NOTES ON HILBERT SPACE 1 DEFINITION: by Prof C-I Tn Deprtment of Physics Brown University A Hilbert spce is n inner product spce which, s metric spce, is complete We will not present n exhustive mthemticl discussion of this subject Rther, by using exmples nd nlogies, hopefully you will feel more t ese with Hilbert spce t the end of this short discussion 11 REMARKS: (1) Consider the spce of functions, {f : [, b] C}, where f is squre-integrble complex-vlued function on the rel intervl [,b], ie, f(x) 2 < (1) This spce will be denoted s L 2 One cn directly verify tht L 2 is complex vector spce, ie, f nd g L 2, then αf + βg L 2 for α nd β, etc We will return to clrify in wht sense one cn visulize function s vector (2) By introducing dot-product, f g f (x)g(x) (2) this complex vector spce becomes n inner product spce This inner product provides us with positive definite norm for ech vector, ( f = 0 if nd only if f = 0) (3) Define distnce between two functions by f f f (3) d(f,g) f g (4) This turns L 2 into metric spce We shll explin the fct tht L 2 is Hilbert spce next 2 I Hilbert Spce is Complete 21 COMMENTS: (1) In deling with rel numbers, it is nturl to first strt working with integers One then finds tht one needs rtionls, nd finlly irrtionls To formlly introduce irrtionls, 1
one cn first introduce notion of distnce, d(x, y), between two numbers x nd y, which of course cn be chosen to be the bsolute vlue x y (With notion of distnce, the spce of numbers becomes metric spce ) Now one cn tlk bout n infinite sequence of numbers, eg, {3, 31, 314, 3141, 31415, } Note tht (i) the differences between ny pirs of numbers further down the sequence become smller nd smller (This is n exmple of Cuchy sequence ) (ii) As finite decimls, every term in the sequence is rtionl number Now, s you cn guess, the limit of this sequence is π, nd it is not rtionl number (2) A metric spce is not complete if it contins Cuchy sequences whose limit points re not contined in the spce As the exmple bove shows, the spce of rtionl numbers, with the usul notion of distnce, is not complete metric spce By including ll irrtionls, the spce of ll rel numbers is complete Mthemticlly, one sttes tht the rel numbers, R, is the completion of the set of rtionl numbers (3) We ll gree irrtionl numbers exist nd they must be included in ny sensible usge of numbers However, for physics, it is lso cler tht, for given problem, it is sufficient to work with rtionls only, eg by greeing to lwys work to the twentieth decimls Another wy of stting this fct is tht one cn lwys pproximte n irrtionl number by rtionl number to ny degree of ccurcy one desires Mthemticlly, one sttes tht the set of rtionl numbers is dense in R (4) It is mthemticl fct tht the spce L 2, defined by 1, s metric spce with distnce between functions f nd g defined by f g, iscomplete Tht is, the limiting function of ny Cuchy sequence of functions in L 2 is lso in L 2 Therefore, L 2 is Hilbert spce, which will lso be denoted by H 1 [Note: The nottion H 2 will be used to denote the Hilbert spce of L 2 functions defined over (, ) ] (5) Let us clrify the sitution by drwing n nlogy with the cse of rel numbers The role of rtionls is now plyed by continuous functions on the intervl [,b] However, the condition of squre-integrbility, 1, is much less stringent thn continuity Mny functions which re not continuous nevertheless stisfy 1 It cn be shown tht these dditionl functions cn lwys be thought of s limits of Cuchy sequences of continuous functions We complete L 2 by dding to the set of continuous functions these limiting points Tht is: L 2 is the completion of the spce of continuous functions on the intervl [,b], with respect to distnce defined by d(f,g) f g (6) Just like the sitution with rel numbers, ny function f L 2 cn lwys be pproximted by continuous functions, to ny desired degree of ccurcy Mthemticlly, one sttes tht the set of continuous functions on [,b] is dense in L 2 (7) Let us end with n exmple using Fourier series Let us consider L 2 defined over the intervl [ π, π] Consider periodic function f(x) Assume tht f(x) is continuous nd differentible except t point x o [ π, π], where f(x) hs finite discontinuity Let us 2
consider the following sequence of functions, {S(x)},N = 0,1,2,, defined by S(x) = 1 2π N n N n e ınx (5) where π n = 1 2π π n e nπx f(x) (6) for ll integers n Note tht ech {S(x)} is continuous t {S(x)} where f(x) hs discontinuity For ny smll number ɛ, we cn lwys choose N lrge enough so tht f {S(x)} <ɛ forn N (7) Tht is, to the ccurcy ɛ, f cn be pproximted by continuous function, {S(x)}, lthough f(x) is not The set of {S(x)} is Cuchy sequence Therefore, the spce of continuous functions is not complete However, the spce L 2 is Indeed, it is precisely in the sense of 7 tht we understnd the sttement tht every periodic function of period 2π cn be represented by Fourier series): with n given by 6 f(x) = 1 2π N n N n e ınx (8) 3 Stndrd Complete Orthonorml Bsis for H 1 31 DEFINITION: A system of functions, {φ n (x)}, defined on [,b] is sid to be n orthonorml system if φ n(x)φ m (x) =δ n,m (9) 32 DEFINITION: A system of functions, {φ n (x)}, defined on [,b] nd belonging to L 2, is sid to be complete if there exists no functions different from zero in L 2 which is orthogonl to ll functions φ n (x) Without loss of generlity, let us now shift the intervl [,b] to [-L/2, L/2] Consider the set of bsis functions, {U n (x)}, <n<, defined by U n (x) 1 L e 2πnx L (10) 3
We hve previously pointed out tht this set forms n orthonorml set: U n(x)u m (x) = 1 L e 2π(n m)x L = δ n,m (11) For ech function F in L 2, consider the Fourier series representtion F(x) = C n U n (x) (12) C n = U n(x)f(x) (13) The fct tht ny function in L 2 cn be expnded in this bsis cn be used to show tht this set is complete By substituting 10 into 12, one finds tht the completeness cn be expressed mthemticlly s 33 COMMENTS: U n(x)u n (x )= 1 L e 2πın(x x) L = δ(x x ) (14) (1) This set of complete orthonorml bsis functions will be referred to s our stndrd bsis In terms of this stndrd bsis, every function f cn be expnded s in 12 Insted of writing out 12 explicitly, we cn lso represent the function f by column vector, with components:,c n,,c 2,C 1,C o,c 1,C 2, C n C 2 C 1 C o C 1 C 2 C n Tht is, in this bsis, we cn ssocite function with vector with infinite mny components In this representtion, one sees function s column vector, just like the sitution with finite dimensionl vector spce (15) 4
(2) To drw closer nlogy with our stndrd representtion for finite dimensionl vector spce, let s re-write the inner-product, 2, nd the norm, 4, s follows: Consider two functions, F (x) nd F (x), described by components F (x) nd F (x) respectively in this stndrd bsis It follows from 11 tht F F F (x)f (x) = f 2 f f = C nc n C n 2 (3) In this representtion, the vector nture of the spce is mnifest! Insted of representing vector by its components, one cn lso represent function, f(x), by n bstrct vector in Dirc nottion, f If we first introduce Dirc nottion for our stndrd bsis vectors: { U n }, <n<, 8 cn be written s: with f = The orthonormlity condition in Dirc nottion becomes (16) C n U n, (17) C n = U n F (18) U n U m = δ n,m (19) We will shortly show tht the completeness condition in Dirc nottion is U n U n = Î (20) You should now compre these representtions with those for finite dimensionl vector spce, nd convince yourself tht these two sets re formlly identicl (4) You should lso convince yourself tht every liner opertor on L 2, ˆM : L 2 L 2, cn be ssocited with n infinite by infinite mtrix s follows: ˆM = M m,n U m U n (21) m= Wht is the form of the mtrix {M m,n } for the momentum opertor ˆp = ı h x? 34 Coordinte Bsis for H 1 It is convenient to formlly define set of bsis vectors, { x }, lbelled by continuous index,x [, b], stisfying Dirc-delt orthonormlity condition s follows: Orthonormlity : x x δ(x x ) (22) 5
In terms of this bsis, it is now possible to express ech bstrct vector f in H 1 s Coordinte Representtion : f F (x) x (23) It follows from 22 nd 23 tht f(x) cn be recovered by dotting br vector x into Eq 23, ie, f(x) = x f (24) Tht is, we hve chieved representtion where the originl function f(x) inl 2 now ppers s components of the bstrct vector f in coordinte bsis In prticulr, the inner product of two vectors, f nd g, through 24 nd 23, is f g f (x)g(x )= f (x)g(x) (25) in greement with the originl definition, 23 Since every vector in cn be expressed by 23, they form complete bsis: Completeness : x x = Î (26) 35 COMMENTS: (1) We clim tht x o is n eigenstte of our position opertor, ˆx, ie, ˆx x o = x o x o (27) for x o [, b] Eq 27 follows if one writes down the components of x o, ie, x o = δ(x x o ) x (28) Tht is, δ(x x o ) is the coordinte wvefunction for the eigenvector of ˆx with n eigenvlue x o, ˆxδ(x x o )=x o δ(x x o ) By pplying 24 to 28, one simply reproduces the orthogonlity condition, 22 (2) In this representtion, every liner opertor cn be expressed s ˆx = ˆM = M(x, x ) x M(x, x ) x = M(x, x ) x x (29) This representtion follows by considering ˆM = Î ˆMÎ nd using 26 twice For instnce, xδ(x x ) x x, ˆp = { ı hδ(x x ) x } x x (30) Whenever M(x, x ) is proportionl to δ(x x ), the opertor ˆM will be clled locl opertor Insted of double-sum, Eq 30, one hs single-sum representtion: ˆx = x x x, ˆp = 6 { ı h } x x (31) x
For locl opertors, one often simply writes down the mtrix M(x, x ) for the opertor M, ˆx xδ(x x ) ˆp { ı hδ(x x ) x } (32) Better yet, one cn drop the δ(x x ) fctor entirely ˆx x ˆp { ı h x } (33) if it is understood tht one is deling with locl opertor! For instnce, insted of writing out ˆp f = explicitly, it is much simpler to ccept the nottion { ı hδ(x x ) x }f(x ) x (34) ˆpf(x) = ı h f(x) (35) x [Convince yourself tht the Hmiltonin opertor Ĥ is locl in coordinte representtion, nd mke sure you understnd the proper interprettion of the expression ı h (x,t) t = Ĥ(x, t) ] (3) Lstly, let s return to our stndrd bsis, {U N (x)} It is cler tht the bstrct vector, U N, is given by U N = U N (x) x U N (x) = x U N (36) With this understnding, by sndwiching 20 between x nd x, it leds to 14, which proves 20, s promised 7