pc iv: molecular spectroscopy / molekülspektroskopie prof ... · 12.05.2007 introduction to...
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12.05.2007 Introduction to Quantum Mechanics
PC IV: MOLECULAR SPECTROSCOPY /
Molekülspektroskopie
Prof. Oleg Vasyutinskii
Summer Semester: from April 9, 2007 till July 20, 2007
Lectures: Thursday 8:00 – 9:30 (PK11.2)Friday 11:30 – 12:15 (PK11.2)
Exercises: Friday 12:15 – 13:00 (PK11.2)
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12.05.2007 Introduction to Quantum Mechanics
URL Internet Skript
Vorlesung: http://www.pci.tu-bs.de/aggericke/PC4Englisch: http://www.pci.tu-bs.de/aggericke/PC4e_osv/
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12.05.2007 Introduction to Quantum Mechanics
Basis Concepts of Classical Mechanics The Newton equations of motions describes the movement of particles along well-defined trajectories which are the result of boundary conditions and the forces between them.
For example, let us concider a Harmonic Oscillator (a massive particle on a spring):V(x) = ½ kx², E = p²/2m + ½ kx² ⇒ E = ½ m (dx/dt)² + ½ kx²Having in mind the energy conservation law, we see that all possible energies are ellipses in the phase space (x ,p): E = p²/2m + ½ kx²
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12.05.2007 Introduction to Quantum Mechanics
Hamilton Equations
dq/dt = ∂H/∂p
dp/dt = − ∂H/∂q
* 4. Aug. 1805 in Dublin, Irland+ 2. Sep. 1865 in Dublin, Irland
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12.05.2007 Introduction to Quantum Mechanics
Problems of Classical Mechanics: Black Body Radiation.
How can be explained the intensity of black body radiation as function of the radiation wavelength?
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12.05.2007 Introduction to Quantum Mechanics
Black Body Radiation In 1900 Max (Karl Ernst Ludwig) Planck first suggested that the radiation energy
cannot have all continuum values. He postulated that the energy is always proportional to a certai very small discret portion of energy which cannot be disintegrated. This elementary portion of energy (quant) is proportional to the radiation frequency ν, E = h ν , where h is the proportionality constant which is now known as Planck constant.
h = 6,626176 . 10-34 J·s
* 23. April 1858 in Kiel, Schleswig-Holstein+ 4. Oktober1947 in Göttingen
Nobelpreis 1918
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12.05.2007 Introduction to Quantum Mechanics
Planck‘s Formula
The Planck‘s formula was found to be in perfect agreement with experiment. However, the Planck‘s postulate about existance of the elementary indivisible portion of energy (quant) resulted at that time to fierce discussions with the adepts of the classical theory.
The maximim of the Planck distribution corresponds to the radiation velocity of :
νmax = 2,8214 kT/hwhich can easily be measured experimentally and thus the Planck constant can be determined. Its value is
k = 1,3806503.10-23 J·K-1
ννπννν ν de
hc
du kTh 18)( 3
2
−=
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12.05.2007 Introduction to Quantum Mechanics
Problems of Classical Mechanics: Photoeffect
Experiment:• No photoelectrons are detected under some light frequency for any light intensity (the threshold effect). • The photoelectron energy does not depend on the light intensity. • The photoelectron energy linearly depends on the light frequency.Classical Interpretation:The electromagnetic field of the incident light E causes oscillation of the free electrons and pulls them out from the metall. However, this model predicts that the output electron flux is proportional to the light intensity and does not explain the threshold effect and why the electron flux is proportional to the light frequency. Quantum mechanical interpretation:The light is absorbed by the metall quant by quant. An electron is bound inside the metall and a certain energy (photoelectric work function) is needed for extracting it out to the vacuum. The rest of the photon energy is realized as the photoelectron kinetic energy. This model perfectly fits all experimental data.
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12.05.2007 Introduction to Quantum Mechanics
Albert Einstein* 14. März 1879 in Ulm, Württemberg+ 18. April 1955 in Princeton, New Jersey, USANobel Prize 1921 für Photoeffekt
He developed the Theory of Photoeffect, the Theory of Light Absorption my Matter, the Special Relativistic Theory, and the General Ralativistic Theory.
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12.05.2007 Introduction to Quantum Mechanics
Albert Einstein
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12.05.2007 Introduction to Quantum Mechanics
Why Quantum mechanics?
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12.05.2007 Introduction to Quantum Mechanics
Why Quantum mechanics?
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12.05.2007 Introduction to Quantum Mechanics
Why Quantum mechanics?
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12.05.2007 Introduction to Quantum Mechanics
Why Quantum mechanics?
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12.05.2007 Introduction to Quantum Mechanics
Why Quantum mechanics?
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12.05.2007 Introduction to Quantum Mechanics
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12.05.2007 Introduction to Quantum Mechanics
Interference of Electromagnetic Waves
φ= A e−iωt+ikx
ω=2πν: Wave Frequencyλ= c / ν : Wavelengthk = 2π/λ : Wavevector
Total Wave Intensity: I = |φ|2
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12.05.2007 Introduction to Quantum Mechanics
Experimente with particles
P12 = P1 + P2
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12.05.2007 Introduction to Quantum Mechanics
de Broglie: a Particle is a Wave
De Broglie's Dissertation “Recherches sur la théorie des quanta“ in 1924 at the firts time gave a relationship between a particle massm, its velocity v, and the corresponding wavelength λ:
Wavelength: λ = h/mv
Louis Victor Pierre Raymond duc de Broglie* 15. Aug. 1892 in Dieppe, France+ 19. März 1987 in Paris, France
Nobelpreis 1929
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12.05.2007 Introduction to Quantum Mechanics
Interference of Matter Waves
φ = A e−iωt+ikx
ω=2π Ε/h: Frequency
k = 2π/λ = p/h : Wavevector
Probability to Detect the Particle I = |φ|2
φ = φ1+φ2I = |φ1+φ2|2
I = I1+I2+2(I1I2)½.cos ∆ϕ
Where ∆ϕ is the phase difference
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12.05.2007 Introduction to Quantum Mechanics
Experiment: a Particle is a Wave
Diffraction of Elecrtons Diffraction of X-Rays
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12.05.2007 Introduction to Quantum Mechanics
Spectrum of Electromagnetic Waves
Spectral Areas
Visible Light
700 600 500 400
Wavelength, nm
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12.05.2007 Introduction to Quantum Mechanics
Electromagnetic Spectrum
Type of Radiation
Frequency Range (Hz)
Wavelength Range Type of Transition
gamma-rays 1020-1024 <1 pm nuclear
X-rays 1017-1020 1 nm-1 pm inner electron
ultraviolet 1015-1017 400 nm-1 nm outer electron
visible 4-7.5x1014 750 nm-400 nm outer electron
near-infrared 1x1014-4x1014 2.5 µm-750 nm outer electron molecular vibrations
infrared 1013-1014 25 µm-2.5 µm molecular vibrations
microwaves 3x1011-1013 1 mm-25 µm molecular rotations, electron spin flips*
radio waves <3x1011 >1 mm nuclear spin flips*
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12.05.2007 Introduction to Quantum Mechanics
Electromagnetic Radiation
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12.05.2007 Introduction to Quantum Mechanics
Light PolarizationE
Unpolarized light
E
Linearly polarized light
Circularly polarized light
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12.05.2007 Introduction to Quantum Mechanics
Diffraction on a Slit: The Uncertainty Principle
Interference: θ = λ/(2 ∆x)
Impulse: ∆px ≈ p • θ = p λ/(2 ∆x)
De Broglie wavelength: p = h/λ
∆x .∆px ≈ ħ
Planck-Constante ħ = h/2π
h = 6,6260755·10-34 J·s
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12.05.2007 Introduction to Quantum Mechanics
Werner Karl Heisenberg
* 5. Dez. 1901 in Würzburg +1. Feb. 1976 in München
Nobelpreis 1932
1927 Unschärferelation
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12.05.2007 Introduction to Quantum Mechanics
A Particle as a Wavepacket
The wavepacket can be presented as a superposition of many harmonic waves with different wavelengths (impulses).
λπ h
h2,)()( ==Φ ∫
∞
∞−
pdpepwxxpi
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12.05.2007 Introduction to Quantum Mechanics
Einstein: Interaction of Light with Radiation
dN2/dt = B12. u(ν) . N1
dN2/dt = − A21 N2dN2/dt = − B21
. u(ν) . N2
A21 / B12 = 8πh ν ³/c³B21 / B12 = g2 / g1
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12.05.2007 Introduction to Quantum Mechanics
Three Main Principles of Quantum Mechanics
1. The probability of an experimental event P is given by the square of a complex number Φwhich is called the probability amplitude, or the wave function:
P = ⏐Φ⏐2 = Φ . Φ*
2. If the event can be realized through indistinguishable ways each described by the probability amplitudes Φ1 , Φ2 , ets., the total probability amplitude Φ can be found as a linear superposition of the amplitudes Φ1 and Φ2 (Superposition Principle):
Φ = a1Φ1 + a2Φ2
P = ⏐Φ⏐2 = ⏐a1Φ1 + a2Φ2⏐2 ← Interference
3. If the experiment allows to determine which alternative is realized, the total probability of the event P is the sum of probabilities P1 and P2 .
P = P1 + P2 = ⏐a1 Φ1⏐2 + ⏐a1 Φ2⏐2 ← no Interference
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12.05.2007 Introduction to Quantum Mechanics
<bra| und |ket> Vectors
< to | from >
The total probability amplitude:
<x|Q> = Σi=1<x|i><i|Q>
Σi=1 |i><i| = 1
The values <i|Q> = ai show the contributions from the slits 1 and 2 to the total amplitude:
<x|Q> = a1<x|1> + a2 <x|2> ψ(x) = a1φ1(x) + a2φ2(x)
|Q> = a1 |1> + a2 |2>
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12.05.2007 Introduction to Quantum Mechanics
Paul Adrien Maurice Dirac
* 8. Aug. 1902 in Bristol, England+ 20. Okt. 1984 in Tallahassee, Florida, USA
Nobelpreis 1933
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12.05.2007 Introduction to Quantum Mechanics
Comparison of the Vector and Amplitude Notations
|Q> = a1 |1> + a2 |2> +... ψ(x) = a1φ1(x) + a2φ2(x)+…
Pi = |ai|2
Σx <i|x> <x|i> = 1 and Σx |x><x| = 1 <i|i> = 1 normalization
<i|k> = 0 (i≠k) orthogonality
Pi = |ai|2
-∞∫∞ |φi(x)|2dx = 1
-∞∫∞
φi(x)*φk(x) dx = 0 (i≠k)
Σi Pi = 1
Σi |ai|2 = 1
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12.05.2007 Introduction to Quantum Mechanics
Calculation of the Probability Amplitude (Wavefunction)
There are three mathematically equivalent ways of calculation of the probability amplitude.1) Matrix algebra
Werner HeisenbergNobelpreis 1932
2) Differential equation, DGLErwin SchrödingerNobelpreis 1933
3) Tragectory integrals Richard Feynman; Nobelpreis 1965
* 12. Aug. 1887 in Erdberg, Wien+ 4. Jan. 1961 in Wien
* 11. Mai 1918 in Far Rockaway, New York+ 15. Feb. 1988 in Los Angeles
* 5. Dez. 1901+ 1. Feb. 1976
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12.05.2007 Introduction to Quantum Mechanics
The Schrödinger Equation
Time-independent Schrödinger Equation:
[− h²/2m∂²/∂x² + V(x)] ψ(x) = E ψ(x) (1 dimension)
[− h²/2m ∆ + V(x,y,z)] ψ(x,y,z) = E ψ(x,y,z) (3 dimensions)
Where ∆ is the Laplace operator: ∆ = ∂²/∂x² + ∂²/∂y² + ∂²/∂z²
Time-dependent Schrödinger Equation:
[− h²/2m∂²/∂x² + V] ψ(x,t) = ih ∂/∂t ψ(x,t)
Normalization
−∞∫+∞
|ψ(x,y,z)|2 dx dy dz = 1
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12.05.2007 Introduction to Quantum Mechanics
Schrödinger Equation for a Free Particle
If V(x)= 0 => − h²/2m∂²/∂x² ψ(x) = E ψ(x)
The solution (plane wave) is:
Ψ(x) = A eikx + B e-ikx where k²=2Em/h²
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12.05.2007 Introduction to Quantum Mechanics
A Particle in a One-Dimensional Box
The Schrödinger Equation
)()(2 2
22xEx
xdd
m nnn ψψ =−h
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12.05.2007 Introduction to Quantum Mechanics
A Particle in a One-Dimensional Box
In conjugated dye molecules the π-electrons behave like a free particles moving along the molecular chain.
The Schrödinger equation:
d²ψ(x)/dx² + 2mE/h² ψ(x) = 0
With boundary conditions: ψ(0)=0 und ψ(L)=0.
The solution is:
En = h²/8mL² n² where n = 1, 2, 3, ...
ψn(x) = (2/L)½ sin (nπx/L)
The factor (2/L)½ results from the wavefunction normalization:
0 ∫L|ψn|2dx = 0 ∫
LC2 sin2(nπ x/L)dx = C2 . ½ L = 1
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12.05.2007 Introduction to Quantum Mechanics
A Particle in a One-Dimensional Box
ψn(x)WallWall
n = 5 2 1 4 3
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
L0
E = n2h2/8mL2ψn(x) = (2/L)1/2 sin(πnx/L)
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12.05.2007 Introduction to Quantum Mechanics
What Happens if a Wall is Not Infinitely High?Tunneling
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12.05.2007 Introduction to Quantum Mechanics
Tunneling
Tunneling Microscope
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12.05.2007 Introduction to Quantum Mechanics
Example: Scanning Tunneling MicroscopeNobel Prize 1986
Gerd Binning and Heihrich Rohrer
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12.05.2007 Introduction to Quantum Mechanics
Scanning Tunneling Microscope
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12.05.2007 Introduction to Quantum Mechanics
Harmonic OscillatorDiatomic molecule
F = - k x
µ = mAmB/(mA+mB))()(
22
2
2
22xExxk
xdd
m nnn ψψ =⎟⎟⎠
⎞⎜⎜⎝
⎛+−
h
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12.05.2007 Introduction to Quantum Mechanics
Harmonic Oscillator: Diatomic Molecule Energy Levels
Ev = (v + ½) ħ ω, ω = (k/µ)½, µ = mAmB/(mA+mB)
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12.05.2007 Introduction to Quantum Mechanics
Harmonic Oscillator: the Wavefunctions41
22 ;;)()(
2
⎟⎟⎠
⎞⎜⎜⎝
⎛=== −
kxyeyHNx y
vvv µα
αψ h
Hv are Hermite polynomials: H0(y) = 1, H1(y) = 2y, H2(y) = 2y2 – 2, H3(y) = 8y2 – 12y, ets.
-4 -2 2 4
-0.75
-0.5
-0.25
0.25
0.5
0.75
1
v = 2 1 0
|ψ(x)|2ψ(x)
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12.05.2007 Introduction to Quantum Mechanics
Hydrogen AtomHamiltonian-
+
reZ
mH e
e 0
22
2
42 επ−∇−=
hkkk EH ψψ =
Polar coordinates2
2
2
2
2
22
zyxe ∂∂
+∂∂
+∂∂
=∇ ( )φθ ,1 222
2 Λ+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
=∇r
rrre
( )θ
θθθθθ
φθ∂∂
∂∂
+∂∂
=Λ sinsin
1sin
1, 2
2
22
Wavefunctions),,( φθψ rk where k ≡ n, l, s, ml, ms
sl msmlnlk rRr χφθφθψ ),()(),,( Υ=
Rn l are Associated Laguerre polynomials and Yl m are Spherical Harmonics
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12.05.2007 Introduction to Quantum Mechanics
Spherical Harmonics for l = 0, 1, 2, 3, 4
Yl,m(θ, φ) = Plm(cos θ) .fm (φ) fm (j) = 1/(2π)½ eimφ
Elektron l m Yl,m(θ, φ) |Yl,m|2
s 0 0 1/(4 π)½ 1/4 π
p11
±10
±(3/8 π)½ sin θ e±i φ
(3/4 π)½ cos θ
3/8 π sin² θ3/4 π cos² θ
d222
±2±10
(15/32 π)½ sin² θ e±2i φ
±(15/8 π)½ sin θ cos θ e±i φ
(5/16 π)½ (3cos² θ - 1)
15/32 π sin4 θ15/8 π sin² θ cos² θ5/16 π(3cos² θ - 1)²
f
3333
±3±2±1 0
(35/64 π)½ sin3 θ e±3i φ(105/32 π)½ sin² θ cos θ e±2i φ
(21/64 π)½ sinJ (5 cos² θ - 1) e±i φ
(7/16 π)½ (5 cos3 θ - 3 cos θ)
35/64 π sin6 θ105/32 π sin4 θ cos² θ
21/64 π sin² θ (5 cos² θ - 1)²7/16 π (5 cos3 θ - 3 cos θ)
g
4 4444
±4 ±3±2±10
(315/512 π)½ sin4 θ e±i4 φ
(315/64 π)½ sin3 θ cos θ e±3i φ
(225/660 π)½ sin² θ (7 cos² θ - 1) e±2i φ
(225/320 π)½ sinJ (7 cos3 θ - 3 cos θ) e±i φ
(9/256 π)½ (35 cos4 θ - 30 cos² θ + 3)
315/512 π sin8 θ315/64 π sin6 θ cos² θ
225/660 π sin4 θ (7 cos3 θ - 1)²225/320 π sin² θ (7 cos3 θ - 3 cos θ)²9/256 π (35 cos4 θ - 30 cos² θ + 3)²
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12.05.2007 Introduction to Quantum Mechanics
Real Wavefunctions Obtained from Linear Combinations
l |ml| Wavefunctions
0 0 s = 1/(4π)½
pz = (3/4π)½cosϑ
px = (3/4π)½sinϑcosϕ
py = (3/4π)½sinϑsinϕ
d3z²−r² = (5/16π)½(3 cos²ϑ − 1)
dxz = (15/4π)½sinϑcosϑcosϕ
dyz = (15/4π)½sinϑcosϑsinϕ
dx²−y² = (15/16π)½sin²ϑcosϕ
dxy = (15/16π)½sin²ϑsinϕ
2
0
1
2
1
0
1
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12.05.2007 Introduction to Quantum Mechanics
Hydrogen Atom Wavefunctions: Angular PartpX orbital pY orbital
l = 1 m = ±1s orbital:
l = 0 m = 0pZ orbital:l = 1 m = 0
dz² orbital: m=0 dx²-y² orbital
dxy orbital dzx orbital dyz orbital
l = 2
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12.05.2007 Introduction to Quantum Mechanics
Hydrogen Atom Wavefunctions: Radial Part
2/30
10
02aeR
ar
−
=
2/30
2
020 22
2 0
a
ear
R
ar
−
⎟⎠⎞⎜
⎝⎛ −
= 2/30
2
021 62
0
a
ear
R
ar
−
=
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12.05.2007 Introduction to Quantum Mechanics
Niels Henrik David Bohr
* 7. Okt. 1885 in Kopenhagen, + 18. Nov. 1962 in Kopenhagen
Nobelpreis 1965
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12.05.2007 Introduction to Quantum Mechanics
Hydrogen Atom Energy Levels
K,3,2,124
12
0
2
0
=−= nna
eEnl πε
ε0 is the Vacuum permittivity; ε0 = 8.854·10-12 J-1 C2 m-1
e is the Elementary charge; e = 1.602·10-19 Ca0 is the Bohr radius; a0 = 4πε0ħ2/mee2 = 5.292·10-11 m