Download Harmonic Oscillator: Understanding Finite Potential Wells and Solutions and more Study notes Quantum Mechanics in PDF only on Docsity! Harmonic Oscillator – Levine Ch 4/Atkins 3.5/R&S 5.1/Dykstra 2.2 Talked about ∞ potential wells and finite ones now ∞ well with sloping sides: V = (1/2) kx2 so V = ∞ only at x = ∞ , since V(x) is finite expect: a. Ψ(x) penetrate walls but then damp to zero on both sides (finite) b. in between walls Ψ(x) must be nonzero for some x, so Ψ(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 Ψ(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. Insert Graph Solution—can read about details in text (Levine 4.2) ==> H (x) = (T + V) (x) = [ -(h2/2m) d2/dx2 +kx2/2] (x) = E (x) -(h2/2m) [d2/dx2 + α2x2]Ψ(x) = E Ψ(x) where α2 = mk/h2 [d2/dx2 +( 2mE/ h2 - α2x2)]Ψ(x) = 0 again recall h = h/2 Recall that solutions Ψ(x) must damp to zero as x increase (pass barrier) One way to do this is multiply oscillating function by exp(-αx2/2) form e-cx_ would damp, but need x2 for ±x sym. (even) (c= α/2 arbitrary) If Ψ(x) = e -αx2/2 f(x) then second derivative above reduces to: [d2/dx2] (x) = exp(-αx2/2) [(α2x2)f(x) + 2(-αx)f(x) (df/dx) - αf(x) + (d2f/dx2)] Now could eliminate exp(-αx2/2) from Schroedinger Eqn since in each term: ( x2)f(x) + 2(- x)f(x) [df/dx] - f(x) + [d2f/dx2] +( 2mE/ h2 - 2x2) f(x) = 0 Results in 2nd order eqn: [d2f/dx2] - 2( x)f(x) [df/dx] + ( 2mE/ h2 - ) f(x) = 0 This can be solved with a power series solution : f(x) = Σ cnxn Solution has parity Ψ(x) = e -ax2/2 [c0 + c1x + c2x2 +c3x3 + . . .] As it must, [Π, 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 v(x) = Nv exp(- x 2/2) Hv(x) = Nv exp(-y 2/2) Hv(y) --where y= 1/2x H0(y) = 1 H4(y) = 16y 4 - 48y2 + 12 even H1(y) = 2y H5(y) = 32y 5 - 160y3 + 120y odd H2(y) = 4y 2 – 2 H6(y) = 64y 6 - 480y4 + 720y2 – 120 even H3(y) = 8y 3 – 12y H7(y) = 128y 7 – 1344 y5 + 3360y3 – 1680y odd INSERT GRAPH See term oscillate in sign—causes oscillation of Ψv (x) v = 0 just exponential decay from 1 at x=0 v = 1 odd, Ψ = 0 at x = 0 then inc + and - and damp with opposite sign curvature v = 2 even - Ψ = -2 at x = 0, then curve positive (E>V), but as x inc. 4y2> |-2| so Ψ becomes (+), opposite curvature then needed for decay--oscillation INSERT GRAPH 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)hν è ν = (1/2π) √k/m or α = 2πνm/h INSERT GRAPH So see steady, uniform increase of E with v With v+1 – Ev = h , where ν is the frequency of the oscilllator v=0 à E0=hν/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 INSERT GRAPH Potential energy, diatomic: U(R) = <Hel(R)> + Vnn(R) where Vnn nuclear repulse Electrons hold atoms together nuclei repell (+ ßà +) 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( R)2 where ∆R=R-Reg “harmonic approximation” -- (note power series expansion):
