Model Problems – 9
Model Problems – 9.5 Atkins / Particle in box all texts
Consider E-M wave 1st
wave: E0ei(kx – ωt) = E0 [cos (kx - ωt) – i sin (kx - ωt)]
[pic]
magnitude: |k| = 2π/λ ω = 2πc/λ =2πν ν = c/λ
moves in space and time – traveling wave
reflect at the node
keeps the wave continuous (if not create an interference)
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
Now think of traveling particle ( 1-D no forces ( V = 0
Hψ = Eψ = Tψ let V = 0, free moving particle
Hψ= -(2/2m d2/dx2 ψ(x)
Solution: need some function that can take derivatives twice and get function back
choices:
a) deax/dx = aeax derivative works d2/dx2 eax = a2eax
(but can’t be negative unless a = iα, need negative to balance -(2/2m and end up with positive Energy)
b) d2/dx2 sin kx = -k2 sin kx
(Note: eiαx = cos αx - i sin αx ( general form wave)
No constraint ( traveling wave (but for particle)
Solve Schroedinger Equation for free particle:
-((2/2m) d2/dx2 ψ = Eψ
if ψ = eiαx ( plug in -((2/2m)(iα)2 eiαx = E eiαx
from ( (b): ) α = k = (2mE)1/2/( (
E = α2(2/2m (all K.E. - positive, not quantized)
no restrictions – free particle, any energy magnitude
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:
pψ = -i((ik) ψ ( Magnitude: |p| = (k
signs ( direction p = (k (motion in +x)
[opposite ψ = e-ikx, (motion -x)]
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 ()
-(2/2m d2/dx2 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 = 2πn/2L = nπ/L (sin nπ = 0)
for ψ(x) ≠ 0, must have: A(0 , n(0 and n=1,2,3,…
(i.e. must be node both sides – integral number waves and be non-zero someplace--non-trivial solution)
forms a standing wave -- quantized (recall - laser)
-(2/2m d2/dx2 (A sin nπx/L) = E (A sin nπx/L)
(2n2π2/2mL2 = En = n2h2/8mL2
Expanding E-levels ~ n2
each has increasing
number of nodes
restricted energy levels
lowest energy ( 0 (particle always moving)
Probability distribution: ψ*ψ dx
[pic]ψ*ψ dx = 1 (if normalize)
but plot ψ*ψ → not uniform in x 2
n = 1 more probable in middle 1
n = 2 zero probability at x = L/2 L/2
as n increases – probability more even → classical
Orthogonal (ψm*ψn dx = 0 if n ( m ((sinx cosx dx=0)
Amplitude? [pic]ψn*ψn dx = 1 from normalization
[pic]A2 sin2(nπx/L) dx = 1 ( A2 (L/2) = 1 ( A = (2/L)
Probability [pic]ψ*ψ dx ( probability between a + b
Use for pib? Great model / see how potential or B.C.
leads to quantization
Application: polyenes [pic]…
π-system delocalize – electrons move through π-bonds
spectra – e could be in different levels
ΔE = En–En1 = hν
n ( n + 1
ΔE = (n+1)2h2/8mL – n2h2/8mL
= (2n + 1) h2/2mL2 = hν
Now see properties
a) bigger n ( more separation – higher frequency
b) bigger m ( less separation (but all me electron)
c) bigger L ( less separation (as square), experimental
Sample dye problem: λmax (Å)
|[pic] |N |Obs |Calc |
| |1 |1600 |870 |
| |2 |2170 |2080 |
| |3 |2600 |3360 |
| |4 |3020 |4650 |
| |5 |3460 |5940 |
| |6 |3690 |7200 |
| |11 |4510 |13700 |
| |12 |4750 |15000 |
| |15 |5040 |18900 |
ΔE = h2/8mL2 (2N + 1) L ( 2.81 N
λ = c/ν = hc/ΔE
= (8mc/h2) (2.81)2 N2/2N + 1
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)
|[pic] |N |Obs |Calc |
| |2 |4250 |3280 |
| |3 |5600 |4540 |
| |4 |6500 |5800 |
| |5 |7600 |7060 |
| |6 |8700 |8330 |
| |7 |9900 |9600 |
Model does better (here use N+1 ( N+2) and use different length, but still λ ~ N2/N type term
Vision: retinal undergoes cis-trans isomerization
[pic]
Butadiene examples
[pic]
Ionization potential – measures energy of the “orbital” – see decrease ethylene ( butadiene
[pic]
2-D box example ( π-system expand energy , difference gets smaller—big box, small energies
poly arene examples:
[pic]
Now what if wall not so “high” or “wide”
[pic]
( high wall wave must shorter wall wave can
have zero amplitude penetrate–thin– go through
ψ*ψ = 0 at wall ( reflect
(-(2/2m d2/dx2 + V)ψ = Eψ
if ψ = eiαx ( [(2a2/2m + (V–E)]ψ = 0 ( α =[pic]
now x < x0 , V = 0 E – V = (+)
ψ = eiαx is complex ( wave
but for : x > x0 , V > E E – V = (–)
so α = i[pic] = iΚ ( ψ' = e-Κx ( decaying function
At wall ψ(x0) = ψ'(x0) i.e. must be continuous
if penetrate: ψ'(x1) = ψ''(x1) (contin. out ψ'' < ψ)
equation 9.10 Atkins: Tunnelling
T ( 16ε (1 - ε) e-2KL where : ε = E/V L = x1 – x0
Look at just the barrier:
HA = -(2/2m d2/dx2 = HC
HB = -(2/2m d2/dx2 + V
solve each region separately:
ψA = Aeikx + Be-ikx k = (2mE/()1/2
ψB = A'eik'x + Be-ik'x k' = [2m (E - V)/(]
ψC = A''eikx + B''e-ik'x k = (2mE/()1/2
Note: if E < V, then k' = imaginary
let k' = iΚ , Κ = [2m (E - V)/(]1/2 ( = real)
ψB = A'e-Κx + B'e+Κx
– exponentially decreasing or increasing function
– no oscillation in barrier
– amplitude: ψ*ψ ( 0 in barrier, thus can tunnel
– damping ~ mass – heavy don’t penetrate – classic low energy don’t penetrate
“tunnelling”
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
ψA(0) = ψB(0) A + B = A' + B'
ψB(ℓ) = ψC(ℓ) A'e-Κℓ + B'e+Κℓ = A''eik ℓ + B''e -ikℓ
and continuous slopes
(ψA/(x(0 = (ψB/(x(0 ikA – ikB = -ΚA' + ΚB'
(ψB/(x(ℓ = (ψC/(x(ℓ -ΚA'eΚℓ+ΚB'eΚℓ = ikA''eik ℓ+ikB''e-ikℓ
Then consider structure as:
B = 0 , A ( 0
then B'' = 0
and (A''(2 ~ transmission
(B(2 ~ reflection
Probability of tunneling: (A''(2 / (A(2
P = 1/(1 + G) G = [pic]
Note: P – non zero , K > 0
E increased, G decreased, P increased
3-D Particle in box – Separation of Variables
V = 0 0 < x L ;y = -x , x V0 levels continuous
Harmonic Oscillator – use pib to “think through”
Consider one ball on a spring
Hook’s law states force
F = -k (x – xe) = -kq
q = Δx ( variable
F = -(V/(q
V = 1/2 kq2 k = force constant
T = [pic] = [pic]
Hψ = [pic] ψ = Eψ
a) like a box with sloped sides
– softer potential expect penetration
b) – still a well expect oscillator
c) – must be integrable – expect damping
Note: ψ ~ ((q)e–αq only work on side
ψ ~ ((q)e–αq2 works both
(α must be positive)
((a) – must be a polynomial
if ((a) ~ constant
but ~q2 – (constant)
Result – see text nice pictures
Wave function – modify variable to simplify
ψυ(y) = Hυ(y)e–y2/2
υ = 0, 1, 2, … (quantize)
recursion formula: Hυ(y) = (-1)υey2 dυ/dyυ (e-y2)
y = αq α = 2π [pic] = [pic]
Hermite polynomials:
ex: H0 = 1 note: – odd - even progression
H1 = 2y – alternate exponents
H2 = 4y2 – 2 – ( number solutions
H3 = 8y3 – 12y – exponential damping
thus: ψυ = (α/αυυ! π1/2)1/2 Hυ(y)e-y2/2 nomenclature
Homework
insert ψυ into Schrödinger equation Hψ = Eψ and get
Eυ = (υ + 1/2) [pic]w w = [pic]
= (υ + 1/2) hϖ ϖ = w/2π
note: – even energy spacing: ΔE = hν
– zero point energy: 1/2 hν
– heavier mass ΔE ( 0 classical
– weaker force constant ΔE ( 0
Probabilities – low υ – high in middle; high υ – high on edge
To describe tow masses on a spring (relate to molecules)
need change variable
q = (x2 - x1) – (x2 - x1)
= r – req
r = x2 – x1 = relative position
in this case use: μ = m1m2/m1 + m2 reduced mass in H
2-mass harmonic oscillator:
(-[pic]2/2μ d2/dq2 + 1/2 kq2)ψ = Eψ
and get Eυ = (υ + 1/2) [pic]w w = [pic]
Can use this to model vibration of a diatomic molecule – low υ
harmonic (ideal) E-levels collapse in real molecule
anharmonic
multiatom 3n - 6 relative coordinate – complex but separable
Two-dimensional Harmonic oscillator:
H = T + V
T = -[pic]2/2m ((2/(x2 + (2/(y2)
V = V(x, y) (expand about ???)
= V(0, 0) + (V/(x(0 x + (V/(y(0 y + 1/2 (2V/(x2(0 (x2) + 1/2 (2V/(y2(0 y2 + (2V/(x(y(0 + …
– more complex potential
– not separated
– V(0, 0) = 0 – arbitrary constant
– (V/(x(0 = (V/(y(0 = 0 – choose x = 0, y = 0 as the minimum
Then V = 1/2 ((2V/(x2)0 x2 + 1/2 ((2V/(y2)0 y2 + 1/2 ((2V/(x(y)0 xy
= 1/2 kx x2 + 1/2 ky y2 + 1/2 kxy xy + …
– so form like harmonic oscillator if can separate
– do exchange of variable x, y ( q1, q2 where q1, q2 chosen so that potential is not coupled
– call this diagonalize potential – use matrix approach can do to arbitrary accuracy
– also works for n-dimensions: (3n - 6) vibration
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