Lecture 9 phys 404 Fourier transforms

PHYS 404

Lecture 9: Fourier Transforms

Dr. Vasileios Lempesis

Introduction-a

•  Frequently in mathematical physics we encounter pairs of functions related by an expression of the following form:

The function is called the integral

transform of by the kernel .

( ) ( ) ( , )

b

a

g α =

∫

f t K α t dt ( )

g α

f t

( )

^{K( , )}α ^t

Introduction-b

Examples of integral transforms

Integral Transforms Kernel Form

Fourier

Laplace Hankel

Mellin

eiat f t e dt( ) ^iat

+∞

−

−∞

∫

e⁻at

( )

tJ atn

t^α−1

1

( )

2 f t tJ( ) ⁿ αt dt π

+∞

−∞

∫

( ) ^at f t e dt

+∞

−

−∞

∫

( ) 1

f t t dt^α

+∞

−

−∞

∫

Introduction-c

Linearity

All these transforms are linear; that is

[

^{1 1 2 2}

]

1 1 2 2

( ) ( ) ( , )

( ) ( , ) ( ) ( , )

b

a

b b

a a

c f t c f t K t dt

c f t K t dt c f t K t dt α

α α

+ =

+

∫

∫ ∫

( ) ( , ) ( ) ( , )

b b

a a

cf t K α t dt c f t K= α t dt

∫ ∫

Introduction-c

The inverse operator

If we adopt for our linear integral

transformation the operator L, we obtain

For all the above transforms we have an inverse operator such that

( ) ( )

g α = Lf t

L⁻1

( ) 1 ( ) f t = L g a⁻

The Fourier Transform

The problem that a Fourier transform answers is the representation of a non-periodic function over the infinite range. A physical example of this is the resolution of a wave packet into sinusoidal waves.

The Fourier transform is defined by

[ ]

( ) ( ) ( ) ^{i t}

g ω F f t f t e ^ω dt

+∞

−

−∞

= =

∫

[ ] ( )

1 1

( ) ( )

2

f t F g ω g ω e di t^ω ω π

+∞

−

−∞

= =

∫

The Fourier Transform

The cosine transform

If the function is even it may be represented by the so called Fourier cosine transform.

g_c(ω⁾ = F f⎡⎣ (t)⎤⎦ = 2 f (t)cos

( )

ω^t

0 +∞

∫

^dt

[ ] ( ) ( )

1

0

( ) _c ( ) 1 _c cos

f t F g ω g ω ωt dω

π

+∞

= − =

∫

The Fourier Transform

The sine transform

If the function is odd it may be represented by the so called Fourier sine transform.

g_s(ω) = F f⎡⎣ (t)⎤⎦ = 2 f (t)sin

( )

ωt

0 +∞

∫

^dt

f (x) = F⁻¹ _⎡⎣g_s(ω ⁾_⎤⎦ = 1

π ^gs

( )

ω ^sin

( )

^ωx

0 +∞

∫

^dω

(If ) g(t − t₀)

g(t)e^iω0t g(at) g(−t)

g^{( )n} (t)

g(x)dx

−∞

t

∫

g(t)dt

−∞

∞

∫ ^{= G(0) = 0}

The Fourier Transform

In three dimensional space

When we move to three dimensional space the Fourier transform becomes:

( ) ( ) ⁱ 3

g ^k =

∫

f ^r e^{− ⋅k r}d x

( )

3 3

( ) 1 ( )

2

f x g e d ki

π

=

∫

^{k k r}⋅

The Fourier Transform

The convolution theorem-a

Some times in the probability theory we need to determine the probability density of two random independent variables f and g. This is given by the so called convolution of the functions;

f t( )^{∗ g t}( ) ^{≡ f x}( )^{g t}( ^{− x})

−∞

∞

∫ ^dt

( ) ( ) ( )^* ( )^{, 1}( )^{* 2} ( ) ³( )^{= 1}( )^{* 2} ( )^{* 3}( )

f t ∗g t = g t f t ⎡⎣ f t f t ⎤⎦∗ f t f t ⎡⎣ f t f t ⎤⎦

( ) ( ) ( ) ( ) ( ) ( ) ( )

1 * 2 3 = 1 * 2 1 * 3

f t ⎡⎣ f t + f t ⎤⎦ f t f t + f t f t

The Fourier Transform

The convolution theorem-b

Let’s assume that J(ω) and G(ω) are the

Fourier transforms of the functions j(t) and g(t). It can be proved that the Fourier inverse transform of a product of Fourier transforms is the convolution of the original functions

[

^{( ) ( )}

]

( ) ( )

F j t ∗ g t = J ω G ω ^{F−1 ⎡}_⎣^J ( ) ( )^{ω G ω ⎤}_⎦ ^{= j t( )∗ g t( )}

The Fourier Transform

The frequency convolution theorem

Let the functions , with

Fourier transforms , . We can prove the frequency convolution theorem:

Let’s assume that is the Fourier

transform of the function f . It can be proved

( )

² ₂¹

( )

²

f t dt F ω dω

π

∞ ∞

−∞ −∞

∫

=

∫

1( )

f t f₂ ( )t

1( )

F ω ^F₂ ⁽ω⁾

( ) ( )

1

1 * 2 2 1( ) ( )2

F⁻ ⎡⎣F ω F ω ⎤⎦ = π f t f t

F

( )

ω

The Fourier Transform

The momentum representation-a

In quantum mechanics the wave function, , which is a solution of Schrödinger

equation, has the following properties:

1.  is the probability of finding the particle between x and x+dx.

2. 

3. 

( )

^x

ψ

( ) ( )

* x x dx

ψ ψ

( ) ( )

* x x dx 1

ψ ψ

+∞

−∞

∫

=

( ) ( )

x ψ * x xψ x dx

+∞

−∞

=

∫

The Fourier Transform

The momentum representation-b

We want a function , that will give the same information for momentum:

1.  is the probability of finding the particle with momentum between p and p+dp.

2. 

3. 

g p

( ) ( ) ( )

g p g p dp*

( ) ( )

* 1

g p g p dp

+∞

−∞

∫

=

( ) ( )

p g p pg p dp* +∞

−∞

=

∫

The Fourier Transform

The momentum representation-c

Such a function is given by the Fourier transform of our space function ,

The corresponding three-dimensional momentum function is

( )

^x

ψ

( )

^/

( ) 1

2

g p ψ x e ipx dx π

∞

−

−∞

= _!

∫

^!

* 1 *

( )

/

( ) 2

g p ψ x eipx dx π

∞

−∞

= _!

∫

^!

(

¹

)

^3/2

( )

^{/ 3}

( ) 2

g ψ e i d x

π

∞

− ⋅

−∞

=

∫ ∫ ∫

^{r p}

p r ^!

!

Mathematical Supplement

The Dirac Delta function -a

•  The Dirac function in the one-dimensional case is defined as:

( ) ( ) ( )

( ) 0, 0

1, (0)

x x

x dx f x x dx f

δ

δ δ

∞ ∞

−∞ −∞

= ≠

= =

∫ ∫

Mathematical Supplement

The Dirac Delta function -b

•  Delta function has also the following properties:

( ) ( )

( )( ) ( ) ( )

1 1

1 2 1 2 1 2

1 ,

/

a x x x x

a

x x x x x x x x x x

δ δ

δ δ δ

− = −

⎡ ⎤

⎣ ⎦

− − = − + − −

⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

x dδ

( )

x

dx = −δ

( )

x ^{, δ'}

( )

^x

−∞

∞

∫

^{f (x)dx = − f '(0)}

F ^⎡_⎣δ

( )

t ^⎤_⎦^{=1, F(eiω0t) = 2πδ ω − ω}

(

⁰

)

The Fourier Transform

…of a periodic function

Periodic functions can also be Fourier

transformed. We can show that if g(t) is a

periodic function with period T then its Fourier transform is given by

Where the coefficients are associated with the corresponding Fourier series representation of the function

F(ω⁾ = 2π ^c_n

n=−ω

∞

∑

^{δ ω}

(

^{− nω0}

)

cⁿ

Mathematical Supplement

Tables of Fourier transforms

In the literature we can find tables with

Fourier transforms of different functions. We give here such a table which also contains

useful trigonometric formulae and integrals.

(Click here).