Hermite Polynomials

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  Quantum Mechanics I

  A Problem Text

  • David DeBruyne, Larry Sorensen(Authors)
  • 2018(Publication Date)
  • Sciendo

    (Publisher)

  The solutions to Hermite’s equation are the Hermite Polynomials. We will solve this differential equation thereby deriving the Hermite Polynomials using a power series solution in part 2 of this chapter. Using equation (1) with the appropriate Hermite polynomial is likely the easiest way to obtain a position space eigenfunction for the quantum mechanical SHO. ψ 0 ( x ) = parenleftBig mω π ¯ h parenrightBig 1 / 4 1 √ 2 0 0! H 0 parenleftbiggradicalbigg mω ¯ h x parenrightbigg e -mωx 2 / 2¯ h = parenleftBig mω π ¯ h parenrightBig 1 / 4 ( 1 ) e -mωx 2 / 2¯ h = parenleftBig mω π ¯ h parenrightBig 1 / 4 e -mωx 2 / 2¯ h , in agreement with our earlier calculation . ψ 1 ( x ) = parenleftBig mω π ¯ h parenrightBig 1 / 4 1 √ 2 1 1! H 1 parenleftbiggradicalbigg mω ¯ h x parenrightbigg e -mωx 2 / 2¯ h = parenleftBig mω π ¯ h parenrightBig 1 / 4 1 √ 2 2 parenleftbiggradicalbigg mω ¯ h x parenrightbigg e -mωx 2 / 2¯ h = parenleftbigg 4 π parenleftBig mω ¯ h parenrightBig 3 parenrightbigg 1 / 4 xe -mωx 2 / 2¯ h , also in agreement with our earlier calculation . 292 Postscript: Hermite Polynomials are not square integrable. The products of a Hermite polynomial and the Gaussian envelope e -ξ 2 / 2 are, however, square integrable over all space so are acceptable wave functions. The constants parenleftBig mω π ¯ h parenrightBig 1 / 4 1 √ 2 n n ! are simply normalization constants. The adjective form of Charles Hermite’s name is Hermitian, as in Hermitian operator. 10–18. Show that H 1 ( ξ ) is orthogonal to H 2 ( ξ ) . The Hermite Polynomials are orthogonal when both are weighted by e -ξ 2 / 2 . Including the weighting functions, the orthogonality condition is integraldisplay ∞ -∞ H 1 ( ξ ) e -ξ 2 / 2 H 2 ( ξ ) e -ξ 2 / 2 dξ = integraldisplay ∞ -∞ 2 ξ ( 4 ξ 2 -2 ) e -ξ 2 dξ = integraldisplay ∞ -∞ ( 8 ξ 3 -4 ξ ) e -ξ 2 dξ = 8 integraldisplay ∞ -∞ ξ 3 e -ξ 2 dξ -4 integraldisplay ∞ -∞ ξe -ξ 2 dξ. The integrands are both odd functions integrated between symmetric limits.

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  Mathematical Problems for Chemistry Students

  • Gyorgy Pota(Author)
  • 2011(Publication Date)
  • Elsevier Science

    (Publisher)

  Chapter 1 Problems 1.1 ALGEBRA 1. The Hermite Polynomials [1, 2, p. 60] play an important role in the description of the vibrational motion of the molecules [3, p. 476]. The first few Hermite Polynomials are given in Table 1.1 ( −∞ < x < ∞ ), and each question in this problem concerns these polynomials. (a) Confirm by direct calculation that the general relationship H n + 1 ( x ) = 2 xH n ( x ) − 2 nH n − 1 ( x ) is valid for the polynomials. (b) Which polynomials are even functions and which are odd ones? (c) Which are the polynomials whose zeros include 0? (d) Which are the polynomials whose real zeros are symmetrical to the origin? (e) Which polynomials can have an even number of real zeros? Why? Table 1.1 The first few Hermite Polynomials n H n ( x ) 0 1 1 2 x 2 4 x 2 − 2 3 4 x (2 x 2 − 3) 4 4(4 x 4 − 12 x 2 + 3) 5 8 x (4 x 4 − 20 x 2 + 15) 6 8(8 x 6 − 60 x 4 + 90 x 2 − 15) 7 16 x (8 x 6 − 84 x 4 + 210 x 2 − 105) 8 16(16 x 8 − 224 x 6 + 840 x 4 − 840 x 2 + 105) 9 32 x (16 x 8 − 288 x 6 + 1512 x 4 − 2520 x 2 + 945) 10 32(32 x 10 − 720 x 8 + 5040 x 6 − 12600 x 4 + 9450 x 2 − 945) 1 2 Chapter 1. Problems Table 1.2 Several associated Legendre functions l m P m l ( x ) 1 1 √ 1 − x 2 2 2 3(1 − x 2 ) 3 2 15 x (1 − x 2 ) 5 2 105 x (1 − x 2 )(3 x 2 − 1) 2 7 2 63 x (1 − x 2 )(143 x 4 − 110 x 2 + 15) 8 10 2 495(1 − x 2 )(4199 x 8 − 6188 x 6 + 2730 x 4 − 364 x 2 + 7) 128 (f) On the basis of Descartes’rule of signs what can be said about the number of positive zeros of the polynomials? (g) Determine the exact zeros of the polynomials H n for n = 1, 2, 3, 4, 5. (h) Determine the approximate zeros of the polynomials not mentioned in paragraph 1g by mathematical/spreadsheet software. 2. The associated Legendre functions P m l ( x ) [2, p. 192, 4] appear in the angular parts of the orbital functions of hydrogenic (one-electron) atomic particles [3, p. 334]. Some associated Legendre functions are included in Table 1.2, and each question in this problem concerns these functions.

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  Special Functions and Orthogonal Polynomials

  • Richard Beals, Roderick Wong(Authors)
  • 2016(Publication Date)
  • Cambridge University Press

    (Publisher)

  Schl¨ afli [355] gave the integral formula (5.2.1) for the Legendre polynomials in 1881. The series expansion (5.5.13) for Legendre polynomials was given by Murphy in 1833 [296]. Hermite Polynomials occur in the work of Laplace on celestial mechanics [242] and on probability [243, 244], and in Chebyshev’s 1859 paper [72], as well as in Hermite’s 1864 paper [181]. Laguerre polynomials for α = 0 were considered by Lagrange [237], Abel [2], Chebyshev [72], Laguerre [239], and, for general α, by Sonine [374]. Jacobi polynomials in full generality were introduced by Jacobi in 1859 [201]. The special case of Chebyshev polynomials was studied by Chebyshev in 1854 [69]. Gegenbauer polynomials were studied by Gegenbauer in [159, 160]. Further remarks on the history are contained in several of the books cited in Section 5.11, and in Szeg˝ o’s article (and Askey’s addendum) [393]. A common approach to classical orthogonal polynomials is to use the generating function formulas (5.3.6), (5.4.6), and (5.5.8) as definitions. The three-term recurrence relations and some other identities are easy consequences, as is orthogonality (once one has selected the correct weight) in the Hermite and Laguerre cases. The fact that the polynomials are the 138 The classical orthogonal polynomials eigenfunctions of a symmetric second-order operator is easily established, but in the generating function approach this fact appears as something of a (fortunate) accident. It does not seem clear, from the generating function approach, why it is these polynomials and no others that have arisen as the “classical” orthogonal polynomials. In our view, the characterization theorem (3.4.1), the connection with the basic equations of mathematical physics in special coordinates (Section 3.6), and the characterization of “recursive” second-order equations (Section 1.1) provide natural explanations for the significance of precisely this set of polynomials.

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  Physics with MAPLE

  The Computer Algebra Resource for Mathematical Methods in Physics

  • Frank Y. Wang(Author)
  • 2008(Publication Date)
  • Wiley-VCH

    (Publisher)

  Although we offer no derivation of the normalization factor one can, in a practical manner, use Maple to find it by integrating the square of any unnormalized wave function. Worksheet 15.4 Hermite Polynomials are defined as in Maple. In this worksheet we plot the energies and normalized wave functions. > > ψ := ( n, ξ ) → e ( − 1 / 2 ξ 2 ) HermiteH( n, ξ ) √ 2 n n ! > e ( − ξ 2 2 ) > √ 2 e ( − ξ 2 2 ) ξ > − 1 4 √ 8 e ( − ξ 2 2 ) + 1 2 √ 8 e ( − ξ 2 2 ) ξ 2 > 1 6 √ 48 e ( − ξ 2 2 ) ξ 3 − 1 4 √ 48 e ( − ξ 2 2 ) ξ > 1 32 √ 384 e ( − ξ 2 2 ) + 1 24 √ 384 e ( − ξ 2 2 ) ξ 4 − 1 8 √ 384 e ( − ξ 2 2 ) ξ 2 > 1 120 √ 3840 e ( − ξ 2 2 ) ξ 5 − 1 24 √ 3840 e ( − ξ 2 2 ) ξ 3 + 1 32 √ 3840 e ( − ξ 2 2 ) ξ > − 1 384 √ 46080 e ( − ξ 2 2 ) + 1 720 √ 46080 e ( − ξ 2 2 ) ξ 6 − 1 96 √ 46080 e ( − ξ 2 2 ) ξ 4 + 1 64 √ 46080 e ( − ξ 2 2 ) ξ 2 440 15 Schrödinger Equation in One Dimension (II) > > > > 0 2 4 6 8 –6 –4 –2 2 4 6 xi We list the first six wave functions for the harmonic oscillator in Table 15.1. Solutions appear in two classes – symmetric and antisymmetric wave functions; this feature is consistent with the symmetric nature of the potential energy, because V ( x ) = V ( − x ) . Table 15.1: The first six wave functions for harmonic oscillator with variable ξ = p ( mω/ ) x . n ψ n / ( mω/π ) 1 / 4 0 e − ξ 2 / 2 1 √ 2 e − ξ 2 / 2 (2 ξ ) 2 1 √ 2 e − ξ 2 / 2 (2 ξ 2 − 1) 3 1 √ 3 e − ξ 2 / 2 (2 ξ 3 − 3 ξ ) 4 1 √ 24 e − ξ 2 / 2 (4 ξ 4 − 12 ξ 2 + 3) 5 1 √ 60 e − ξ 2 / 2 (4 ξ 5 − 20 ξ 3 + 15 ξ ) 15.6 Homogeneous Field 441 15.6 Homogeneous Field We consider this function for the potential energy, V ( x ) = F | x | , (15.39) which might represent the case of a particle subject to a constant force F . This potential serves as an excellent model for a quark and an antiquark in a bound system. 2 The Schrödinger equation for this problem is − 2 2 m d 2 ψ dx 2 + F | x | ψ = Eψ. (15.40) Because V ( x ) = V ( − x ) , we take advantage of the symmetry property discussed in Sec-tion 15.3.

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  Special Functions

  A Graduate Text

  • Richard Beals, Roderick Wong(Authors)
  • 2010(Publication Date)
  • Cambridge University Press

    (Publisher)

  4.4 Hermite Polynomials 107 D H n = 2 3n(n−1)/2 n  j =1 j j ; (4.3.20) D (α) n = n  j =1 j j −2n+2 ( j + α) j −1 ; D (α,β) n = 1 2 n(n−1) n  j =1 j j −2n+2 ( j + α) j −1 ( j + β) j −1 (n + j + α + β) j −n . For a proof, see for example Section 6.71 of [279]. The next three sections contain additional results and some alternate deriva- tions for these three classical cases. 4.4 Hermite Polynomials The Hermite Polynomials  H n  are orthogonal polynomials associated with the weight e −x 2 on the line R = (−∞, ∞). They are eigenfunctions H  n (x ) − 2x H  n (x ) + 2n H n (x ) = 0, (4.4.1) satisfy the derivative relation H  n (x ) = 2n H n−1 (x ), and can be defined by the Rodrigues formula H n (x ) = (−1) n e x 2 d n dx n  e −x 2  =  2x − d dx  n {1}. It follows that the leading coefficient is 2 n and that H  n (x ) − 2x H n (x ) = −H n+1 (x ). (4.4.2) They are limits H n (x ) = lim α→+∞ 2 n n! α n/2 P (α,α) n  x √ α  . The three-term recurrence relation (4.3.6) may also be derived as follows. It is easily shown by induction that H n is even if n is even and odd if n is odd: H n (−x ) = (−1) n H n (x ). 108 Orthogonal polynomials Therefore the relation must have the form x H n (x ) = a n H n+1 (x ) + b n H n−1 (x ). (4.4.3) Identities (4.4.2) and (4.4.3) imply that H  n = 2b n H n−1 . Comparing leading coefficients, we see that a n = 1 2 and b n = n: x H n (x ) = 1 2 H n+1 (x ) + n H n−1 (x ). (4.4.4) If we write H n (x ) = ∑ n k =0 a k x k , then equation (4.4.1) implies the relation (k + 2)(k + 1) a k +2 = 2(k − n) a k . Since a n = 2 n and a n−1 = 0, this recursion gives H n (x ) =  2 j ≤n (−1) j n ! j ! (n − 2 j ) ! (2x ) n−2 j . (4.4.5) The first six of the H n are H 0 (x ) = 1; H 1 (x ) = 2x ; H 2 (x ) = 4x 2 − 2; H 3 (x ) = 8x 3 − 12x ; H 4 (x ) = 16x 4 − 48x 2 + 12; H 5 (x ) = 32x 5 − 160x 3 + 120x .

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