A 2 sin 2 (n x/l) dx = 1 A 2 (L/2) = 1

Transcription

1 VI 15 Model Problems 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 t) = E 0 [cos (kx - t) i sin (kx - t)] magnitude: k = / = c/ = = c/ moves in space and time traveling wave, vector k direction reflect at node (same both ways) keeps the wave continuous (if not it creates an interference) is other cycle, at t=/c 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) Picture: free space E-M wave, all frequencies possible (like bb) Restrict space, like cavity, then restrict wavelengths (laser) Think of traveling particle 1-D no forces V = 0 = E = T let V = 0, free moving particle = (- /m) d /dx [(x)] Solution: need some (wave)function that we can take derivative twice and get function back - eigenfunction 15 node

2 VI 16 choices: a) de ax /dx = ae ax derivative works d /dx e ax = a e ax but energy, = (ħ /m)d /dx, must be positive, term uses negative of nd deriv., so use: a = i e ix wavefunct. complex, sum of real: cos x + i sin x b) d /dx sin kx = -k sin kx (note: same for cos kx) (Note: e ix = cos x - i sin x general form wave) No constraint traveling wave (but for particle) Solve Schroedinger Equation for free particle: -( /m) d /dx = E if = e ix plug in -( /m)(i) e ix = E e ix from (b): = k = (me) 1/ / E = /m (all K.E. - positive, not quantized) no restrictions free particle, any energy or wavelength Note effect of momentum: (for = e ikx ) (need complex) p = -i(ik)= k Magnitude: p = k (now real) signs direction p = k (motion in +x direction) [ompare to: = e -ikx, (opposite motion in -x)] Boundary onditions Restrictions on wave function so it will fit postulates B.. relate to continuous and finite properties wave/fct properties - both sides boundary--must match 16

3 VI 17 Particle in a box introduce simplest potential energy in box V = 0 outside V = but not in SchEqn For finite E: particle must be in box, or definite E B.. (think of as F = -dv/dx, force at wall ) - /m d /dx = E try = A sin x + B cos x Looks same but now: B.. (0) = 0 restrict: B = 0 (since cos 0 = 1) (L) = 0 restrict: = n/l (since sin n = 0) idea continuous w/f, zero probability outside: out = 0 for (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- A0 non-trivial solution) forms a standing wave -- quantized (recall - laser) - /m d /dx (A sin nx/l) = E (A sin nx/l) E n = (- /m)(- n /L ) E n = n h /8mL = h/ Expanding E-levels ~ n each increases number of nodes curvature restricted energy levels lowest energy 0 (particle always moving) 17

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

5 VI 19 Now see properties of p.i.b-like molecule E-levels/spectra a) bigger n more separation higher frequency - h b) bigger m less separation (but all same m e - electron) c) bigger L less separation (as square), experimental Sample dye problem: Polyene N max (Å) N Obs alc E = (N + 1) h /8mL L 0.81 N (in nm) = c/ = hc/e = (8m e c/h)(0.81x10-9 m) N /(N + 1) units! m [to do: insert m e (for elect), h, c; then convert to nm, 10-9 m] Note: trend is as expected N increase, 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) R e f : J o c h e n A u t s c h b a c h, J.hem. Ed. 84, ( 007) 19

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 ~ N /N type term (linear) Bio-connect -Vision: retinal undergoes cis-trans isomerization N (trans) OMO LUMO difference Decrease with length alculated with full QM Transition wavelength (=c/) increase with box length fit 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. E, fluor. dec. E 1

8 VI -D box example -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) -D wavefct for rectangle box (Atkins), a is n 1 =n =1, uniform, b & c related by rotation, n 1 =1,n = or v.v. poly arene examples (in wavelength, go to right, lower energy 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 example - need to solve atoms & molecules write: = m x V = 0 0 < x <a 0 < y < b 0 < z < c V = outside the box y z = E = -me/ 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 = 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 divide by XYZ: me 1 X X x 1 Y Y y 1 Z Z z XYZ each term must be a constant since independent i.e. 1/X X/x = etc. + + = -me/ These individual 1-D equations are p.i.b solutions again: (x,y,z) = 8 abc sin n x n sin y n z sin x a 3 y b z c

10 VI 4 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 shorter wall wave can have zero amplitude penetrate also thin wall * = 0 at wall reflect can go through or tunnel (- /m d /dx + V) = E if = e ix [ /m + (V E)] = 0 = m (E V) Explore behavior in different regimes: now x < x 0, V = 0 E V = (+) = e ix is complex exponential fct. wave but for : x > x 0, V > E E V = ( ) complex m (V E) so = i = i '(x) = e -x real, decaying fct. 4

11 VI 5 At wall, x-x 0 (x 0 ) = '(x 0 ) i.e. must be continuous If non-zero in wall, then must decay as move +x On other side: '(x 1 ) = ''(x 1 ) (contin. go out: '' < equation 8.19b Atkins: Tunneling probability, T T 16 (1 - ) e -L where : = E/V E or V T =[m(v-e)] 1/ / and L=x 1 x 0 inc. V or L, T Solution (extra): Look at just the barrier: A = - /m d /dx = B = - /m d /dx + V solve each region separately: A = Ae ikx + Be -ikx k = (me/) 1/ B = A'e ik'x + Be -ik'x k' = [m (E - V)/] 1/ (in barrier) = A''e ik x + B''e -ik x k = (me/) 1/ = k Note: if E < V, then k' = imaginary let k' = i, = [m (E - V)/] 1/ ( = real) Inside the barrier: B = A'e -x + B'e +x exponentially decreasing or increasing function no oscillation in barrier amplitude: * 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 A (0) = B (0) A + B = A' + B' B (l) = (l) A'e -l + B'e +l = A''e ik l + B''e -ikl 5 A B

12 VI 6 and continuous slopes A /x 0 = B /x 0 B /x l = /x l ika ikb = -A' + B' -A'e l +B'e l = 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 e - ) 4 (E / V) (1 E / V) Note: P non zero, K > 0 E increased, G decreased, P increased 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) : Solution to this more complex but have new property (0) & (L) 0 -- since V hence w/f non zero inside wall -- from B.. (turns out to be exponential e -y, i.e. decay function where y = x L, x > L ;y = -x, x<0) 6

13 VI 7 Imagine boxes side by side: as (L M) 0 wave functions will overlap, then * 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 Particle on a ring: (Atkins, p ) ircumference = r (fixed distance) B.. () = ( + ) (vary angle) single value (continuous no wall) = mr r unit length ~ r d = E dr = Ae i + Be -i r=xi+yj, x +y =1 x= r cos y= r sinorcos -1 (x/ r ) B.. e i = e i( + ) e i() = 1 = n = 0, 1,, nd term (B-dependent) redundant, let, same fct. E n = n /mr = n / I I = mr moment of inertia Note: levels degenerate for n no zero-point E E 0 = 0, unknown on ring spacing ~n same pattern (OK with uncertainty) bigger ring lower E n 7

14 VI 8 Angular Momentum J = r x p in general J z = r p (ring: 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/ = r/n (integ. # waves on ring) E n = p n /m = (h/) n /mr = E n from above E n = n /I E = J z /I J z = n relate to Bohr get quantized solution for Energy and angular momentum why? [,L] = 0, commuting operators, simultaneous eigenfct. This form works for molecular rotation / atom, add 3 rd dimension 8

15 9 VI 9