Theoretical Physics Prof Ruiz, UNC Asheville, doctorphys on YouTube Chapter P Notes Fourier Transforms P Fourier Series with Exponentials Here is a summary of our last chapter, where we express a periodic wave as a Fourier series a + [ am cos( mx) + bm sin( mx) ] m a + π + π an cos( nx) + π bn sin( nx) Our goal is to replace the above series with an integral so we will be able to represent functions which are not periodic Think of this as including sine waves with frequencies that fall in between the harmonic frequencies But first we need to exp the interval We will achieve our goal in four steps We use the Euler relation Our Four Steps Introducing Exponentials Exping the Interval 3 Transforming to an Integral 4 Aiming for Infinity ix e cos x + i sin x Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
We can write cosines sines in terms of exponentials PP (Practice Problem) Refresh your memory of working with Euler's relation to be sure you can easily arrive at these "backward" Euler relations Using these, we can write ix ix e + e cos x sin x e e i ix ix as a f x a nx b nx ( ) + n cos( ) + n sin( ) n [ ] n c e n inx PP (Practice Problem) Show the following c ( an ibn ) for n > ( an + ibn ) for n < We can calculate the "c" coefficients from our familiar formulas: a a + π, + π an cos( nx), + π bn sin( nx) Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
The results are worked out below The equations a c a + π lead to the following c + π Our equation ( an ibn ) for n > with + π an cos( nx) + π bn sin( nx) gives + π [ cos( nx) i sin( nx) ] for n > + π inx e The equation ( an + ibn ) for n < is + π inx e All of these can be written for all n as follows + π inx e Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
Summary n c e n inx + π inx e P Exping the Interval For notational purposes, rewrite the above with a new z-variable g(z) inz g( z) e n + π inz g( z) e dz Then exp the interval by a transformation of variables z + π x + This leads us to z x π, ie, z π x π dz We arrive at π g( x) e i x n + π i x π g( x) e Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
We can write these as follows n c e n i x + i x e P3 Transforming to an Integral It is time for some "theoretical physics" magic Note that this chapter is not meant to be super mathematically rigorous Our focus here is trying to underst where the Fourier transform comes from rather than just giving it to you We would like to transform the series to an integral We note that since n are integers, then n We then write i x e n n Now we introduce a new variable, one which we intend to promote to a continuous variable Therefore, et c c( k) n k π k n n k π With the new variable k c( k) e dk The variable k has become a continuous variable we have replaced the sum with an integral The three things we did: ) replace delta k with dk, ) "rip off" the n from the Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
c introduce c(k), 3) turn the summation sign into a "snake" where we integrate over all k since our sum did that for the discrete case What about our other equation? With our new variable we have Summary: + i x e + c( k) e c( k) e dk + c( k) e Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
P4 Aiming for Infinity Now comes a "questionable" mathematical step Can we extend the limits of integration to infinity? + c( k) e Doesn't c( k) we get nonsense? But what about n c( k) c in this limit? For now, let's hope c( k ) is finite the write the following pair of equations c( k) e dk + c( k) e Well, if we substitute c( k ) in the f ( x ) integral using some x' for the integration to get the c( k ), we should get f ( x ) back again It is important to use something like x' because we have an "x" on the left side c( k) e dk Now substitute c k f x e + ' ( ) ( ') ' see what we get + ' f ( x ') e ' Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
We are going to do the k integral first Note x' k are independent of each other π + ' f ( x ') e e ' dk f x f x + ik ( x x') ( ) ( ') ' Do you see it? Do you recognize what is in the brackets? Here is a reminder below δ ( x) + π + ik ( x x') x x δ ( ') This means we have, f ( x') δ ( x x ') ' which is a true statement Everything checks out Summary The following are consistent c( k) e dk + c( k) e Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
P5 The Fourier Transform We will the following convention for defining the Fourier Transform Our convention will involve a symmetric definition, but you do not have to do this Some authors go have π as a unit The convention we will use to define Then, our two equations π c( k) F( k) c( k) e dk + c( k) e become c( k) F( k) π F( k) e π + The result is the symmetric equations below F( k) π + F( k) e π The function F( k ) is called the Fourier transform of f ( x ) f ( x ) is the inverse Fourier transform of F( k ) Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
Some authors write the Fourier transform with the following notation + I{ } F( k) e π The inverse Fourier transform is then I { F( k)} F( k) π P6 Parseval's Theorem In quantum mechanics the probability distribution is given by In terms of the Fourier transform, P( x) f *( x) F( k) π f x F k π ik ' x *( ) *( ') ' Note our careful use of two integration variables k k' since we plan on multiplying these together Remember, the k k' are integration variables similar to our summation variations we encountered earlier ik ' x P( x) F *( k ') ' F( k) π π P x F k F k dk π i( k k ') x ( ) *( ') ( ) ' Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense
et's integrate the probability distribution over all x π P( x) i( k k ') x P( x) F *( k ') F( k) dk ' Now, watch this simplification Rearrange things plan on doing the x integration first π i( k k ') x P( x) F *( k ') F( k) e dkdk ' Note that π i( k k ') x e δ ( k k ') Then P( x) F *( k ') F( k) δ ( k k ') dkdk ', which means we can do one of the remaining two integrals quickly due to the delta function But this tells us P( x) F *( k) F( k) dk f *( x) F *( k) F( k) dk This is Parseval's Theorem If f ( x ) is normalized, so is F( k ) Michael J Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3 Unported icense