Hermite polynomials in Quantum Harmonic Oscillator

1 Hermite polynomials in Quantum HarmonicOscillatorChristos T. AravanisChristos T. Aravanisis a senior majoring inMathematics and Theoretical Physics at the Uni-versity of Athens, graduation heplans to attend graduate school where he will studyMathematics. The content of this article reflects hisinterest in the applications of Mathematics to Quantum mechanics and in other branches of physics, it is common to ap-proach physical problems using algebraic and analytic methods. Examplesinclude the use of differential equations for many interesting models, the use ofquantum groups in Quantum physics, and of differential geometry in relativitytheory. In this article, we discuss the Hermite polynomials , some of their prop-erties and a brief description of their applications to the Quantum PolynomialsHermite polynomials , named after the French mathematician Charles Hermite ,are orthogonal polynomials , in a sense to be described below, of the formHn(x) = ( 1)nex2dndxne x2(1)forn= 0,1,2,3.

2 The first few Hermite polynomials are forn= 0 we haveH0(x) = 1 forn= 1 we haveH1(x) = 2x forn= 2 we haveH2(x) = 4x2 Undergraduate Mathematics Exchange, Vol. 7, No. 1 (Fall 2010)27 Definition N, we define Hermite polynomialsHn(x) by n=0Hn(x)n!rn=e2xr r2,for|r|< .(2)To findHn(x), expand the right hand side of (2) as a Maclaurin series inrand equate coefficients. From Equation (2) we derive the closed expressionHn(x) =bn/2c k=0( 1)kn!k!(n 2k)!(2x)n 2k(3)wherebxcdenotes the largest integer less than or equal tox. Checking withn= 0,1,2,.., we find that (3) yields the expected Hermite polynomials . Toprove that (3) holds in general, one can use induction (see [2]).Recurrence RelationsNext we discuss recurrence relations that Hermite polynomials satisfy. We startwithHn+1(x) 2xHn(x) + 2nHn 1(x) = 0, n= 1,2,..,(4)which follows from the fact that the generating functionw(x,r) =e2xr r2satisfies the differential equation w/ r (2x 2r)w= next recurrence relation connectsH n(x) andHn 1(x).

3 From (2), wehaveH n(x) = 2nHn 1(x), n= 1,2,..(5)Now, after a moment s thought, and combining the above two recurrence rela-tions we have another relationH n(x) 2xH n(x) + 2nHn(x) = 0, n= 1,2,..(6)From a mathematician s viewpoint, relation (6) is a second-order linear differen-tial equation, which is calledHermite s differential equation. From a physicist spoint of view, differential equation (6) plays a central role in one of the mostimportant physical models, namely the one-dimenisionalQuantum HarmonicOscillator. For both mathematicians and physicists, solutions of (6) are theHermite , we turn to a common topic for polynomials such as Hermite polynomials ,theorthogonality. Our goal is to prove that the family of Hermite polynomials {Hn}forn= 0,1,2,..is orthogonal with respect to theweighte x2. In otherwords, we will prove Undergraduate Mathematics Exchange, Vol.

4 7, No. 1 (Fall 2010) + e x2Hn(x)Hm(x)dx= 0(7)form6=n. To prove this we will use a technique from [2]. It is easy to showby direct differentiation and using (6) thatun=e x22Hn(x)andum=e x22Hm(x)satisfyu n+ (2n+ 1 x2)un= 0(8)andu m+ (2m+ 1 x2)um= 0.(9)Multiplying (8) byumand (9) byuntransforms each intoumu n+ (2n+ 1 x2)umun= 0(10)andunu m+ (2m+ 1 x2)unum= 0,(11)respectively. Subtracting (11) from (10) we have(umu n unu m) + 2(n m)umun= 0.(12)Finally, integrating (12) from to + shows that(n m) + e x2Hn(x)Hm(x)dx= , (7) follows ifm6= , we have + e x2Hn(x)Hm(x)dx= 2nn! ,a result which takes more work than them6=ncase for a detailed proof, see[2].Connection with Quantum Harmonic OscillatorIn this final part of our paper, we will show the connection of Hermite Poly-nomials with the Quantum Harmonic Oscillator . First of all, the analogue ofthe classical Harmonic Oscillator in Quantum Mechanics is described by theSchr odingerequation +2m~2(E V(y)) = 0, Undergraduate Mathematics Exchange, Vol.

5 7, No. 1 (Fall 2010)29where is thestateof a particle of massmin the potentialV(y), with will suppose that the potential has the formV(y) =y2, and thereforewe consider the following equation +2m~2(E y2) = 0.(13)In order to simplify this equation, we make a change of variabley= (13) is transformed to 2mk4~2x2 = 2mk2~2E ,where the differentiation is now with respect to the new variablex. Choosingthe constantkappropriately our equation becomes x2 = ,(14)where := 2m~ (14) is a second order differential equation with variable solve this equation, we first notice that (x) =e x2/2is a solution ofthe differential equation x2 = . Using the method of variation ofparameters (see [3]) we find the following solutions of Equation (14): n(x) = (x)H(x).(15)Here, (x) is the solution defined above, andH(x) is a function to be de-termined.

6 To find the form ofH(x), we substitute n(x) given by (15) intoequation (14) and we obtain the following equation:H 2xH + ( 1)H= 0.(16)Setting 1 = 2n = n:= 2n+ 1 in (16) we obtain none other thanthe Hermite differential equation (6) whose solutions areH(x) :=Hn(x), theHermite [1] L. D. Faddeev and Yakubovskii, Lectures on Quantum Mechanics forMathematics Students, AMS, 2009.[2] N. N. Lebedev, Special functions & their applications, Dover, 1972.[3] J. D. Logan, Applied Mathematics, Wiley, Undergraduate Mathematics Exchange, Vol. 7, No. 1 (Fall 2010)