Model Problems update Particle in box - all texts plus Tunneling, barriers, free particle Atkins (p ), House Ch 3

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Abraham Wheeler
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1 VI 15 Model Problems update 01 - Particle in box - all texts plus Tunneling, barriers, free particle Atkins (p ), ouse h 3 onsider E-M wave first (complex function, learn e ix form) E 0 e i(kx ωt) = E 0 [cos (kx - ωt) i sin (kx - ωt)] magnitude: k = π/λ ω = πc/λ =πν ν = c/λ moves in space and time traveling wave, vector k direction reflect at the node keeps the wave continuous (if not create an interference) is other cycle, at Δt=λ/c node if trap wave like violin string tied down at end standing wave (principle of laser light trap in cavity specific frequency / phase amplified) restriction- number wavelengths is integral divisor of length integer representation of frequencies not continuous (analogy to Bohr orbit, wavelength must match path) Think of traveling particle 1-D no forces V = 0 ψ = Eψ = Tψ let V = 0, free moving particle ψ= (-h /m) d /dx [ψ(x)] 15

2 16 VI 16 Solution: need some (wave)function that we can take derivative twice and get function back - eigenfunction choices: a) de ax /dx = ae ax derivative works d /dx e ax = a e ax (but energy, ψ= (ħ /m)d ψ/dx, must be positive, nd deriv. must be negative, so use: a = iα, e iαx wavefunction complex, sum: cos αx & sin αx) b) d /dx sin kx = -k sin kx (note: same for cos kx) (Note: e iαx = cos αx - i sin αx general form wave) No constraint traveling wave (but for particle) Solve Schroedinger Equation for free particle: -(h /m) d /dx ψ = Eψ if ψ = e iαx plug in -(h /m)(iα) e iαx = E e iαx from (b): α = k = (me) 1/ /h E = α h /m (all K.E. - positive, not quantized) no restrictions free particle, any energy or wavelength Note effect of momentum: (for ψ = e ikx ) pψ = -ih(ik)ψ = hk ψ Magnitude: p = hk signs direction p = hk (motion in +x direction) [ompare to: ψ = e -ikx, (opposite motion in -x)] Boundary onditions Restrictions on wave function to let fit postulates B.. relate to continuous and finite properties wave/fct properties - both sides boundary--must match

3 Particle in a box introduce simplest potential energy in box V = 0 outside V = but not in SchEqn 17 VI 17 For finite E: particle must be in box, or definite E B.. (think of as F = -dv/dx, force at wall ) -h /m d /dx ψ = Eψ try ψ = A sin αx + B cos βx Looks same but now: B.. ψ(0) = 0 restrict: B = 0 (since cos 0 = 1) ψ(l) = 0 restrict: α = nπ/l (since sin nπ = 0) idea continuous w/f, zero probability outside: ψ out = 0 for ψ(x) 0, must have: A 0, n 0 and n = 1,,3, (i.e. node both sides integral number internal lobes and be non-zero someplace- A 0 non-trivial solution) forms a standing wave -- quantized (recall - laser) -h /m d /dx (A sin nπx/l) = E (A sin nπx/l) E n = (-h /m)(- n π /L ) E n = n h /8mL h=h/π Expanding E-levels ~ n each increases number of nodes curvature restricted energy levels lowest energy 0 (particle always moving)

4 VI 18 Probability distribution: ψ*ψ dx (this is Born interpretation) ψ*ψ dx = 1 to normalize, replace A by N: ψ = Nφ = Nsin(nπx/L) L 0 or 1/N L = 0 L φ*φ dx = 0 sin nπx/l, let y=nπx/l, dy=(nπ/l)dx = (L/nπ)[ ½ y+ ¼ sin y nπ 0 = L/ N=(/L) 1/ =A but plot ψ*ψ not uniform in x n = 1 more probable in middle n = zero probability at x = L/ 1 L/ as n inc. probability evens classical Orthogonal ψ m *ψ n dx = 0 if n m L/ sin(nπx/l) sin(mπx/l) dx = 0 easiest seen graphically Amplitude? A = (/L) 1/ see normalization L A sin (nπx/l) dx = 1 A (L/) = 1 0 Probability b a ψ*ψ dx probability between a + b From above: P ab =(/L)(L/nπ)[½ y+¼ sin y nπa/l nπb/l 18 (y=nπx/l) Use for pib? Great model / see how B.. quantization Application: polyenes π-system delocalize electrons move through π-bonds ( e - per level know) spectra e - could be in different levels ΔE = E n+1 E n = hν n n + 1 ΔE = (n+1) h /8mL n h /8mL = (n + 1) h /ml = hν

5 19 VI 19 Now see properties of p.i.b-like molecule E-levels/spectra a) bigger n more separation higher frequency - hν b) bigger m less separation (but all same m e - electron) c) bigger L less separation (as square), experimental Sample dye problem: Polyene N λ max (Å) N Obs alc ΔE = (N + 1) h /8mL L 0.81 N (in nm) λ = c/ν = hc/δe = (8m e c/h)(0.81x10-9 m) N /(N + 1) units! m [to do: insert m e (for elect), h, c; when done convert to nm, 10-9 m] Note: trend is as expected N increase, λ increase (big boxes lower energy states) values off calc. change much faster than exper. -- box length approximate -- and evenness of V (real potential vary over bonds) (if use oscillating V(x) - potential in box, answer fits data) Ref: Jochen Autschbach, J.hem. Ed. 84, 1840 (007)

VI 0 Dye S N 5 N 5 N + S N Obs alc 450 380 3 5600 4540 4 6500 5800 5 7600 7060 6 8700 8330 7 9900 9600 Model does better (here use N+1 N+) and use different length, but still λ ~ N /N type term

6 VI 0 Dye S N 5 N 5 N + S N Obs alc Model does better (here use N+1 N+) and use different length, but still λ ~ N /N type term (linear) Bio-connect -Vision: retinal undergoes cis-trans isomerization N (trans) OMO LUMO difference Transition wavelength (λ=c/ν) Decrease with length increase with box length fit alculated with full QM with oscillating V(x) potential 0

7 VI 1 Ionization potential measures energy of the orbital OMO dec. from ethylene butadiene (left peak lowest) Butadiene examples real spectra shift with length structure from vibrations, absorp. inc. ΔE, fluor. dec. ΔE 1

VI -D box example π-system expand energy,

3) poly arene examples (in wavelength, so going

8 VI -D box example π-system expand energy, difference gets smaller big box, small energies Problems worked out in many books (Atkins p , Engel h 14.4, ouse h. 3) poly arene examples (in wavelength, so going to right, lower energy due to larger D boxes): 1 ring rings 3 rings 4 rings

9 VI 3 3-D Particle in box Separation of Variables (ouse h.3) ritical Method - needed to solve atoms & molecules write: ψ = h m x V = 0 0 < x <a 0 < y < b 0 < z < c V = outside the box + y + z ψ = Eψ ψ = -me/h ψ Note: a) / x only operate on x-dependent function b) is a sum of terms each depend on 1 variable IN GENERAL can find solution -- product function form Ψ = X(x) Y(y) Z(z) where X(x) is only fct. of x, etc. AND energy also a sum: E = E 1 + E + E 3 Substitute: X ( x) Y ( y) Z( z) me XYZ = YZ + XZ + XY = x y z h XYZ divide by XYZ: me h = 1 X X x + 1 Y Y y + 1 Z Z z each term must be a constant since independent i.e. 1/X X/ x = α etc. α + β + γ = -me/h 3

10 VI 4 These individual 1-D equations are p.i.b solutions again: ψ (x,y,z) = 8 abc sin n x n sin y n sin z π x a π y b π z c E = h n n x y n + 8m a b c z + = E 1 + E + E 3 Lowest state n x = n y = n z = 1 But 3 ways for next state n x =, n y = n z = 1, etc. Each of these could have different energies owever, if a=b=c, then each has same energy degeneracy from symmetry Barriers (Atkins p.38-9, Engel h.14.9) goal get concept, not derivation Now what if wall not so high or wide high wall wave must have zero amplitude ψ*ψ = 0 at wall reflect shorter wall wave can penetrate also thin wall can go through or tunnel (-h /m d /dx + V)ψ = Eψ if ψ = e iαx [h α /m + (V E)]ψ = 0 α = m (E V) h 4

11 5 VI 5 Explore behavior in different regimes: now x < x 0, V = 0 E V = (+) ψ = e iαx is complex exponential fct. wave but for : x > x 0, V > E E V = ( ) α complex so α = i = iκ ψ'(x) = e -Κx real, decaying fct. m (V E) h At wall, x-x 0 ψ(x 0 ) = ψ'(x 0 ) i.e. must be continuous If non-zero in wall, then ψ must decay as move +x On other side: ψ'(x 1 ) = ψ''(x 1 ) (contin. go out: ψ'' < ψ) equation 9.11 Atkins: Tunneling probability, T T 16ε (1 - ε) e -ΚL where : ε = E/V E or V T Κ=[m(V-E)] 1/ /h and L=x 1 x 0 inc. V or L, T Solution (extra-repeat): Look at just the barrier: A = -h /m d /dx = B = -h /m d /dx + V solve each region separately: ψ A = Ae ikx + Be -ikx k = (me/h) 1/ ψ B = A'e ik'x + Be -ik'x k' = [m (E - V)/h] 1/ (in barrier) ψ = A''e ik x + B''e -ik x k = (me/h) 1/ = k Note: if E < V, then k' = imaginary let k' = iκ, Κ = [m (E - V)/h] 1/ ( = real) Inside the barrier: ψ B = A'e -Κx + B'e +Κx

12 6 VI 6 exponentially decreasing or increasing function no oscillation in barrier amplitude: ψ*ψ 0 in barrier, thus can tunnel probability non-zero of in and other side barrier damping ~ mass heavy don t penetrate classic low energy don t penetrate tunnelling --skip, read i.e. w/f okay if bound in area of wall must be thin to solve for A, B s must set up simultaneous equation based on: boundary constraints ψ A (0) = ψ B (0) A + B = A' + B' ψ B (l) = ψ (l) A'e -Κl + B'e +Κl = A''e ik l + B''e -ikl and continuous slopes ψ A / x 0 = ψ B / x 0 ika ikb = -ΚA' + ΚB' ψ B / x l = ψ / x l -ΚA'e Κl +ΚB'e Κl = ika''e ik l +ikb''e -ikl Then consider structure as: B = 0, A 0 (come from left) then B'' = 0 and A'' ~ transmission B ~ reflection Probability of tunneling: A'' / A P = 1/(1 + G) G = (e Κl e -Κl ) 4 (E / V) (1 E / V) Note: P non zero, K > 0 E increased, G decreased, P increased

13 VI 7 onsider particle in box - short side (finite well) (Engel 14.5): V = 0 0 < x < L; V = V 0 0 < x < L E n : Energy no longer ~n (spacing will get closer with n) ψ: Solution to this more complex but have new property ψ(0) & ψ(l) 0 -- since V hence w/f non zero inside wall -- from B.. (turns out to be exponential e -βy, i.e. decay function where y = x L, x > L ;y = -x, x<0) Imagine boxes side by side: as (L M) 0 wave functions will overlap, then ψ*ψ will be non zero in other box and particle will tunnel this is seen as inversion, e.g. N 3, -bond Additional property as E V 0, levels must close in together E > V 0 levels continuous 7

14 Particle on a ring: (Atkins, p ) ircumference = πr (fixed distance) B.. ψ(φ) = ψ(φ + π) (vary angle) continuous but not zero (no wall) r 8 unit length VI 8 h ψ = ψ = Eψ mr φ dr ~ r dφ ψ = Ae iαφ + Be -iβφ r=xi+yj, x +y =1 x= r cosφ y= r sinφ or φ = cos -1 (x/ r ) B.. e iαφ = e iα(φ + π) e iα(π) = 1 α = n = 0, ±1, ±, nd term (B-dependent) redundant, let α = β, same fct. E n = h n /mr = h n / I I = mr moment of inertia Note: levels degenerate for ±n no zero-point E E 0 = 0, φ unknown on ring spacing ~n same pattern (OK with uncertainty) bigger ring lower E n Angular Momentum J = r x p in general J z = r p (1-D z out of plane) I = mr moment of inertia E = p /m = J z /mr = J z /I from de Broglie p = h/λ λ = πr/n (integ. # waves on ring) E n = p n /m = (h/π) n /mr = E n from above E n = n h /I E = J z /I J z = nh get quantized solution for Energy and angular momentum why? [,L] = 0, commuting operators, simul. eigenfct. This form works for molecular rotation / atom, add dimen.

Model Problems update Particle in box - all texts plus Tunneling, barriers, free particle - Tinoco (pp455-63), House Ch 3

Model Problems update Particle in box - all texts plus Tunneling, barriers, free particle - Tinoco (pp455-63), House Ch 3 VI 15 Model Problems update 010 - Particle in box - all texts plus Tunneling, barriers, free particle - Tinoco (pp455-63), House Ch 3 Consider E-M wave 1st wave: E 0 e i(kx ωt) = E 0 [cos (kx - ωt) i sin