Model Problems 09 - Ch.14 - Engel/ Particle in box - all texts. Consider E-M wave 1st wave: E 0 e i(kx ωt) = E 0 [cos (kx - ωt) i sin (kx - ωt)]

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1 VI 15 Model Problems 09 - Ch.14 - Engel/ Particle in box - all texts Consider E-M wave 1st wave: 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 reflect at the node keeps the wave continuous (if not create an interference) is other cycle, at Δt=λ/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 integral divisor of length integer representation of frequencies not continuous Now think of traveling particle 1-D no forces V = 0 Hψ = Eψ = Tψ let V = 0, free moving particle Hψ= -h /m d /dx ψ(x) 15

2 VI 16 Solution: need some function that can take derivatives twice and get function back choices: a) de ax /dx = ae ax derivative works d /dx e ax = a e ax (but Energy must be positive, so nd deriv. must be negative, so need a = iα, e iαx wavefunction complex) b) d /dx sin kx = -k sin 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 Boundary Conditions Restrictions must fit postulates B.C. relate to continuous and finite properties of wavefct., etc. relates to properties of wave/fct on both sides of boundary--must match Note effect of momentum: (for ψ = e ikx ) pψ = -ih(ik) ψ Magnitude: p = hk signs direction p = hk (motion in +x) [opposite: ψ = e -ikx, (motion -x)] 16

3 VI 17 Particle in a box in box V = 0 outside V = For E to be finite: particle must be in box (need definite E-state also think of as F = -dv/dx, force at wall is ) -h /m d /dx ψ = Eψ try ψ = A sin αx + B cos βx B.C. ψ(0) = 0 restrict: B = 0 (since cos 0 = 1) ψ(l) = 0 restrict: α = nπ/l (since sin nπ = 0) for ψ(x) 0, must have: A 0, n 0 and n=1,,3, (i.e. must be node both sides integral number modes and be non-zero someplace--non-trivial solution) forms a standing wave -- quantized (recall - laser) -h /m d /dx (A sin nπx/l) = E (A sin nπx/l) (-h /m)(- n π /L ) = E n = n h /8mL Expanding E-levels ~ n each has increasing number of nodes restricted energy levels lowest energy 0 (particle always moving) 17

4 VI 18 Probability distribution: ψ*ψ dx L ψ*ψ dx = 1 (if normalize) 0 but plot ψ*ψ not uniform in x 1 n = 1 more probable in middle n = zero probability at x = L/ L/ as n increases probability more even classical Orthogonal ψ m *ψ n dx = 0 if n m sin(nπx/l) sin(mπx/l) dx=0 easiest seen graphically Amplitude? L 0 L 0 ψn*ψ n dx = 1 from normalization A sin (nπx/l) dx = 1 A (L/) = 1 A = (/L) 1/ Probability b a ψ*ψ dx probability between a + b Use for pib? Great model / see how potential or B.C. leads to quantization Application: polyenes π-system delocalize electrons move through π-bonds 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 18

5 VI 19 Now see properties 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: λ max (Å) Polyene N Obs Calc H H C C H H C C H C N C C H H ΔE = (N + 1) h /8mL L 0.81 N (in nm) λ = c/ν = hc/δe = (8mc/h) (0.81x10-9 m) N /N + 1 units! 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) 19

VI 0 Dye S C N C H 5 C H H C C N H C C H 5 N + S N Obs Calc 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

6 VI 0 Dye S C N C H 5 C H H C C N H C C H 5 N + S N Obs Calc 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) Butadiene examples real spectra shift with length 0

7 VI 1 Ionization potential measures energy of the orbital see decrease ethylene butadiene (left peak lowest) 1

8 VI -D box example π-system expand energy, difference gets smaller big box, small energies Problems worked out most books (Engel Ch 14.4) poly arene examples (in wavelength, so going to right, lower energy): 1 ring rings 3 rings 4 rings

9 VI 3 3-D Particle in box Separation of Variables (Engel 14.4) Method we will need to solve atoms & molecules write: Hψ = 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) H 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 are pib solutions again: n ψ (x,y,z) = sin x n π sin y n y sin z π z 8 abc π x a b 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 However, if a=b=c, then each has same energy degeneracy from symmetry Barriers (Engel Ch.14.9) 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 be penetrated also thin wall go through or tunneling (-h /m d /dx + V)ψ = Eψ if ψ = e iαx [h α /m + (V E)]ψ = 0 α = now x < x 0, V = 0 E V = (+) ψ = e iαx is complex wave 4 m (E V) h

11 but for : x > x 0, V > E E V = ( ) so α = i m (V E) h VI 5 = iκ ψ' = e -Κx real, decaying function At wall 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 9.10 Atkins: Tunneling probility, T T 16ε (1 - ε) e -ΚL where : ε = E/V L = x 1 x 0 Κ=[m(V-E)] 1/ /h Solution (extra-repeat): Look at just the barrier: H A = -h /m d /dx = H C H 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] (in the barrier) ψ C = 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) ψ B = A'e -Κx + B'e +Κx exponentially decreasing or increasing function no oscillation in barrier 5

12 VI 6 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 C (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 C / 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 6

13 Particle on a ring: Circumference = πr B.C. ψ(φ) = ψ(φ + π) continuous but not zero (no wall) h Hψ = ψ = Eψ mr φ r dr 7 unit length ~ r dφ ψ = Ae iαφ + Be -iβφ r=xi+yj, x +y =1 x= r cosφ y= r sinφ B.C. e iαφ = e iα(φ + π) e iα(π) = 1 α = n = 0, ±1, ±, nd term (B-dependent) redundant VI 7 E n = h n /mr Note: levels degenerate for ±n no zero point E E 0 = 0, φ unknown on ring spacing ~n same pattern (OK uncert.) 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 (int. # waves 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 This form works for molecular rotation / atom, add dimen.

14 VI 8 Now consider if particle in box with short side (finite well) (Engel 14.5): V = 0 V = V 0 0 < x < L 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.C. (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 Additional property as E V 0, levels must close in together E > V 0 levels continuous 8

if trap wave like violin string tied down at end standing wave

if trap wave like violin string tied down at end standing wave VI 15 Model Problems 9.5 Atkins / Particle in box all texts onsider E-M wave 1st wave: 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