5.61 Fall, 013 Lecture #8 Page 1 Last time Lecture #8: Quantum Mechanical Harmonic Oscillator Classical Mechanical Harmonic Oscillator * V(x) = 1 kx (leading term in power series expansion of most V(x) potential energy functions) * x is displacement from equilibrium (x = 0 at equilibrium) * angular frequency ω = [k μ] 1/ m 1 m * μ = reduced mass m 1 + m d x k From F = ma we get = x [we get x(t) from this, not from ψ(x)] dt m x(t) = Asin ωt + Bcos ωt = C sin (ωt + φ) get A,B or C,φ from initial conditions of pluck E 1/ turning point x ± (E) = ± from E = V (x ± (E)) k ν,ω,τ T(t),T (kinetic energy) V (t),v (potential energy) Today * simplify Schrödinger Equation to get rid of constant factors * solution: Gaussian envelope Hermite polynomials * pictures * semiclassical interpretation (not in most texts): combination of classical mechanics with λ(x) = h/p(x) (a unique source of insight) * vibrational transition intensities and selection rules Quantum Mechanical Harmonic Oscillator (McQuarrie, Chapters 5.5, 5.8-10) pˆ H = T + V = + kxˆ μ ħ 1 = + kxˆ μ x 1 We can clean up this equation by making the substitution
5.61 Fall, 013 Lecture #8 Page ξ=α 1/ x where α = kμ ħ (ξ is dimensionless, ħ ħ 1/ ħ k ħ = α = = ω μ x μ ξ μ ξ ξ ξ because = x ξ x ax α 1/ 1 1 1 kx = k ξ α 1 k 1/ 1 = ħ ξ = ħ ωξ μ ( ) 1/ which makes the equation universal ) ħ ω H = +ξ ( much simpler form) ξ H ψ(ξ) = Eψ ξ E (entire differential equation, except 0 = +ξ ħ ξ ψ ω ψ, is dimensionless) One can convert this into the Hermite differential equation by making the substitution ψ ( ξ) = e ξ f ( ξ ) and finding a new differential equation for f(ξ). The reason for doing this is ξ that e ensures that ψ 0 as ξ. Note that letting ξ means that E/ ħ ω is negligible with respect to ξ. What is the solution to the differential equation if we ignore the E term? ħ ω 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 ( ) d H n dξ dh ξ n + nh n = 0. (Hermite equation) dξ There are very convenient recursion relations that relate the H n 1 to the H n, etc. dh n = nh n 1 ( ξ ). dξ This looks like the effect of pˆ on H n (because the leading term in H n is ξ n ).
5.61 Fall, 013 Lecture #8 Page 3 H n+1 ( ξ ) = ξh n ( ξ ) nh n 1 ( ξ ), rearranging 1 ξh n (ξ) = H n+1 (ξ) + nh n 1(ξ). This looks like the effect of xˆ on H n. We will use this second recursion relation to compute integrals of the form d ξ ψ* nξ m ψ p. (n, m, p are integers). These recursion relations enable us to evaluate all integrals of the form m ψ v (xˆn pˆ ) ψ v+fdx (we will postpone the actual evaluation until next lecture) and we derive the selection rule for nonzero integrals f = n + m, n + m, (n + m) There is also a general expression (Rodrigues formula) for the H n ( ) 1 d n n ξ ξ H n ξ = ( ) e dξ e n The Hermite equation is a well known (to mathematicians) differential equation. The solutions of the Schrödinger equation are
5.61 Fall, 013 Lecture #8 Page 4 E v = ħ ω(v + ½) 1/ kμ α = Gaussian! ħ envelope ensures ξ = α 1/ x ψ 0 as s x o± 1 α 1/4 ξ ψ v (x) = 1/ H ( v ξ ) e ( v v!) π a x!######### " vibrational quantum number v = 0, 1,, N v Normalization * Normalized * ψ v (± ) = 0 * ψ v (0) = 0 for odd-v (odd function) dψ * v = 0 for even-v (even function), to be worked out dx x=0 Hermite polynomials What do we know about orthogonality? Based on results derivable from postulates? Non-degenerate eigenvalues. dxψ * v ψ v = δ vv orthonormal Semi-Classical Picture applicable to all 1-dimensional problems for insight, prediction, and obtaining ψ(x) without solving any differential equation. Classical: T(x) = E V(x) = p(x) /μ p(x) = [m(e V(x))] 1/ p(x) is momentum in classical mechanics but just a convenient function in quantum mechanics. Quantum: h de Broglie λ= p valid not just for free particle or a piecewise constant potential
5.61 Fall, 013 Lecture #8 Page 5 = Semi-classical: λ(x) h p classical (x) Qualitative Shapes of ψ v (x): * pair of nodes nearest to x are spaced by λ(x)/. * exponentially damped envelope, extending into non-classical 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 6, Fig. 5.10 right side] * envelope within classically allowed region resembles what you expect from classical mechanics dx ψ *(x)ψ(x)dx (slow speed high probability) v classical 1 1/ v classical = p classical μ = { μ[e V (x)]} μ To get 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 v classical ( x ) = = time (x to x + ) to right between x and x + dx τ / 1 π μ 1/ τ= = = π v = ω/π, ω =[k / μ] 1/ ν ω k ψ *(x)ψ(x)dx = k /π 1/ ψ *(x)ψ(x) = E V (x) 1/ 1/ dx dx 1 μ k /π ) = 1/ = dx ( v classical )(τ π k E V(x) [E V (t)] μ gives the classical average of ψ *(x)ψ(x) near 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 k / π 1/ ψ (x)ψ(x) envelope is. E V (x)
5.61 Fall, 013 Lecture #8 Page 6 * node spacing h (1) recall λ=, use classical ideas to qualitatively locate nodes, p(x) () nodes are closest together when p is largest (near x = 0), envelope and node spacing allows you to sketch ψ (x)ψ(x) without solving a differential equation. Δ x 1 (3) or compute a phase integral = (want to find value of Δx that is λ(x) equal to λ/, the distance between nodes) replace λ(x) by h p(x). We get h x+λ/ h p(x)δ x = as the distance, Δx, between nodes more accurately, dxp(x) = x x phase integral pe (x)dx tells us how many nodes there are between x 1 and x h x 1 at energy E. This is the same as knowing how many bound energy levels lie at or below E.
5.61 Fall, 013 Lecture #8 Page 7 McQuarrie, Donald A. Quantum Chemistry. nd ed. Sausalito, CA: University Science Books, 007. Used with permission. * non-classical tails [(extend into region where E < V(x)] * x + lobe positive by convention * lobes nearest x and x + largest * nodes closest together near x = 0 * no zero crossings in classically forbidden region What about pictures of ψ *(x)ψ (x)? Non-Lecture What do we do with these HO wavefunctions? 1. calculate relative intensities of vibrational transitions
5.61 Fall, 013 Lecture #8 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 Spectral intensities E v = ωr (v +1/ ) ω x (v +1/ ) ( ~ means cm 1 units) hc anharmonicity, comes mostly from x 3 and x 4 terms in V(x) I fi dx ψ* v f μ( x )ψ v i hν v f electric dipole moment (HCl vs. H, N, O, Cl ) (think of radio antenna) v i Δv = 0 pure rotation spectrum Δv = ±1 Δv = 0, ± vibrational vibrational fundamental overtone Two contributions to vibrational overtone transitions * mechanical anharmonicity (Morse potential) * electronic anharmonicity (higher derivatives of μ(x))
5.61 Fall, 013 Lecture #8 Page 9 In the following we will look only at the electronic anharmonicity contributions. * 1 dxψ μ 0 + μ 1 x + μ x ψ = μ 0 dx ψ * v f v i v f ψ v i orthonormal δ v f v i * μ + * μ1 dx ψ xψ + dx ψ v x f v i v f ψ v i j jj see recursion relationship H n+1(ξ) = ξ H n(ξ) nh n 1 ( ξ) 1 n n+1 ξ n 1 ψ ψ ξh (ξ) = H (ξ) + nh ( ) for x term (evaluate in two steps) xψ v v+1 v 1 selection rule: Δv = ±1 1 ξ H n = ξh n+1 + nξh n 1 1 1 1 = H + (n +1)H + n n+ n H n + (n 1)H n Δ v = 0, ± Next time: a, a treatment
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