

f(x) = F

−1

[f(k)] =

1

2π

Z

∞

−∞

e

ikx

˜

f(k) dk.

This is a really dodgy argument. We first took the limit as

L → ∞

to turn

the integral from

R

L

−L

to

R

∞

−∞

. Then we take the limit as ∆

k →

0. However,

we can’t really do this, since ∆

k

is defined to be 2

π/L

, and both limits have

to be taken at the same time. So we should really just take this as a heuristic

argument for why this works, instead of a formal proof.

Nevertheless, note that the inverse Fourier transform looks very similar to

the Fourier transform itself. The only differences are a factor of

1

2π

, and the sign

of the exponent. We can get rid of the first difference by evenly distributing the

factor among the Fourier transform and inverse Fourier transform. For example,

we can instead define

F[f(x)] =

1

√

2π

Z

∞

−∞

e

−ikx

f(x) dx.

Then F

−1

looks just like F apart from having the wrong sign.

However, we will stick to our original definition so that we don’t have to deal

with messy square roots. Hence we have the duality

F[f(x)] = F

−1

[f(−x)](2π).

This is useful because it means we can use our knowledge of Fourier transforms

to compute inverse Fourier transform.

Note that this does not occur in the case of the Fourier series. In the Fourier

series, we obtain the coefficients by evaluating an integral, and restore the

original function by taking a discrete sum. These operations are not symmetric.

Example. Recall that we defined the convolution as

f ∗ g(x) =

Z

∞

−∞

f(x − y)g(y) dy.

We then computed

F[f ∗ g(x)] =

˜

f(k)

˜

f(g).

It follows by definition of the inverse that

F

−1

[

˜

f(k)˜g(k)] = f ∗ g(x).

Therefore, we know that

F[f(x)g(x)] = 2π

Z

∞

−∞

˜

f(k −`)˜g(`) d` = 2π

˜

f ∗ ˜g(k).

Example. Last time we saw that if ∇

2

φ − m

2

φ = ρ(x), then

˜

φ(k) = −

˜ρ(k)

|k|

2

+ m

2

.

Hence we can retrieve our φ as

φ(x) = F

−1

[

˜

φ(k)] = −

1

(2π)

n

Z

R

n

e

ik·x

˜ρ(k)

|k|

2

+ m

2

d

n

x.

Note that we have (2

π

)

n

instead of 2

π

since we get a factor of 2

π

for each

dimension (and the negative sign was just brought down from the original

expression).

Since we are taking the inverse Fourier transform of a product, we know that

the result is just a convolution of F

−1

[˜ρ(k)] = ρ(x) and F

−1

1

|k|

2

+m

2

.

Recall that when we worked with Green’s function, if we have a forcing term,

then the final solution is the convolution of our Green’s function with the forcing

term. Here we get a similar expression in higher dimensions.