Transcript Quantum Physics

Quantum Physical Phenomena
in Life (and Medical) Sciences
Quantum Physics
Experiments and principles beyond the capacity
of the classical (Newtonian) physics
Péter Maróti
Professor of Biophysics, University of Szeged, Hungary
Suggested text:
S. Damjanovich, J. Fidy and J. Szőlősi: Medical Biophysics, Semmelweis, Budapest 2006
New (quantum) physics is needed
to describe and to understand
• the chemical-physical nature of electrons in
atoms and molecules (bonds, interactions, etc.),
• the behavior of nucleus (radioactivity, nuclear
reactions, radiation therapy, etc.),
• the energetic changes in nuclea, atoms and
molecules (spectroscopy) and
• the image formation from human body and
organs (iconography),
because classical physics fails to work in these fields of the microworld.
What is new that quantum physics offers?
1. Some physical quantities are quantized
- the energy of electromagnetic fields (black body radiation)
- the energy of the atoms (absorption and emission spectra of atoms, Franck-Hertz
experiment)
- the magnetic moments of particles, the spin (Einstein-de Haas experiment)
2. Wave-particle dualism
- Wave properties of particles (Davisson-Germer experiment, Tunneling)
- Particle properties of waves (photoelectric (Hallwachs) effect)
3. „Unusual” distribution
- Spatial distribution: description using the term of probability (wave function)
- Deviation from equipartition (temperature-dependence of the molar heat capacity of
solids)
4. Uncertainty principle after Heisenberg
Complementer quantities cannot be measured simultaneously with arbitrarily small
precision.
1. The energy is quantized: the electromagnetic
field and the atomic oscillators
1. Black body radiation and the UV catastrophe
explained by quantized energy levels for light and emitting oscillators (Planck).
Light emitted by a black body has a spectrum that
depends on temperature. The “ideal black body” is a box
with black walls, opening to the exterior through a small
hole. Radiation escaping through the hole will be in
thermal equilibrium with the temperature of the walls.
Classical physics attempted to explain the shape of the curve of power (or energy
density, ρ) as a function of wavelength through four laws.
1. Kirchhoff’s law: e/a = E(λ,T),
for any substances, where e: emission and
a: absorption. For black body: a = 1.
2. Wien’s displacement law:
T·λmax= 2896 μm·K (constant)
T is the absolute temperature and λmax is
the peak of the curve.
3. Stefan-Boltzmann law: Etotal = σ·T4
Etotal is the emittance (total power per unit
area), and σ = 56.7 nW·m-2·K-4. This is why
bulb filaments are run hot!
4. Rayleigh-Jeans law:

8kT
4
The first three laws worked fine, but not the Rayleigh-Jeans law that led to
„UV catastrophe”
In the UV, the
Rayleigh-Jeans
curve diverges to
infinity
Ultraviolet catastrophe
Rayleigh-Jeans curve

Experiment
black body
radiation
8kT
4
All energy levels allowed, so fraction
of energy density with high energy
increases as temperature increases.
The radiation escaping from a black body is
determined by the energy loss from vibration of
molecules (the electromagnetic oscillators) in the
walls. These increase with T. The energy of the
light emitted is determined by the oscillators.
Planck’s curve
Planck proposed that the energy of the electromagnetic
oscillators was limited to discrete values, rather than
continuous. Planck’s famous equation is E = nhn,
where n = 0, 1, 2, …, n (= c/) is the frequency, and h
is Planck’s constant, 6.626 10-34 Js.

8 hc 
1



4   e hc / kT  1 
The energy levels are quantized. At any temperature,
the transitions at higher energy levels are less likely to
occur, so contribution to energy density falls off in
high energy range (UV). As the energy (E = hn =
hc/) approaches kT, the term in brackets approaches
1, and the Planck equation becomes the same as the
Rayleigh-Jeans equation.
Conclusion: the energy of the oscillators that make up the electromagnetic field are
QUANTIZED
2. Temperature dependence of molar heat capacities of solids
Upon lowering the
temperature, the observed
heat capacity did not remain
constant but approached to
zero.
The heat capacity (specific heat) is the
proportionality factor relating
temperature rise, DT, to the heat
applied: q = C·DT. The classical view
was that C was related to the oscillation
of atoms about their mean position,
which increased as heat was applied. If
the atoms could be excited to any
energy, then a value of
C = 3R = 25 J·K-1
was expected, and this value, proposed
by Dulong and Petit, was observed for
many systems at ambient temperature.
However, the expected behavior was a
constant value as a function of T, and
this was not seen, and some elements
(diamond) had values way off.
Temperature dependence of molar heat capacities of solids
The Dulong and Petit’s law
CV, m
 U m 

  3 R
 T V
comes from the equipartition
theorem of the classical
physics.
Significant deviations were
observed at low temperatures.
Einstein and Debye formula:
The atoms oscillate with the
same frequency (Einstein) or
with the range of frequencies
(Debye).
Temperature dependence of molar heat capacities of solids
Einstein showed that if Planck’s
hypothesis of quantized energy
levels was applied, the equation:
 2hkTn 

hn  e
2
C  3Rf , where f 


kT  hkTn 1 
e

provided a good fit. This was
later improved by Debye who
allowed a range of values for n.
Conclusion: the energy of atomic oscillators
in matter are also QUANTIZED.
Conclusions from Planck’s solution to the UV catastrophe and
Einstein’s solution to the heat capacity problem
1. Heating a body leads to oscillations in the structure at the atomic
level that generate light as electromagnetic waves over a broad
region of the spectrum. This is an idea from classical physics.
2. From Planck’s hypothesis, the properties can be understood if the
energy of the oscillations (and hence of the light) are constrained
to discrete values, - 0, hn, 2hn, 3hn, etc. At any frequency value,
the intensity of the light is a function of the number of quanta, n,
at a fixed energy, determined by hn. This is in contrast to the
classical view in which energy levels were assumed to be
continuous, and intensity at any frequency was dependent on the
amplitude of the wave.
3. A similar conclusion comes from Einstein's treatment of the heat
capacity, but here the effect is seen from the oscillations
generated in the substance on application of heat. Absorption of
energy is therefore also quantized.
1. The energy is quantized: the atoms
When a high voltage is discharged
through a gas, the atoms or
molecules absorb energy and from
collision with the electrons, and reemit the energy as light. It is found
that the light is not a continuous
range at all frequencies, but is
constrained to a few narrow lines
(right). Explanations of the
emission spectrum of atomic
hydrogen played a critical role in
development of a quantum
mechanical understanding of the
structure of atoms.
Lines similar to those in the
hydrogen emission spectrum were
seen as absorption lines in the light
from stars.
Lines of hydrogen emission spectrum
Balmer studied the emission spectrum in the near UV-visible region, and
noticed that the distribution of the lines along the wavelength scale
showed an interesting pattern that could be described by the Balmer
formula:
1
1
1
~
n   109678 2  2 

n 
2
where n is 3, 4, 5,… Later work revealed a more extensive set of lines in
the UV and IR, which all followed the general pattern given by:
 1
1
1 
~

n   2  2
n


 L nH 
where  is the Rydberg constant, and nL and nH are lower and higher
value integers.
Term scheme of the H-atom
Balmer series:
Franck-Hertz experiment
One of the most demonstrative experiments for the quantization of energy of atoms.
The wavelegth of the emitted light is

hc
e U
At the first step of U = 4.9 V, λ = 253.7 nm
The steps of the currentvoltage characteristics
describe the set of discrete
energy levels (term scheme)
of the mercury atom.
1. The magnetic moment is quantized:
Spin; The Einstein- de Haas Experiment
An iron bar (S) is hung from a string (R) in
such a way that it can rotate around its axis.
From the torsion vibrations of the string, the
angular momentum N, which the bar gets as a
whole when it is magnetized by external
currents through the spoils (E), can be
measured. Suppose that n electrons with their
elementary angular momentum j contribute to
the magnetization.
m: mass of the electron,
q =-e: elementary charge of the
electron,
c: speed of the light,
μB: Bohr magneton
 q 
M bar  n  M electron  ng 
 j
 2mc 
As n  j  N  0
 q 
 e 
 
M bar   g 
  N   g
  N   g B   N
 2mc 
 2mc 
  
By measuring the macroscopic quantities N and Mbar, we can determine the
gyromagnetic ratio g of the elementary magnetic moments (the Landé factor): g = 2.
The magnitude of two Bohr magnetons for the magnetic moment excludes the orbital
magnetic moment as the source of the ferromagnetism and can be only explained by the
existence of the electron spin.
Deflection of Ag atom beam by a
magnetic field: the Stern-Gerlach
experiment
Splitting of Na D-line emission by
a magnetic field: the Zeeman
effect.
Classical physics:
continuum of deflections
No magnetic field,
no splitting
N
S
Quantum physics: two
sharp peaks separated
The atomic beam passes through an
inhomogeneous magnetic field. One
observes the splitting of the beam into
two components. Consequences:
1) Experimental demonstration of
directional quantisation and
2) The direct measurement of the magnetic
moments of atoms.
Some electron energy levels could be
split by a magnetic field. This is the
Zeeman effect.
The electron spin quantum number, and Pauli’s exclusion principle.
In order to account for the magnetic
splitting of atomic lines, it was
necessary to understand how the
magnetic properties of the electron
could be included in the grand
scheme of the quantum physics.
The interpretation of the spin by spinning the
electron around the axis is a fiction only that
makes the imagination easier (but it is not true)!
spin up or a
spin down or b
Since the electron has a negative charge, it is clear that any electron moving in an
orbital should also generate a magnetic field. Pauli realized that the problem could
be inverted, - why do all atoms not show a magnetic response? The answer he
proposed was that electrons normally came in pairs, so that their magnetic effects
cancelled out. This would be the case if, in addition to movement in an “orbit”, the
electrons were also spinning on their axes. Each electron would then be a magnet. A
pair of electrons in the same orbit would cancel out, but only if they had opposite
spins. More formally, Pauli’s exclusion principle states that two, but no more than
two, electrons can occupy any orbit. As a result, the number of electron orbitals was
doubled, by adding the fourth quantum number, ms, which can have one of two
values, +½ and -½.
With the addition of ms, the properties of elements in the Periodic Table
could be accounted for. In addition to the atomic number, and the matching
of electronic to protonic charge, the reactivities of the elements could be
explained in terms of the need to fill the electronic orbitals by sharing
electrons, thus explaining the valence properties, and their periodic pattern.
Filling of orbital shells, - the
building-up (Aufbau) principle
Spin-orbit coupling
In addition to the angular momentum and
magnetic properties arising from its spin (S), an
electron in a molecular orbital has an orbital
angular momentum, which also generates a
magnetic field (L). The interaction between these
two magnetic fields is called spin-orbit coupling.
The antiparallel arrangement (bottom, right) has
the lower energy, and is therefore favored. The
electron magnetic field depends on the total
angular momentum, J. In general, both electrons
and nucleons spin-couple with other magnets in
their environment. This coupling to neighboring
spins can be measured in pulsed EPR and NMR
applications which look at the kinetics of decay
of spin states populated by a pulse of excitation.
Analysis of the results provides structural
information, - distance and types of atoms close
enough to couple.
Some nuclear magnets of importance in biological studies
The gyromagnetic ratio (γ) determines the energy at
which electromagnetic radiation will flip the spin of a
nuclear magnet. A flip occurs when the energy
matches (or is in resonance with) the energy of the
transition. The energy needed (expressed in terms of
frequency) depends on the applied magnetic field (the
strength of the magnet), -4.7 Tesla in this case.
What is the difference between the electron and the nucleus as magnets?
1. The spin quantum numbers for the electron, proton and neutron, all have
the same value of ½. This is an intrinsic angular momentum, - every
electron, proton and neutron has this property.
2. Because of the relation between angular momentum and mass, a spin ½
body has a frequency of rotation determined by the mass. Because
nuclear particles (nucleons) have masses 1,836  that of the electron, the
nucleus spins much more slowly.
3. Because the magnetic field depends on the rate of rotation, electrons
have magnetic fields ~2,000  that of protons, which have higher fields
than more massive nuclei. Hence magnetic resonance occurs at much
higher energies for EPR than NMR.
2. Wave-Particle Dualism
1. Wave character of particles
Diffraction of
X-rays
electron beams
DavissonGermer
experiment
Beams of
accelerated
electrons were
directed through
thin metal foil and
arrival of fast
electrons were
detected on a
photographic plate.
The diffraction pattern seen when the electron beam was accelerated to give a
wavelength of 0.5 Å gave a pattern similar to that seen when a beam of X-rays of
similar wavelength (0.71 Å) was used.
barrier
Application: Quantummechanical tunneling
The barrier can be defeated by
1) thermal activation (Arrhenius) and
ΔE
A particle approaches
a potential barrier
2) tunneling (Gamow)
Δx
k=k0·exp(-ΔE/kBT)
ΔE
T
1.0
Transmittance (Gamow’s expression)


2
8

m


T  exp   2
D
E

D
x


h2


The temperature-dependence of the rates of the two
processes are highly different:
1)
Arrhenius: strong temperature-dependence
2)
Tunneling: no temperature-dependence
Estimating the tunneling probability
Estimate the relative probabilities that a proton and a deuteron can tunnel through the
same barrier of height 1.0 eV and a length 100 pm when their energy is 0.9 eV!
The mass of the proton is m = 1.673·10-27 kg, the Planck’s constant is h = 6.626·10-34
J·s
the energy difference to overcome by tunneling is ΔE = 0.1 eV and the length of the
barrier is Δx = 100 pm.
Apply the Gamow-relationship of transmittance via tunneling:


2
8

m


T  exp   2
D
E

D
x


h2


Conclusions: 1) the tunneling probability of a proton (in the system specified) is
Tp = 9.56·10-7 and is much (about 300 times) greater than that of a deuteron:
Td = 3.07·10-9.
2) The ratio of the tunneling probabilities will be much greater when the barrier is twice
as long and the other conditions remain unchanged: Tp/Td = 9·104.
Quantum tunneling effect:
Scanning Tunneling Microscope (STM)
The electron
passes
through the
gap between
the surface of
the sample
and the tip of
the scanning
electrode via
tunneling.
By moving the
electrode
above the
surface, the
distance d can
be measured
and therefore
the surface
can be
mapped.
Application of tunneling in life sciences:
Scanning Tunneling Microscope
1. Ultra-precise electromechanical instrumentation
(piezoelectric scanner, precise electromechanic
components for sample positioning)
2. Quantummechanical tunneling-effect: When
approaching two biased conductors, the current
increases exponentially by a factor of 10 for each
0.1nm reduction in distance before they finally make
contact.
Cross-section through tip and sample in STM
(contour-lines indicate constant charge density)
Because of the sharp increase in tunneling current
with decreasing distance, even tips with moderate
sharpness yield atomic resolution easily, because
the front atom carries the lion´s share of the
tunneling current!
Electron tunneling in reaction center protein of
photosynthetic bacteria
LIGHT
Upon lowering the temperature from
ambient temperature, all electron transfer
reaction rates are decreasing and they
freeze finally except of the first and
fastest reaction (3 ps halftime) which
increases by a factor of about 3 at 4 K.
Rate of electron transfer
This clearly indicates for tunneling
pathway instead of a temperature
activated reaction of electron transfer
between P and HA.
Tunneling (e.g. P→BA→HA)
k
Thermal reactions (e.g. QA→QB)
High temperature
1/T
Low temperature
Arrangement of the Rb. sphaeroides reaction center cofactors, representation of electron transfer pathways
(arrows) and the corresponding time constants. Notations P: bacteriochlorophyll (BChl) dimer, B: monomeric BChl,
H: bacteriopheophytine, Q: quinone, Fe: non-heme iron atom, A: photoactive branch, B: photoinactive branch. The
circles in the middle of chlorins indicate the Mg atom in BChl.
2. Particle character of waves
The photoelectric (Hallwachs) effect
The limiting wavelengths at the red
ends of the photoelectric spectra are
indicated in brackets.
The photoelectric
effect
When UV light shines on a metal surface, it induces the
release of electrons, which can be detected as a current in a
circuit such as that on the left. The released electrons are
attracted by an applied voltage to an anode, and the
resulting current detected, and used to measure the rate of
electron release. The characteristics of this effect are as
follows:
1. No electrons are ejected, regardless of intensity, unless the
light is sufficiently energetic. In terms of Planck’s
equation, they have to have a high enough frequency. The
actual value (the work function) depends on the metal.
2. The kinetic energy of the ejected electrons varies linearly
with the frequency of the incident light, but is independent
of intensity.
3. Even at low intensity, electrons are ejected immediately if
the frequency is high enough.
According to the classical view, the energy of radiation should be proportional to
the amplitude squared. It should therefore be related to intensity, which is in
contradiction to the result observed.
Einstein suggested a solution to this dilemma, by
invoking Planck’s hypothesis. The electron is
ejected if it picks up enough energy from
collision with a photon. However, the energy of
the photon is given by the Planck equation, and
so is proportional to frequency, and quantized.
From the 1st law of thermodynamics, energy has to
be conserved. We can therefore write an equation in
which the kinetic energy of the electron is equal to
the energy picked up from the photon, minus the
energy needed to dislodge the electron (the work
function, f):
½mev2 = hn – f
This is Einstein’s photoelectric law.
Dual nature of matter – wave and particle
properties apply to subatomic particles
Einstein’s relation between energy and mass and the de Broglie equation
Einstein suggested in the context of special relativity that energy and mass are
equivalent, and related through the famous E = mc2. De Broglie realized that, if all
matter was quantized, this implied a general relation between the momentum of a
particle and its energy as expressed in terms of frequency. By combining the Planck
equation, E = hn = hc/, and the relationship for the momentum of an electromagnetic
wave (given by p = mc), p = E/c, we get p = h/λ or, rearranging:

h
h

p mv
The de Broglie relationship implies that any particle of mass m moving with velocity v
will possess wavelike properties. In view of the value of the Planck constant, the effect
will be appreciable for particles of low mass.
Conclusion: on atomic scale, the concept of particle and wave melts together,
particles take on the characteristics of waves, and waves the characteristics of
particles.
Direct application of the de Broglie relation
to resolution limit of microscopes
The resolution limit of an optical microscope is defined as the shortest distance between
two points on a specimen that can still be distinguished by the observer or camera
system as separate entities: 0.61·λ/(n·sinα), where n·sinα is the numerical aperture.
The wavelength is given by the de Broglie relationship:

h
mv
The resolution limit of the microscope is inversely proportional to the mass and velocity
of the particle.
The higher is the speed of the particle, the smaller will be the resolution limit. E.g. for
electron of v = c/50 in the electronmicroscope, the limit is λ = 0.12 nm (X-ray).
The larger is the mass of the particle, the smaller will be the resolution limit. E.g. the
(hypothetical) neutronmicroscope would offer about 2000 times smaller resolution limit
than the electronmicroscope with the same velocities of the particles because
mneutron ≈ 2000·melectron
The atomic de Broglie microscope is an imaging system which is expected to provide
resolution at the nanometer scale using neutral He atoms as probe particles.
3. Unusual spatial distribution and
discrete energy levels
Examples: Particles in potential wells
• Electron in hyperbolic potential well
(spectroscopy of H-type atoms)
• Electrons in rectangular potential well
(spectroscopy of dyes in the visible range)
• Atoms/molecules in parabolic potential well
(spectroscopy of vibrations in the infrared
spectral range)
1. Electron in hyperbolic potential well
(quantization of the H atom)
Semiclassical treatment: the forms of the
wave functions and the corresponding energy
leveles can be anticipated in terms of the de
Broglie relation („standing waves”).
The Coulomb potential around the nucleus
U (r ) 
1
4 0

e
r
where ε0 is the dielectric constant of the vacuum,
e is the elementary charge of the nucleus (proton),
The potential energy of the electron
Epot
r is the distance of the electron from the nucleus.
e2


4 0 r
1
The shape of the function (potential barrier) is hyperbolic.
The bound electron should be inside the hyperbolic walls.
The electron is free to move inside the segment but reflected at the boundary. The
hyperbolic wall is the constrain that will quantize the energy of the electron. In lack of
this condition (i.e. the electron is not bound, it is free), the energy of the electron can
take any values, it is continuous.
Each wavefunction is
- a standing wave,
- fits into the segment and
- successive functions must possess
one more half-wavelength.
The permitted wavelengths
n
2
 n  2  rn
where n  1,2,3,...
The average speed of the electron can be estimated by the de Broglie relation
vn 
h
h

n
n  me 4 rn me
Here me is the mass of the electron. The electron has only kinetic energy inside the well
thus the kinetic energy of the electron
Ekin
1
h2
2
2
 me v n 

n
2
32  rn2 me
The total energy of the electron as a function of r
En (r )  Ekin,n  Epot,n
h2
2
e2

n 

2
4  0 rn
32  rn me
where
n  1,2,3,...
1
The permitted energies are
En  
me  e 4
2
2
 02
h

2
1
n
2
The rigorous solution from the Schrödinger equation (see below)
En  
me  e 4
8  02
h
2

1
n
2
where
n  1,2,3,...
It differs from the semiclassical solution by a factor of
π2/4 ≈ 2 only.
Conclusions: 1)The energy of the electron in the H atom is
quantized, 2) the quantization arises from the boundary
conditions that ψ must satisfy and 3) the subsequent energy
levels relate as the inverse squares of the neighboring integer
numbers: E1 : E2 : E3: ... = 1 : 2-2 : 3-2: ...
The Table on the right shows
wave functions, in x, y and z
coordinates, for hydrogenic
atoms (H, He+, Li2+, etc., atoms with only 1 electron;
Z is the atomic number,
giving the charge of the
nucleus).
The Figures below show the
“shapes” (boundary
surfaces) of the
corresponding orbitals.
s
The subshells are named s, p, d, f, g, and h for electrons with l = 0, 1, 2, 3, 4 and 5.
Molecular orbitals
The energy due to
atomic interaction
can be calculated
as a function of R.
The equilibrium
bond length is at
the minimum.
Formation of the bonds in N2. The 2pz orbitals
overlap and coalesce into a s molecular
orbital; the px and py orbitals form  orbitals
perpendicular to each other.
1s orbitals of H-atoms
coalesce to form a s bond of
H2, with cylindrical symmetry
Bond wave functions are formed by summing
atomic wave functions. Bonds formed by
combing antiparallel spins are favored. For H2,
the molecular orbital has a lower energy than
the atomic orbitals, so is stable.
Bonds of carbon
Figs. from Atkins, The Elements of Physical Chemistry
In methane, each sp3 forms a s
bond with a 1s atom of H
Carbon has one empty, and two 1-electron filled 2p
orbitals, giving four valence electrons. In order to
explain the symmetrical chemical behavior of
molecules like methane, we have to have four
identical orbitals in tetrahedral symmetry. In order to
provide these, a mixing of orbitals occurs, to give
four sp3 hybrid orbitals (left). This involves
promotion of 1 2s electron to the vacant 2p orbital,
hence sp3.
In ethylene (ethene), the double-bond is made
up of a s and a  bond.
Take-home message on bonds.
1. Bonding orbitals can take up quite complicated shapes. Extended -bonding
occurs in molecules like cytochromes, chlorophylls, flavins, nucleic acid
bases, tryptophan, etc. The electrons in these extended orbitals roam over
the entire coordinated  system (see later the quantumphysical treatment of
these systems).
2. Transitions between energy levels in molecules can occur between orbitals
of different type. The change in electron distribution results in an electrical
dipole difference between the ground and excited states. When the orbital is
asymmetric, as in an extended  system, the excited state can remain in the
-orbital, but will have a different eigenfunction and eigenvalue, and so a
different orbital “shape”. This also gives rise to an excited state dipole.
3. Normally, molecular orbitals are at lowest energy when they are filled by
two electrons of antiparallel spin. If an electron is removed (for example, by
oxidation), or an extra electron added (for example, by reduction), the lone
electron is not spin-coupled, and therefore acts as a magnet. The magnetic
effect arises from the unpaired spin, and the magnetic dipole results from
the angular momentum of the electron in its orbital.
2. Electrons in rectangular potential well
(electronic energy levels)
2π-electrons in conjugated chains
Metallic model for the π-electrons:
They can move freely along the
chain (in the well) but cannot
escape (from the well).
The shape of the potential well is
rectangular and infinitely deep.
The double and single bonds
between neighboring carbon atoms
are alternating.
The length of an elementary group is l (≈ 3 Å)
Bond
Length
(Å)
Energy
(kcal/mole)
C-C
1.54
83
C=C
1.34
122
C-H
1.08
101
the number of elementary groups is N,
each group has 2 π-electrons,
the whole chain has 2N π-electrons and
the total length of the chain is L = N·l
Electrons in the box: rectangular potential well
There are 2·N π-electrons in the well. Because of the
bundary conditions, the electrons cannot have arbitrary
(continuous set of) energies but well defined discrete
values only. We are looking for the possible energy levels
using 1) semiclassical physics (based on de Broglie’s
standing waves of the electrons) and 2) quantum physics
(Schrödinger equations).
Semiclassical treatment
Only those energy levels are allowed along which standing

n   L where n  1,2,3,... waves can be created with nodal points at the potential
2
walls.
h
The speeds of the electrons are determined from the de
vn 
Broglie’s relationship.
m
vn 
nh
2mL
2
1
h
2
En  m v 2n 

n
2
8 m L2
Replacement of the wavelength in the expression of speed.
The total energy consists of kinetic energy term only.
Energies of electrons in the box:
rectangular potential well
The energy of the electrons are quantized:
- The lowest energy (zero-point energy) is not zero. It
comes from the uncertainty principle that requires a
particle to possess kinetic energy if it is confined to a
finite region.
- The energy levels are dense on the bottom and
become more diluted upon getting out of the well.
The heights of the energy levels are related as the
squares of the integer numbers.
- According to the Pauli principle, there can be two
electrons with antiparallel spins at every level of
energy. As there are N groups, the 2·N π-electrons
occupy the lowest N energy levels.
En 
h2
8 m L2
 n2
- The separation between adjacent energy levels
decreases as the length of the container (L)
increases and becomes very small when the
where n  1,2,3,...container has macroscopic dimensions. In
laboratory-sized vessels the energy is not quantized.
The lowest energy band (the most red part)
of the absorption spectrum
The absorption band with the longest
wavelength (smallest energy, the center of the
farest red absorption band)
h n red abs  EN 1  EN
n red abs
1  ( N  1) 2 h 2 N 2 h 2 
 


2
2
h  8 m L
8 m L 
n red abs 
If N >>1, then
n red abs
h
1


2 N
4ml
h
8ml
2

2N  1
N2
or red abs
The wavelength of the absorption band is proportional to N.
4 m c l2

N
h
Discussion
4 m c l2
 red abs
N
h
The wavelength of the most red
absorption band is proportional
to N, the length of the conjugated
chain. With elongation of the
chain, its absorption band shifts
toward the longer wavelengths:
It shifts to red („red shift”).
N
(length of chain)
λred abs
spectral
range
3
445 nm
visible
4
593 nm
visible
5
741 nm
near IR
6
889 nm
near IR
molar absorption
3. Particles (atoms and molecules)
in parabolic potential well
The energy profile of a harmonic oscillator is parabolic.
The particle is freely moving inside the parabolic
potential well.
Semiclassic treatment
The total energy of the particle is E  1 k x02  1 m v 2
2
2
L  2  x0
which is equal to the kinetic energy between the walls.
Those energy levels are allowed which can be covered
with standing waves
n
n
h
Possible velocities of the particle
Possible energies of the particle
1
n2 h2
k
2
E n  m vn 

2
8m 2En
E n
mn

4
n  hn

nh
nh
k


4 x0 m 4 m 2 E n
vn 
where
n  0,1,2,3,...
2
L
Energy levels of the harmonic oscillator
Uniform ladder of spacing h·ν
Solution from the Schrödinger equation:
En = h·ν(n + ½), where n = 0, 1, 2, ...
Conclusions:
(a) The energy levels of the oscillator are
equidistant: ΔE = h·ν.
(b) The selection rule: Δn = ± 1.
Transitions are allowed between adjacent
states only (in first order). Transitions
among not neigboring levels are forbidden.
(c) The minimum energy of the oscillator
(n = 0) differs from zero: E0 = ½ hν. The
zero-point energy is not zero, because of
For typical molecular oscillator, the zerothe uncertainty principle: the particle is
point energy is about
confined, its position is not completely
E0 = 3·10-20 J = 100 meV
uncertain, and therefore its momentum,
The Boltzmann-energy at room temperature and hence its kinetic energy, cannot be
exactly zero.
is ½ kBT = 25 meV
4. Heisenberg’s uncertainty principle
Simultaneous estimation of the precision of the space coordinate and the impuls of a particle
Wave packet
The uncertainty (Δx) of the position (x) of the
particle is characterized by the extension of the wave
packet:
Dx  n    n 
h
p
λ
n·λ
The precision of the determination of the impuls (p) depends on the precision of the
determination of the wavelength. The longer is the wave packet (i.e. the larger is the number n),
the more precise is the determination of the wavelength:
Dp  Dx  h
Dp D 1


p
 n
From the two
equations:
The product of the uncertainties of the simultanous determination of the space and impuls
coordinates cannot be smaller than a definit limit determined by the Planck’s constant. The
measuring device inherently perturbes the state of the microsystem under physical observation.
The perturbation causes the uncertainty: the energy and impuls of the absorbed and/or emitted
photon are commeasureable with those of the observed (micro)particle.
Similar complementer quantity-pair is the time (t) and energy (E). The product of the
uncertainties of the lifetime (Δt) and the excited state (ΔE) in simultaneous measurement is
limited by the Planck’s constant:
DE  Dt  h
Problems for Seminar
1. What is the resolution power of the electron microscope using electrons
accelerated to one hundreds of the light speed in vacuum (c/100)?
2. What is the ratio of protons at room temperature that are able to absorb
photon from the 1 GHz microwave radiation to provide NMR signal in the
spectrometer?
3. What is the magnetic induction in the 1 GHz NMR speψctrometer? Make
reasonable estimate for the required current!
4. How many 1000 nm photons does a 1 mW monochromatic infrared
rangefinder emit in 0.1 s?
5. Based on the spectral lines of mercury (Hg) emission, construct the
approximate current-voltage characteristics of the Franck-Hertz experiment!
The most intense spectral lines (in nm): 184.45, 253.7, 300, 312, 334, 365.4,
404.656, 435.835, 546.074 and 579.065
Problems for Seminar
6. Calculate the wavelength of an electron in a 10 MeV particle accelerator.
7. The wave function of an electron in the lowest energy state of a hydrogen
atom is ψ =(π·a03)-1/2·exp(-r/a0), with a0 = 52.9 pm, and r the distance from
the nucleus. Calculate the probabilities of finding the electron inside a small
volume of magnitude 1 pm3 located at a) the nucleus, (b) a distance a0 from
the nucleus. The total volume of a hydrogen atom is of the order of 107 pm3.
How many times are more probable to find the electron at the nucleus than
in the same volume element located at a distance a0 from the nucleus?
8. Determine the surface temperature of the star Sirius when the wavelength of
the radiation of the most intense electromagnetic radiation emitted from the
surface is 263.27 nm.
9. A glow-worm of mass 5.0 g emits red light (650 nm) with a power of 0.10 W
entirely in the backward direction. To what speed will it have accelerated
after 10 years if released into free space and assumed to live?
10. In an X-ray photoelectron experiment, a photon of wavelength 150 pm
ejects an electron from the inner shell of an atom and it emerges with a
speed of 2.14·107 m/s. Calculate the binding energy of the electron.
Appendix
Introduction of advanced studies
in quantum physics
Somewhat advanced level
Precaution!
The Schrödinger equation
but not detrimental to your
health
The refinement of the Bohr model by de Broglie (we will see later in discussion of the principles
of spectroscopy of H-atom) paved the way for the more formal description by Schrödinger.
What Schrödinger added was a more powerful formalism for the description of the wave
function. Unfortunately, this involves some elegant math, - a thing of beauty to the physicist,
but perhaps not to life scientists. Fortunately, we don’t need to understand the math in detail
to appreciate the results, so the approach here will be non-mathematical.
It will help to appreciate a few points:
1. The new approach makes it possible to describe wave functions that are three-dimensional, and
more complex in shape, while maintaining the constraints required by quantization, and the
need to form a “standing wave”. This made it possible to make the Schrödinger equation
time-independent.


E  H 
Here H is the Hamiltonian (energy term operator) of the system, ψ is the wave function and
E denote the allowed (actual) energy values of the system.
2. Heisenberg had introduced his Uncertainty Principle, which showed that the momentum and
position of the electron could not be determined simultaneously with certainty. Schrödinger
therefore used a term (wave function) related to the probability of the electron occupying a
particular volume.
Example: the Schrödinger equation for H-like atoms
The Hamiltonian for H-like atoms consists of kinetic and potential energy terms:

h2
2
Ze
H   2 2 
4 0r
8 m
and the Schrödinger equation:
h2
2
Ze
E   2  2 

4 0r
8 m
Let’s compare this to the classical expression for the energy of the electron:
1
Ze2
2
E  me v 
2
40r
The similarity arises from the need to consider the two balancing forces that determine
the electron energy, - the kinetic term (the “centrifugal force” if we consider a planetary
model), and the constraining electrostatic term for the potential energy. The first term on
the right (the kinetic term) is now quantized, and all terms are modified by the wave
function, .
What is , the wave function and what is that odd symbol, 2 („del squared” or nabla
operator)?
h2
Ze 2
2
E   2   

8 m
40 r
Wave
function of an
electron in Htype atom
Probability of
finding the
electron
between x
and x+ δx
Radial distribution of the electron
Born’s interpretation of the wave function
It’s difficult to form a solid picture of what 
represents, but 2 is easier to grasp; it represents the
probability of finding the electron at a particular point
in space. This is more formally described by the Born
approximation, which, for a one-dimensional system,
states that the probability of finding the particle
between x and x + dx is proportional to 2dx. In three
dimensions the probability is proportional to 2dV.
The term2 is a probability density, and we get the
probability by multiplying the density by the volume
of the region of interest.
Examples of a particular wave function, , in onedimension, and its square, 2, plotted against distance
are shown on the left. Note that for 2, all values are
positive even when  goes negative. The shaded bar
at the bottom is gray-scale encoded to show
probability.
The Laplace-operator
The odd symbol, 2 (pronounced del-squared) is a Laplace operator.
In the one-dimensional case (particle-in-a-box treatment, see later), the term 2 is
replaced by the second derivative of  with distance
 
2
d2
dx 2
For the three dimensional case, in its simplest form as applied to Cartesian coordinates,
2 is an abbreviation for the term
2
2
2
d
d
d 
2   2  2  2 
dz 
 dx dy
It tells us that we have to perform the second derivative operation on the wave function
in a three dimensional coordinate system. It takes a more complicated appearance for
other coordinate systems, but serves the same function. Because trigonometric functions
provide a rich field for investigation of potential wave functions, 2 is often used in the
context of spherical polar coordinates.
1  2 
1
2
1
 
 
  2 r

sin

 2 2


r r  r  r sin  f 2 r 2 sin   
 
2
The energy operators
h2
Ze 2
2
 2 
  E
8 m
40 r
As noted above, the three operators on  all have energy units. They can be
expressed in various forms, which makes for some confusion, but basically they
represent the same energies as appear in the classical equation. The form of the
equation above is that usually presented (this is just the previous equation
rearranged). The Table below shows the equation for the 1-D case in different forms.
To cut the clutter, Planck’s
constant, h, has been replace
by ħ, defined by ħ=h/2π.
Note that in the last line, the
kinetic and potential energy
terms have been lumped
together into a single
operator. This term also has
energy units, and is called a
Hamiltonian operator, H.
p x2
2m,
The kinetic energy operator, Tx is shown in classical form, Tx = ½mv2 =
and in
the quantized form above. The potential energy operator is represented by U(x).
Finding appropriate forms for 
Solution of the Schrödinger equation involves the second derivative of the wave
function, . There are many solutions that can be found, and the problem is to know
which ones are appropriate. We have to recognize constraints that relate to the
physical reality of the system. These can be summarized as follows:
1. Acceptable solutions must be single-valued at all points in space. This is because
there cannot be more than one answer to the question “What is the probability of
finding the particle at a particular point in space?”.
2. Similarly,  must be continuous and finite. This constraint is necessary to provide
the “standing-wave” character of the function.
3. The total value of 2 summed over all space must be 1. This is because the particle
has to be somewhere, but there’s only one of them.
Because energy levels are quantized, the wave function can take different forms,
depending on the quantum number.
The different quantum numbers in the Bohr model represent different energy levels for
orbitals in the ground state and excited states of the H atom. The effect of this last
property has interesting connotations when we consider the shapes of the orbitals
defined by the wave function (see later the next lecture).
1. Applying the Schrödinger equation to the hydrogen atom
h2
The time-independent Schrödinger equation is:  2 2  V  E
8 m
e2
where the potential energy term is:
40 r
V
After substitution, we bring all terms to one side and rearrange them.
h2
e2
2
 2  
  E
8 m
40 r
We substitute into the leftmost term using 2 in its spherical polar coordinate form
because of the spherical symmetry of the H atom:
 2 2
1
 
 
1
 2 8 2m
e2


 2 (E 
)  0
 sin 

dr 2
rr r 2 sin   
  r 2 sin 2  f 2
h
40r
Fortunately, this fearsome-looking equation has already been solved, so we need be
concerned only with the result.
The solutions are given in a form appropriate for the spherical polar coordinates
chosen. The solutions requires that we use three quantum numbers to describe
each energy level. The nomenclature for these is not very helpful.
The three quantum numbers are:
n, the principal quantum number;
l, the angular (azimuthal) momentum
quantum number;
ml, the magnetic quantum number.
n = 1, 2, 3, …n
l = 0, 1, 2, …(n-1)
ml = 0, ±1, ±2, …, ±l
The resulting orbitals are grouped into
subshells and shells, as nicely
summarized on the left.
The solutions of the equation give the energy of the electrons in the orbitals, and
their distribution in space. The energy of the electron is given by:
mee4 1
En   2 2 2 n  1, 2, 3, ...
8h  0 n
Not surprisingly (because it worked), this is the same as the equations Bohr and de
Broglie developed to account for the emission lines for the H-atom.
Where have the l and ml quantum numbers gone?
The answer is straightforward. All three quantum numbers characterize the electron
distribution, but for any shell, all the wave functions use the same value for radius.
The radius enters into the energy through the potential energy function, - the
coulombic force holding the electron to the nucleus, which depends only on r, and
is spherically symmetrical. Since this balances the centrifugal energy represented
by the angular momentum of the electron, all electrons in the shell have the same
potential energy function, determined by n, as in the Bohr equation.
This simple case applies only to the H-atom.
Because n uniquely determines the energy, the value of E for a particular n is called
its characteristic energy, or eigenvalue. The wave function for a particular n is
called the characteristic wave function, or eigenfunction.
The relation between  and probability of distribution depends on the quantum
number. Choice of polar coordinates gives  as a function of r, , and f, so that
( r, , f )  R( r )Q( )F (f )
The terms on the right are called the radial part, (R(r)), and the angular part,
(Q()F(f)), of this expression. As the equation shows, the radial part is dependent
on r, so we can think of n as characterizing the distance dependence, with each
shell having a different radius. The solution for each orbital differs, depending on
the values of the quantum numbers. For orbitals with l and m equal to zero, the
solution depends only on r, the radius. As a consequence, these orbits are spherical.
For non-zero l, and for m quantum numbers,
the “shape” of the orbit is not spherical, and the
wave function is determined by x, y and z, the
parameters of its position in Cartesian space.
For example, of the four electrons in the L
shell, one has l = 0, and hence m is also zero,
and the orbital is spherical. The other three are
arranged along the x, y, and z axes, and are not
spherical.
2. Particle in a rectangular box, - the Schrödinger equation in 1-D
If we constrain our electron to a one-dimensional box,
we can simplify things. In 1-D, we don’t have to worry
about the potential energy term of the planetary model, the electron is constrained by the walls of the box, not by
the attraction of the nucleus, so we only have to worry
about the kinetic energy. And we don’t have to worry
about 2, because we’re in 1-D. The Schrödinger
equation then becomes:
h2 d 2
 2
  E
2
8 m dx
To solve this equation, we have to find an appropriate wave function, . The
wavelength of a harmonic oscillator in one-dimension is a sine function, and this
provides an appropriate starting point for our search for a wave function. Let’s try
 = A·sin kx
where x is distance. We will take a look at what happens if we set k = n/L, with L as
the length of our box. Because sin  = sin 2 = sin 3 = 0, we expect this function to
generate a curve that oscillates. In the next slide this function (with A =1) is plotted
against distance (x) with different values of n, and with distance scaled to units of L.
n=1
n=2
n=3
n=4
2
1.0
(sin(nx/L))
sin(nx/L)
1.2
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
-1.2
n=1
n=2
n=3
n=4
0.8
0.6
0.4
0.2
nodes
0.0
0.0
0.2
0.4
0.6
x/L
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
x/L
Note that if we change the value of n in this equation, we change the number of
beats per unit distance as measured by the number of times the curve crosses the
zero-line at a node. The nodes define the distance at which we will form a
“standing wave”. At distance x=L, for n = 1 we have 1 node; for n = 2, 2 nodes; n =
3, 3 nodes; etc.
The plot on the right shows the square of the function versus the same distance
scale. Each node generates a peak, so there are n peaks for each curve. The curves
are all positive, so (remembering the 2 gives probability) they look suitable.
Our choice of wave function looks pretty good, - it has the sorts of property we’re
looking for. We now need to plug it into our Schrödinger equation to see if it
satisfies all the constraints. Starting with  = Asinkx, the critical steps are:
1. Find the second derivative of this function (it’s given by -k2)
 8 mE 
2. Substitute  from this into the Schrödinger equation to find that k  
, 
2
1
 h

 8 2 mE  2
and that, by back substitution,   A sin  2  x.
2

h

3. Satisfy the boundary condition that  is zero at the walls. As we saw in previous
slides, sin n = 0, so this is satisfied when x =1 0 or L, the length of our box.
From this, our function can work if
substitution,
 n  A sin
n
x
L
 8 2mE  2 n

 
2
L
 h

and
(n is integer). Then by back
n2h2
En 
8mL2
4. Satisfy the constraint that the probability is 1. For this we need to adjust the
amplitude of our sine wave, using the factor A. Tables of integrals give us a
normalization constant such that A 
2
L
. We finally end up with:
2
n
n 
sin
x
L
L
1
2
We have already seen plots of this function, and a version is shown on the left
below, in the context of energy levels. We can now say that the energy of an
electron in a one-dimensional box has different levels that are quantized, and that
at each level the wave function, , is a sine wave, with the number of nodes
equal to the quantum number. The probability is given by 2, and this is shown
in the right panel below.
3. Solution of the Schrödinger equation for
harmonic oscillator
 2 d 2 1 2

 k x   E
2
2 m dx
2
The Scrödinger equation
The wavefunction
n 
8
4
mk
h / 2  2n n!
 H n e y
2
/2
The permitted energy levels
mk
where y  4
2
x
1
En  (n  )  hn
2
with n  0,1,2,... and n 
1
2
The factor Hn is a Hermite polynomial, it
depends on the vibrational quantum
number n and some of which are listed in
the Table.
n
Hn(y)
0
1
1
2y
The wavefunction of the ground state (the
lowest energy state) is (n = 0)
2
4y2 – 2
3
8y3 – 12y
4
16y4 -28y2 +12
5
32y5 – 160y3 + 120y
0 
and the probability
density is a bell-shaped
Gaussian function
 02

8
mk
4
h/2
4
mk
h/2
 e y
2
 e y
/2
2
The Hermite polynomials
k
m