Fourier Series, Fourier Transforms and the Delta Function

Transcription

1 Fourier Series, Fourier Trasforms ad the Delta Fuctio Michael Fowler, UVa 9/4/06 Itroductio We begi with a brief review of Fourier series Ay periodic fuctio of iterest i physics ca be expressed as a series i sies ad cosies we have already see that the quatum wave fuctio of a particle i a box is precisely of this form The importat questio i practice is, for a arbitrary wave fuctio, how good a approximatio is give if we stop summig the series after terms We establish here that the sum after terms, f ( θ, ca be writte as a covolutio of the origial fuctio with the fuctio that is, ( ( ( x = / si( + x / si x, f ( θ = ( θ θ f( θ dθ f θ, gives a good ituitive guide to how good a approximatio the sum over terms is goig to be f θ I particular, it turs out that step discotiuities are ever hadled The structure of the fuctio ( x (plotted below, whe put together with the fuctio ( for a give fuctio ( perfectly, o matter how may terms are icluded Fortuately, true step discotiuities ever occur i physics, but this is a warig that it is of course ecessary to sum up to some where the sies ad cosies oscillate substatially more rapidly tha ay sudde chage i the fuctio beig represeted We go o to the Fourier trasform, i which a fuctio o the ifiite lie is expressed as a ikx itegral over a cotiuum of sies ad cosies (or equivaletly expoetials e It turs out that argumets aalogous to those that led to x x such that ( ow give a fuctio ( f ( x = ( x x f( x dx Cofroted with this, oe might well woder what is the poit of a fuctio ( x which o covolutio with f(x gives back the same fuctio f(x The relevace of ( x will become evidet later i the course, whe states of a quatum particle are represeted by wave fuctios o the ifiite lie, like f(x, ad operatios o them ivolve itegral operators similar to the covolutio above Workig with operatios o these fuctios is the cotiuum geeralizatio x is the ifiite-dimesioal of matrices actig o vectors i a fiite-dimesioal space, ad ( represetatio of the uit matrix Just as i matrix algebra the eigestates of the uit matrix are a set of vectors that spa the space, ad the uit matrix elemets determie the set of dot products of these basis vectors, the delta fuctio determies the geeralized ier product of a cotiuum

2 basis of states It plays a essetial role i the stadard formalism for cotiuum states, ad you eed to be familiar with it! Fourier Series Ay reasoably smooth real fuctio f ( θ defied i the iterval < θ ca be expaded i a Fourier series, A0 f ( θ = + ( A cos θ + B si θ = where the coefficiets ca be foud usig the orthogoality coditio, ad the same coditio for the si θ 's to give: cos mθ cos θdθ = m, ( θ cos θ d A = f θdθ B = f( si θ θ ote that for a eve fuctio, oly the A are ozero, for a odd fuctio oly the B are ozero How Smooth is Reasoably Smooth? The umber of terms of the series ecessary to give a good approximatio to a fuctio depeds o how rapidly the fuctio chages To get a idea of what goes wrog whe a fuctio is ot smooth, it is istructive to fid the Fourier sie series for the step fuctio f ( f ( θ = for < θ 0, θ = for 0 < θ Usig the expressio for B above it is easy to fid: 4 si 3θ si 5θ f ( θ = si θ Takig the first half doze terms i the series gives:

3 As we iclude more ad more terms, the fuctio becomes smoother but, surprisigly, the iitial overshoot at the step stays at a fiite fractio of the step height However, the fuctio recovers more ad more rapidly, that is to say, the overshoot ad rigig at the step take up less ad less space This overshoot is called Gibbs pheomeo, ad oly occurs i fuctios with discotiuities How the Sum over Terms is Related to the Complete Fuctio To get a clearer idea of how a Fourier series coverges to the fuctio it represets, it is useful to f θ, teds towards stop the series at terms ad examie how that sum, which we deote ( f ( θ So, substitutig the values of the coefficiets i the series ( θ cos ( θ si d A = f θdθ B = f θ θ A 0 = + + = f ( θ ( A cos θ B si θ gives

4 4 f ( θ = f( θ dθ (cosθ cosθ siθsi θ f( θ dθ + + = = f( θ dθ cos ( θ θ f( θ dθ + We ca ow use the trigoometric idetity = si( + x cos x = = si x to fid where si( + ( θ θ f ( θ = f( θ dθ ( θ θ f( θ dθ = si ( θ θ si( + x ( x = si x (ote that provig the trigoometric idetity is straightforward: write cos x = z + z, ad sum the geometric progressios ( z ix = e, so Goig backwards for a momet ad writig ( x = cos x + = it is easy to check that ( xdx = To help visualize ( θ, here is = 0:

5 5 We have just established that the total area uder the curve =, ad it is clear from the diagram that almost all this area is uder the cetral peak, sice the areas away from the ceter are almost equally positive ad egative The width of the cetral peak is / ( +, its height ( + / Exercise: for large, approximately how far dow does it dip o the first oscillatio? ( / For fuctios varyig slowly compared with the oscillatios, the covolutio itegral f ( θ = ( θ θ f( θ dθ will give f ( θ close to f ( θ, ad for these fuctios f ( θ will ted to ( f θ as icreases It is also clear why covolutig this curve with a step fuctio gives a overshoot ad oscillatios Suppose the fuctio f ( θ is a step, jumpig from 0 to at θ = 0 From the covolutioary form of the itegral, you should be able to covice yourself that the value of f ( θ at a poit θ is the total area uder the curve ( θ to the left of that poit (area below zero that is, below the x-axis of course coutig egative For θ = 0, this must be exactly f θ at 05 (sice all the area uder ( θ adds to But if we wat the value of (

6 θ = /( 6 + (that is, the first poit to the right of the origi where the curve cuts through the x-axis, we must add all the area to the left of θ /( = +, which actually adds up to a total area greater tha oe, sice the leftover area to the right of that poit is overall egative That gives the overshoot A Fourier Series i Quatum Mechaics: Electro i a Box The time-idepedet Schrödiger wave fuctios for a electro i a box (here a oedimesioal square well with ifiite walls are just the sie ad cosie series determied by the boudary coditios Therefore, ay reasoably smooth iitial wavefuctio describig the electro ca be represeted as a Fourier series The time developmet ca the be foud be multiplyig each term i the series by the appropriate time-depedet phase factor Importat Exercise: prove that for a fuctio ( i f θ = ae θ, with the a i geeral complex, = ( f θ dθ = a ( The physical relevace of this result is as follows: for a electro cofied to the circumferece of a rig of uit radius, θ is the positio of the electro A orthoormal basis of states of the i / e θ with a iteger, a correctly ormalized electro o this rig is the set of fuctios ( superpositio of these states must have = a = =, so that the total probability of fidig the electro i some state is uity But this must also mea that the total probability of fidig the electro aywhere o the rig is uity ad that s the left-had side of the above equatio the 'scacel Expoetial Fourier Series I the previous lecture, we discussed briefly how a Gaussia wave packet i x-space could be represeted as a cotiuous liear superpositio of plae waves that tured out to be aother Gaussia wave packet, this time i k-space The pla here is to demostrate how we ca arrive at that represetatio by carefully takig the limit of the well-defied Fourier series, goig from the fiite iterval (, to the whole lie, ad to outlie some of the mathematical problems that arise, ad how to hadle them The first step is a trivial oe: we eed to geeralize from real fuctios to complex fuctios, to iclude wave fuctios havig ovaishig curret A smooth complex fuctio ca be writte i a Fourier series simply by allowig A ad B to be complex, but i this case a more atural iθ iθ expasio would be i powers of e, e We write:

7 7 ad retracig the above steps iθ iθ f ( θ = ae with a = e f( θ dθ = i( θ θ ( = ( = f θ e f θ dθ = f ( θ dθ cos ( θ θ f( θ dθ + = exactly the same expressio as before, therefore givig the same ( i i i i because usig cos ( e θ e θ, si i ( e θ θ θ = + θ = e simply be rearraged to a sum over i e θ terms for θ This is t surprisig, the first terms i A, B ca Electro out of the Box: the Fourier Trasform To break dow a wave packet ito its plae wave compoets, we eed to exted the rage of itegratio from the (, used above to (, We do this by first rescalig from (, to ( /, / ad the takig the limit Scalig the iterval from to (i the complex represetatio gives: / f ( x = a e where a = f x e ix / ix / ( dx = the sum i beig over all itegers This is a expressio for f (x i terms of plae waves e ikx where the allowed k s are /, with = 0, ±, ±, Retracig the steps above i the derivatio of the fuctio ( x, we fid the equivalet fuctio to be ( x / (( + x si ( / x si / = cos + = = x Studyig the expressio o the right, it is evidet that provided is much greater tha, this has x we cosidered earlier But we are the same peaked-at-the-origi behavior as the ( iterested i the limit, ad there for fixed this fuctio ( x is low ad flat Therefore, we must take the limit goig to ifiity before takig goig to ifiity

8 8 This is what we do i the rest of this sectio Provided is fiite, we still have a Fourier series, represetig a fuctio of period Our mai iterest i takig ifiite is that we would like to represet a operiodic fuctio, for example a localized wave packet, i terms of plae-wave compoets Suppose we have such a wave packet, say of legth, by which we mea the wave is exactly zero outside a stretch of the axis of legth Why ot just express it i terms of a ifiite Fourier series based o some large iterval ( /,/ provided the wave packet legth is completely iside this iterval? The poit is that such a aalysis would ideed accurately reproduce the wave packet iside the iterval, but the same sum of plae waves evaluated over all the x-axis would reveal a ifiite strig of idetical wave packets apart! This is ot what we wat As a prelimiary to takig to ifiity, let us write the expoetial plae wave terms i the stadard k-otatio, ix / ikx e = e So we are summig over a (ifiite set of plae waves havig wave umber values k = /, = 0, ±, ±,, a set of equally-spaced k s with separatio Δ k = / Cosider ow what happes if we double the basic iterval from ( /, / to (, The ew allowed k values are k = /, = 0, ±, ±,, so the separatio is ow Δ k = /, half of what it was before It is evidet that as we icrease, the spacig betwee successive k values gets less ad less Goig back to the iterval of legth, writig a = a( k, k = / we have ix / ( f ( x = a ik x e = a k e = Recall that the Riema itegral ca be defied by = with k = Δ k, = 0, ±, ±, ( lim ( f kdk= f k Δk Δk o The expressio o the right-had side of the equatio for f(x has the same form as the right-had side of the Riema itegral defiitio, ad here Δ k = / That is to say,

9 9 f ( x = a k e = a k e Δ k = a( k e ikx ikx ikx ( ( dk = = i the limit Δk 0, or equivaletly We are of course assumig here that the fuctio a( k, which we have oly defied (for a give o the set of poits k, teds to a cotiuous fuctio a( k i the limit It follows that i the ifiite limit, we have the Fourier trasform equatios: / ikx ix/ ikx ( / f ( x = a( k e dk ad a( k = lima = lim f x e dx = f ( x e dx Dirac s Delta Fuctio ow we have take both ad to ifiity, what has happeed to our fuctio ( x Remember that our procedure for fidig f ( θ i terms of f ( θ gave the equatio? ad from this we foud ( θ f ( θ = f( θ dθ cos ( θ θ f( θ dθ + = Followig the same formal procedure with the ( = Fourier trasforms, we are forced to take ifiite (recall the procedure oly made sese if was take to ifiity before, so i place f i terms of f, f x i terms of itself! et s of a equatio for ( θ ( θ we get a equatio for ( write it dow first ad thik afterwards: f x = f x e dx e dk e f x dx = ikx ikx dk ik ( x x ( ( ( ( I other words, where f ( x = ( x x f( x dx + dk ikx ( x = e This is the Dirac delta fuctio This had-wavig approach has give a result which is ot clearly defied This itegral over x is liearly diverget at the origi, ad has fiite oscillatory behavior everywhere else To make ay progress, we must provide some form of cutoff i k-

10 0 space, the perhaps we ca fid a meaigful limit by placig the cutoff further ad further away by first takig to ifiity, the That is to say, si (( + x/ ( x = lim ( lim ( x = lim lim si ( x/ From our argumets above, we should be able to recover ( x as a limit of ( x A way to uderstad this limit is to write ( M = + / ad let M go to ifiity before (This meas as we take large o its way to ifiity, we re takig far larger! So the umerator is just si Mx I the limit of ifiite, for ay fiite x the deomiator is just x, sice siθ = θ i the limit of small θ From this, ( x si Mx = lim M x This is still a rather pathological fuctio, i that it is oscillatig more ad more quickly as the ifiite limit is take This comes about from the abrupt cutoff i the sum at the frequecy To see how this relates to the (also ill-defied ikx ( x = ( dk / e, recall ( x from the series ( x ( + si + x/ x = = cos si x/ + = came Expressig the cosie i terms of expoetials, the replacig the sum by a itegral i the large limit, i the same way we did earlier, writig k = /, so the iterval betwee successive k 's is k /, Δ = so ( ( ( f k dk / f k : / ix/ dk ikx ( x e e = / ( x si / = = x

11 So it is clear that we re defiig the ( x abruptly cut off at the large values / ( ikx as a limit of the itegral dk / e, which is ( / ± / I fact, this is ot very physical: a much more realistic sceario for a real wave packet would be a gradual dimiutio i cotributios from high frequecy (or short wavelegth modes that is to say, a getle cutoff i the itegral over k that was used to replace the sum over For example, a reasoable cutoff procedure would be to multiply the itegrad by exp Δ k, the take the limit of small Δ ( Therefore a more reasoable defiitio of the delta fuctio, from a physicist s poit of view, would be + dk ikx Δ k x /4Δ / Δ 0 Δ 0 Δ 0 ( x = im e e = im e = im ( x, say ( 4Δ Δ That is to say, the delta fuctio ca be defied as the arrow limit of a Gaussia wave packet Δ x has o oscillatig sidebads, thaks to our with total area Ulike the fuctio ( θ, ( smoothig out of the upper k-space cutoff, so step discotiuities do ot geerate Gibbs pheomeo overshoot istead, a step will be smoothed out over a distace of order Δ Properties of the Delta Fuctio It is straightforward to verify the followig properties from the defiitio as a limit of a Gaussia wavepacket: ( x dx =, ( x = 0 for x 0 ( x = ( x, ( ax = ( x, a ( a x ( x b dx = ( a b Yet Aother Defiitio, ad a Coectio with the Pricipal Value Itegral There is o uique way to defie the delta fuctio, ad other cutoff procedures ca give useful isights For example, the k-space itegral ca be split ito two ad simple expoetial cutoffs applied to the two halves, that is, we could take the defiitio to be Evaluatig the itegrals, 0 dk + ikx εk dk ikx εk ( x = lim e e + e e ε 0 0

12 ε ( x = lim = lim ε 0 ix ε ix ε ε 0 + x + ε It is easy to check that this fuctio is correctly ormalized by makig the chage of variable x = ε taθ ad itegratig from / to / This represetatio of the delta fuctio will prove to be useful later ote that regarded as a fuctio of a complex variable, the delta fuctio has two poles o the pure imagiary axis at z = ± iε The stadard defiitio of the pricipal value itegral is: D D ε D P f( x f( x f ( x dx= lim dx+ dx x ε 0 x x D ε It is ot difficult to see that for a cotiuous differetiable fuctio f(x this is equivalet to D D D ( P lim ( x f x dx= f x dx x x + ε ε 0 D Therefore the pricipal value operator ca be writte symbolically: P x = lim = lim 0 + x ε x + ε ε 0 x+ iε x iε Puttig this together with the similar represetatio of the delta fuctio above, ad takig the limit of ε 0 to be uderstood, we have the useful result: Importat Exercises: Prove Parseval s Theorem: P i ( x x iε = ± x dk ikx dk If f ( x = a( k e, the f ( x dx a( k = Prove the rule for the Fourier Trasform of a covolutio of two fuctios: dk ikx dk ik x If f ( x = a( k e, g( x = b( k e, dk ikx the f ( x x g( x dx = a( k b( k e

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