Transcript Quantum mechanics – an introduction

Introduction to
Quantum mechanics and Molecular Spectra
Ka-Lok Ng
Asia University
Contents
• The postulates of quantum mechanics (QM)
The wave equation – Schrodinger equation
• Quantum mechanical operators
• Eigenvalues of QM operators
• Wave functions
• The particle in a 1D box
• Physical methods of determining the 3D structure of proteins
References
• House J.E. Fundamentals of quantum chemistry, 2nd ed. Elsevier
2004
• Whitford D. Proteins: structure and function. J. Wiley 2005.
• http://www.spaceandmotion.com/Physics-Erwin-Schrodinger.htm
• Molecular spectra, see
http://spiff.rit.edu/classes/phys315/lectures/lect_14/lect_14.html
http://cref.if.ufrgs.br/hiperfisica/hbase/molecule/molec.html#c2
The postulates of quantum mechanics (QM)
Postulate I
For any possible state of a system, there is a function y of the coordinates of
the parts of the system and time that completely describes the system.
  ( x, y, z, t )
 Is called a wave function. For two particles system,
  ( x1 , y1 , z1 , x2 , y2 , z2 , t )
The wave function square 2 is proportional to probability. Since  may be
complex, we are interested in *, where * is the complex conjugate (i  -i) of .
The quantity *dt is proportional to the probability of finding the particles of the
system in the volume element, dt = dxdydz.
*

 dt  1
all _ space
that is the probability of finding the particle in the universe is 1  normalization condition.
The postulates of quantum mechanics (QM)
Orthogonal of two wave functions
  *ydt  0
y *dt  0
Example: sinq and cosq are orthogonal functions.

 sin q cos qdq  0
0
Fourier series expansion – sin(nq) and cos(nq) orthogonal functions
The Wave Equation
•
•
In 1924 de Brogile shown that a moving particle has a wave character. This idea
was demonstrated in 1927 by Davisson and Germer when an electron beam was
diffracted by a nickel crystal.
According to the de Brogile relationship, there is a wavelength associate with a
moving particle which is given by
h

mv
where , h, m and v denote the wavelength, Planck’s constant, mass and velocity.
Erwin Schrodinger adapted the wave model to the problem of the hydrogen atom and
propose the Schrodinger equation. The model needs to describe a three-dimensional
wave.
Classical physics – the flooded planet problem = the waveforms that would result form a
disturbance of a sphere that is covered with water
The classical 3D wave equation is
 2  2  2 1  2
 2  2  2 2
2
 x  y  z v t
where  is the amplitude function and v is the phase velocity of the wave.
Schrodinger equation
The Wave Equation
The Schrodinger wave equation
 2  2  2  4 2 m2 2( E  V )
 2  2 

2
2
 x  y  z
h
m
The 1D wave equation solution
http://www-solar.mcs.stand.ac.uk/~alan/MT2003/PDE/node12.html
The 2D wave equation solution
http://www.math.harvard.edu/archive/21b_fall_03/waves/index.html
Operators
Postulate II
With every physical observable q there is associated an operator Q, which
when operating upon the wavefunction associated with a definite value of that
observable will yield that value times the wavefunction F, i.e. QF = qF.
H
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qmoper.html
Operators
(1) The operators are linear, which means that
O(1  2) = O1 + O2
• The linear character of the operator is related to the superposition of states
and waves reinforcing each other in the process
(2) The second property of the operators is that they are Hermitian (the 19th
century French mathematician Charles Hermite).
• Hermitian matrix is defined as the transpose of the complex conjugate (*) of a
matrix is equal to itself, i.e. (M*)T = M
x  iy 
 a

M  
x

iy
c


x  iy 
 a

M *  
x

iy
c


x  iy 
 a
  M
( M * )T  
x

iy
c


In QM, the operator O is Hermitian if
*
* *
F
O

d
t


O

 F dt
C. Hermite
http://commons.wikimedia.org/wiki/Image:Charles_Hermite_circa_1887.jpg
Eigenvalues of QM operator
Postulate III
• The permissible values that a dynamical variable may have are
those given by OF = aF, where F is the eigenfunction of the QM
operator (Hermitian) O that corresponds to the observable whose
permissible real values are a.
• The is postulate can be stated in the form of an equation as
O
operator
F
wave function
= a
eignevalue
F
wave function
Example Let F = e2x and O=d/dx
 dF/dx = d(e2x)/dx = 2 e2x
 F is an eigenfunction of the operator d/dx with an
eigenvalue of 2.
Eigenvalues of QM operator
• Eigenvalues of QM operator must be real !
• Example
b  ic 
 a

M  
d 
 b  ic
solve  Mx  x
 (a   )(d   )  (b  ic)(b  ic)  0
 2  (a  d )  ad  b 2  c 2  0
The two values for  are real
Wave functions
Postulate IV
• The state function  is given as a solution of
Hˆ   E
•
Schrodinger equation
where Hˆ is the total energy operator, that is the Hamiltonian operator.
The hamiltonian function is the total energy, T+V, where T is the kinetic energy and
V is the potential energy. In operator form
Hˆ  Tˆ  Vˆ
where Tˆ is the operator for kinetic energy and Vˆ is the operator for potential energy.
In differential operator form, the time dependent Schrodinger equation is
h2
2
2
2
h 

(



V
(
q
,
t
))

(
q
,
t
)


(qi , t )
i
i
2
2
2
2
8m x y
z
2i t
where qi is the generalized coordinates, m is the mass of the particle.
The particle in a one-dimensional box
•
•
•
•
•
We treat the behavior of a particle that is confined to
motion in a box
The coordinate system for this problem is show at the
right
The Hamiltonian operator, H, is H = T + V = p2/2m + V
where p is the momentum, mass is the mass of the
particle, and V is the potential energy
Outside the box V = ∞, so the Schrodinger equation
Hˆ   E
h2 d 2

    E
8m 2 dx2
For the equation to be valid, y must be 0 
Boundary condition = the probability of finding the
particle outside the box is zero
Inside the box, V = 0, so the Schrodinger equation
becomes
h2 d 2

  E
2
2
8m dx
http://www.everyscience.com/Chemistry/Physical/Quantum_Mechanics/a.1128.php
The particle in a one-dimensional box
d 2
2

k
0
2
dx
Where k2 = 82mE/h2. This is a linear differential equation
with constant coefficients, which have a solution fo the
form  = A cos(kx) + B sin(kx).
The constant A and B must be evaluated using the
boundary conditions. Boundary conditions are those
requirments that must be met becase of the the physical
limits fo the system.
For the probability of finding the particle to vanish at the
walls of the box, that is  = 0 both at x≦0 and x≧L.
At x≦0
 = 0 = A + B(0)  A = 0
At x≧L
 = 0 = B sin kL
 Since B ≠ 0, otherwise the complete wavefunction = 0 !
 sin kL = 0 that is kL = n  quantization condition !
The particle in a one-dimensional box
•
•
•
quantization condition  kL = n
Recall k2 = 82mE/h2
k2 L2 = 82mE/h2 L2 = n22, where n = 1, 2 ……. is the quantum number
n2h2
E
8m L2
• Zero-point energy
• One quantum number arise from a 1D system
E ~ n2
E ~ 1/L2
E~m
To determine the wavefunction , one uses the normalization condition
L
  dt   B
*
all _ space
  ( x) 
2
sin (n / L) xdx  1
2
n = 1, 2, 3
0
2
nx
sin(
)
L
L
http://www.everyscience.com/Chemistry/Physical/Quantum_Mechanics/a.1128.php
The particle in a one-dimensional box
• Consider a carbon chain like
C=C-C=C-C
• as an arrangement where the  electrons can move along the
chain. If we take an average bond length of 1.40 Angstrom, the
entire chain would be 5.60 Angstrom length, Therefore, the
energy difference between the n=1 and n=2 state would be
3h 2
3(6.631027 erg  s) 2
E

2
8m L 8(9.101028 g )(5.60108 cm) 2
E  5.781012 erg
The energy corresponds to a wavelength of 344 nm, and the maximum in the
absorption spectrum of 1,3-pentadiene is found at 224 nm. Although this not
close agreement, the simple model does predict absorption in the UV region of
the spectrum.
Molecular Spectra
Three types of energy levels in molecules
• electronic: large energy separations (200-400 kJ/mol)  optical or UV
• vibrational: medium energy separations (10-40 kJ/mol)  Infrared
• rotational: small energy separations (10-40 J/mol)  microwave
• All the energy levels are quantized
Molecular Spectra
• For a spectral line of 6000 Angstrom, which is in the visible light region,
the corresponding energy is E = hc/ = 3.3x10-12 erg
 a molar quantity multiply by Avogadro’s number  E = 200 kJ/mol
• Diatmoic molecule can be viewed as if they are held together by bonds
that have some stretching and bending (vibrational) capability, and the
whole molecule can rotate as a unit.
http://cref.if.ufrgs.br/hiperfisica/hbase/molecule/molec.html#c2
Molecular Spectra
• Normal mode of vibration for the CO2 molecule
behaves much like a simple harmonic
oscillator
• The vibrational energies can therefore be
described by the relation (+1/2)*h, where ,
the vibrational quantum number = 0,1,2,3….
and =the classical frequency
• the symmetric stretch mode
• the asymmetric stretch mode
• the bending mode
http://www.phy.davidson.edu/StuHome/shmeidt/JuniorLab/CO2Laser/Theory.htm
Vibrational and Rotational energy levels transition spectra
• The CO2 molecule is free to rotate. The energies of the rotational
modes (E = h2/(82I), where I is the moment of inertia)) are smaller
than for vibrational modes. Hence, the energy levels for two
vibrational states with the rotational divisions looks like:
Vibrational and Rotational energy levels transition spectra for HCl
http://universe-review.ca/F12-molecule.htm
Applications of the Vibrational Energy Levels
•
•
•
•
determination of bond lengths
determination of bond force constants
determination of bond dissociation energy
qualitative and quantitative chemical analysis
Selection rules for energy level transitions
• Selection rules are divided into high probability or allowed transitions
and Forbidden transitions of much lower probability
• Forbidden transition – symmetry-forbidden and spin-forbidden
transitions
• Spin-forbidden transitions involve a change in spin multiplicity defined
as 2S+1 where S is the electron spin number. Spin multiplicity
reflects electron pairing (see Table). For a favourable transition there
is no change in multiplicity (DS=0)
Number of unpair Electron
electrons
spin S
2S+1
Multiplicity
0
0
1
Singlet
1
½
2
Doublet
2
1
3
Triplet
3
3/2
4
quartet
Selection rules for energy level transitions
• Symmetry-forbidden transitions reflect redistributin of charge during
transitions in a quantity called the transition dipole moment.
• Differences in dipole moment arise from the different electron
distributions of ground and excited states
• For an allowing transition it requires a change in dipole moment
• EM radiation can induce Rotational transitions only in molecule with
a permanent dipole moment.
• Not all molecules have dipole moments!
– (1) only polar molecules can absorb and emit electromagnetic photons
– (2) non-polar molecules : H2 ,CO2 ,CH4
– (3) energy transfer can take place during collisions
• The intensity of the signals in a rotational spectrum increase with the
molecular dipole moment.
Fluorescence Spectroscopy
•
•
•
•
•
•
•
Fluorescence – excited molecules decay to the
ground state via the emission of a photon with
DS = 0 ( no change in spin multiplicity, S1  S0)
Emission is occurs at longer wavelengths than
the corresponding absorbance band
Quantum yield of Fluorescence emission
= photons emitted/number of photons
absorbed
maximum value of quantum yield is 1
Photophysical properties of a fluorophore can
be used to obtain information on its immediate
molecular environment. Relaxation of a
fluorophore from its excited state can be
accelerated by fluorescence resonance energy
transfer (FRET).
FRETcan be used to characterize proteinprotein interactions as observed in signaling
complexes of ion channel proteins.
http://www.physiologie.uni-freiburg.de/fluorscence.html
Raman Spectroscopy
•
•
•
•
Chandrasekhara Venkata Raman (1888-1970) who discovered in 1928 that
light interacts with molecules vai absorbance, transmission or scattering
the first Indian Nobel Laureate in physics
Raman made many major scientific discoveries in acoustics, ultrasonic, optics,
magnetism and crystal physics
Scattering can occur at the same wavelength when it is known as Rayleigh
scattering (elastic, 0) or it can occur at altered frequency (change in the
colour of the scattered light) when it is the Raman effect
Figure. See http://www.search.com/reference/Raman_spectroscopy
Figure. See http://www.inphotonics.com/raman.htm
http://www.vigyanprasar.gov.in/dream/feb2002/article1.htm
Raman Spectroscopy
• some weaker bands of shifted frequency are detected. Moreover,
while most of the shifted bands are of lower frequency 0 - i, there
are some at higher frequency, 0 + i. By analogy to fluorescence
spectrometry, the former are called Stokes bands and the latter antiStokes bands. The Stokes and anti-Stokes bands are equally
displaced about the Rayleigh band; however, the intensity of the
anti-Stokes bands is much weaker than the Stokes bands and they
are seldom observed.
http://www.gfz-potsdam.de/pb4/pg2/equipment/raman/raman.html
Raman Spectroscopy Application
• commonly used in chemistry
• provides a fingerprint by which the molecule can be
identified. The fingerprint region of organic molecules is
in the range 500-2000 cm-1.
• to study changes in chemical bonding, e.g. when a
substrate is added to an enzyme.
• Raman gas analyzers have many practical applications,
for instance they are used in medicine for real-time
monitoring of anaesthetic (麻醉的) and respiratory gas
mixtures during surgery.
• In solid state physics, spontaneous Raman spectroscopy
is used to, among other things, characterize materials,
measure temperature, and find the crystallographic
orientation of a sample.
http://www.search.com/reference/Raman_spectroscopy
Nuclear Magnetic Resonance Spectroscopy
In 1945, the NMR phenomenon was given by
F. Bloch and E. M. Purcell (both share the
1952 Nobel Prize)
NMR spectra are observed upon the pulse
absorption of a photon (radio frequency) of
energy and the transition of nuclear spins
from ground to excited states
Bloch F (1905-1983)
Purcell E.M. (1912-1997)
http://nobelprize.org/nobel_prizes/physics/laureates/1952/
http://cancer.stanford.edu/research/milestones/
http://www.pulseblaster.com/gallery/1.html
Nuclear Magnetic Resonance Spectroscopy
For 1H there are two orientations. In one orientation the protons are
aligned with the external magnetic field (north pole of the nucleus
aligned with the south pole of the magnet and south pole of the
nucleus with the north pole of the magnet) and in the other where
the nuclei are aligned against the field (north with north, south with
south)
= aligned against
= aligned with
A spinning nucleus is
equivalent to a magnet
http://www.brynmawr.edu/Acads/Chem/mnerzsto/The_Basics_Nuclear_Magnetic_Resonance%20_Spectroscopy_2.htm
http://vam.anest.ufl.edu/simulations/nuclearmagneticresonance.php
Nuclear Magnetic Resonance Spectroscopy
Nuclei possessing angular moment (also called spin)
have an associated magnetic moment (current
generate magnetic field). Certain atomic nuclei,
such as 1H, 13C, 15N and 31P have spin S=½ and
2H, 14N have spin S=1, 18O has S=5/2).
For nuclei such as 12C is the most common isotope is
NMR silent, that is not magnetic. If a nucleus is
not magnetic, it can't be studied by nuclear
magnetic resonance spectroscopy. For the
purposes, biomolecular NMR spectroscopy
requires proteins enriched with 1H, 13C or 15N or
ideally all nuclei.
Nuclear transitions differed in frequency from one
nucleus to another but also showed subtle
differences according to the nature of the
chemical group (chemical shift effect).
Methyl protons resonating at a frequency ≠
amide proton ≠ a-carbon proton ≠ b-carbon
proton  The chemical environment of such
nuclei are different
Probe by NMR and this technique can be exploited to
give information on the distances between atoms
in a molecules. These distances can then be
used to derive a 3D model of the molecule.
f 
H 0
2
f  frequency_ of _ absorption
  gyrom agnetic _ ratio _ of _ the _ nucleus
H 0  external_ m agnetic_ field
The frequency range needed to excite
protons is relatively high. It ranges from
300 MHz to 900 MHz.
Nuclear Magnetic Resonance Spectroscopy
Why do we see peaks ?
• When the excited nuclei in the beta orientation start to relax back down to the alpha
orientation, a fluctuating magnetic field is created. This fluctuating field generates a
current in a receiver coil that is around the sample. The current is electronically
converted into a peak.
Why do we see peaks at different positions?
• because nuclei that are not in identical structural situations do not experience the
external magnetic field to the same extent. The nuclei are shielded or deshielded due
to small local fields generated by circulating s,  and lone pair electrons.
NMR spectra
Solid-state 900 MHz (21.1 tesla) NMR spectrometer
at the Canadian National Ultrahigh-field NMR Facility
for Solids.
http://www.answers.com/topic/solid-state-nuclear-magnetic-resonance
http://www.scielo.br/scielo.php?pid=S0100-41582002000500017&script=sci_arttext
Nuclear Magnetic Resonance Spectroscopy
Limitations for NMR methods
1. For small proteins with size < 100 kD
2. Require highly concentrated protein solutions on the
order of 1-2 mM.
3. pH of solution < 6.