A Fourier series can be defined for any function over the interval 0 x 2L

1

0 sincos2

)(n

nn L

xnb

L

xna

axf

where dxL

xnxf

La

L

n

2

0cos)(

1

dxL

xnxf

Lb

L

n

2

0sin)(

1

Ofteneasiestto treat

n=0 casesseparately

Compute the Fourier series of the SQUARE WAVE function f given by

)(xf2,1

0,1

x

x

2

Note: f(x) is an odd function ( i.e. f(-x) = -f(x) )

so f(x) cos nx will be as well, while f(x) sin nx will be even.

dxL

xnxf

La

L

n

2

0cos)(

1)(xf

2,1

0,1

x

x

dxxfa 0cos)(1 2

00

dxdx 0cos)1(0cos11 2

0

0

dxnxdxnxan

2

0cos)1(cos1

1

dxnnxdxnx ( )coscos1

00

dxnxdxnx

00coscos

1

change of variables: x x' = x-

periodicity: cos(X+n) = (-1)ncosX

for n = 1, 3, 5,…

dxL

xnxf

La

L

n

2

0cos)(

1)(xf

2,1

0,1

x

x

00 a

dxnxan

0cos

2for n = 1, 3, 5,…

0na for n = 2, 4, 6,…

change of variables: x x' = nx

dxxn

an

n

0cos

2 0

IF f(x) is odd, all an vanish!

dxL

xnxf

Lb

L

n

2

0sin)(

1)(xf

2,1

0,1

x

x

00sin)(1 2

00 dxxfb

dxnxdxnxbn

2

0sinsin

1

dxnnxdxnx ( )sinsin1

00

periodicity: sin(X±n) = (-1)nsinX

dxnxdxnx

00sinsin

1

for n = 1, 3, 5,… and vanishing for n = 2, 4, 6,…

change of variables: x xnew = xold-

)(xf2,1

0,1

x

x

00 b

dxnxbn

0sin

2for n = 1, 3, 5,…

0nb for n = 2, 4, 6,…

change of variables: x x' = nx

dxxn

n

0sin

2

dxL

xnxf

Lb

L

n

2

0sin)(

1

dxxn

0sin

2

for odd n

nxn

40cos

2 for n = 1, 3, 5,…

)5

5sin

3

3sin

1

sin(

4)( xxx

xf

1

2x

y N = 1

N= 5

http://www.jhu.edu/~signals/fourier2/

http://www.phy.ntnu.edu.tw/java/sound/sound.html

http://mathforum.org/key/nucalc/fourier.html

http://www.falstad.com/fourier/

Leads you through a qualitative argument in building a square wave

Add terms one by one (or as many as you want) to build fourier series approximation to a selection of periodic functions

Build Fourier series approximation to assorted periodic functionsand listen to an audio playing the wave forms

Customize your own sound synthesizer

Fourier transformsof one another

Two waves of slightly different wavelength and frequency produce beats.

x

x

1k

k = 2

NOTE: The spatial distribution depends on the particular frequencies involved

Many waves of slightly different wavelength can produce “wave packets.”

Adding together many frequencies that are bunched closely together

…better yet…

integrating over a range of frequencies

forms a tightly defined, concentrated “wave packet”

http://phys.educ.ksu.edu/vqm/html/wpe.html

A staccato blast from a whistle cannot be formed by a single pure frequencybut a composite of many frequencies close to the average (note) you recognize

You can try building wave packets at

The broader the spectrum offrequencies (or wave number)

…the shorterthe wave packet!

The narrower the spectrum offrequencies (or wave number)

…the longerthe wave packet!

Fourier Transforms Generalization of ordinary “Fourier expansion” or “Fourier series”

de)(g2

1)t(f ti

tdetf2

1g tiω)()(

Note how this pairs “canonically conjugate” variables and t.

Whose product must be dimensionless (otherwise eit makes no sense!)

Conjugate variablestime & frequency: t,

f

2

2

What about

coordinate position & ???? r or x

inverse distance??wave number,

p2

In fact through the deBroglie relation, you can write:ph /

/iEte

/hchEc

hhp

/x

ixpe

x0

For a well-localized particle (i.e., one with a precisely known position at x = x0 )

we could write:

)(0

xx

Dirac -function

a near discontinuous spike at x=x0,(essentially zero everywhere except x=x0 )

x0

with

1)(

0dxxx

such that

dxxxxfdxxxxf )()()()(

000

f(x)≈ f(x0), ≈constant over xx, x+x

x

1x

For a well-localized particle (i.e., one with a precisely known position at x = x0 )

we could write: )(0

xx

In Quantum Mechanics we learn that the spatial wave function (x) can be complemented by the momentum spectrum of the state, found through the Fourier transform:

dxexp ixp /)(

2

1)(

Here that’s

//

00

2

1)(

2

1)(

pixixp edxexxp

Notice that the probability of measuring any single momentum value, p, is:

2

1

2

1)(

//2

2 00

pixpixeep

What’sTHAT mean?

The probability is CONSTANT – equal for ALL momenta! All momenta equally likely!The isolated, perfectly localized single packet must be comprised of an infinite range of momenta!

(k) (x)

11

2

kx

k0

2

2

2

kx

(x)(k)

k0

Remember:

…and, recall, even the most general whether confined by some potential OR free actually has some spatial spread within some range of boundaries!

Fourier transforms do allow an explicit “closed” analytic form for

the Dirac delta function

de2

1)t( )t(i

Area within1 68.26%1.28 80.00% 1.64 90.00%1.96 95.00%2 95.44%2.58 99.00%3 99.46%4 99.99%

-2 -1 +1 +2

2

2

2

)x(

e2

1x

Let’s assume a wave packet tailored to be something like aGaussian (or “Normal”) distribution

A single “damped”pulse bounded tightlywithin a few of its

mean postion, μ.

For well-behaved (continuous) functions (bounded at infiinity)

like f(x)=e-x2/22

dxexfkF ikx)(2

1)(

Starting from:

f(x) g'(x) g(x)= e+ikxik

dxxgx'fxgxf )()()()(

2

1

dxek

ix'fe

k

xif ikxikx )()(

2

1

f(x) is

boundedoscillates in thecomplex plane

over-all amplitude is damped at ±

we can integrate this “by parts”

dxex'fk

ikF ikx)(

2

1)(

)()(2

1kikFdxex'f ikx

Similarly, starting from:

dkekFxf ikx)(2

1)(

)()(2

1xixfdkek'F ikx

And so, specifically for a normal distribution: f(x)=ex2/22

differentiating: )()(2

xfx

xfdx

d

from the relation just derived: kdekF

ixf

dx

d xki ~)

~(

2

1)(

~

2 '

Let’s Fourier transform THIS statement

i.e., apply: dxeikx

21

on both sides!

dxei

kikF ikx 2

1)( 2 1

2 F'(k)e-ikxdk~ ~~

kdkFi ~

)~

(2 ' ei(k-k)xdx

~ 1 2

(k – k)~

kdkFi

kikF~

)~

()( 2 ' ei(k-k)xdx

~ 1 2

(k – k)~

)()( 2 kFi

kikF ' selecting out k=k

~

rewriting as: 2

)(

/)( kkF

dkkdF

0

k

0

k

dk''

''dk'

22

2

1)0(ln)(ln kFkF

2221

)0(

)( ke

F

kF 22

21

)0()(k

eFkF

2221

)0()(k

eFkF

22 2/)( xexf Fourier transforms

of one anotherGaussian distribution

about the origin

dxexfkF ikx)(2

1)(

Now, since:

dxxfF )(2

1)0(

we expect:

10 xie

22

1)0(

22 2/

dxeF x

2221

2)(k

ekF

22 2/)( xexf Both are of the form

of a Gaussian!

x k 1/

x k 1

orgiving physical interpretation to the new variable

x px h