aa r X i v : . [ m a t h . G M ] M a y Bessel Type Orthogonality For Hermite Polynomials
Omid Hamidi
Department of Physics, Shahid Bahonar University of Kerman, Kerman, IRAN.(e-mail: [email protected])
Abstract
It is shown that Hermite polynomials satisfy a Bessel type orthogonality relationbased on the zeros of a single indexed Hermite polynomial and with a finite inte-gration interval. Because of the role of non-symmetric zeros in the final relation, itsapplicability covers Hermite polynomials P n ( x ) with n ≥ Keywords:
Orthogonality, Hermite Polynomials, Bessel Functions. Introduction
Among the orthogonal functions (and even the polynomials) appearing in mathematicalphysics the Bessel functions J ν ( x ) have a different type of orthogonality relation [1], Z a dx xJ ν ( α νm xa ) J ν ( α νn xa ) = 0 . (1)As seen, the two Bessel functions under the integral have the same function index ν andthe orthogonality is relative to the location of the zeros α νk satisfying J ν ( α νk ) = 0. Alsothe interval, [0 , a ] ( a is arbitrary), over which the integral is carried out is only part ofthe domain over which J ν ( x ) is defined. In this short note it is shown that the Hermitepolynomials H n ( x ) which satisfy the standard orthogonality relation Z ∞−∞ dx e − x H n ( x ) H m ( x ) = 0 , n = m (2)also satisfy an orthogonality relation similar in spirit to the orthogonality relation of theBessel functions. In the next section the details of work are presented. To obtain the desired orthogonality relation consider the Hermite differential equation, H ′′ n ( x ) − xH ′ n ( x ) + 2 nH n ( x ) = 0 , −∞ < x < ∞ (3)where as usual prime indicates differentiaion with respect to the argument. Let us firstchange the independent variable to y according to x = k n y . Equation (3), upon multiply-ing by k n , then transforms into, d H n ( k n y ) dy − k n y dH n ( k n y ) dy + 2 nk n H n ( k n y ) = 0 (4)Now in (4), a transformation of the dependent variable according to H n ( k n y ) = e k n y / ψ n ( k n y ) (5)2ill transform it into d ψ n ( k n y ) dy + (cid:2) k n (1 + 2 n ) − k n y (cid:3) ψ n ( k n y ) = 0 (6)where the common exponential factor has been dropped. From (5) it is noticed that ψ n ( y )is either even or odd like H n ( y ) and the location of zeros of the ψ n ( y ) are symmetric withrespect to the origin and on the finite x axis coincide with the zeros of the Hermitepolynomail H n ( y ). Now let us choose k n as follows k n = α nj b (7)where α nj indicates the location of the j th zero of ψ n ( y ), that is ψ n ( α nj ) = 0. The nextstep is to write (6) for two different values of k n , i.e., d ψ n ( α ni yb ) dy = (cid:20) α ni y b − α ni (1 + 2 n ) b (cid:21) ψ n ( α ni yb ) d ψ n ( α nj yb ) dy = (cid:20) α nj y b − α nj (1 + 2 n ) b (cid:21) ψ n ( α nj yb ) (8)Now multiply the first equation by ψ n ( α nj yb ) and the second one by ψ n ( α ni yb ) and subtractthe second one from the first and integrate over y from y = − b to y = b . Once done, onthe Left Side (LS) (Right Side (RS)), one would getLS = Z b − b dy (cid:20) ψ n ( α nj yb ) d ψ n ( α ni yb ) dy − ψ n ( α ni yb ) d ψ n ( α nj yb ) dy (cid:21) (9)RS = ( α ni − α nj ) b Z b − b dy (cid:20) ( α ni + α nj ) y b − (1 + 2 n ) (cid:21) ψ n ( α nj yb ) ψ n ( α ni yb ) (10)Due to the formal self-adjointness of the differential operator d dy , the expression LS equalsa boundary term given byLS = (cid:20) ψ n ( α nj yb ) dψ n ( α ni yb ) dy − ψ n ( α ni yb ) dψ n ( α nj yb ) dy (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) y = by = − b = 0 (11)where the fact ψ n ( α nj ) = ψ n ( − α nj ) = 0 (the same also for j → i ) has been used. Onthe right side RS, the coefficeint in front, i.e., ( α ni − α nj ), will be non-zero if α ni and3 nj are non-symmetric zeros. Therefore, the result bellow holds for n ≥ α ni = α nj , the equality LS=RSwill yield 0 = Z b − b dy (cid:20) ( α ni + α nj ) y b − (1 + 2 n ) (cid:21) ψ n ( α nj yb ) ψ n ( α ni yb ) (12)One last step in (12) is to use (5) to express ψ n ( α nj yb ) in terms of the correspondingHermite polynomial H n ( α nj yb ) and will yield0 = Z b − b dy (cid:20) ( α ni + α nj ) y b − (1 + 2 n ) (cid:21) e − ( α ni + α nj ) y / b H n ( α ni yb ) H n ( α nj yb ) (13)which is the desired orthogonality relation. As it may have been noticed the interval ofintegration could have been chosen [0 , b ] in which case LS in (11) would be zero again,LS = (cid:20) ψ n ( α nj yb ) dψ n ( α ni yb ) dy − ψ n ( α ni yb ) dψ n ( α nj yb ) dy (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) y = by =0 = 0 (14)where the vanishing of the expression at lower end of the interval is due to the vanishingof ψ n ( y ) or its derivative at y = 0 (depending on n being odd or even, respectively). Soone would get0 = Z b dy (cid:20) ( α ni + α nj ) y b − (1 + 2 n ) (cid:21) e − ( α ni + α nj ) y / b H n ( α ni yb ) H n ( α nj yb ) (15)Having shown the new orthogonality relations (13) and (15) it is a question in which, ifany, occasion would such a type of orthogonality play a role. Looking back at (6) onenotices that it is the Schr¨odinger equation for the simple harmonic oscillator problem. Inthat equation k n (1 + 2 n ) plays the role of the eigenvalue while k n y is the potential energy.Because of the multiplicative factor k n (which also appears in the energy eigenvalue), thepotential energy depends on the energy eigenvalue. Such type of potentials are not trivialin proving the orthogonality of eigenfunction. Problems of this sort have been discussedin the literature [2, 3, 4]. References [1] G. B. Arfken, H. J. Weber:
Mathematical Methods For Physicists , 5th ed., AcademicPress, London 2001. 42] H. Jallouli, H. Sazdjian, Ann. Phys. (1997) 376.[3] R Yekken, R J Lombard J. Phys. A: Math. Theor. (2010) 125301 (17pp)[4] J. Garca-Martnez, J. Garca-Ravelo , J.J. Pea , A. Schulze-Halberg , Physics LettersA373