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.