Mathematical Physics Homework 10

Transcription

1 Georgia Institute of Technology Mathematical Physics Homework Conner Herndon November, 5 Several types of orthogonal polynomials frequently occur in various physics problems. For instance, Hermite polynomials H n (x) arise in the quantum harmonic oscillator problem. We will try to find them here via an eigenvalue problem for the following differential operator ˆL d x d. (.) This operator is defined on a space of functions (in our case polynomials) such that ˆLΨ (x) Ψ (x) xψ (x) for any function Ψ (x). The Hermite polynomials are known to be eigenvectors (or more accurately eigenfunctions) of this operator. We want to find a few of these eigenfunctions and the corresponding eigenvalues by calculating the matrix elements of this operator and then diagonalizing it. (a) For the formalism to work we have to ensure that the space of polynomials is a vector space and the operator is linear. Please prove that this is indeed the case. This operator is infinite in dimension, so the corresponding matrix will also be infinite in dimension. We don t know how to diagonalize matrices of infinite dimension. To circumvent this problem let us restrict our attention to a subspace of low order polynomials (up to order three). As our first step, let s introduce a basis: ê ê x ê 3 x ê 4 x 3. (.) (b) Calculate the matrix elements L mn of ˆL in this basis. The easiest way to do this is by simply expanding ˆLê n in the basis {ê k }, i.e., ˆLê n k L knê k. Alternatively, you can use the scalar product

2 L mn ê m, ˆLê n ê m ˆLên w (x), (.3) with the weight w (x) e x. (c) Find the eigenvalues and eigenvectors of L mn and thus determine the coefficients of the first four Hermite polynomials. If you are feeling adventurous, you can find quite a few more higher order polynomials this way, but the procedure quickly becomes too cumbersome due to the necessity of evaluating the determinant and solving algebraic equations of high order. The next problem explores an alternative route for generating sets of orthogonal polynomials. Consider the space of polynomials P. A polynomial of degree n may be written may be written as v :, (a n ) (.4) where the a C are coefficients to the variable x raised to corresponding power x. Then for v, w, u P and µ, ν C, we have the properties commutativity: v + w + b β x β b β x β + β β w + v. (.5) associativity p (v + w) + u + b β x β + c γ x γ β + b β x β + γ β γ v + (w + u). p c γ x γ (.6) identity v + + v, (.7)

3 and + v + v. (.8) inverses and v + ( v) ( ) +, ( ) ( v) + v +. + (.9) (.) scalar associativity ( ) µ (νv) µ ν ( ) µν (µν) v (.) scalar distribution (µ + ν) v (µ + ν) µ + ν µv + νv (.)

4 vector distribution µ (v + w) µ + b β x β β µ + µ b β x β β µv + µw. (.3) scalar identity and v v v, v. (.4) (.5) Thus, the space of polynomials is a vector space over the field of complex numbers. If we act with this operator ˆL on this space, we have ˆL (µv + νu) x d ) µ + ν b β x β so the operator ˆL is linear. µ x d ) µˆlv + ν ˆLu, (b) The operator ˆL acting on each basis vector yields β + ν x d ) m b β x β β (.6)

5 ˆLê x d ) ˆLê x d ) x x ˆLê 3 x d ) x 4x ˆLê 4 x d ) x 3 6x 6x 3. (.7) Then L L L 3 L 4 L L L 3 L 4 L 3 L 3 L 33 L 34 ên (.8) L 4 L 4 L 43 L 44 in the basis of polynomials means that the coefficient for each order polynomial is described by the corresponding row. So since ˆLê, then all rows in the first column are zero. Since ˆLê x, the second column has rows of zero except the second which corresponds to polynomials of degree. Following this logic, we receive the resulting operator ˆL 6 4. (.9) 6 (c) From the matrix form of ˆL we have the eigenvalues ) det (ˆL λi λ λ 6 4 λ, 6 λ (.) which has eigenvalues λ 6 λ 4 λ 3 λ 4. (.) The eigenvectors are then

6 6 v () 6 v () v (3) 6 6 v (4) 6v () + v (3) v () + 6v (4) v () v () v (3) v (4) v (3) 6v (4) 3. v () v () v (3) v (4) 6v () 6v () 6v (3) 6v (4) repeating this process for all eigenvalues yields the corresponding eigenvectors v 3 3 v 5 v 3 v 4. (.) (.3) Since the index of each eigenvector corresponds to a power of x, the Hermite polynomials are (starting from λ 4 to λ because I apparently named them backward) H (x) H (x) x H 3 (x) x H 4 (x) 3x + x 3. (.4) These polynomials are different than what wikipedia says they are, and that s because the eigenvectors may be scaled by some scalar. Orthogonal polynomials can be very conveniently generated by using the Gram-Schmidt orthogonalization procedure. For instance, the Leguerre polynomials L n (x) arise in solutions of the Schrödinger equation for the hydrogen atom. The polynomials with different indices are orthogonal with respect to the scalar product defined (Φ, Ψ) : e x Φ (x) Ψ (x), (.) where w (x) e x is the weight function. Start with a set of non-orthogonal polynomials

7 ê ê a + x ê 3 b + cx + x ê 4 d + fx + gx + x 3 (.).. where a, b, c, d, f, and g are constants, and orthogonalize them (with respect to the scalar product defined above) sequentially beginning with ê, ê 3, and so on to generate the first four Leguerre polynomials. Start with ẽ ê. The next basis vector may be found by Gram-Schmidt orthogonalization ẽ x (xẽ, ẽ ) ẽ (ẽ, ẽ ) xe x x e x x. The third is found by the same process ẽ 3 x (xẽ, ẽ ) (ẽ, ẽ ) And finally the fourth is ẽ 4 x (xẽ 3, ẽ 3 ) (ẽ 3, ẽ 3 ) x x 4 x ẽ e x x (x ) e x (x ) (x 3) (x ) x 4x +. ẽ 3 (ẽ3, ẽ 3 ) (ẽ, ẽ ) e x x ( x 4x + ) e x (x 4x + ) (x 4x + ) x 3 9x + 8x 6. 4 ẽ (ẽ, ẽ ) ẽ (ẽ, ẽ ) e x (x ) (x ) (x 4x + ) e x ( x 4x + ) e x (x ) (x ) (x ) (.3) (.4) (.5) These generated orthogonal polynomials are within a factor of the first four Leguerre polynomials.