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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