SIMPLE QUANTUM SYSTEMS Chapters 14, 18 "ceiiinosssttuu" (anagram in Latin which Hooke published in 1676 in his "Description of Helioscopes") and deciphered as "ut tensio sic vis" (elongation of any spring is proportional to the force applied published in "Lectures de Potentai Restitutiva, orofspring" in 1678) Robert Hooke, 1676 Overview of Harmonic Oscillator, (14.3, 18.3) Asystem of reduced mass µ subject to the potential: V = 1 2 kx2 where k is the force constant Hooke s law: F x = kx Bohr's Correspondence Principle: The behavior of systems described by quantum mechanics reproduces classical physics in the limit of large quantum numbers HAMILTONIAN Ĥ = ˆT + ˆV = ˆp2 x 2µ + 1 2 k ˆx2 = h2 2µ dx 2 + 1 2 kx2 The above oscillations of the probability density has practical consequences only if it is an observable. As the quantum number becomes larger the distance between peaks becomes smaller. At high enough υ, the uncertainty principle will prevent resolving the gaps without altering the physical state of the oscillator. The exponential tails into the classically forbidden region decrease with increasing υ, so the quantum oscillator looks more and more like the classical one. GENERAL SOLUTION multiply through by 2µ/h 2 Ĥψ (x) = h2 2µ let α = µk/h 2,(text, bottom of p. 349) Limitations of the Harmonic Oscillator Model 1. Equal energy spacing => all transitions occur at same frequency (single line spectrum). Experimentally many lines can be observed (called overtones, e.g., υ = 0 > υ = 2) 2. Model does not predict bond dissociation. dx 2 + 1 2 kx2 ψ (x) = Eψ (x) (14.35) dx kµ 2 h x2 2 ψ (x) = 2µ Eψ (x) h2 dx 2 α 2 x 2 ψ (x) = 2µ Eψ (x) h2 The general solution is a gaussian times a Hermite polynomial, H υ ( α x) Morse Potential (text, Fig. 18.6) Explicitly includes the effects of bond breaking and accounts for the anharmonicity of real bonds giving rise to overtone and combination bands. ψ υ (x) = N υ H υ ( α x) e α x2 /2 for n = 0, 1, 2,... (14.36) and E υ = υ + 1 2 hν 0 where ν 0 = 1 1/2 k (oscillator frequency -even classically) (14.39, 14.40) 2π µ Note: even (constant) energy level spacing: E = hν 0 zero point energy: ½ hν 0 heavier mass, ν and E 0 also weaker force constant, ν and E 0 } classically
NORMALIZATION ψ *(x)ψ (x)dx = -2- ORTHOGONALITY OF THE HERMITE POLYNOMIALS, H υ (y) H υ(y)h υ (y) e y2 dy = 2 υ υ! π 0 if υ = υ if υ υ N υ H υ ( α x)e α x2/2 N υ H υ ( α x)e α x2/2 dx let y = α x, dy = α dx = N 2 1 υ α H 2 υ (y)e y2 dy WAVEFUNCTION = N 2 1 υ α 2υ υ! π = 1 => N υ = α 1 π 1/4 2 υ υ! 1/2 (14.37) Note: odd - even progression number of solutions, = 0, 1, add exponential term, exp[-y 2 /2], to get damping since wavefunction must remain bounded (14.38) tunneling
-3- EX 1. Calculate the frequency and wavelength of the radiation absorbed when a quantum oscillator whose characteristic frequency, ν 0, is 3.15 10 13 s -1 makes a transition from the υ = 2 to the υ = 3 state. CAREFUL! There are two frequencies here - the frequency you are to calculate for the photon and the fundamental frequency of the harmonic oscillator. E photon = hν = E H.O. = hν 0. And the wavelength is obtained in the usual way: νλ = c. [ν = 3.15 10 13 s -1, λ = 9.52 10-6 m] EX 2. The vibrational frequency of 35 Cl 2 is 1. 68 10 13 s 1. What is the force constant for this molecule? Solve for k from the equation for the frequency of the oscillator, Eq. (14.40). You can use dimensional analysis to determine the units of the force constant since F = - kx = ma. [324 kg s -2 (unit is also N m -1 where the unit for force is a newton, N)] EX 3. The force constant for 35 Cl 2 is 323 N m 1.Determine its vibrational zero point energy. From Eq. (14.39) one obtains the energy for any level υ. For the zero point energy υ = 0 so Ε = ½ hν with the frequency ν again being given by (Eq. 14.40). [5.56 10-21 J]
EX 4. The strongest infrared band of 1 H 35 Cl occurs at ~ ν = 2991 cm -1. Find the force constant in SI units. [516.3 N m -1 ] You are not responsible for the specifics of this section but do know why the Morse oscillator is a more realistic potential. Anharmonic Oscillator (18.3) In real molecules nuclei repel each other upon approach. Also, a bond can break. A more realistic potential is the Morse. V = D e (1 e b x ) 2 (18.4) Anharmonicity increases with υ, decreasing spacing between vibrational levels and increasing the bond length. E υ = ( υ + 1 / 2 ) ν x e ν e ( υ + 1 / 2 ) 2 (18.5) where x e ν e is the anharmonicity constant. The "correction" is quadratic in υ and lowers the energy. Comparison of the harmonic and Morse potentials (from p. 1 of these notes and text Fig. 18.6). Comparison of Harmonic and Anharmonic Oscillator Models for HCl Vibrational Energies transition ν obs (cm -1 ) ν calc (cm -1 ) harmonic oscillator Morse oscillator 0 > 1 2 885.9 2 885.9 2 885.70 0 > 2 5 668.0 5 771.8 5 668.20 0 > 3 8 347.0 8 657.7 8 347.50 0 > 4 10 923.1 11 543.6 10 923.6 0 > 5 13 396.5 14 429.5 13 396.5
You are not responsible for this section. It is only to make contact with PChem lab. Rigid Rotor/Harmonic Oscillator (18.4) For a molecule which is both vibrating and rotating E υ,j = ( υ + 1 / 2 ) ~ ν + B ~ J(J + 1) the infrared spectrum for the υ = 0 υ = 1 transition is not merely one peak. A separate peak is observed for each rotational level in υ = 0 which is allowed to go to a higher or lower value of J in the excited υ = 1 vibrational state. Those going to a higher J value in the excited state have more energy than the υ = 0 υ = 1 transition excluding rotation and those going to a lower J value have an energy less than the υ = 0 υ = 1 transition excluding rotation. Rotating harmonic oscillator energy levels and transitions (text Fig. 18.8). The observed vibration/rotation spectrum consists of a vibrational level which is generally split into a series of two sets of peaks: 1) a P branch where ΔJ = 1 an) an R branch where ΔJ = +1. Then for the fundamental transition ( υ = 0 υ = 1) the wavenumbers of the branches are: ~ E J J 1 ~ ~ νp = = ν 2BJ hc ~ E J J + 1 ~ ν R = = ~ ν 2B( J + 1) hc Rotational/Vibrational infrared spectrum of carbon monoxide.