5.61 Lecture 8: Quantum Mechanical Harmonic Oscillator

Lecture #8 Page 1 Lecture #8: Quantum Mechanical Harmonic Oscillator Last time Classical Mechanical Harmonic Oscillator 1 * V (x) = kx 2 (leading term in power series expansion of most V(x) potential energy 2 functions) * x is displacement from equilibrium (x = 0 at equilibrium) 1/2 * angular frequency ω = [ k µ] mm * µ = 1 2 reduced mass m1 + m2 d2x k From F = ma we get 2 = − x dt m Classical Mechanics] [we get x(t) from this, not from ψ(x) because this is x(t) = Asinωt + B cosωt = C sin ( ωt + φ ) : general solution get (A,B) or (C,φ) from initial conditions of “pluck” 1/2 ⎛ 2E ⎞ turning points x± (E) = ± ⎜ from E = V ( x± (E)) ⎝ k ⎟⎠ ν,ω,τ definitions and inter-relationships T (t),T (kinetic energy) ⎫ ⎬ overbar means average value V (t),V (potential energy) ⎪⎭ Today * * * * * simplify Schrödinger Equation to get rid of constant factors solution: Gaussian envelope × Hermite polynomial pictures: at least as important as the mathematical form semiclassical interpretation (not in most texts): combination of classical mechanics with quantum mechanics via λ(x) = h/p(x) (a unique and never-ending source of insight) vibrational transition intensities and “selection rules” Quantum Mechanical Harmonic Oscillator (McQuarrie, Chapters 5.5, 5.8-10) 2  = T + V = p̂ + 1 kx̂ 2 H 2µ 2 =− 2 ∂ 1 + kx̂ 2 2 2 µ ∂x 2 Lecture #8 Page 2 We can “clean up” this equation by making the substitution where α = ( kµ) 1/2 ξ = α1/2 x  (ξ is dimensionless, which makes the equation “universal”)  2 ∂2  2 ∂2 ⎛ k⎞ = α 2= ⎜ ⎟ 2 2 µ ∂x 2 µ ∂ξ 2 ⎝ µ⎠ 1/2 ∂2  ∂2 = ω ∂ξ 2 2 ∂ξ 2 ∂ ∂ ∂ξ ⎤ ⎡ because = ⎢ ∂x ∂ξ  ∂x ⎥ ⎢⎣ ⎥ α1/2 ⎦ 1 2 1 ⎛ 1⎞ 2 kx = k ⎜ ⎟ ξ 2 2 ⎝ α⎠ 1⎛ k⎞ = ⎜ ⎟ 2 ⎝ µ⎠ 1/2 ξ 2 = 1 ωξ 2 2 2 ! = "ω ⎡ − ∂ + ξ2 ⎤ H ⎥ 2 ⎢⎣ ∂ξ2 ⎦ ( much simpler form ) ! ψ(ξ) = Eψ ( ξ ) H "ω ⎞ ⎛ ⎜⎝ rearrange and divide by ⎟ 2 ⎠ ⎡ ∂2 2E ⎤ 0 = ⎢ − 2 + ξ2 − ψ ( ξ) !ω ⎥⎦ ⎣ ∂ξ (entire differential equation, except ψ, is dimensionless) One can convert this into the Hermite differential equation by making the substitution ψ ( ξ ) = e− ξ 2 2 f ( ξ ) and finding a new differential equation for f(ξ). The reason for doing this is that the e− ξ 2 factor ensures that ψ → 0 as |ξ| → ∞. Note that letting ξ → ∞ means that 2E/hω is negligible with respect to ξ . What is the solution to the differential equation if we ignore the 2E term? ω 2 2 2 This is a very clean form of the Schrödinger equation because all of the k,µ-specific factors are absorbed into a dimensionless ξ variable. Why would we want this? The Hermite polynomials (in integer powers of ξ) are solutions to the differential equation d2Hn dH n − 2ξ + 2nH n = 0 . 2 dξ dξ (Hermite equation. Hn is a polynomial in ξ.) There are two very convenient “recursion relations” that relate the Hn–1 to the Hn, etc. Lecture #8 Page 3 dH n = 2nH n−1 ( ξ ) . dξ This looks like the effect of p̂ on Hn (because the leading term in Hn is ξn). 2. H n+1 ( ξ ) = 2ξH n ( ξ ) − 2nH n−1 ( ξ ) , rearranging 1. 1 H (ξ) + nH n−1 (ξ). 2 n+1 This looks like the effect of x̂ on Hn. We will use this second recursion relation to compute integrals of the form ξH n (ξ) = ∫ dξ ψ *n ξ m ψ p . (n, m, p are integers). These recursion relations enable us to evaluate all integrals of the form ∫ ψ v ( x̂ n p̂ m ) ψ v+ dx . (We will postpone the actual evaluation until next lecture when we will also derive the “selection rule” for nonzero integrals): l = n + m, n + m – 2, … – (n + m). l is an integer and goes from its maximum value of n + m down to its minimum value in steps of 2. There is also a general expression (Rodrigues formula) for all of the Hn H n ( ξ ) = ( −1)n eξ 2 d n − ξ2 e dξ n The Hermite equation is a well known (to mathematicians) differential equation. The solutions of the harmonic oscillator Schrödinger equation are Lecture #8 Page 4 ⇠ = ↵1/2 x Gaussian envelope ensures ! 0 as x!±1 z }| { ⇣ ↵ ⌘1/4 ? ⇠ 2 /2 1 Hv (⇠) e v (x) = | {z } (2v v!)1/2 ⇡ 6 | {z } 6 Nv vibrational quantum Normalization Hermite polynomial number v = 0, 1, 2, … (vee, not nu!) * Normalized * ψv(±∞) = 0 * ψv(0) = 0 for odd-v (odd function) dψ v * = 0 for even-v (even function), to be worked out dx x=0 Ev = hω(v + ½) What do we know about orthogonality? Based on results derivable from postulates? For non-degenerate eigenvalues: ∫ dxψ v*ψ v′ = δ vv′ orthonormal Semi-Classical Picture – applicable to all 1-dimensional problems as a basis for insight, prediction, and obtaining ψ(x) without solving any differential equation. Classical: T(x) = E – V(x) = p(x)2/2µ p(x) = [2m(E – V(x))]1/2 p(x) is momentum in classical mechanics but just a convenient function in quantum mechanics. ⊕ Quantum: h p valid not just for free particle or a piecewise constant potential de Broglie λ = = Lecture #8 Semi-classical: λ(x) ≡ Page 5 h pclassical (x) * pair of nodes nearest to x are spaced by λ(x)/2. Qualitative Shapes of ψv(x): * * * * * exponentially damped envelope, extending into non-classical (E < V(x)) regions (!!!!) oscillations within classically allowed region with number of internal nodes equal to the quantum number even v, even function, antinode at x = 0 outer lobes (near x+ and x–) are largest [see McQuarrie, page 226, Fig. 5.10 right side] envelope within classically allowed region resembles what you expect from classical mechanics dx (slow speed↔high probability) ψ * (x)ψ(x)dx ∝ v classical 1 1/2 vclassical = pclassical µ = {2 µ[ E − V (x)]} µ To derive the proportionality constant, consider the fraction of time the particle is found between x and x + dx: time ( x → x + dx ) ⎧⎪ probability of finding particle moving ⎫⎪ dx vclassical ( x ) =⎨ ⎬= time ( x− to x+ ) to right between x and x + dx τ/2 ⎪⎩ ⎪⎭ τ is the quantum number independent period of the harmonic oscillator. 1 2π ⎛ τ= = = 2π ⎜ ⎝ ν ω ψ *(x)ψ(x)dx = µ⎞ ⎟ k⎠ 1/2 [ v = ω / 2π, ω = [k / µ] ] 1/2 dx ( vclassical ) ( τ 2 ) 1/2 * = dx 1/2 ⎧2 ⎫ ⎨ [ E − V (t)]⎬ ⎩µ ⎭ 1⎛ ⎜ π⎝ µ⎞ ⎟ k⎠ −1/2 1/2 ⎡ k / 2π 2 ⎤ =⎢ ⎥ dx ⎣ E − V (x) ⎦ gives the classical (i.e. nodeless) ⎡ k / 2π 2 ⎤ ψ *(x)ψ(x) = ⎢ ⎥ average of ψ *(x)ψ(x) near x ⎣ E − V (x) ⎦ (but not the phase). [To get the classical envelope, assume that the maximum value of ψ*ψ is twice the average value. [This is always a good approximation for a rapidly oscillating always positive function.] Thus the envelope of 1/2 ⎡ 2k / π 2 ⎤ ⎤ ψ*(x)ψ(x) is ⎢ ⎥ ⎥ . The envelope is the smooth curve that connects all ⎣ E − V (x) ⎦ ⎥⎦ of the maxima of ψ*(x)ψ(x) between x– and x+. node spacing Lecture #8 (1) (2) (3) Page 6 h , use classical ideas to qualitatively locate nodes, p(x) nodes are closest together when p is largest (near x = 0). Knowledge of the envelope and node spacing allow you to sketch ψ∗(x)ψ(x) without solving a differential equation. ∆x 1 = (want to find value of ∆x that is or compute a “phase integral” λ(x) 2 equal to λ/2, the distance between nodes) recall λ = replace λ(x) by h p(x). We get p(x)∆ x = h as the distance, ∆x, between nodes 2 ⎡ ⎢⎣ more accurately, ∫ x+λ/2 x h⎤ dx p(x) = ⎥ . 2⎦ 2 x2 pE (x)dx tells us how many nodes there are between x1 h ∫x1 and x2 at energy E. This is the same as knowing how many bound energy levels lie at or below E. The phase integral Lecture #8 What about pictures of ψ *(x)ψ(x) ? Figure removed due to copyright restrictions. See Figure 5.1 in: Merzbacher, Eugen. Quantum Mechanics. Wiley, 1997. ISBN: 9780471887027. * non-classical tails [(extend into region where E < V(x)] * x+ lobe positive by convention (x– lobes alternate sign as –1v) * lobes nearest x– and x+ largest * nodes closest together near x = 0 * no zero crossings in classically forbidden region Non-Lecture What do we do with these HO wavefunctions? 1. calculate relative intensities of vibrational transitions Page 7 Lecture #8 2. Page 8 Use perturbation theory (Lectures #14, #15 and #18) to compute consequences of higher than quadratic terms in V(x) e.g. for Morse oscillator Ev  ( v + 1 / 2 )2 = ω ( v + 1 / 2 ) − ωx hc (~ means cm–1 units) “anharmonicity”, comes mostly from x3 and x4 terms in V(x) Spectral intensities I fi ∝ ∫ dx ψ µ( x ) ψ vi vvf f 2 * vf hν electric dipole moment (HCl vs. H2, N2, O2, Cl2) (think of radio antenna) ∆v = 0 pure rotation spectrum spectrum ∆v = ±1 vibrational fundamental Two contributions to vibrational overtone transitions * mechanical anharmonicity (Morse potential) * electronic anharmonicity (higher derivatives of µ(x)) ∆v = 0, ±2 vibrational overtone vvi i Lecture #8 Page 9 In the following we will look only at the electronic anharmonicity contributions. ∫ 1 dxψ v*f ⎡⎢ µ0 + µ1 x + µ2 x 2 ⎤⎥ψ vi = µ0 ∫ dx ψ v*f ψ vi  ⎣ ⎦ 2 orthonormal δv v f i µ + µ1 ∫ dx ψ v*f xψ vi + 2    2 ∫ dx ψ v*f x 2ψ vi see recursion relationship Recursion Relationships H n+1 (ξ) = 2ξH n (ξ) − 2nH n−1 ( ξ ) 1 ξH n (ξ) =  H n+1 (ξ) + n  H n−1 ( ξ )  2 ψ ψ xψ v v+1 v−1 selection rule: ∆v = ±1 for x2 term selection rules (evaluate in two steps) 1 ξH n+1 + nξH n−1 2 1⎛ 1 ⎞ ⎛1 ⎞ = ⎜ H n+2 + (n + 1)H n ⎟ + n ⎜ H n + (n − 1)H n−2 ⎟ ⎠ ⎝2 ⎠ 2⎝ 2 ∆ v = 0, ±2 ξ2 H n = Next time: a†, a treatment