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Model Problems – 9.5 Atkins / Particle in box all texts
Consider E-M wave 1st
wave: E0ei(kx – wt) = E0 [cos (kx - wt) – i sin (kx - wt)]
magnitude: |k| = 2p/l w = 2pc/l =2pn n = c/l
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 à light trap in cavity
– specific frequency / phase amplified)
restriction - number wavelengths integral divisor of length
Now think of traveling particle à 1-D no forces Þ V = 0
Hy = Ey = Ty let V = 0, free moving particle
Hy= -2/2m d2/dx2 y(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 = ia, need negative to balance -2/2m and end up with positive Energy)
b) d2/dx2 sin kx = -k2 sin kx
(Note: eiax = cos ax - i sin ax à general form wave)
No constraint ® traveling wave (but for particle)
Solve Schroedinger Equation for free particle:
-(2/2m) d2/dx2 y = Ey
if y = eiax à plug in -(2/2m)(ia)2 eiax = E eiax
from (b): a = k = (2mE)1/2/ à
E = a22/2m (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. à relates to properties of wave/fct on both sides of boundary--must match
Note effect of momentum:
py = -i(ik) y à Magnitude: |p| = k
signs à direction p = k (motion in +x)
[opposite y = 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 y = Ey try y = A sin ax + B cos bx
B.C. y(0) = 0 Þ restrict: B = 0 (since cos 0 = 1)
y(L) = 0 Þ restrict: a = 2pn/2L = np/L (sin np = 0)
for y(x) ≠ 0, must have: A¹0 , n¹0 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 npx/L) = E (A sin npx/L)
2n2p2/2mL2 = En = n2h2/8mL2
Expanding E-levels ~ n2
each has increasing
number of nodes
restricted energy levels
lowest energy ¹ 0 (particle always moving)
Probability distribution: y*y dx
y*y dx = 1 (if normalize)
but plot y*y → 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 òym*yn dx = 0 if n ¹ m (òsinx cosx dx=0)
Amplitude? yn*yn dx = 1 from normalization
A2 sin2(npx/L) dx = 1 Þ A2 (L/2) = 1 Þ A = (2/L)
Probability y*y dx Þ probability between a + b
Use for pib? Great model / see how potential or B.C.
leads to quantization
Application: polyenes …
p-system delocalize – electrons move through p-bonds
spectra – e could be in different levels
DE = En–En1 = hn
n ® n + 1
DE = (n+1)2h2/8mL – n2h2/8mL
= (2n + 1) h2/2mL2 = hn
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: lmax (Å)
/ N / Obs / Calc1 / 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
DE = h2/8mL2 (2N + 1) L » 2.81 N
l = c/n = hc/DE
= (8mc/h2) (2.81)2 N2/2N + 1
Note: trend is as expected N ® increase, l ® 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)
/ N / Obs / Calc2 / 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 l ~ N2/N type term
Vision: retinal undergoes cis-trans isomerization
Butadiene examples
Ionization potential – measures energy of the “orbital” – see decrease ethylene ® butadiene
2-D box example ® p-system expand energy , difference gets smaller—big box, small energies
poly arene examples:
Now what if wall not so “high” or “wide”
¥ high wall wave must shorter wall wave can
have zero amplitude penetrate–thin– go through
y*y = 0 at wall Þ reflect
(-2/2m d2/dx2 + V)y = Ey
if y = eiax Þ [2a2/2m + (V–E)]y = 0 Þ a =
now x < x0 , V = 0 E – V = (+)
y = eiax is complex ® wave
but for : x > x0 , V > E E – V = (–)
so a = i = iK à y' = e-Kx ® decaying function
At wall y(x0) = y'(x0) i.e. must be continuous
if penetrate: y'(x1) = y''(x1) (contin. out y'' < y)
equation 9.10 Atkins: Tunnelling
T @ 16e (1 - e) e-2KL where : e = 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:
yA = Aeikx + Be-ikx k = (2mE/)1/2
yB = A'eik'x + Be-ik'x k' = [2m (E - V)/]
yC = A''eikx + B''e-ik'x k = (2mE/)1/2
Note: if E < V, then k' = imaginary
let k' = iK , K = [2m (E - V)/]1/2 ( = real)
yB = A'e-Kx + B'e+Kx
– exponentially decreasing or increasing function
– no oscillation in barrier
– amplitude: y*y ¹ 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
yA(0) = yB(0) A + B = A' + B'
yB(ℓ) = yC(ℓ) A'e-Kℓ + B'e+Kℓ = A''eik ℓ + B''e -ikℓ
and continuous slopes
¶yA/¶xï0 = ¶yB/¶xï0 ikA – ikB = -KA' + KB'
¶yB/¶xïℓ = ¶yC/¶xïℓ -KA'eKℓ+KB'eKℓ = ikA''eik ℓ+ikB''e-ikℓ
Then consider structure as:
B = 0 , A ¹ 0
then B'' = 0
and ïA''ï2 ~ transmission
ïBï2 ~ reflection
Probability of tunneling: ïA''ï2 / ïAï2
P = 1/(1 + G) G =
Note: P – non zero , K > 0
E increased, G decreased, P increased
3-D Particle in box – Separation of Variables
V = 0 0 < x <a
0 < y < b
0 < z < c
V = ¥ outside the box
write: Hy = y = Ey
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 a solution product form
Y = X(x) Y(y) Z(z) / q1(x) q2(y) q3(z) X(x) only on x, etc.
AND ® energy also a sum: E = E1 + E2 + E3
substitute
divide by xyz
Þ each term must be a constant – since independent
i.e. 1/x ¶2x/¶x2 = a etc. a + b + g = -2mE/
These are pib solutions again:
y = sin sin sin
E = = E1 + E2 + E3
Lowest state nx = ny = nz = 1
But 3 ways for next state ® nx = 2, ny = h2 = 1, etc.
Particle on a ring
Circumference = 2pr
B.C. y(0) = y(f + 2p)
continuous but not zero (no edge)
Hy = y = Ef
y = Aeiaf + Be-ibf
B.C. eiaf = eia(f + 2p) Þ eia(2p) = 1 Þ a = n = 1, ±1, ±2,
2nd term redundant
En = 2n2/2mr2
Note: – levels degenerate ±n
– no zero – point E
E0 = 0 , f unknown
– spacing ~na – same point
– bigger ring – lower En
Angular Momentum
J = r x p in general ® Jz = rp here (1-D)
z out of plan
I = mr2 moment of inertia
E = p2/2m = dz2/2mr2 = Jz2/2I
from de Broglie p = h/l l = 2pr/n (integer number waves rin)
En = pn2/2m = (h/2p)2 n2/2mr2 = En from above
En = n22/2I Þ E = Jz2/2I Þ Jz = n
so get solution for Energy and angular momentum
This form works for other (molecular rotation / atom) , add dimension
Now consider if particle in box with short sides:
V = 0 0 < x < L
V = V0 0 < x < L
En: Energy no longer ~n2 (spacing will get closer)
y: Solution to this more complex but have new property – y(0) + y(L) ¹ 0
hence w/f non zero inside wall (turns out to be exponential e-by
where y = x – L , x > L ;y = -x , x<0)
i.e. decay function y(¦)
Imagine 2 boxes side by side:
as L – M ® 0 the wave functions will overlap and stick there, then y*y will be non zero in other box and say that particle tunneled
Additional property – as E ® V0 , levels must get close together ® E > 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 = Dx ¬ variable
F = -¶V/¶q
V = 1/2 kq2 k = force constant
T = =
Hy = y = Ey
a) like a box with sloped sides
– softer potential expect penetration
b) – still a well expect oscillator
c) – must be integrable – expect damping
Note: y ~ ¦(q)e–aq only work on side
y ~ ¦(q)e–aq2 works both
(a 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
yu(y) = Hu(y)e–y2/2
u = 0, 1, 2, … (quantize)
recursion formula: Hu(y) = (-1)uey2 du/dyu (e-y2)
y = aq a = 2p =
Hermite polynomials:
ex: H0 = 1 note: – odd - even progression
H1 = 2y – alternate exponents
H2 = 4y2 – 2 – ¥ number solutions
H3 = 8y3 – 12y – exponential damping
thus: yu = (a/auu! p1/2)1/2 Hu(y)e-y2/2 nomenclature
Homework
insert yu into Schrödinger equation Hy = Ey and get
Eu = (u + 1/2) w w =
= (u + 1/2) hv v = w/2p
note: – even energy spacing: DE = hn
– zero point energy: 1/2 hn
– heavier mass DE ® 0 classical
– weaker force constant DE ® 0
Probabilities – low u – high in middle; high u – 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: m = m1m2/m1 + m2 reduced mass in H
2-mass harmonic oscillator:
(-2/2m d2/dq2 + 1/2 kq2)y = Ey
and get Eu = (u + 1/2) w w =
Can use this to model vibration of a diatomic molecule – low u
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 = -2/2m (¶2/¶x2 + ¶2/¶y2)
V = V(x, y) (expand about ???)
= V(0, 0) + ¶V/¶xï0 x + ¶V/¶yï0 y + 1/2 ¶2V/¶x2ï0 (x2) + 1/2 ¶2V/¶y2ï0 y2 + ¶2V/¶x¶yï0 + …
– more complex potential
– not separated
– V(0, 0) = 0 – arbitrary constant
– ¶V/¶xï0 = ¶V/¶yï0 = 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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