A Fourier series can be defined for any function over the interval 0  x  2L where Often easiest to treat n=0 cases separately

Compute the Fourier series of the SQUARE WAVE function f given by  2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.

change of variables: x  x' = x-  periodicity: cos ( X+n  ) = (-1) n cosX for n = 1, 3, 5,…

for n = 2, 4, 6,… change of variables: x  x' = nx IF f(x) is odd, all a n vanish!

periodicity: sin ( X±n  ) = (-1) n sinX for n = 1, 3, 5,… and vanishing for n = 2, 4, 6,… change of variables: x  x new = x old - 

for n = 1, 3, 5,… for n = 2, 4, 6,… change of variables: x  x' = nx for odd n for n = 1, 3, 5,…

1 22 x y N = 1 N= 5

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 functions and listen to an audio playing the wave forms Customize your own sound synthesizer

Fourier transforms of one another

Two waves of slightly different wavelength and frequency produce beats.  x   x x 1k1k 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” A staccato blast from a whistle cannot be formed by a single pure frequency but a composite of many frequencies close to the average (note) you recognize You can try building wave packets at

The broader the spectrum of frequencies (or wave number) …the shorter the wave packet! The narrower the spectrum of frequencies (or wave number) …the longer the wave packet!

Fourier Transforms Generalization of ordinary “Fourier expansion” or “Fourier series” Note how this pairs “canonically conjugate” variables  and t. Whose product must be dimensionless (otherwise e  i  t makes no sense!)

Conjugate variables time & frequency: t,  What about coordinate position & ???? r or x inverse distance?? wave number,  In fact through the deBroglie relation, you can write:

x0 x0 For a well-localized particle (i.e., one with a precisely known position at x = x 0 ) we could write: Dirac  -function a near discontinuous spike at x=x 0, (essentially zero everywhere except x=x 0 ) x0 x0 with such that f(x)≈ f(x 0 ), ≈constant over x  x, x+  x xx 1x1x

For a well-localized particle (i.e., one with a precisely known position at x = x 0 ) we could write: 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: Here that’s Notice that the probability of measuring any single momentum value, p, is: What’s THAT 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)(k) (x)(x) k0k0 (x)(x) (k)(k) k0k0 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

Area within  1  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   Let’s assume a wave packet tailored to be something like a Gaussian (or “Normal”) distribution A single “damped” pulse bounded tightly within a few  of its mean postion, μ.

For well-behaved (continuous) functions (bounded at infiinity) like f(x)=e -x 2 /2  2 Starting from: f(x)f(x) g'(x)g'(x)g(x)= e +ikx ikik f(x) is bounded oscillates in the complex plane over-all amplitude is damped at ±  we can integrate this “by parts”

Similarly, starting from:

And so, specifically for a normal distribution: f(x)=e  x 2 /2  2 differentiating: from the relation just derived: Let’s Fourier transform THIS statement i.e., apply:on both sides! 1  2  F'(k)e -ikx dk ~ ~ ~ e i(k-k)x dx ~ 1 2   (k – k) ~

e i(k-k)x dx ~ 1 2   (k – k) ~ selecting out k=k ~ rewriting as: 0 k 0 k dk' ' ' '

Fourier transforms of one another Gaussian distribution about the origin Now, since: we expect: Both are of the form of a Gaussian!  x    k  1/ 

 x  k  1 or giving physical interpretation to the new variable  x  p x  h