Harmonic Oscillator — Levine Ch 4/Atkins 3.5/R&S 5.1/Dykstra 2.2

Talked about ƒ potential wells and finite ones now • well with sloping sides:

V = (1/2) kx2 so V = ƒ only at x = ƒ , since V(x) is finite expect:

a. Y(x) penetrate walls but then damp to zero on both sides (finite)

b. in between walls Y(x) must be nonzero for some x,

so Y(x) must have curvature è T ‚ 0 è zero point E

c. solutions must oscillate with # nodes increasing with increasing E

d. as k inc è V and T inc à zero point E will increase

as k decrease à Y(x) becomes continuum state (V~ const., flat)

e. as v n(q.n.) increase è nodes inc à T inc à V inc , results in more time at turning points, contrast to particle in box, more like classic harmonic oscillator.

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Solution–can read about details in text (Levine 4.2)

==> HY(x) = (T + V) Y(x) = [ -(h2/2m) d2/dx2 +kx2/2]Y(x) = E Y(x)

-(h2/2m) [d2/dx2 + a2x2]Y(x) = E Y(x) where a2 = mk/h2

[d2/dx2 +( 2mE/ h2 - a2x2)]Y(x) = 0 again recall h = h/2p

Recall that solutions Y(x) must damp to zero as x increase (pass barrier)

One way to do this is multiply oscillating function by exp(-ax2/2)

form e-cx_ would damp, but need x2 for ±x sym. (even) (c= a/2 arbitrary)

If Y(x) = e -ax2/2 f(x) then second derivative above reduces to: [d2/dx2]Y(x) =

exp(-ax2/2) [(a2x2)f(x) + 2(-ax)f(x) (df/dx) - af(x) + (d2f/dx2)]

Now could eliminate exp(-ax2/2) from Schroedinger Eqn since in each term:

(ax2)f(x) + 2(-ax)f(x) [df/dx] - af(x) + [d2f/dx2] +( 2mE/ h2 - a2x2) f(x) = 0

Results in 2nd order eqn: [d2f/dx2] - 2(ax)f(x) [df/dx] + ( 2mE/ h2 - a) f(x) = 0

This can be solved with a power series solution : f(x) = S cnxn

Solution has parity Y(x) = e -ax2/2 [c0 + c1x + c2x2 +c3x3 + . . .]

As it must, [P, H] = 0 since V = (1/2) kx2 è even

Oscillation will develop from the relative signs of c’s for powers of x

Solution known as Hermite polynomials to this form differential eqn

Yv(x) = Nv exp(-ax2/2) Hv(x) = Nv exp(-y2/2) Hv(y) --where y=a1/2x

H0(y) = 1 H4(y) = 16y4 - 48y2 + 12 even

H1(y) = 2y H5(y) = 32y5 - 160y3 + 120y odd

H2(y) = 4y2 — 2 H6(y) = 64y6 - 480y4 + 720y2 — 120 even

H3(y) = 8y3 — 12y H7(y) = 128y7 — 1344 y5 + 3360y3 — 1680y odd

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See term oscillate in sign–causes oscillation of Yv (x)

v = 0 just exponential decay from 1 at x=0

v = 1 odd, Y = 0 at x = 0 then inc + and - and damp with opposite sign curvature

v = 2 even - Y = -2 at x = 0, then curve positive (E>V), but as x inc. 4y2> |-2|

so Y becomes (+), opposite curvature then needed for decay--oscillation

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As v inc amplitude grows when E ~ V (i.e. T~0) called turning point, classically,

in q.m. this is due to higher powers increasing faster than lower ones.

Solving the equation: Eu = (v+1/2)hn è n = (1/2p) ÷k/m or a = 2pnm/h

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So see steady, uniform increase of E with v

With DE = Ev+1 — Ev = hn , where n is the frequency of the oscilllator

v=0 à E0=hn/2 — zero point energy, continually oscillating

or T ‚ 0 so <| p2|>‚ 0 so need <| x2|>‚ 0 which fits the uncertainty principle

Balls on springs or pendulums are nice, but can we use this for chemistry?

Harmonic Oscillator is a good model for vibration of molecules

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Potential energy, diatomic: U(R) = <Hel(R)> + Vnn(R) where Vnn nuclear repulse

Electrons hold atoms together nuclei repell (+ ß à +) keep atoms from collapse

At long distance this approaches Energy of 2 atoms- arbitrary zero

At short distance, repulsion dominate - Ueq à lowest energy at Req

Approximate bottom of well as parabola U(R) ~ (1/2)k(DR)2 where DR=R-Reg

"harmonic approximation" -- (note power series expansion):

U(R) = U(R) + dU/dR|Re(R-Re) +(1/2) d2U/dR2|Re(R-Re)2 + . . .

First term const, second zero at minimum, so the third tem is lowest order non-zero, non-constant term gives rise to harmonic motion

Higher terms eg. d3U/dR3|Re(R-Re)3 etc --Anharmonic corrections

Kinetic energy must be modified since vibration is relative motion

not interested in translation so reduce coordinates to q= DR = R-Re 1-D

And mass term modifies to: m = (m1m2/m1+m2) -- which is reduced mass

HvibYv(q) = [ -(h2/2m) d2/dq2 +kq2/2 Yv(q) = E Yv(q)

Solution just as before Eo(v +1/2)hn

extent U(R)~ kq2/2 get even spacing of levels DE = hn, where n freq vib

If differ from this, get anharmonic correction, DE < hn, spacings collapse

This happens at top of potential where curve bends to allow dissociation

IR spectroscopy (handwaving insight only - details Chem 543)

electric field of light leads to a resonant oscillator when nlight = nvib

if molecule has a change (dm/dq ‚ 0) in electric dipole moment (m)

classically this causes molecule to have larger amplitude

— in q.m. this leads to a change in quantum number, v, and energy Ev

Harmonic oscillator: Dv = ± 1 = v - v' only + à absorption, - à emission

DE = (v + 1/2)hn - (v' + 1/2)hn = hn -- all allowed transitions overlap

IR spectrum v = 0 à v' = 1 (fundamental) yields the frequency of vibration for dipolar oscillator -- frequency vary like ÷k/m

-- strong bonds higher freq C-C ~1000 cm-1, C=C ~1600 cm-1, C=C ~2200 cm-1

-- higher mass lower freq HCl ~ 2800 cm-1 , DCl ~ 2000 cm-1

*Also responsible for kinetic isotope effects: deuterate a position and if H-motion involved in reaction coordinate, the rate will decrease due to difference in zero point energy of ground and excited (transition) state INSERT GRAPH

*Also affect thermodynamics: Boltzmann Ni/Nj = gi/gj exp[-(Ei-Ej)/kT]

this gives the relative population (Ni/Nj ) of Ei + Ei energy levels, so changing (eg. lowering) frequency changes (increases) population of (higher quantum number) E- levels and increases entropy (corresponds to the # states populated)