harmonic oscillator in quantum mechanics

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1 Physics 400 Spring 016 harmonic oscillator in quantum mechanics lecture notes, spring semester rozman/ourses/p400_17s/ Last modified: May 19, 017 Dimensionless Schrödinger s equation In quantum mechanics a harmonic oscillator with mass m and frequency ω is described by the following Schrödinger s equation: d ψ m dx + 1 mω x ψ(x) = Eψ(x). (1) The solution of Eq. (1) supply both the energy spectrum of the oscillator E = E n and its wave function, ψ = ψ n (x); ψ(x) is a probability density to find the oscillator at the position x. Since the probability to find the oscillator somewhere is one, the following normalization conditil supplements the linear equation (1): ψ(x) dx = 1. () As a first step in solving Eq. (1) we switch to dimensionless units: ω has the dimension of energy, hence E ω is dimensionless. Therefore, we introduce the parameter ε, ε E ω. (3) Page 1 of 7

2 We divide Eq. (1) by ω : d ψ mω dx + mω x ψ(x) = εψ(x). (4) The only dimensional parameter combination that remains in Eq. (4), mω, has the dimension of [length]. Therefore, the new variable u, mω u x (5) is dimensionless. or mω x = u, d mω dx = d du. (6) d ψ du + u ψ(x) = εψ(x), (7) d ψ du + ( ε u ) ψ = 0. (8) Asymptotics of the wave function as u ± How ψ(u) behave when u ±? Let s search for the solution of Eq. (8) in the following form: ψ = e S, (9) where S(u) is a new unknown function. de S du = S e S, d e S du = S e S + S e S, (10) where S = ds du. Substituting Eq. (10) into Eq. (8), arrive at the following nonlinear differential equation: S + S + ε u = 0. (11) In the limit u ±, ε u. In addition, as we see shortly, S S. (1) Therefore, Eq. (11) can be simplified to S u = 0 S = u S = u. (13) Page of 7

3 The choice of minus sign in Eq. (13) is the only one consistent with the requirement Eq. (). The solution Eq. (13) is also consistent with the assumption Eq. (1). Thus, u lim ψ(u) = e. (14) u ± Hermite differential equation Based on the result Eq. (14) we are going to search for a solution of Eq. (8) in the following form: ψ(u) = v(u)e u. (15) ψ = v e u v u e u (16) ψ = v e u v u e u v e u + v u e u. (17) Substituting Eqs. (15), (17) into Eq. (8) and simplifying, we arrive at the following equation: For the later convenience, we introduce the notation v uv + (ε 1)v = 0. (18) ε 1 n. (19) The equation is called Hermite equation. v uv + nv = 0 (0) Solutions of Hermite equation Let s search for the solution of Hermite equation in the following integral form, v(u) = e ut Y (t)dt, (1) where the contour integral in the complex plane is taken over yet unspecified contour and Y (t) is a yet unknown function. Page 3 of 7

4 The first and the second derivatives of v(u), Eq. (1), are as following: v = e ut t Y (t)dt, () v = e ut t Y (t)dt. (3) uv = t Y (t)(ue ut dt) = t Y (t)de ut = t Y (t)e ut B where A and B denote the end points of the contour. Let s impose the following restriction on the contour : A e ut d (t Y (t)), (4) dt In this case, uv = t Y (t)e ut B = 0. (5) A e ut d (t Y (t)). (6) dt Substituting Eqs. (1), (3), and (6) into Eq. (0): e ( ut d ) dt (t Y (t)) + (t + n)y (t) dt = 0. (7) Hence, ( ) d t dt (t Y ) + + n Y = 0. (8) Equation (8) is a first order ordinary differential equation that can be integrated separating variables. To simplify expressions, let s introduce the following notation: and write Eq. (8) in the following form: Z(t) = t Y (t), Y (t) = 1 Z(t), (9) t dz dt + ( t + n t Page 4 of 7 ) Z = 0. (30)

5 Separating variables in Eq. (30), Finally, and v(u) = ( dz t Z = + n ) dt lnz = t t 4 nlnt. (31) Z(t) = 1 t n e t 4 Y (t) = 1 t e 4 tn+1, (3) 1 t eut 4 tn+1 dt, ψ(u) = e u 1 t eut 4 tn+1 dt. (33) Let s accept, for now without a proof, that Eq. (33) describes a physically acceptable (that is normalizable per Eq. ()) wave function only if n is a non-negative integer. In particular that means that the energy spectrum of a harmonic oscillator where we used Eqs. (3) and (19). E = ω ( ε = ω n + 1 ), (34) If n is an integer we can chose an arbitrary closed contour that encircles t = 0 as the integration contour in Eq. (33). Hermite polynomials Let s have a closer look at v(u). 1 t v(u) = eut 4 tn+1 dt = 1 t n+1 eut t 4 u +u dt = e u 1 t n+1 e (u t ) dt. (35) Introducing a new integration variable, z, and dropping an irrelevant constant factor, obtain: v(u) = e u z = u t, (36) e z dz, (37) (z u) n+1 Page 5 of 7

6 where is an closed contour encircling the point z = u. Using auchy s formula for derivatives of analytic functions, d k f (u) du k = k! πi f (z) dz, (z u) k+1 the expression Eq. (37) can be rewritten as following: dn v(u) = e u du n e u. (38) We can see that v(u) is actually a polynomial of order n. The first few non-normalized wave functions are as following: n v n (u) ψ n (u) 0 1 e u 1 u ue u 4u (4u )e u 3 8u 3 + 1u ( 8u 3 + 1u)e u Hermite s polynomials define with a factor of ( 1) n to keep positive the coefficient next to the highest power of the argument: dn H n (u) = ( 1) n e u du n e u. (39) For the reference, the explicit expression for Hermite polynomials is as following: n H n (u) = ( 1) m n m n! m!(n m)! un m. (40) m=0 The coefficient next to the highest power of the argument is n. Summarizing, the un-normalized wave function of a harmonic oscillator can be expressed as following: ψ n (u) = e u Hn (u). (41) Page 6 of 7

7 Normalized wave function To find the normalized wave function, let s calculate the normalization integral: N = ψ n du = e u H n(u)du = ( 1) n H n (u) where in the last equality we substituted Eq. (39) for H n. Integrating by parts in the last integral n times, we get Using Eq. (40), therefore The normalized wave function, N = [ d n du n e u ] du, (4) e u dn H n du. (43) dun d n H n du n = n n!, (44) N = n n! e u du = n n! π. (45) φ(u) = 1 1 ψ n (u) = N n n! u e Hn (u). (46) π References [1] L. D. Landau and E. M. Lifshitz, Quantum Mechanics Non-Relativistic Theory, vol. III of ourse of Theoretical Physics. Butterworth-Heinemann, 3 ed., Page 7 of 7