Linear Operators / Functions as Vectors

Transcription

1 ecture 4 Phys 3750 iear Operators / Fuctios as Vectors Overview ad otivatio: We first itroduce the cocept of liear operators o a vector space. We the look at some more vector-space examples, icludig a space where the vectors are fuctios. Key athematics: ore vector-space math! I. iear Operators. Defiitio ad Examples The essetial ature of a liear operator is cotaied i its ame. The operator part of the ame meas that a liear operator operates o ay vector u (i the space of iterest) ad produces aother vector v i the space. That is, if u is a vector, the v = u () is also vector. The liear part of liear operator meas that ( a u + b v) = a u b v () + is satisfied for all scalars a ad b ad all vectors u ad v i the space. iear operators come i may differet forms. The oes of iterest for ay give vector space deped upo the problem beig solved. Whe dealig with a vector space of fiite dimesio, we ca always use stadard liear-algebra otatio to represet the vectors as colum matrices of legth ad the liear operators as square matrices of size. For = 3, for example, Eq. () ca be writte as v v v 3 = u u u 3, (3) where v i, u i, ad ij are the (scalar) compoets of v, u, ad, respectively, i some particular orthoormal basis. s we shall see below, sometimes we are iterested i a vector space where the vectors are fuctios. I that case the liear operators of iterest may be liear For example, if dealig with vectors i real space, the elemets i a colum vector are ofte the scalar compoets (also kow as Cartesia coordiates) of that vector i the xˆ, ŷ, ẑ basis. D Riffe -- /8/03

2 ecture 4 Phys 3750 differetial operators. example of a liear differetial operator o a vector space of fuctios of x is d dx. I this case Eq. () looks like d dx g ( x) = f ( x), (4) where f ( x) ad g ( x) are vectors i the space ad d dx is the liear operator. B. Eigevalue Problems importat vector-space problem is the eigevalue problem. We already have some experiece with this problem as part of the process of fidig the ormal modes of the coupled oscillators. Simply stated, the eigevalue problem is this: for a give liear operator, what are the vectors u ad scalars λ such that u = λu (5) is satisfied? These vectors u ad scalars λ are obviously special to the operator: whe operated o by, these vectors oly chage by the scale factor λ. These special vectors u are kow as eigevectors ad the values of λ are kow as eigevalues. Each eigevector u has associated with it a particular eigevalue λ. For a vector space of dimesios (where we are usig stadard liear algebra otatio) the eigevalues are solutios of the characteristic equatio ( λ I ) 0, (8) det = where I is the idetity matrix. s we did whe solvig the = ad = 3 (homework) coupled oscillator problems, substitutig the eigevalues (oe at a time!) back ito Eq. (5) allows us to fid the eigevectors u. If the (fiite dimesio) vector space is complex the Eq. (8) always has solutios. ow here is the cool thig. If the operator is self-adjoit (also kow as Hermitia), which meas that its matrix elemets satisfy ji = ij, the (i) its eigevalues are real, (ii) its eigevectors spa the space, ad (iii) the eigevectors with distict eigevalues are orthogoal. Thus, if the operator is self-adjoit ad all eigevalues are distict, the those eigevectors form a orthogoal basis for the space. If the eigevalues are ot This result is kow as the fudametal theorem of algebra. D Riffe -- /8/03

3 ecture 4 Phys 3750 distict, a orthogoal basis ca still be formed from the eigevectors. (it just takes a little bit of work.) Ofte, however, the eigevalue problem of iterest is o a real vector space. I this case, if is symmetric (that is, the matrix elemets of satisfy ij = ji ), the Eq. (8) will have real solutios ad, agai, the associated eigevectors u ca be used to form a basis for the vector space. famous eigevalue problem from quatum mechaics is oe other tha the timeidepedet Schrödiger equatio H ψ = Eψ, (6) which is a eigevalue problem o a vector space of fuctios. Here the vectors are the fuctios ψ ( x, y, z) ; the operator is the differetial operator H h = V m x y z, ( x, y z) ; (7) ad the eigevalues are specific values of E. 3 This is perhaps the most importat equatio i quatum mechaics because the (ormalized) eigevectors describe the (spatial part of the) states of the system with a defiite value of eergy, ad the eigevalues E are the eergies of those states. II. The Coupled Oscillator Problem Redux et's revisit the coupled oscillator problem to see how that problem fits ito our discussio of vector spaces. We first review the associated eigevalue problem that we solved whe fidig the ormal modes, ad the we make some remarks about the iitial-value problem.. The Eigevalue Problem Recall, i that problem we started with equatios of motio (oe for each object) ( q q + q ) = 0 ~ q& ω, (9) j j j j+ ( j =, K, ), where q j () t is the time-depedet displacemet of the j th oscillator. We the looked for ormal-mode solutios 3 The fuctio V ( x y, z), is the classical potetial eergy a particle of mass m. D Riffe -3- /8/03

4 ecture 4 Phys 3750 q j iωt () t = q 0, e, (0) j where all objects oscillate at the same frequecy Ω. By assumig that the solutios had the form of Eq. (0), the coupled ordiary differetial equatios became the coupled algebraic equatios ( q q + q ) 0 ~ q + ω, () Ω = 0, j 0, j 0, j 0, j+ which we rewrote as ~ ω 0 0 ~ ω 0 0 ~ ω ~ ω 0 0 ~ ω q q q q0, 0, 0, 0,3 = Ω q q q q 0, 0, 0,3 0, () otice that this is exactly of the form of Eq. (5) (the eigevalue problem) where the vectors are row colum matrices, the liear operator is a matrix, ad the eigevalues λ are the squared frequecies Ω. s we previously discovered i solvig that problem there are eigevectors, q q q q0, 0, 0, 0,3 si si = si si π ( ) + π ( ) + π ( 3) + ( ) π + (3) ( =, K, ), ad the th eigevector has the eigevalue Ω = ~ ω si ( + ). (4) 4 π otice that the eigevalues are real, as they should be for a symmetric operator. lso, because the eigevalues are distict, the eigevectors from a orthogoal basis for the space. D Riffe -4- /8/03

5 ecture 4 Phys 3750 B. The Iitial Value Problem s part of solvig the iitial-value problem for this system, we eded up with the equatio 4 q q q q 3 ( 0) ( 0) ( 0) ( 0) = = Re ( a ) si si si si π ( ) + π ( ) + π ( 3) + π ( ) +, (5) which we eeded to solve for the coefficiets Re ( a ). et's ow place the previous solutio of Eq. (5) for the coefficiets Re ( a ) withi the cotext of the curret discussio of vector spaces. 5 s we talked about i the last lecture, if we write a vector v as a liear combiatio of orthogoal vectors u v = v = u, (6) the the coefficiets v i are give by v ( u, v) =. (7) ( u, u ) For the example at had, Eq. (5) is equivalet to Eq. (6), but i order apply Eq. (7) to fid the coefficiets Re ( a ) i Eq. (5), we eed the defiitio of the ier product of two vectors for this vector space. For ay dimesioal vector space the ier product betwee two vectors w ad v ca be writte as (See Exercise 4.5) v v w, (8) v (, v) = ( w w K w ) where w ad v are the compoets of the vectors w ad v i the same basis. otice that the elemets of the row matrix i Eq. (8) are the complex cojugates of 4 This is Eq. (0) of the ecture 0 otes. 5 ote that Eq. (5) says, at its most basic level, that the eigevectors [Eq. (3)] form a basis for the space of iitial displacemets of the objects (which ca be ay set of real umbers). D Riffe -5- /8/03

6 ecture 4 Phys 3750 the elemets of u. Of course, if we are dealig with real vector, the the complex cojugate is simply the elemet itself. ote that Eq. (8) ca be writte i more compact form as ( ) = w w, v j v j. (9) j= Usig the form of the ier product i Eq. (9), the applicatio of Eq. (7) to the coupled oscillator problem is thus 6 Re ( a ) j= = si j= ( π j) q ( 0) si + j π ( j) +. (0) III. Vectors Spaces ad Fourier Series The last vector-space example is Fourier Series. Recall, the complex Fourier-series represetatio of a fuctio f ( x) defied o the iterval to is iπ x f ( x) = c e, (a) = where the coefficiets c are give by c = f ( x) e iπx dx. (b) If you have bee payig attetio to this poit (i.e, if you are still awake), the you should be thikig "h ha! Equatio () says that we ca write the fuctio f ( x) as i x a liear combiatio of the (basis!) fuctios e π with coefficiets c. ooks like a vector space to me! d ah ha, agai! It seems that somehow Eq. (b) is the equivalet of Eq. (7), where the coefficiets are expressed i terms of ier products o this space." But likely you are ow asleep ad thikig about other thigs. But if you were awake, you would be etirely correct. et's see that this is the case. The vectors i this space are ideed fuctios o the iterval to, ad oe set of 6 Equatio (0) is Eq. (8) of the ecture 0 otes. D Riffe -6- /8/03

7 ecture 4 Phys 3750 iπx basis vectors u is ideed the set of fuctios u( x) = e, < <. So what is the ier product o this space that makes these basis vectors orthogoal? You actually saw the ier product back i ecture before you kew it was a ier product, so let me remid you. Deotig, for example, a fuctio f ( x) as the vector f, we defie the ier product ( g, f ) i this space as ( g,f ) = g ( x) f ( x)dx. () gai, ote the complex cojugate i the defiitio. lso otice the similarity of Eqs. (9) ad (). Usig Eq. (), Eq. () ca be writte i vector-space otatio as = f = c u, (3a) c ( u, f ) =. (3b) ( u, u ) astly, let's revisit the idea of a orthoormal basis withi the cotext of Fourier series. Recall, a ormalized (or uit) vector û is defied by u ˆ = ( uˆ, uˆ ) =, ad we ca ormalize ay vector u via u uˆ =. (5) ( u, u) iπx et's fid the ormalized versio of the basis fuctios u( x) = e. Calculatig ( u, u ) we have iπx iπx ( u, u ) e e dx =. (6) = We ca thus tur our orthogoal basis ito a orthoormal basis by usig the ormalized vectors iπx e u ˆ =. (7) D Riffe -7- /8/03

8 ecture 4 Phys 3750 If we ow write a vector i this space as a liear combiatio of these ormalized basis vectors, = f = c uˆ, (8a) c = ( u,f ) (8b) ˆ the the fuctioal expressio of Eq. (8) results i the Fourier series beig writte as f ( ) x = = c e iπx, (9a) c = f ( x) e iπx dx. (9b) To some, the Fourier Series writte as Eq. (9) is more appealig because it has a certai symmetry that Eq. () lacks. Exercises i j 4. Cosider the operator = o a two dimesioal vector space. Show k l u v that for ay two scalars a ad b ad ay two vectors = v = that this operator is liear, i.e., that it satisfies Eq. (). u u ad v 4. Show that for ay two scalars a ad b ad ay two fuctios f ( x) ad ( x) that the differetial operator d i is liear, i.e., that it satisfies Eq. (). dx 4.3 I solvig the = 3 coupled oscillator problem, we foud the three eigevectors to the associated eigevalue problem, which ca be writte as g, D Riffe -8- /8/03

9 ecture 4 Phys u =, u = 0, ad û of each of these vectors. = u. Fid the ormalized versios û, û, ad 4.4 Cosider the time idepedet Schrödiger equatio eigevalue problem H ψ = Eψ, where H is the operator h + kx. This is the (D) quatum m x mechaical harmoic-oscillator problem. The solutios (eigevectors ad eigevalues) of this problem ca be writte as ( = 0,,, K) ax ψ ( x) = ax e, where a = mk h ad d dx h E h ~ = + ω, where = k m (a) For = 0 (the groud state), show that ψ 0 ( x) is a solutio to H ψ = Eψ with the appropriate eigevalue. (b) For this vector space, the ier product of two vectors ψ ad φ is defied as ( φ,ψ) = ( x) ψ ( x)dx. Show that the = 0 ad = states are orthogoal. ϕ (c) Fid the orm of the = 0 state. Thus costruct the ormalized eigevector correspodig to this state. (d) Give that H ψ 0 = E0ψ 0 ad H ψ = Eψ, fid H ϕ, where ϕ = C 0ψ 0 + Cψ (Here C 0 0 ad C 0 are two costats.) Thus argue that the wave fuctio ϕ is ot a eigevector of H (for ay value of E ). 4.5 Ier product. Cosider two vectors writte i terms of some orthoormal basis, v = v u, = w m = = m w u. (a) Usig Eq. (0) of the ecture 3 otes, show that the ier product ( w, v) ca be expressed i terms of the compoets of the two vectors as ( w, v) = w v. = (b) What is the orm of the vector v expressed i terms of its compoets? m D Riffe -9- /8/03

10 ecture 4 Phys Ier Product ad Fourier Series. Cosider two fuctios expressed as their ormalized Fourier Series represetatios, f ( ) x = = c e iπx, ( ) g x = = m d e imπx (a) Startig with these expressios, show that the ier product ( g,f ) = g ( x) f ( x)dx ca be expressed i terms of the Fourier coefficiets c ad d as ( ) g,f = d c. = (b) What is the orm of f ( x) i terms of its Fourier coefficiets? otice the similarity of these results ad those of Exercise 4.5. Cool, eh?. D Riffe -0- /8/03