Quantum Harmonic Oscillator

Quantum Harmonic Oscillator Chapter 13 P. J. Grandinetti Chem. 4300 Oct 20, 2017 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 1 / 26

Kinetic and Potential Energy Operators Harmonic oscillator well is model for small vibrations of atoms about bond as well as other systems in physics and chemistry. m 1 m 2 r e Model bond as spring that acts as restoring force whenever two atoms are squeezed together or pulled away from equilibrium position. Kinetic and potential energy operators are x = r r e and μ is reduced mass K = p2 2μ and V(x) = 1 2 κ f x 2 1 μ = 1 m 1 + 1 m 2 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 2 / 26

Force constants for selected diatomic molecules Bond κ f /(N/m) μ 10 28 kg ν/cm 1 Bond length/pm H 2 570 8.367664 4401 74.1 D 2 527 16.72247 2990 74.1 H 35 Cl 478 16.26652 2886 127.5 H 79 Br 408 16.52430 2630 141.4 H 127 I 291 16.60347 2230 160.9 35 Cl 35 Cl 319 290.3357 554 198.8 79 Br 79 Br 240 655.2349 323 228.4 127 I 127 I 170 1053.649 213 266.7 16 O 16 O 1142 132.8009 1556 120.7 14 N 14 N 2243 116.2633 2331 109.4 12 C 16 O 1857 113.8500 2143 112.8 14 N 16 O 1550 123.9830 1876 115.1 23 Na 23 Na 17 190.8770 158 307.8 23 Na 35 Cl 117 230.3282 378 236.1 39 K 35 Cl 84 306.0237 278 266.7 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 3 / 26

Schrödinger equation for harmonic oscillator Insert Kinetic and Potential Energy operators ħ2 d 2 ψ(x) + 1 2μ dx 2 2 κ fx 2 ψ(x) = Eψ(x) and defining so Schrödinger equation becomes k 2 = 2μE ħ 2 and α 4 = μκ f ħ 2 d 2 ψ(x) dx 2 + (k 2 α 4 x 2 )ψ(x) = 0 This ODE doesn t have simple solutions like particle in infinite well. Harmonic oscillator potential becomes infinitely high as x goes to Wave function is continuous and single valued over x = to. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 4 / 26

Schrödinger equation for harmonic oscillator Notice at large values of x one can approximate ODE as With solutions d 2 ψ(x) dx 2 α 4 x 2 ψ(x) 0 ψ(x) Ae ±α2 x 2 2 but only accept ψ(x) Ae α2 x 2 2 as physical But these are NOT solutions for all x. However, in light of this asymptotic solution we further define ξ = αx and ψ(x)dx = χ(ξ)dξ to transform Schrödinger equation into d 2 ( ) χ(ξ) k 2 + dξ 2 α 2 ξ2 χ(ξ) = 0 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 5 / 26

Schrödinger equation for harmonic oscillator We expect solution to this ODE to have asymptotic limits We propose a general solution Substituting into ODE gives d 2 H(ξ) dξ 2 2ξ dh(ξ) dξ lim χ(ξ) = 2 ξ Ae ξ2 χ(ξ) = AH(ξ)e ξ2 2 + 2nH(ξ) = 0 where n = k2 α 2 1 9 out of 10 math majors recognize this as Hermite s differential equation Its solutions are the Hermite polynomials. Only solutions with n = 0, 1, 2, are physically acceptable for harmonic oscillator. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 6 / 26

First seven Hermite polynomials and approximate roots. n H n (y) Roots 0 1 1 2y 0 2 4y 2 2 ±0.707107 3 8y 3 12y 0 ±1.224745 4 16y 4 48y 2 + 12 ±0.5246476 ±1.650680 5 32y 5 + 160y 3 + 120y 0 ±0.958572 ±2.020183 6 64y 6 480y 4 + 720y 2 120 ±0.436077 ±1.335850 ±2.350605 7 128x 7 1344x 5 + 3360x 3 1680x 0 ±0.816288 ±1.673552 ±2.65196 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 7 / 26

Harmonic Oscillator Wave Function Normalized solutions to Schrödinger equation for harmonic oscillator are χ n (ξ) = A n H n (ξ)e ξ2 2, where A n 1 2 n n! π 1 2 Condition that n only be integers leads to harmonic oscillator energy levels E n = ħω 0 (n + 1 2), n = 0, 1, 2, where ω 0 = κ f μ Energy levels are equally spaced at intervals of ħω 0. In spectroscopy vibrational frequencies are given in terms of the spectroscopic wavenumber, ν = ω 0 2πc 0 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 8 / 26

Harmonic Oscillator Energy Levels -2-1 0 1 2 Ground state with n = 0 has zero point energy of 1 2 ħω 0. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 9 / 26

Maximum displacement of classical harmonic oscillator 10 8 6 4 2 0-1.0-0.5 0.0 0.5 1.0 Maximum displacement of classical harmonic oscillator in terms of energy x max = 1 ω 0 2E m Combined with E n = ħω 0 (n + 1 2) we obtain corresponding x max : The classical turning point for each oscillator state ξ max n = αx max n = 2n + 1 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 10 / 26

Harmonic Oscillator Wave Functions 1.0 0.5 1.0 0.5 nth oscillator state has n nodes. 0.0-0.5-1.0 1.0 0.5 0.0-0.5 0.0-0.5-1.0 1.0 0.5 0.0-0.5 approximate roots of the Hermite polynomials in earlier slide. Note similarities and differences with infinite well. -1.0 1.0 0.5-1.0 1.0 0.5 nth oscillator state corresponds to (n + 1)th infinite well state. 0.0-0.5-1.0 1.0 0.5 0.0-0.5-1.0 1.0 0.5 Tiny horizontal arrows represent classical oscillator displacement range for same energy. 0.0-0.5-1.0 1.0 0.5 0.0-0.5-1.0-4 -2 0 2 4 0.0-0.5-1.0 1.0 0.5 0.0-0.5-1.0-4 -2 0 2 4 Gray represents classically excluded region, ξ > ξ max n Finite potential leads to wave function penetration into classically excluded region. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 11 / 26

Classical Oscillator Turning Points As n increases probability density function approaches that of classical harmonic oscillator displacement probability (gray line) shown with the n = 112 oscillator 1.0 0.8 classical limit 0.6 0.4 0.2 0.0-15 -10-5 0 5 10 15 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 12 / 26

Web App - 1D QM simulation of single bound particle Link here: 1D Quantum Wells Solves Schrödinger equation and shows solutions. Default is infinite square well (zero everywhere inside, infinite at edges). Top is graph of potential and horizontal lines show energy levels. Below is probability distribution of particle s position, oscillating back and forth in a combination of two states. Below particle s position is graph of momentum. Bottom set of phasors show magnitude and phase of lower-energy states. To view state, move mouse over energy level on potential graph. To select a single state, click on it. Select single state by picking one phasor at bottom and double-clicking. Click on phasor and drag value to modify magnitude and phase to create combination of states. Select different potentials from Setup menu at top right. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 13 / 26

Harmonic Oscillator Wave Functions χ n (ξ) = A n H n (ξ)e ξ2 2, where A n 1 2 n n! π 1 2 Rewriting in terms of x gives ψ n (x) = N n H n (αx)e (αx)2 2 where N n α αa n = 2 n n!π 1 2 Lowest energy wave function has form of Gaussian function α 2 ψ 0 (x) = π 1 2 e (αx)2 and probability distribution that is Gaussian ψ 0 (x)ψ 0(x) = α π e (αx)2 with standard deviation of Δx = 1 (α 2). P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 14 / 26

Integrals involving Hermite polynomials Hermite polynomials with even n are even functions while those with odd n are odd functions. Keep this in mind when evaluating integrals. Example Calculate x for harmonic oscillator wave function. Starting with integral x = ψ n (x) xψ n(x)dx = xψ n (x)ψ n(x)dx, Note that ψ n are real, so ψ n (x) = ψ n(x). Since ψ n (x) is either even or odd depending on whether n, then product, ψ n (x)ψ n(x) = ψ n (x)ψ n (x) is always even. Therefore xψn 2 (x) is always odd and we obtain x = 0. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 15 / 26

Integrals involving Hermite polynomials Example Calculate p for harmonic oscillator wave function. Calculate p = ψ n (x) pψ n(x)dx = iħ ψ n (x) ψ n(x) dx. x The derivative of an odd function is even The derivative of an even function is odd Integrand is product of even and odd functions Thus, integrand is odd function and therefore integral is zero, that is, p = 0. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 16 / 26

Integrals involving Hermite polynomials A useful integral involving the Hermite polynomials is A m A n H m (ξ)h n (ξ)e ξ2 dξ = δ m,n Hermite polynomials also obey two useful recursion relations H n+1 (ξ) 2ξH n (ξ) + 2nH n 1 (ξ) = 0 and dh n (ξ) dξ = 2nξH n 1 (ξ) P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 17 / 26

Example Using the recursion relations above show that x = 0. Since ξ = αx this is identical to ξ = 0. Start with ξ = (H n e ξ2 2 )ξ(h n e ξ2 2 )dξ = H n ξh n e ξ2 dξ Using recursion relation we find ξh n = 1 2 H n+1 + nh n 1, Substituting back into our integral we obtain ( ) 1 ξ = H n 2 H n+1 + nh n 1 e ξ2 dξ = 1 2 H n H n+1 e ξ2 dξ + n H n H n 1 e ξ2 dξ Since n n ± 1, both integrals are zero and thus ξ = 0. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 18 / 26

Creation and Annihilation Operators Nearly every potential well can be approximated as harmonic oscillator Describes situations from molecular vibration to nuclear structure. Quantum field theory starting point basis of quantum theory of light. Consider harmonic oscillator Hamiltonian written in form = ħ2 d 2 2μ dx + 1 2 μω2 2 0 x2 We now define two non-hermitian operators ( ) ( ) μω0 p μω0 p â + = x i and â 2ħ μω = x + i 0 2ħ μω 0 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 19 / 26

Creation and Annihilation Operators Calculate product â â + and rearrange to â + â = μω 0 2ħ â + â = 1 ħω 0 ( ) x 2 i x p i p x + + p2 μω 0 μ 2 ω 2 ( ) p 2 μω2 0 x2 + 2μ 2 + i ( x p p x) 2ħ [ x, p]=iħ Recognizing the commutator in the last term we obtain and obtain â + â = ħω 0 + i 2ħ (iħ) = ( = ħω 0 â + â + 1 ) 2 ħω 0 1 2 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 20 / 26

Creation and Annihilation Operators Both and â â + have same ψ n (x) as eigenstates. We further define the number operator as N = â + â where Nψ n (x) = nψ n (x) and = ħω 0 ( N + 1 2 ) Similarly you can show that â â + = and identify â â + = N + 1 + 1 ħω 0 2 or = ħω 0 (â â + 1 ) 2 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 21 / 26

Creation and Annihilation Operators Look at effect of applying â + on eigenstates of. ( If ψ = â + ψ n then ψ = (â + ψ n ) = ħω 0 â + â + 1 ) (â 2 + ψ n ) and ( ψ = ħω 0 â + â â + + 1 ) ( ψ n = â + ħω 0 â â + + 1 ) ( ) ψ 2â+ 2 n = â + + ħω 0 ψn +1 ħω 0 Then we can write ψ = (â + ψ n ) = â + ( + ħω 0 ) ψn = â + ( E + ħω0 ) ψn = ( E + ħω 0 ) (â+ ψ n ). We learn that energy of â + ψ n is ħω 0 higher than E, the energy of ψ n. Effect of â + on ψ n is to change it into ψ n+1 with E n+1 = E n + ħω 0. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 22 / 26

Creation (Raising) and Annihilation (Lowering) operators Similarly, one can show that â applied to ψ n turns it into ψ n 1. â + and â are called Creation and Annihilation operators, respectively. â + and â are also called Raising and Lowering operators, respectively. Without proof, coefficients that maintain normalization of the wave functions when applying â ± are â + ψ n = (n + 1)ψ n+1 and â ψ n = nψ n 1 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 23 / 26

Fun trick with Creation and Annihilation operators As we can t go any lower than n = 0 we must have â ψ 0 = 0 We can use this to determine ψ 0 Since ( μω0 p x + i 2ħ μω 0 Expanding and rearranging gives Integrating gives dψ 0 dx = μω 0 ħ xψ 0 ) ψ 0 = 0 dψ 0 = μω 0 ψ 0 ħ xdx ψ 0 = A 0 e (μω 0 ħ)x2 2 Recalling (μω 0 ħ) = μκ f ħ 2 = α 2 gives ψ 0 = A 0 e α2 x 2 2 = A 0 e ξ2 2 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 24 / 26

Fun trick with Creation and Annihilation operators Normalizing ψ 0 with integral ψ 0 (x)ψ 0(x)dx = A 0 2 e α2 x 2 dx = 1 gives A 0 2 = α π so we have ψ 0 = α1 2 π 1 4 e α2 x 2 2 From ψ 0 we can use â + to generate all higher energy eigenstates. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 25 / 26

Example Use â + to generate ψ 1 (x) from ψ 0 (x). [ ) α ψ 1 = â + ψ 0 = ( x ] p α i 1 2 [ ] 2 μω π 1 4 e α2 x 2 2 = α3 2 1 + ħα2 xe α2 x 2 2 2π 1 4 μω Check that ħα 2 = ħ ( μκf ) 1 2 (μκ = f ) 1 2 μω 0 μω 0 ħ 2 = (μκ ( ) 1 2 f )1 2 μ = 1 μω 0 μ κ f Thus we obtain ψ 1 (x) = α3 2 2xe α2 x 2 2 = 2π 1 4 α1 2 2π 1 4 A 1 (2αx) H 1 (αx) e α2 x 2 2 Although tedious you can find all Hermite polynomials this way. P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 26 / 26