Quantum Theory - Particle Physics Group

CHAPTER 2. WAVE MECHANICS AND THE SCHRÖDINGER EQUATION 34To each value N = 0,1,2,... will then correspond an even function v(ξ) which is a polynomial oforder 2N in ξ, and an even, physically acceptable, wave function u(ξ), which is given by (??).In a similar way we obtain the odd states by using the power seriesv(ξ) =∞∑b ν ξ 2ν+1 , (2.121)ν=0which contains only odd powers of ξ. We again substitute the ansatz into the Hermite equationand obtain a recursion relation for the coefficients b ν . We now see, that the series terminatesfor the discrete valuesλ = 4N + 3, N = 0, 1, 2,... . (2.122)To each value N = 0,1,2,... will then correspond an odd function v(ξ) which is a polynomial oforder 2N+1 in ξ, and an odd, physically acceptable wave function u(ξ) given by (??). Puttingtogether the results we see that the eigenvalue λ must take on one of the discrete valuesλ = 2n + 1, n = 0, 1, 2,... (2.123)where the quantum number n is a non-negative integer. Inserting into (2.109) we therefore findthat the energy spectrum of the linear harmonic oscillator is given byE n =(n + 1 )ω 0 , n = 0, 1, 2,... (2.124)2We see that, in contrast to classical mechanics, the quantum mechanical energy spectrum of thelinear harmonic oscillator consists of an infinite sequence of discrete levels. The eigenvalues arenon-degenerate since for each value of the quantum number n there exists only one eigenfunction(apart from an arbitrary multiplicative constant) and the energy of the lowest state, the zeropoint-energyE 0 = ω 0 /2, is positive. This can be understood as a consequence of theuncertainty relation ∆X ∆P ≥ /2 because 〈P 〉 = 〈X〉 = 0 so thatE = 12m (∆P)2 + mω2 02 (∆X)2 , (2.125)which has the positive minimum E 0 . Since the wave function factors v n (ξ) are solutions of theHermite equation and polynomials of order n, they are proportional to the Hermite polynomialsH n (ξ), which can be definded asThis leads to the explicit formulaH n (ξ) = (−1) n e ξ2 dn e −ξ2(2.126)dξ(n= e ξ2 /2ξ −dξ) d ne −ξ2 /2 . (2.127)H 2n =n∑n−k (2n)!(2ξ)2k(−1)(n − k)!(2k)! , H 2n+1 = 2ξk=0n∑(−1)k(2n + 1)!(2ξ)2n−2kk!(2n − 2k + 1)!k=0(2.128)