Lecture 10. Infinite Dimensional Vector Spaces
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1 Lecture 10 Infinite Dimensional Vector Spaces October 6, 2010 Lecture 10
2 Infinite Dimensional Space Consider the set of continuous, single-variable functions, f(x), on the interval L/2 x L/2. If we specify these functions by giving there values at n equally-spaced, discreet points, x i, they form vectors, f, in an n-dimensional vector space. We can choose the orthonormal basis vectors to those specifying each of the discreet point. i = with i j = δ ij and i i = I The components of the vector f are then the values of the function at the discreet points f = i f i with i f = f(x i ) Lecture 10 1
3 Infinite Dimensional Space (cont.) The norm of f is given by f f = f (x i )f(x i ) = f(x i ) 2 As n increases, the norm of f increases essentially proportionally to n. For the continuous case where n the norm goes to infinity. This isn t good. Let s redefine the components of f so that the norm remains finite and essentially independent of n. f = i f i now with i f = f(x i ) L where L = L n Since L varies inversely with n, the norm is essential independent of n f f = f (x i )f(x i ) L = f(x i ) 2 L Lecture 10 2
4 Infinite Dimensional Space (cont.) Let s now redefine the basis vectors. x i = 1 L i such that x i f = f(x i ) Note that these basis vectors are still orthoganal x i x j = 0 for i j but they are not normalized to unity. In fact x i x i as n The completeness relation in term of this basis is then x i x i L = I f = f f = x i f x i L = f x i x i f L = f(x i ) x i L f (x i )f(x i ) L Lecture 10 3
5 Infinite Dimension Space (cont.) We re now ready to let n and go to the continuous case. The sum over n will go an integral over x and L will go to dx L/2 L/2 x x dx = I f = L/2 x f x dx = L/2 f(x ) x dx L/2 L/2 f f = L/2 f x x f dx = L/2 f (x )f(x )dx L/2 L/2 f g = L/2 f x x g dx = L/2 f (x )g(x )dx L/2 L/2 normalization: x x = δ(x x ) Lecture 10 4
6 Operators in Infinite Dimension Space An operator, Ω, in an infinite dimensional space transforms a vector, f, in the vector space to another vector, f, in the space. In the x basis Ω f = g x Ω f = x g = g(x) Ω is represented by an infinite dimension matrix. Ω xx = x Ω x x g = x Ω x x f dx Formally, we can write Ωf(x) = g(x) What this really means though is x Ω x x f dx = g(x) Lecture 10 5
7 The Position Operator Consider the operator, ˆx, that when operating on a function, f(x), has the effect of multiplying the function by x. x ˆx f = x x f = xf(x) x ˆx x x f dx = xf(x) x ˆx x = x δ(x x ) The eigenvectors of ˆx are the basis vectors, x. x ˆx x 0 = x 0 δ(x x 0 ) = x 0 x x 0 As we ve noted before, these eigenvectors have a delta function rather than unity normalization. Lecture 10 6
8 The Derivative Operator Another natural operator to consider is the operation of taking a derivative of a function. x ˆd f = f (x) x ˆd f = x ˆd x x f dx = d dx f(x) What are the matrix elements x ˆd x? Lecture 10 7
9 The Derivative of the Delta Function The matrix elements of D are given by the derivative of the delta function x ˆd x = δ (x x ) = d dx (δ(x x )) δ (x x )f(x )dx = d dx (δ(x x )) f(x )dx = d dx (δ(x x )) f(x )dx = δ(x x ) ( ) d dx f(x ) dx + [ ] f(x )δ(x x ) = δ(x x ) ( ) d dx f(x ) dx = d dx f(x) Lecture 10 8
10 The ˆk Operator The operator ˆd is not Hermitian. d xx = δ (x x ) = d dx δ(x x ) = δ(x x ) d dx d xx = d x x = D x x = δ (x x) = d dx δ(x x) = δ(x x) d dx = δ(x x ) d dx = d xx Define the operator ˆk = i ˆd k xx = iδ (x x ) k xx = kx x = k x x = iδ (x x) = iδ (x x ) = k xx For an infinite dimension operator it is not sufficient that x ˆk x = x ˆk x We must also show that g ˆk f = f ˆk g Lecture 10 9
11 Hermitivity of an Infinite Dimension Operator g ˆk f = f ˆk g b b a a g x x K x x f dxdx = ( b b a a f x x K x x g dxdx ) i b a g (x) ( ) df(x) b dx = i dx a ( ) dg (x) f(x)dx dx but the left-hand side is ig (x)f(x) b a + i b a ( ) dg (x) f(x)dx dx Hermitivity requires that f(x) or g(x) go to zero at the limits a and b. Lecture 10 10
12 Eigenvectors of ˆk Let k be an eignevector of ˆk with eigenvalue k. ˆk k = k k Let s work in the x basis. x ˆk k = k x k = kk(x) x ˆk x x k dx = kk(x) i d dx ψ k(x) = kk(x) Solution: k(x) = Ae ikx Lecture 10 11
13 Normalization Now let s normalize k. k k = k x x k dx = A 2 e i(k k )x dx = A 2 2πδ(k k ) Choose: A = 1 2π k k = δ(k k ) As for the eigenvectors of ˆx, these have delta function normalization rather than unity normalization. Lecture 10 12
14 Fourier Transforms Let s find the vector f in the k-basis. f(k) = k f = k x x f dx = 1 2π e ikx f(x)dx Similarly f(x) = x f = x k k f dk = 1 2π e ikx f(k)dk f(k) and f(x) are the Fourier transforms of each other. Lecture 10 13
15 Canonical Commutation Relation The operators ˆx and ˆk do not commute. x ˆx f = xf(x) x ˆk f = i df(x) dx x ˆxˆk f = ix df(x) dx x ˆkˆx f = ix d ( ) xf(x) dx x [ˆx, ˆk] f = x [ˆx, ˆk] f x [ˆk, ˆx] f = ix df(x) dx df(x) + ix dx + if = if = i x I f [ˆx, ˆk] = ii This means that there are no vectors that are eigenvectors of both ˆx and ˆk. As you know and as we ll see, this leads to one of the most important and basic features of quantum mechanics. Lecture 10 14
16 Reciprocity Between x and k Representations x x dx = I k k dk = I f = x f(x ) dx = k f(k ) dk x x = δ(x x ) k k = δ(k k ) x ˆx x = xδ(x x ) x ˆk x = iδ (x x ) k ˆx k = iδ (k k ) k ˆk k = kδ(k k ) x ˆx f = xf(x) x ˆk f = if (x) k ˆx f = i f (k) k ˆk f = k f(k)) x k = 1 2π e ikx k x = 1 2π e ikx f(x) = x f = x k f(k ) dk = 1 2π e ik x f(k ) dk Lecture 10 15
17 The Momentum Operator The momentum operator is simply the k operator multiplied by h. ˆp = hˆk x ˆp f = i hf (x) [ˆx, ˆp] = i hi Lecture 10 16
18 Correspondence with the Wavefunction Formalism ψ = x x ψ dx = x ψ(x) dx ψ(x) gives the components of ψ in the x basis. ψ = p p ψ dp = p ψ(p) dp ψ(p) gives the components of ψ in the p basis. x p are the matrix elements of the unitary transformation from the p basis to the x basis. ψ(x) = x ψ = x p ψ(p) dp = 1 2π h e ipx/ h ψ(p) dp Lecture 10 17
19 Hilbert Space A Hilbert space is an infinite dimensional complex vector space of normalizable vectors. f f = f(x) 2 dx is finite The state of a quantum system is an element ψ of a Hilbert space ψ(x) 2 dx = ψ(p) 2 dp are finite Note that the basis vectors x and p are not elements of the Hilbert space since: x x = δ(0) and p p = δ(0) We also require that the components of vectors corresponding to physical states be given by a continuous function. x ψ = ψ(x) is a continuous functuon This is a subset of a true Hilbert space and is called a rigged or physics Hilbert space. Lecture 10 18
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