6Fourier transforms
IB Methods
6.1 The Fourier transform
So far, we have considered writing a function
f
:
S
1
→ C
(or a periodic function)
as a Fourier sum
f(θ) =
X
n∈Z
ˆ
f
n
e
inθ
,
ˆ
f
n
=
1
2π
Z
π
−π
e
−inθ
f(θ) dθ.
What if we don’t have a periodic function, but just an arbitrary function
f : R → C? Can we still do Fourier analysis? Yes!
To start with, instead of
ˆ
f
n
=
1
2π
R
π
−π
e
−inθ
f
(
θ
) d
θ
, we define the Fourier
transform as follows:
Definition (Fourier transform). The Fourier transform of an (absolutely inte
grable) function f : R → C is defined as
˜
f(k) =
Z
∞
−∞
e
−ikx
f(x) dx
for all k ∈ R. We will also write
˜
f(k) = F[f(x)].
Note that for any k, we have

˜
f(k) =
Z
∞
−∞
e
−ikx
f(x) dx
≤
Z
∞
−∞
e
−ikx
f(x) dx =
Z
∞
−∞
f(x) dx.
Since we have assumed that our function is absolutely integrable, this is finite,
and the definition makes sense.
We can look at a few basic properties of the Fourier transform.
(i)
Linearity: if
f, g
:
R → C
are absolutely integrable and
c
1
, c
2
are constants,
then
F[c
1
f(x) + c
2
g(x)] = c
1
F[f(x)] + c
2
F[g(x)].
This follows directly from the linearity of the integral.
(ii) Translation:
F[f(x − a)] =
Z
R
e
−ikx
f(x − a) dx
Let y = x − a. Then we have
=
Z
R
e
−ik(y+a)
f(y) dy
= e
−ika
Z
R
e
−iky
f(y) dy
= e
−ika
F[f(x)],
So a translation in xspace becomes a rephasing in kspace.
(iii) Rephasing:
F[e
−i`x
f(x)] =
Z
∞
−∞
e
−ikx
e
−i`x
f(x) dx
=
Z
∞
−∞
e
−i(k+`)x
f(x) dx
=
˜
f(k + `),
where ` ∈ R and
˜
f(x) = F[f(x)].
(iv) Scaling:
F[f(cx)] =
Z
∞
−∞
e
−ikx
f(cx) dx
=
Z
∞
−∞
e
−iky/c
f(y)
dy
c
=
1
c
˜
f
k
c
.
Note that we have to take the absolute value of
c
, since if we replace
x
by
y/c
and
c
is negative, then we will flip the bounds of the integral. The
extra minus sign then turns c into −c = c.
(v) Convolutions: for functions f, g : R → C, we define the convolution as
f ∗ g(x) =
Z
∞
−∞
f(x − y)g(y) dy.
We then have
F[f ∗ g(x)] =
Z
∞
−∞
e
−ikx
Z
∞
−∞
f(x − y)g(y) dy
dx
=
Z
R
2
e
ik(x−y)
f(x − y)e
−iky
g(y) dy dx
=
Z
R
e
−iku
f(u) du
Z
R
e
−iky
g(y) dy
= F[f]F[g],
where
u
=
x − y
. So the Fourier transform of a convolution is the product
of individual Fourier transforms.
(vi)
The most useful property of the Fourier transform is that it “turns differ
entiation into multiplication”. Integrating by parts, we have
F[f
0
(x)] =
Z
∞
−∞
e
−ikx
df
dx
dx
=
Z
∞
−∞
−
d
dx
(e
−ikx
)f(x) dx
Note that we don’t have any boundary terms since for the function to be
absolutely integrable, it has to decay to zero as we go to infinity. So we
have
= ik
Z
∞
−∞
e
−ikx
f(x) dx
= ikF[f(x)]
Conversely,
F[xf(x)] =
Z
∞
−∞
e
−ikx
xf(x) dx = i
d
dk
Z
∞
−∞
e
−ikx
f(x) dx = i
˜
f
0
(k).
This are useful results, since if we have a complicated function to find the Fourier
transform of, we can use these to break it apart into smaller pieces and hopefully
make our lives easier.
Suppose we have a differential equation
L(∂)y = f,
where
L(∂) =
p
X
r=0
c
r
d
r
dx
r
is a differential operator of pth order with constant coefficients.
Taking the Fourier transform of both sides of the equation, we find
F[L(∂)y] = F[f(x)] =
˜
f(k).
The interesting part is the left hand side, since the Fourier transform turns
differentiation into multiplication. The left hand side becomes
c
0
˜y(k) + c
1
ik˜y(k) + c
2
(ik)
2
˜y(k) + ··· + c
p
(ik)
p
˜y(k) = L(ik)˜y(k).
Here
L
(
ik
) is a polynomial in
ik
. Thus taking the Fourier transform has changed
our ordinary differential equation into the algebraic equation
L(ik)˜y(k) =
˜
f(k).
Since L(ik) is just multiplication by a polynomial, we can immediately get
˜y(k) =
˜
f(k)
L(ik)
.
Example. For φ : R
n
→ C, suppose we have the equation
∇
2
φ − m
2
φ = ρ(x),
where
∇
2
=
n
X
i=1
d
2
dx
2
i
is the Laplacian on R
n
.
We define the n dimensional Fourier transform by
˜
φ(k) =
Z
R
n
e
−ik·x
f(x) dx = F[φ(x)]
where now k ∈ R
n
is an ndimensional vector. So we get
F[∇
2
φ − m
2
φ] = F[ρ] = ˜ρ(k).
The first term is
Z
R
n
e
−ik·x
∇ · ∇φ d
n
x = −
Z
R
n
∇(e
−ik·x
) · ∇φ d
n
x
=
Z
R
n
∇
2
(e
−ik·x
)φ d
n
x
= −k · k
Z
R
n
e
−ik·x
φ(x) d
n
x
= −k · k
˜
φ(k).
So our equation becomes
−k · k
˜
φ(k) − m
2
˜
φ(k) = ˜ρ(k).
So we get
˜
φ(k) = −
˜ρ(k)
k
2
+ m
2
.
So differential equations are trivial in
k
space. The problem is, of course,
that we want our solution in
x
space, not
k
space. So we need to find a way to
restore our function back to x space.