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Scientific Methods
•
•
•
•
•
Observations (experiments)
Hypothesis to explain (modified and refined)
Models and Theory
Laws (general behaviors)
Assumptions/conditions, time-space scopes,
definitions - limitations
Atkins / Paula
Physical Chemistry, 9th Edition
Chapter 7 (Ch.8 of 8th Ed.)
Quantum theory :
introduction and principles
http://ebooks.bfwpub.com/pchemoup.php
http://ebooks.bfwpub.com/pchemoup
Atkins & Paula ,“Physical Chemistry,” 9th Ed., Oxford, 2010
The origins of quantum mechanics
7.1 The failures of classical physics and
Energy quantization
7.2 Wave-particle duality
The dynamics of microscopic systems
7.3 The Schrödinger equation
7.4 The Born interpretation of the wavefunction
Quantum mechanical principles
7.5 The information in a wavefunction
7.6 The uncertainty principle
7.7 The postulates of quantum mechanics
P.249
Objectives - From Classic to Quantum Mechanics
• Introduction of Quantum Mechanics
• Understand the difference of classical theory
and experimental observations of quantum
mechanics
(Physical Chemistry, 2nd Ed., Thomas Engel, Philip Reid)
Outline
1. Why Study Quantum Mechanics?
2. Quantum Mechanics Arose Out of the Interplay
of Experiments and Theory
3. Blackbody Radiation
4. The Photoelectric Effect
5. Particles Exhibit Wave-Like Behavior
6. Diffraction by a Double Slit
7. Atomic Spectra and the Bohr Model of the
Hydrogen Atom
Why Study Quantum Mechanics?
•
Quantum mechanics predicts that atoms and
molecules can only have discrete (分離的, 離散
的)energies.
•
Quantum mechanical calculations of chemical
properties of molecules are reasonably
accurate.
Quantum Mechanics Arose Out of the Interplay of
Experiments and Theory
•
Two key properties are used to distinguish
classical and quantum physics.
1. Quantization (量子化)- Energy at the atomic
level is not a continuous variable, but in
discrete packets called quanta (量子) .
2. Wave-particle duality (波粒二象性)- At the
atomic level, light waves have particle-like
properties, while atoms and subatomic particles
have wave-like properties.
The origins of quantum mechanics
λν=c
(7.1)
ν – frequency
c - speed of light
Wavenumber (unit: cm-1):
ν 1
~
ν = =
(7.2)
c λ
Fig.7.1 The wavelength, λ, of a wave is
the peak-to-peak distance. (b) The wave is
shown travelling to the right at a speed c.
At a given location, the instantaneous
amplitude of the wave changes through a
complete cycle (the four dots show half a
cycle). The frequency, ν, is the number of
cycles per second that occur at a given
Chapter 7. Quantum theory:
point.
introduction and principles
P.250
Fig.7.2 The electromagnetic spectrum and the classification of the spectral regions
Chapter 7. Quantum theory: introduction and principles
P.250
7.1 Energy quantization - Key points
•
•
•
•
(a) The classical approach to the description of
black-body radiation results in the ultraviolet
catastrophe.
(b) To avoid this catastrophe, Planck proposed
that the electromagnetic field could take up energy
only in discrete amounts.
(c) The thermal properties of solids, specifically
their heat capacities, also provide evidence that
the vibrations of atoms can take up energy only in
discrete amounts.
(d) Atomic and molecular spectra show that atoms
and molecules can take up energy only in discrete
amounts.
Blackbody Radiation (黑體輻射)
•
•
An ideal blackbody is a cubical solid at a high
temperature emits photons from an interior
spherical surface.
The reflected photons ensure that the radiation
is in thermal equilibrium (熱平衡) with the solid.
Black-body Radiation (黑體輻射, cont’d)
A black body is an object of
emitting and absorbing all
frequencies of radiation
uniformly.
Fig.7.4 An experimental
representation of a black-body is a
pinhole in an otherwise closed
container. The radiation is reflected
many times within the container and
comes to thermal equilibrium with
the walls at a temperature T.
Radiation leaking out through the
pinhole is characteristic of the
radiation within the container.
Chapter 7. Quantum theory: introduction and principles
P.251
Black-body Radiation (黑體輻射, cont’d)
Energy density (Jm-3):
dε(λ,T) = ρ(λ,T) dλ
(7.3)
Total energy density in a region:
∞
(7.4)
ε (T ) = ρ (λ, T )dλ
∫
0
Total energy:
E(T) = V ε(T)
(7.5)
Fig.7.3 The energy distribution in a black-body
cavity at several temperatures. Note how the
energy density increases in the region of
shorter wavelengths as the temperature is
raised, and how the peak shifts to shorter
wavelengths. The total energy density (the area
under the curve) increases as the temperature is
increased (as T4).
Chapter 7. Quantum theory: introduction and principles
P.251
Black-body Radiation (黑體輻射, cont’d)
Fig.7.5 The electromagnetic vacuum
can be regarded as able to support
oscillations of the electromagnetic
field. When a high frequency, short
wavelength oscillator (a) is excited,
that frequency of radiation is present.
The presence of low frequency, long
wavelength radiation (b) signifies that
an oscillator of the corresponding
frequency has been excited.
Chapter 7. Quantum theory: introduction and principles
P.252
Black-body Radiation (黑體輻射, cont’d)
Energy density (Jm-3):
dε(λ,T) = ρ(λ,T) dλ
(7.3)
Rayleigh-Jeans law - density of
states (Jm-4) :
ρ (λ,T) =8πkT/ λ4
(7.6)
k – Boltzmann’s constant:
1.381 × 10-23 JK-1
Fig.7.6 The Rayleigh–Jeans law
(eqn 7.6) predicts an infinite
energy density at short
wavelengths. This approach to
infinity is called the ultraviolet
catastrophe.
Chapter 7. Quantum theory:
introduction and principles
P.252
Blackbody Radiation (黑體輻射, cont’d)
•
The limitation of energy to discrete values is called the
quantization of energy.
•
E=nhν
(n=0,1,2,…)
(7.7)
h- Planck’s constant 6.626x10-34Js
Planck distribution:
8πhc
dE= ρ d λ
(7.8)
ρ (λ , T ) =
λ5 (e hc / λkT − 1)
hc / λ kT
→ ∞ faster than λ5->0
For short wavelengthλ, hc/λkT» 1, e
ρ →0
as
λ →0
For long wavelengthλ, hc/λkT«1,
or
ν →∞
hc
hc


e hc / λkT − 1 = 1 +
+ ...  − 1 ≈
λkT
 λkT

Taylor series: f(x) = f(x0) + f’(x0)(x-x0)/1! + f”(x0)(x-x0)2/2! +…
Blackbody Radiation (cont’d)
•
Max Planck derived the agreement between
theory and experiment on radiation energy.
E = nhv
where h = Planck’s constant
n = a positive integer (n 0, 1, 2, . . . )
•
The theory states that the energies radiated by
a blackbody are not continuous, but can take
discrete values for each frequency.
Blackbody Radiation (黑體輻射, cont’d)
energy density
∞
8πhc
d λ = aT 4
ε (T) = ∫ 5 hc / λkT
0 λ e
−1
(
with
)
8π 5 k 4
a=
15(hc) 3
(7.9 )
Fig.7.7 The Planck distribution (eqn 7.8)
accounts very well for the experimentally
determined distribution of black-body
radiation. Planck ’s quantization
hypothesis essentially quenches the
contributions of high frequency, short
wavelength oscillators. The distribution
coincides with the Rayleigh–Jeans
distribution at long wavelengths.
Chapter 7. Quantum theory: introduction and principles
P.246
Blackbody Radiation (cont’d)
•
Spectral density (譜密度))is the
energy stored in the
electromagnetic field
of the blackbody radiator.
Heat Capacity
•
•
•
•
Molar heat capacities Cv=(∂U/∂T)V of all monatomic
solids ~25JK-1mol-1 (Dulong & Petit, early 19th century)
Equipartition principle – mean energy of an atom as it
oscillates about its mean position in a solid is kT for
each direction of displacement
Cv,m=3NA K=3R= 24.9 JK-1mol-1
(7.10)
It was found that the molar heat capacities of all
monatomic solids are lower than 3R at low
temperatures, and that the values approaches zero as
T→0
Einstein assumed that each atom oscillated about its
equilibrium position with a single frequency νin 1905
Heat capacities (cont’d)
molar internal energy :
U m = 3N AkT = 3RT
(7 .10 a )
molar constant v olume heat capacity :
 ∂U m 
−1
−1
CV , m = 
(7.10b)
 = 3 R = 24.9 JK mol
 ∂T V
Einstein formula : CV ,m (T ) = 3 Rf E (T )
Fig.7.8 Experimental low-temperature
molar heat capacities and the
temperature dependence predicted on
the basis of Einstein ’s theory. His
equation (eqn 7.11) accounts for the
dependence fairly well, but is
everywhere too low.
2

θ   e

f E (T ) =  E   θ E / T
T
e
−
1
  

Einstein temperatur e θ E = hν / k
1) At high temperatur es (T >> θ E ) :
θ E / 2T
2
(7.11)
2
 θ   1 +θ E/ 2T + L 
f E (T ) =  E  
 ≈ 1 (7.12a);
 T   (1 +θ E/ T + L) − 1 
2) At low temperatur es (T << θ E ) :
2
θ
 θE   e
2
 θ 
f E (T ) ≈    θE / T  =  E  e -θE / T (7.12b)
Chapter 7. Quantum theory: introduction
P.248
 T  and
 e principles
 T 
2
E
/ 2T
2
Heat capacities (cont’d)
By averaging over all the
frequencies present from zero up
to a maximum value νD:
Debye formula : CV ,m (T ) = 3Rf D (T )
3
 T  θ D / T x 4e x
f D (T ) = 3  ∫
dx ( 7.13)
2
0
x
θ
e −1
 D
Debye temperature : θ D = h ν D / k
(
)
Fig.7.9 Debye’s modification of Einstein’s
calculation (eqn 7.13) gives very good
agreement with experiment. For copper,
T/θD=2 corresponds to about 170 K, so the
detection of deviations from Dulong and
Petit’s law had to await advances in lowtemperature physics.
Chapter 7. Quantum theory: introduction and principles
P.249
The Photoelectric Effect (光電效應)
•
•
The electrons emitted by
the surface upon illumination
are incident on the collector,
which is at an appropriate
electrical potential to
attract them.
This is called the
photoelectric effect.
The Photoelectric Effect (cont’d)
•
Albert Einstein states that the energy of light,
E = βv
•
where β = constant
v = frequency
From energy conservation the energy of the
electron, Ee, is
Ee = βv − φ
(7.15)
where Ф = work function, characteristic of the metal,
the energy required to remove an electron from the
metal to infinity, the analogue of the Ionization Energy
of an atom or molecule.
The Photoelectric Effect (cont’d)
•
The results of β is identical to Planck’s constant,
h, thus
E = hv
The Photoelectric Effect (cont’d)
1) No electrons are ejected, regardless of the
intensity of the radiation, unless its frequency
exceeds a threshold value characteristic of the
metal.
2) The kinetic energy of the ejected electrons
increases linearly with the frequency of the
incident radiation but is independent of the
intensity of the radiation.
3) Even at low light intensities, electrons are
ejected immediately if the frequency is above the
threshold.
Fig.7.13 In the photoelectric effect, it is found
that no electrons are ejected when the incident
radiation has a frequency below a value
characteristic of the metal and, above that
value, the kinetic energy of the photoelectrons
varies linearly with the frequency of the
incident radiation.
Chapter 7. Quantum theory: introduction and principles
P.251
The Photoelectric Effect (cont’d)
1
mev 2 = hν − Φ (eqn 7.15) provides a technique for
2
the determination of Planck’s constant, for the slopes of
the lines in Fig.7.13 are all equal to h.
1)
2)
3)
Photoejection cannot occur if hν < Φ
because the photon brings insufficient energy;
1
me v2 = hν − Φ
2
(Eq.7.15) predicts that the kinetic
energy of an ejected electron should increase
linearly with frequency;
When a photon collides with an electron, it
gives up all its energy, so we should expect
electrons to appear as soon as the collisions
begin, provided the photons have sufficient
energy.
Fig.7.14 The photoelectric effect can be
explained if it is supposed that the incident
radiation is composed of photons that have
energy proportional to the frequency of the
radiation. (a) The energy of the photon is
insufficient to drive an electron out of the metal.
(b) The energy of the photon is more than
enough to eject an electron, and the excess
energy is carried away as the kinetic energy of
the photoelectron (the ejected electron).
Chapter 7. Quantum theory: introduction and principles
P.251
Example 1
Light with a wavelength of 300 nm is incident on a
potassium surface for which the work function, φ ,
is 2.26 eV. Calculate the kinetic energy and speed
of the ejected electrons.
Solution
We write Ee = hv − φ = (hc / λ ) − φ and convert the units
of φ from electron-volts to joules:
φ = (2.26eV )(1.602 ×10 −19 J / eV ) = 3.62 ×10−19 J
Electrons will only be ejected if the photon energy,
hv, is greater than φ . The photon energy is
calculated to be
hc
λ
(
6.626 ×10 )(2.998 ×10 )
=
= 6.62 ×10
34
8
−9
300 ×10
which is sufficient to eject electrons.
−19
J
Solution (cont’d)
We can obtain Ee = (hc / λ ) − φ = 2.99 ×10−19 J .
Using
Ee = 1 / 2mv 2
, we calculate that
(
)
2 Ee
2 2.99 ×10 −19 J
5
v=
=
=
8
.
10
×
10
m/ s
−31
m
9.109 ×10
Example 2 Calculating the number of photons
- The particle character of electromagnetic radiation
• Calculate # of photons emitted by a 100 W yellow lamp in 1.0s.
(wavelength of yellow light ~ 560 nm, assume 100% efficiency).
• Method:
Each photon has an energy hν, so the total # of photons needed to
produce energy E is E/hν.
The frequency of the radiation (ν = c/λ=?) and the total energy
(?) emitted by the lamp E =?.
E = P∆t, P - the power (in watts), ∆t - the time interval for which
the lamp is turned on.
• Answer: The number of photons is
E
P∆t
λP∆t
N=
=
=
h ν h (c λ )
hc
(
5.60×10
= (6.626×10
−7
)(
)
Js )×(2.998×10
m × 100 Js −1 ×(1.0 s )
−34
20
=
2
.
8
×
10
8
ms−1 )
• Note: it would take ~40 min to produce 1 mol of these photons.
Self Test - The particle character of electromagnetic radiation
• How many photons does a monochromatic
(single frequency) infrared rangefinder of
power 1 mW and wavelength 1000 nm emit in
0.1 s?
• Answer:
[5×1014]
Particles Exhibit Wave-Like Behavior
•
•
In 1924, Louis de Broglie suggested a relationship between
momentum and wavelength for light applying to particles - any
particle, not only photons, travelling with a linear momentum p
should have (in some sense) a wavelength given by the
de Broglie relation:
h
λ=
p
where p = mv (particle momentum) (7.16)
Not only has electromagnetic radiation the character classically
ascribed to particles, but electrons (and all other particles) have
the characteristics classically ascribed to waves. This joint
particle and wave character of matter and radiation is called
wave–particle duality.
7.2 Wave-particle duality
• Key points:
(a) The photoelectric effect establishes the view
that electromagnetic radiation, regarded in
classical physics as wave-like, consists of
particles (photons).
(b) The diffraction of electrons establishes the
view that electrons, regarded in classical
physics as particles, are wave-like with a
wavelength given by the de Broglie relation.
Fig.7.16 An illustration of the
de Broglie relation between
momentum and wavelength.
The wave is associated with a
particle (shortly this wave will
be seen to be the wavefunction
of the particle). A particle with
high momentum has a
wavefunction with a short
wavelength, and vice versa.
Chapter 7. Quantum theory: introduction and principles
P.253
Example 7.3 Estimating the de Broglie wavelength
• Estimate the wavelength of electrons that have been
accelerated from rest through a potential difference
of 40 kV.
• Method:
λ=h/p
(7.16)
The linear momentum of the electrons, p=?
E=? The energy acquired by an electron accelerated
through a potential difference ∆Φ is e ∆Φ, where e is
the magnitude of its charge.
At the end of the period of acceleration, all the
acquired energy is in the form of kinetic energy,
EK = 1/2mev2=p2/2me
Let p2/2me = e∆Φ → p
Carry through the calculation algebraically before
substituting the data.
Example 7.3 Estimating the de Broglie wavelength
• Answer: p2/2me = e ∆Φ
p = (2mee ∆Φ)1/2
de Broglie relation λ = h/p
h
λ=
1
(2me e∆φ ) 2
λ=
{2 × (9.109 ×10
6.626 ×10 − 34 J ⋅ s
− 31
) (
) (
kg × 1.602 × 10 −19 C × 4.0 × 10 4V
)}
1/ 2
= 6.1×10 −12 m
• 1 V C = 1 J and 1 J = 1 kg m2 s−2.
• The wavelength of 6.1 pm is shorter than typical bond lengths in
molecules (about 100 pm). Electrons accelerated in this way are
used in the technique of electron diffraction for the determination
of molecular structure
• (see Section 23.3 structure of solid surface p.885).
Self Test 7.2
• Calculate: (neutron mass: mn=1.675×10-27kg
Planck’s const.: h=6.626×10-34Js
Boltzmann’s const.: k=1.381×10-23J/K)
• (a) the wavelength of a neutron with a translational
kinetic energy equal to kT at 300 K,
• (b) a tennis ball of mass 57 g travelling at 80 km/h.
• Answer:
• (a) 178 pm,
• (b) 5.2 × 10−34 m
Example 4
Electrons are used to determine the structure of
crystal surfaces. To have diffraction, the
wavelength of the electrons should be on the order
of the lattice constant, which is typically 0.30 nm.
What energy do such electrons have, expressed in
electron-volts and joules?
Solution:
Using E=p2/2m for the kinetic energy, we obtain
E=
(
)
− 34 2
p
h
6.626 ×10
−18
=
=
=
2
.
7
×
10
or 17eV
2
−31
−10
2 m 2 mλ
2 9.109 × 10
3.0 ×10
2
2
(
)(
)
Diffraction by a Double Slit (雙縫衍射)
Diffraction of Light
•
Diffraction is a phenomenon that can occur with
any waves, including sound waves, water
waves, and electromagnetic (light) waves.
Chapter 7. Quantum theory: introduction and principles
P.252
Diffraction by a Double Slit (cont’d)
•
For diffraction of light from a thin slit, b >> a.
Diffraction by a Double Slit (cont’d)
•
Maxima and minima arise as a result of a path
difference between the sources of the
cylindrical waves and the screen.
Diffraction by a Double Slit (cont’d)
•
The condition that the minima satisfy is
nλ
sin θ =
, n = ±1,±2,±3,±.....
a
where λ = wavelength
Diffraction by a Double Slit (cont’d)
•
For double-slit diffraction experiment,
One slit or the
other blocked
both slits open
Physical Chemistry Fundamentals: Figure 7.15
Fig. 7.15 The Davisson–Germer experiment. The scattering of an electron beam
from a nickel crystal shows a variation of intensity characteristic of a diffraction
experiment in which waves interfere constructively and destructively in different
directions.
Atomic Spectra and the Bohr Model of the Hydrogen Atom
•
Light is only observed at certain discrete
wavelengths, which is quantized.
For the emission spectra, the inverse of the
wavelength, 1 / λ = v~ of all lines in an atomic
hydrogen spectrum is given by
•
1
−1
−1  1
~
v cm = RH cm  2 − 2 , n > n1
 n1 n 
(
)
(
)
Physical Chemistry Fundamentals: Figure 7.10
Atomic spectra
Bohr frequency condition:
∆E = hν
(7.14)
Fig. 7.10 A region of the
spectrum of radiation emitted
by excited iron atoms consists
of radiation at a series of
discrete wavelengths (or
frequencies).
Physical Chemistry Fundamentals: Figure 7.11
Molecular spectra
Fig. 7.11 When a
molecule changes its state,
it does so by absorbing
radiation at definite
frequencies. This
spectrum is part of that
due to the electronic,
vibrational, and rotational
excitation of sulfur
dioxide (SO2) molecules.
This observation suggests
that molecules can possess
only discrete energies, not
an arbitrary energy.
Physical Chemistry Fundamentals: Figure 7.12
The failures of classical physics - Atomic and molecular spectra
Fig. 7.12 Spectroscopic
transitions, such as those
shown above, can be
accounted for if we
assume that a molecule
emits a photon as it
changes between
discrete energy levels.
Note that high-frequency
radiation is emitted
when the energy change
is large.
SUMMARY: Energy quantization - Key points
•
•
•
•
(a) The classical approach to the description of
black-body radiation results in the ultraviolet
catastrophe.
(b) To avoid this catastrophe, Planck proposed
that the electromagnetic field could take up energy
only in discrete amounts.
(c) The thermal properties of solids, specifically
their heat capacities, also provide evidence that
the vibrations of atoms can take up energy only in
discrete amounts.
(d) Atomic and molecular spectra show that atoms
and molecules can take up energy only in discrete
amounts.
Electron microscopy (電子顯微鏡)
• Traditional light microscopy - illuminating a small area of a sample and
collecting light with a microscope.
• The resolution of a microscope, the minimum distance between two
objects that leads to two distinct images, is on the order of the
wavelength of light used as a probe.
• Conventional microscopes employing visible light have resolutions in the
micrometre range and are blind to features on a scale of nanometres.
• Electron microscopy, in which a beam of electrons with a well defined
de Broglie wavelength replaces the lamp found in traditional light
microscopes. Instead of glass or quartz lenses, magnetic fields are used
to focus the beam.
• In transmission electron microscopy (透射電子顯微鏡 TEM), the
electron beam passes through the specimen and the image is collected
on a screen.
• In scanning electron microscopy (掃描電子顯微鏡 SEM), electrons
scattered back from a small irradiated area of the sample are detected
and the electrical signal is sent to a video screen. An image of the
surface is then obtained by scanning the electron beam across the
sample.
Electron microscopy (電子顯微鏡cont’d)
• The wavelength of and the ability to focus the incident beam—
electrons—govern the resolution.
• Electron wavelengths in typical electron microscopes can be as
short as 10 pm, but it is not possible to focus electrons well with
magnetic lenses so. Typical resolutions of TEM and SEM
instruments are about 2 nm and 50 nm, respectively.
• Electron microscopes cannot resolve individual atoms, whose
diameters ~ 0.2 nm=2Å.
• Only certain samples can be observed under certain conditions.
The measurements must be conducted under high vacuum.
• For TEM observations, the samples must be very thin crosssections of a specimen and SEM observations must be made on
dry samples. Neither technique can be used to study living cells.
Physical Chemistry Fundamentals: Figure 7.17
Electron microscopy is very useful in studies of the internal structure of cells
Fig. 7.17 A TEM image of a cross-section of a plant cell showing chloroplasts,
organelles responsible for the reactions of photosynthesis. Chloroplasts are typically 5
µm long. (Brian Bowes)
Objectives - The Schrödinger Equation
• Key concepts of operators, eigenfunctions,
wave functions, and eigenvalues.
Outline
1. What Determines If a System Needs to Be
Described Using Quantum Mechanics?
2. Classical Waves and the Nondispersive(非色
散) Wave Equation
3. Waves Are Conveniently Represented as
Complex Functions
4. Quantum Mechanical Waves and the
Schrödinger Equation
Outline
1. Solving the Schrödinger Equation: Operators,
Observables, Eigenfunctions, and
Eigenvalues
2. The Eigenfunctions of a Quantum Mechanical
Operator Are Orthogonal
3. The Eigenfunctions of a Quantum Mechanical
Operator Form a Complete Set
4. Summing Up the New Concepts
What Determines If a System Needs to Be Described Using
Quantum Mechanics?
•
•
Particles and waves in quantum mechanics are
not separate and distinct entities.
Waves can show particle-like properties and
particles can also show wave-like properties.
What Determines If a System Needs to Be Described Using
Quantum Mechanics? (cont’d)
•
•
In a quantum mechanical system, only certain values of
the energy are allowed, and such system has a
discrete energy spectrum.
Thus, Boltzmann distribution is used.
ni g i − [ei − e j ]/ kT
=
e
nj g j
where ni = number of atoms or molecules
εi = energy of atoms or molecules
gi = degeneracy (簡並度) at energy level i
(energy level vs. energy state)
Example 5
Consider a system of 1000 particles that can only have
two energies, ∆ε = ε 2 − ε 1, with ε1 and ε 2 . The difference
in the energy between these two values is ε 2 > ε1 .
Assume that g1=g2=1.
a. Graph the number of particles, n1 and n2, in states
ε1 and ε 2 as a function of kT / ∆ε . Explain your
result.
b. At what value of kT / ∆ε do 750 of the particles have
the energy ε 1 ?
Solution
We can write down the following two equations:
n2 / n1 = e − ∆ε / kT and n1 + n2 = 1000
Solve these two equations for n2 and n1 to obtain
1000 e − ∆ε / kT
n2 =
1 + e − ∆ε / kT
1000
n1 =
1 + e − ∆ε / kT
Solution
Part (b) is solved graphically. The parameter n1 is
shown as a function of kT / ∆ε on an expanded
scale on the right side of the preceding graphs,
which shows that n1=750 for kT / ∆ε = 0.91 .
Classical Waves and the Nondispersive Wave Equation
•
Transverse(橫向 ), Longitudinal(縱向 ),
and Surface Waves
•
A wave can be represented pictorially by a
succession of wave fronts, where the
amplitude has a
maximum or minimum value.
transverse wave (橫波
橫波)
橫波
• A transverse wave is a moving wave that
consists of oscillations occurring perpendicular
to the direction of energy transfer.
Transverse plane wave
Longitudinal wave (縱波
縱波)
縱波
• Longitudinal waves are waves that have the same direction of
oscillation or vibration along their direction of travel, which means
that the oscillation of the medium (particle) is in the same direction
or opposite direction as the motion of the wave.
Plane pressure wave
• Longitudinal harmonic Sound waves:
  x 
y(x, t) = y0 sin ω  t − 
  c 
Where y is the displacement of the point on the traveling sound wave; x is
the distance the point has traveled from the wave's source; t is the time
elapsed; y0 is the amplitude of the oscillations, c is the speed and ω is the
angular frequency of the wave. The quantity x/c is the time that the wave
takes to travel the distance x.
Classical Waves and the Nondispersive Wave Equation
•
The wave amplitude ψ is:
x t 
Ψ ( x, t ) = A sin 2π  − 
λ T 
•
It is convenient to combine
constants and variables to
write the wave amplitude as
Ψ ( x, t ) = A sin (kx − wt )
where k = 2π/λ (wave vector)
ω = 2πv (angular frequency)
p = hk ,
h≡
h
2π
Classical Waves and the Wave Equation
•
Interference of Two Traveling Waves
•
For wave propagation in a medium where
frequencies have the same velocity (a
nondispersive medium), we can write
∂Ψ ( x , t ) 1 ∂ 2 Ψ ( x , t )
= 2
2
∂x
v
∂t 2
where v = velocity at which the wave propagates
Example 6
The nondispersive wave equation in one dimension
is given by
∂ 2ψ ( x, t ) 1 ∂ 2ψ (x, t )
= 2
2
∂x
v
∂t 2
Show that the traveling wave ψ (x, t ) = A sin (kx − ωt + φ )
is a solution of the nondispersive wave equation.
How is the velocity of the wave related to k and ω?
Solution:
We have
∂ 2ψ (x, t ) 1 ∂ 2ψ (x, t )
= 2
2
∂x
v
∂t 2
∂ 2 A sin (kx − ω t + φ )
2
=
−
k
A sin (kx − ωt + φ )
2
∂x
∂ 2 A sin (kx − ω t + φ ) − ω 2
= 2 A sin (kx − ωt + φ )
2
2
v ∂t
v
Equating these two results gives v = ω / k
Waves Are Conveniently Represented as Complex Functions
•
It is easier to work with the whole complex
function knowing as we can extract the real
part of wave function.
Example 7
a. Express the complex number (4+4i) in the
form reiθ = r (cosθ + i sin θ )
π
3
e
b. Express the complex number
in the
form (a+ib)
i3 / 2
Solution
a. The magnitude of 4+4i is [(4 + 4i )(4 − 4i)]1/ 2 = 4
The phase is given by
cos θ =
4
4 2
=
1
1
π
or θ = cos −1
=
2
2 4
Therefore, 4+4i can be written
4 2ei (π / 4)
2
Solution
b. Using the relation e iα
can be written
= exp(iα ) = cos α + i sin α , 3ei 3π / 2
3π
3π 

3 cos
+ i sin
 = 3(0 − i ) = −3i
2
2 

The dynamics of microscopic systems
• Quantum mechanics acknowledges the wave–
particle duality of matter by supposing that,
rather than travelling along a definite path, a
particle is distributed through space like a
wave.
• The mathematical representation of the wave
that in quantum mechanics replaces the
classical concept of trajectory is called a
wavefunction, ψ (psi).
7.3 Quantum Mechanical Waves
and The Schrödinger equation
• Key point: Schrödinger equation is a second-order
differential(微分) eqn used to calculate the
wavefunction of a system
• (1926) The time-independent Schrödinger
equation for a particle of mass m moving in one
dimension with energy E is
h 2 d 2ψ ( x)
−
+ V ( x)ψ ( x) = Eψ ( x )
2
2m dx
(7.17)
• Study the stationary states of quantum mechanical
systems.
Quantum Mechanical Waves and the Schrödinger Equation
•
An analogous quantum mechanical form of
time-dependent classical nondispersive wave
equation is the time-dependent Schrödinger
equation, given as
∂Ψ ( x, t )
h 2 ∂ 2 Ψ( x, t )
ih
=−
+ V ( x, t )Ψ ( x, t )
2
∂t
2m ∂ x
•
where V(x,t) = potential energy function
This equation relates the temporal and spatial
derivatives of ψ(x,t) and applied in systems
where energy changes with time.
Quantum Mechanical Waves and the Schrödinger
Equation
•
•
For stationary states of a quantum mechanical
system, we have
∂Ψ( x, t )
ih
= E Ψ( x, t )
∂t
Since
, we can show that that wave
functions whose energy is independent of time have
the form of
Ψ ( x, t ) = ψ ( x)e Et / ih = ψ ( x)e − i( E / h ) t
Using the Schrödinger equation to develop the de Broglie relation
d 2ψ
2m
= − 2 (E − V )ψ
2
dx
h
ψ = cos kx = cos( 2πx / λ )
(harmonic
 2m(E − V )
 2mEk 
k =
 = 2 
2
h


 h 
Ek = k 2h 2 / 2 m = p 2 / 2 m
1/ 2
2π
h
h
p = kh =
×
=
λ 2π λ
1/ 2
wave)
Born interpretation of the wavefunction
• Key points: According to Born interpretation,
the probability density is proportional to the
square of the wavefunction.
• (a) A wavefunction is normalized if the integral
of its square is equal to 1;
• (b) The quantization of energy stems from the
constraints that an acceptable wavefunction
must satisfy.
Born interpretation of the wavefunction (cont’d)
• If the wavefunction of a particle has the value ψ at
some point x, then the probability of finding the
particle between x and x + dx is proportional to
|ψ|2dx.
• |ψ|2 is the probability density. The wavefunction
ψ itself is called the probability amplitude.
• If the wavefunction of a particle has the value ψ at
some point r, then the probability of finding the
particle in an infinitesimal volume dτ = dxdydz at
that point is proportional to |ψ|2dτ.
Born interpretation of the wavefunction (cont’d)
Fig. 7.18 The
wavefunction ψ is a
probability amplitude in
the sense that its square
modulus (ψ*ψ or
|ψ|2) is a probability
density. The probability
of finding a particle in
the region dx located at
x is proportional to
|ψ|2dx. We represent
the probability density
by the density of
shading in the
superimposed band.
Born interpretation of the wavefunction (cont’d)
For a particle free to move in 3-dimensions (e.g., an electron near a nucleus in an atom), the
wave-function depends on the point dr with coordinates (x, y, z)
Fig. 7.19 The Born
interpretation of the
wavefunction in
three-dimensional
space implies that
the probability of
finding the particle
in the volume
element dτ =
dxdydz at some
location r is
proportional to the
product of dτ and
the value of |ψ|2 at
that location.
Physical Chemistry Fundamentals: Figure 7.20
Fig. 7.20 The sign of
a wavefunction has no
direct physical
significance: the
positive and negative
regions of this
wavefunction both
correspond to the
same probability
distribution (as given
by the square modulus
of ψ and depicted by
the density of
shading).
Example 7.3 Interpreting a wavefunction
• The wavefunction of an electron in the lowest energy
state of a hydrogen atom is proportional to
exp(−r/a0), with a0 a constant (Bohr radius) and r the
distance from the nucleus. Calculate the relative
probabilities of finding the electron inside a region of
volume 1.0 pm3, located at (a) the nucleus, (b) a
distance a0 from the nucleus.
• Method The region of interest is small on the scale
of the atom. Ignore the variation of ψ within it and
write the probability, P, as proportional to the
probability density (ψ2) evaluated at the point of
interest multiplied by the volume of interest, δV.
i.e., P ∝ ψ2δV, with ψ2 ∝ exp(−2r/a0).
Example 7.3 Interpreting a wavefunction (cont’d)
• Answer In each case δV = 1.0 pm3.
• (a) At the nucleus, r = 0,
(
)
(
P ∝ e 0 × 1.0 pm3 = (1.0 )× 1.0 pm3
)
• (b) At a distance r = a0 in an arbitrary direction,
P ∝ e−2 × (1.0 pm3 ) = (0.14 ) × (1.0 pm3 )
The ratio of probabilities is 1.0/0.14 = 7.1.
Note: it is more probable (by a factor of 7) that the
electron will be found at the nucleus than in a volume
element of the same size located at a distance a0 from
the nucleus. The negatively charged electron is attracted
to the positively charged nucleus, and is likely to be found
close to it.
Test 7.3
• The wavefunction for the electron in its lowest energy state in the
ion He+ is proportional to exp(−2r/a0). Repeat the calculation for
this ion. Any comment?
• Answer:
55; more compact wavefucntion
A note on good practice:
The square of a wavefunction is not a probability: it is a
probability density, and (in 3-dimensions) has the dimensions of
1/length3. It becomes a (unitless) probability when multiplied by a
volume. In general, we have to take into account the variation of
the amplitude of the wavefunction over the volume of interest,
but here we are supposing that the volume is so small that the
variation of ψ in the region can be ignored.
Normalization of wave-function
N - normalization constant
∞
N 2 ∫ ψ ∗ψdx = 1
(7 .18)
−∞
1
N=
1/ 2
∞
 ψ ∗ψdx 
 ∫− ∞

(7 .19 )
A normalized wavefunction satisfies:
In 1 − Dimension:
∫
∞
ψ ∗ψdx = 1
−∞
(7. 20a)
In 3 − Dimensions :
∞
∞
∞
ψ ∗ψdxdydz = 1
∫ ∫ ∫
∫ψ ψdτ = 1
−∞ −∞ −∞
∗
where
dτ = dxdydz
(7. 20b)
(7.20c)
Spherical polar coordinates for system with spherical symmetry
x=rsinθcosφ, y=rsinθsinφ,
z=rcosθ
r – radius [0,∞)
θ - colatitude(餘緯度), [0, π]
φ - azimuth(方位), [0,2π]
dτ=r2sinθdrdθdφ
∞
π
0
0
∫ ∫ ∫
2π
0
ψ ∗ψ r 2dr sin θdθdφ = 1
(7.20d )
Fig. 7.21 The spherical polar
coordinates used for discussing
systems with spherical
symmetry.
Physical Chemistry Fundamentals: Figure 7.22
Fig. 7.22 The
surface of a sphere
is covered by
allowing θ to
range from 0 to π,
and then sweeping
that arc around a
complete circle by
allowing φ to
range from 0 to
2π.
The Eigenfunctions of a Quantum Mechanical Operator
Are Orthogonal
•
•
3-D system is importance to us is the atom.
Atomic wave functions are best described by
spherical coordinates.
Example 7.4 Normalizing a wavefunction
• Normalize the wavefunction for the hydrogen atom
in Ex.7.3.
• Method:
Find the normalization factor N in eqn 7.20c. For a
spherical system is, use spherical coordinates and
carry out the integrations in eqn 7.20d. Note: the
limits on the 1st integral sign - r, 2nd - θ, the 3rd φ.
A useful integral:
∫
∞
0
n!
x e dx = n+1
a
n − ax
where n! denotes a factorial: n! = n(n − 1)(n − 2)...1.
Answer to Ex.7.4 Normalizing a wavefunction:
The integration required is the product of three factors:
∫ψ ψdτ = N ∫
∗
2
∞
0
2 −2 r / a 0
r e
π
2π
0
0
dr ∫ sin θdθ ∫ dπ
1
= N 2 • a03 • 2 • 2π = πa03 N 2 = 1
4
• and the normalized wavefunction is
 1 
N =  3 
 πa0 
1/ 2
1/ 2
 1  − r / a0
ψ =  3  e
 πa0 
• Note: a0 is a length, the dimensions of ψ are 1/length3/2 and
those of ψ2 are 1/length3 - probability density.
• Repeating Ex.7.3, actual probabilities of finding the electron in
the volume element at each location (a0=52.9pm):
(a) 2.2×10−6, ~ 1 chance in about 500,000 inspections of
finding the electron in the test volume;
(b) 2.9 × 10−7, ~ 1 chance in 3.4 million.
Example
−r
Normalize the function e over the interval
0 ≤ r ≤ ∞; 0 ≤ θ ≤ π ; 0 ≤ φ ≤ 2π
Solution:
Volume element in spherical coordinates is
r 2 sin θdrdθdφ , thus
2π
π
∞
0
0
0
N 2 ∫ dφ ∫ sin θdθ ∫ r 2e −2 r dr = 1
∞
4πN 2 ∫ r 2 e−2 r dr = 1
0
Solution
e− r
Using the standard integral ,
∫
∞
0
x n e − ax dx = n!/ a n +1 (a > 0, n is a positive integer)
2
π
4
N
we obtain
2!
1
=
1
so
that
N
=
23
π
The normalized wave function is
1
π
e −r
Note that the integration of any function involving
r, even if it does not explicitly involve φ or θ ,
requires integration over all three variables.
Test 7.4 Normalize the wavefunction
given in test 7.3. He+ g.s.~ exp(−2r/a0).
• Correct Answer: N = (8/πa03)1/2
• 7.4 Born interpretation of wavefunction (b) Quantization
• ψ must be continuous, have a continuous slope (the 2nd
derivative of ψ must be well-defined), be single-valued, and be
square-integrable - The wavefunction ψ must not be infinite
anywhere (the wavefunction must not be infinite over any finite
region).
• An acceptable wavefunction cannot be zero everywhere, because
the particle it describes must be somewhere.
• These are such severe restrictions that acceptable solutions of the
Schrödinger eqn do not in general exist for arbitrary values of the
energy E. In other words, a particle may possess only certain
energies, for otherwise its wavefunction would be physically
unacceptable. i.e., the energy of a particle is quantized.
Physical Chemistry Fundamentals: Figure 7.23
Fig. 7.23
The wavefunction
must satisfy stringent
conditions for it to
be acceptable.
(a) Unacceptable
because it is not
continuous;
(b) unacceptable
because its slope is
discontinuous;
(c) unacceptable
because it is not
single-valued;
(d) unacceptable
because it is infinite
over a finite region.
Quantum Mechanical principles
- 7.5 The information in a wavefunction
• Key points:
(a) The wavefunction (波函數) of a free particle with a specific linear
momentum corresponds to a uniform probability density (均勻概率密度).
• (b) The Schrodinger eqn is an eigen value eqn (本徵方程) in which the
wavefunction is an eigenfunction (本徵函數) of the Hamiltonian
operator(算符).
• (c) Observables (可觀測量) are represented by operators (算符); the
value of an observable is an eigenvalue (本徵值) of the corresponding
operator constructed from the operators for position and linear
momentum (線性動量).
• (d) All operators that correspond to observables are hermitian; their
eigenvalues are real and their eigenfunctions are mutually orthogonal (正
交). Sets of functions that are normalized and mutually orthogonal are
called orthonormal (正交歸一).
• (e) When the system is not described by an eigenfunction of an
operator, it may be expressed as a superposition (線性疊加) of such
eigenfunctions. The mean value (平均值) of a series of observations is
given by the expectation value (期望值) of the corresponding operator.
QM principles - 7.5 The information in a wavefunction (cont’d)
h 2 d 2ψ
−
= Eψ
2
2m dx
( 7.21)
Solutions :ψ = Ae + Be
B =0:
ψ = Ae ikx
ik x
(
−ik x
,
k 2h 2
E=
2m
) (Ae ) = (A e )(Ae ) = A
ψ = Ae
ikx ∗
A = B:
ψ = A(e ikx + e −ikx ) = 2Acoskx
2
ikx
∗ −ikx
ikx
(7.22)
(7.23)
2
(7.24)
(7.25)
probabilit y density :
∗
ψ = (2Acoskx) (2Acoskx) = 4 A cos 2 kx
2
node
2
(7.26)
Fig.7.24 (a) The square modulus of a wavefunction
corresponding to a definite state of linear momentum is a
constant; so it corresponds to a uniform probability of
finding the particle anywhere. (b) The probability
distribution corresponding to the superposition of states
of equal magnitude of linear momentum but opposite
direction of travel.
Chapter 7. Quantum theory: introduction and principles
P.267
7.5 The information in a wavefunction
(b) Operators, eigenvalues, and eigenfunctions
Schr&o&dinger eqn (in operator form) :
Hamiltonia n operator (in one - dimension) :
Hˆ ψ = E ψ
h2 d 2
ˆ
H =−
+ V ( x)
2 m dx 2
(Operator)(function) = (constant factor) × (same function) (7.28a)
ˆ ψ = ωψ
Eigenvalue equation : Ω
(7.28b)
(Operator)(eigenfunction) = (eigenvalu e) × (eigenfunc tion) (7.28c)
( 7.27 a)
( 7.27 b)
Solving the Schrödinger Equation: Operators, Observables,
Eigenfunctions, and Eigenvalues
•
•
We would need to use operators, observables,
eigenfunctions, and eigenvalues for quantum
mechanical wave equation.
The time-independent Schrödinger equation is
an eigenvalue equation for the total energy, E
 h 2 ∂2

+ V ( x )ψ n ( x ) = E n ψ n ( x )
−
2
 2 m ∂x

•
)
where {} = total energy operator or H
It can be simplified as Hˆ ψ n = Enψ n
Example
2
2
d
/
dx
and
d
/
dx
Consider the operators
. Is the
function ψ (x ) = Aeikx + Be −ikx
an eigenfunction of
these operators? If so, what are the eigenvalues?
Note that A, B, and k are real numbers.
Solution
To test if a function is an eigenfunction of an
operator, we carry out the operation and see if the
result is the same function multiplied by a
constant:
(
)
d Ae ikx + Be − ikx
= ikAeikx − ikB −ikx = ik Ae ikx − Be −ikx
dx
(
)
In this case, the result is not ψ (x) multiplied by a
constant, so ψ (x) is not an eigenfunction of the
operator d/dx unless either A or B is zero.
Solution
This equation shows that ψ (x ) is an eigenfunction
of the operator d 2 / dx 2 with the eigenvalue k2.
(
)
d 2 Aeikx + Be − ikx
2
2 −ikx
ikx
=
(
ik
)
Ae
+
(
−
ik
)
B
2
dx
= −k 2 Ae ikx + Be −ikx
(
= −k 2ψ (x )
)
Example 7.5 Identifying an eigenfunction (p.268)
• Show that eax is an eigenfunction of the operator
d/dx ; find the corresponding eigenvalue. Show that
exp(ax2) is not an eigenfunction of d/dx.
• Method: Operate on the function with the operator
and check whether the result is a constant factor
times the original function.
• Answer:
d ax
ˆ
Ωψ = e = ae ax = aψ
dx
d ax2
ax 2
ˆ
Ωψ = e = 2axe = 2ax ×ψ
dx
• Test 7.5: Is the function cos ax an eigenfunction of
(a) d/dx, (b) d2/dx2?
• Correct Answer: (a) No, (b) yes
(c) The construction of operators (p.268)
ˆ,
Observables, Ω, are represente d by operators, Ω
built from the following position and momentum operators :
(Energy operator)ψ = (energy) ×ψ
(Operator correspond ing to an observable)ψ = (value of observable) ×ψ
Specificat ioon of operators : x̂ = x × ,
h d
p̂ x =
i dx
[7.29]
Ex.7.6 Determining the value of an observable (p.269)
•
What is the linear momentum of a particle described by the wavefunction in eqn 7.22:
with (a) B = 0, (b) A = 0?
k 2 h2
ψ = Ae ikx + Be −ikx , E =
•
•
2m
Method Operate on ψ with the operator corresponding to linear momentum eqn 7.29:
Answer:
p̂ x =
(a ) B = 0 :
h d
i dx
h d ψ h deikx h
pˆ xψ =
= A
= A × ikeikx = kh Aeikx = khψ
i dx i
dx
i
p x = +kh
(a ) A = 0 :
h d ψ h de−ikx h
pˆ xψ =
= B
= B × (−ik )e −ikx = − kh Be−ikx = −khψ
i dx i
dx
i
p x = −kh
•
•
In (a) the particle is travelling to the right (positive x ) but in (b) it is travelling to the left
(negative x ).
Test 8.6 The operator for the angular momentum of a particle travelling in a circle in the
xy -plane is ˆ
d , where φ is its angular position. What is the angular momentum
lz = (h / i ) φ
of a particle described by the wavefunction e−2iφ?
• Correct Answer:
l z = −2h
(c) The construction of operators (cont’d, p.270)
1 2
V̂ = kx ×
2
1  h d  h d 
h2 d 2
Ê k =


=−
2 m  i dx  i dx 
2m dx 2
(7.30)
(7.31)
2
2
h
d
Hamiltonia n operator : Ĥ = Ê k + Vˆ = −
+ Vˆ (7.32)
2
2m dx
Physical Chemistry Fundamentals: Figure 7.25
Fig. 7.25 Even if a
wavefunction does not
have the form of a
periodic wave, it is still
possible to infer from it
the average kinetic energy
of a particle by noting its
average curvature. This
illustration shows two
wavefunctions: the
sharply curved function
corresponds to a higher
kinetic energy than the
less sharply curved
function. (p.270)
Physical Chemistry Fundamentals: Figure 7.26
Fig. 7.26 The
observed kinetic
energy of a particle is
an average of
contributions from
the entire space
covered by the
wavefunction.
Sharply curved
regions contribute a
high kinetic energy to
the average; slightly
curved regions
contribute only a
small kinetic energy.
Physical Chemistry Fundamentals: Figure 7.27
Fig.7.27 The wavefunction
of a particle in a potential
decreasing towards the right
and hence subjected to a
constant force to the right.
Only the real part of the
wavefunction is shown, the
imaginary part is similar,
but displaced to the right.
(p.271)
(d) Hermitian operators
• All the quantum mechanical operators that correspond to
observables have a very special mathematical property:
they are ‘hermitian’. An hermitian operator is one for
which the following relation is true:
{
ˆ ψ j d τ = ψ j∗Ω
Hermiticity : ∫ψ Ω
∫ ˆ ψ i dτ
∗
i
∫
∞
−∞
∞
ψ xψ j dτ = ∫ ψ j xψ dτ =
∗
i
∗
i
−∞
}
∗
{∫ ψ xψ dτ }
∞
−∞
∗
j
[7.33]
∗
i
J8.2 The hermiticity of the linear momentum operator :
∫
∞
−∞
∗
i
ψ pˆ xψ j dx =
{∫ ψ pˆ ψ dx}
∞
−∞
∗
j
∗
x
i
integration by parts :
dg
df
f
dx
=
fg
−
g
∫ dx
∫ dxdx
Test 7.7 Confirm that the operator d 2 /dx 2 is hermitian.
The reality of eigenvalues
• Two properties of Hermitian operators: their
eigenvalues are real, and their eigenfunctions are
‘orthogonal’. All observables have real values, and so
are represented by hermitian operators.
• Justification 8.3 The reality of eigenvalues
• For a wavefunction ψ that is normalized and is an
eigenfunction of an hermitian operator with
eigenvalue ω,
∗ ˆ
∗
∗
Ω
d
=
d
=
ψ
ψ
τ
ψ
ωψ
τ
ω
ψ
∫
∫
∫ ψdτ = ω
{
}
∗
ˆ
ˆ ψdτ = ω
ω = ∫ψ Ωψ dτ = ∫ψ ∗ Ω
∗
∗
hermiticity
• ω* = ω confirms that ω is real.
The Eigenfunctions of a Quantum Mechanical Operator
Are Orthogonal
•
•
Orthogonality is a concept of vector space.
3-D Cartesian coordinate space is defined by
x• y = x • z = y • z = 0
•
In function space, the analogous expression
that defines orthogonality (正交 ) is
J7.4 Orthogonality of wavefunctions (p.272)
• Two different functions ψi and ψj are orthogonal if the integral (over all
space) of their product is zero:
Orthogonal ity :
∗
ψ
i
∫ ψ j dτ = 0 for i ≠ j
(7.34)
• Wavefunctions corresponding to different eigenvalues of an hermitian
operators are orthogonal.
sin( a − b ) x sin( a + b) x
−
+ constant (a 2 ≠ b 2 )
2( a − b )
2(a + b)
cos(α ± β ) = cos α cos β m sin α sin β
− 2 sin α sin β = cos(α + β ) − cos(α − β )
∫ sin ax sin bxdx =
∫
2π
0
sin x sin 2 xdx = 0 (a = 1, b = 2)
• Test 8.8 Confirm that the functions sin x and sin 3x are mutually
orthogonal.
∫
2π
0
sin x sin 3 xdx = 0
Example
Show graphically that sin x and cos 3x are
orthogonal functions. Also show graphically that
∞
∫ (sin mx)(sin nx)dx ≠ 0 for n = m = 1
−∞
Solution
The functions are shown in the following
graphs. The vertical axes have been offset to
avoid overlap and the horizontal line indicates
the zero for each plot.
Because the functions are periodic, we can
draw conclusions about their behaviour in an
infinite interval by considering their behaviour
in any interval that is an integral multiple of the
period.
Solution
Solution
The integral of these functions equals the sum
of the areas between the curves and the zero
line. Areas above and below the line contribute
with positive and negative signs, respectively,
∞
and indicate that ∫−∞ (sin mx )(cos 3x )dx = 0 and
∫ (sin x )(sin x )dx > 0 . By similar means, we could
show that any two functions of the type sin mx
and sin nx or cos mx and cos nx are orthogonal
unless n=m. Are the functions cos mx and sin
mx(m=n) orthogonal?
∞
−∞
Physical Chemistry Fundamentals: Figure 7.28
Fig. 7.28 The integral of
the function f(x) = sin x
sin 2x is equal to the area
(tinted) below the green
curve, and is zero, as can
be inferred by symmetry.
The function—and the
value of the integral—
repeats itself for all
replications of the section
between 0 and 2π, so the
integral from −∞ to ∞ is
zero.
7.5 The information in a wavefunction
(e) Superpositions and expectation values
ψ = A(e ikx + e − ikx ) = 2 A cos kx (eqn.7.22 with A = B)
• When the wavefunction of a particle is not an eigenfunction of an
operator, the property to which the operator corresponds does not
have a definite value.
• The momentum not completely indefinite - cosine wavefunction is a
linear combination, or sum, of eikx and e−ikx - definite momentum
states.
• The total wavefunction is a superposition of more than one
wavefunction:
Linear combination of basis functions
• Any wavefunction as a linear combination of eigenfunctions of an
operator - Suppose the wavefunction is known to be a
superposition of many different linear momentum eigenfunctions
•
•
•
•
(7.36)
where the ck are numerical (possibly complex) coefficients and the
ψk correspond to different momentum states. The functions ψk
form a complete set in the sense that any arbitrary function can
be expressed as a linear combination of them.
1. When the momentum is measured, in a single observation one
of the eigenvalues corresponding to the ψk that contribute to the
superposition will be found.
2. The probability of measuring a particular eigenvalue in a series
of observations is proportional to the square modulus (|ck|2) of
the corresponding coefficient in the linear combination.
3. The average value of a large number of observations is given
by the expectation value, <Ω> , of the operator corresponding
to the observable of interest.
The Eigenfunctions of a Quantum Mechanical Operator
Form a Complete Set
•
•
The eigenfunctions of a quantum mechanical
operator form a complete set.
This means that any well-behaved wave
function, f (x) can be expanded in the
eigenfunctions of any of the quantum
mechanical operators.
The expectation value of an operator
• The expectation value of an operator is defined as
(7.37)
(for normalized wavefunctions)
An expectation value is the weighted average of a large number of
observations of a property.
J7.5 The expectation value of an operator
ˆ with eigenvalue ω : Ω
ˆ ψ = ωψ
If ψ is an eigenfunct ion of Ω
• the expectation value is
• A wavefunction is not an eigenfunction of the operator of interest:
can be written as a linear combination of eigenfunctions.
• e.g. the wavefunction is the sum of two eigenfunctions
<Ω> = |c1|2ω1 + |c2|2ω2
Example 7.7 Calculating an expectation value
• Calculate the average value of the distance of an electron from the nucleus
in the hydrogen atom in its state of lowest energy.
• Method The average radius is the expectation value of the operator
corresponding to the distance from the nucleus, x r. To evaluate <r>, we
need the normalized wavefunction (Ex. 7.4):
1/ 2
 1 
ψ =  3  e − r / a0 (E x.7.4)
 πa0 
Bohr radius a 0 = 52.9pm
ˆ ψ dτ
Ω = ψ ∗Ω
[7.37]
∫
• Test 7.9 Evaluate the root mean square distance, <r2>1/2, of the
electron from the nucleus in the hydrogen atom.
• Correct Answer: 31/2a0 = 91.6 pm
Mean kinetic energy of a particle in one dimension
• The mean kinetic energy of a particle in one
dimension is the expectation value of the operator
given in eqn 7.31.
• The kinetic energy is a kind of average over the
curvature of the wavefunction: a large contribution
to the observed value from regions where the
wavefunction is sharply curved (d2ψ/dx2 is large)
and the wavefunction itself is large (ψ* is large).
7.6 The uncertainty principle
• Key points: The uncertainty principle restricts the precision with
which complementary observables may be specified and
measured. Complementary observables are observables for which
the corresponding operators do not commute.
• If the wavefunction is Aeikx , then the particle it describes has a
definite state of linear momentum, namely travelling to the right
with momentum px = +kh/2π . However, the position of the
particle described by this wavefunction is completely
unpredictable. In other words, if the momentum is specified
precisely, it is impossible to predict the location of the particle.
This statement is one-half of a special case of the Heisenberg
uncertainty principle:
• It is impossible to specify simultaneously, with arbitrary
precision, both the momentum and the position of a
particle.
• The other half: if we know the position of a particle exactly, then
we can say nothing about its momentum.
Physical Chemistry Fundamentals: Figure 7.29
A wavefunction
as a superposition
of eigenfunction
Fig. 7.29 The
wavefunction for a
particle at a welldefined location is a
sharply spiked
function that has
zero amplitude
everywhere except at
the particle ’s
position.
Physical Chemistry Fundamentals: Figure 7.30
Fig. 7.30 The wavefunction for a
particle with an ill-defined location
can be regarded as the superposition
of several wavefunctions of definite
wavelength that interfere
constructively in one place but
destructively elsewhere. As more
waves are used in the superposition
(as given by the numbers attached to
the curves), the location becomes
more precise at the expense of
uncertainty in the particle’s
momentum. An infinite number of
waves is needed to construct the
wavefunction of a perfectly localized
particle.
Physical Chemistry Fundamentals: Table 7.2
Example 7.8 Using the uncertainty principle
• Suppose the speed of a projectile of mass 1.0 g is known to within 1 µm s−1.
Calculate the minimum uncertainty in its position.
• Method Estimate ∆p from m ∆v, where ∆v is the uncertainty in the speed; then
use eqn 7.39a to estimate the minimum uncertainty in position, ∆ q.
• Answer The minimum uncertainty in position is
(1 J = 1 kg m2 s−2). The uncertainty is completely negligible for all practical
purposes concerning macroscopic objects. However, if the mass is that of an
electron, then the same uncertainty in speed implies an uncertainty in position far
larger than the diameter of an atom (the analogous calculation gives ∆q = 60 m);
so the concept of a trajectory, the simultaneous possession of a precise position
and momentum, is untenable.
• Test 7.10 Estimate the minimum uncertainty in the speed of an electron in a onedimensional region of length 2a0.
• Correct Answer: 547 km s−1
Complementary observables
• General Heisenberg uncertainty principle: It applies to any pair of
observables called complementary observables, which are
defined in terms of the properties of their operators. Specifically,
two observables Ω1 and Ω2 are complementary if
(7.40)
• When the effect of two operators depends on their order, they do
not commute.
• The commutator of the two operators is defined as
(7.41)
• The commutator of the operators for position and linear
momentum is
J 7.6 The commutator of position and momentum
• Show that the operators for position and momentum do not commute (and
hence are complementary observables)
h
xˆpˆ xψ − pˆ x xˆψ = − ψ = ihψ
i
→ (7.42)
• The Heisenberg uncertainty principle in the most general form - For any two
pairs of observables, Ω1 and Ω2, the uncertainties (to be precise, the root
mean square deviations of their values from the mean) in simultaneous
determinations are related by:
Postulates of Quantum Mechanics
• Postulate (假定)1:
All information that can be obtained about the
state of a mechanical (physical) system is
contained in a wave function Ψ, which is a
continuous, finite, and single-valued function
of time and of the coordinates of the particles
of the system.
(Mortimer 3e)
Postulate 1 of Quantum Mechanics (cont)
• This postulate implies that there is a one-to-one
relationship between the state of the system and a wave
function.
• i.e., each possible state corresponds to one wave
function, and each possible wave function corresponds to
one state.
• The terms “state function” and “wave function” are often
used interchangeably. Information about values of
mechanical variables such as energy and momentum
must be obtained from the wave function, instead of from
values of coordinates and velocities as in classical
mechanics.
• The 4th postulate will provide the method for obtaining
this information.
Postulate 2
• The wave function Ψ obeys the timedependent Schrodinger equation
∂
Ψ
Hˆ Ψ = ih
∂t
(1)
where Ĥ is the Hamiltonian operator of the
system.
Postulate 2 (cont’d)
• The time-independent Schrodinger equation can be
derived from the time-dependent equation by assuming
that the wave function is a product of a coordinate factor
and a time factor:
Ψ(q, t) = ψ(q)η(t)
(2)
where q stands for all of the coordinates of the particles
in the system and where the coordinate wave function ψ
satisfies the time-independent Schrodinger equation.
• Not all wave functions consist of the two factors in Eq.(2),
but all wave functions must obey the time-dependent
Schrodinger equation.
Postulate 3 Mathematical Operators
and Mechanical Variables
• There is a linear hermitian mathematical
operator in one-to-one correspondence with
every mechanical variable.
• This postulate states that for each operator there is one
and only one variable, and
• for each variable there is one and only one mathematical
operator.
• A mathematical operator is a symbol that stands for
performing one or more mathematical operations.
• Usually denote an operator by a letter with a caret (︿)
over it.
Postulate 4 Expectation Values
• (a) If a mechanical variable A is measured without
experimental error, the only possible
outcomes of the measurement are the eigenvalues of the
operator  that corresponds to A.
• (b) The expectation value for the error-free measurement
of a mechanical variable A if given by the formula
A =
∗ ˆ
Ψ
∫ AΨdτ
∫ Ψ Ψdτ
∗
=
Ψ Aˆ Ψ
Ψ Ψ
where  is the operator corresponding to the variable
A, and where ψ=ψ(q, t) is the wave function
corresponding to the state of the system at the time of
the measurement.
Postulate 5. Measurements and
the Determination of the State of a System
• Immediately after an error-free measurement
of the mechanical variable A in which the
outcome was the eigenvalue aj , the state of
the system corresponds to a wave function
that is an eigenfunction of  with eigenvalue
equal to aj .
Postulate 5 (cont’d)
• This postulate says very little about the state of the
system prior to a single measurement of the variable A,
because the act of measurement can change the state of
the system.
• How a measurement can change the state of a system?
Consider the determination of the position of a particle by
the scattering of electromagnetic radiation:
• When an airplane reflects a radar wave, the effect on the
airplane is negligible because of the large mass of the
airplane.
• When an object of small mass such as an electron
scatters ultraviolet light or X-rays, the effect is not
negligible.
Sum 7.7 The postulates of quantum mechanics (p.279)
• The wavefunction. All dynamical information is contained in the wavefunction
ψ for the system, which is a mathematical function found by solving the
Schrödinger equation for the system. In one dimension:
• The Born interpretation. If the wavefunction of a particle has the value ψ at
some point r, then the probability of finding the particle in an infinitesimal
volume dτ = dxd ydz at that point is proportional to |ψ|2dτ.
• Acceptable wavefunctions. An acceptable wavefunction must be continuous,
have a continuous first derivative, be single-valued, and be square-integrable.
• Observables. Observables, Ω, are represented by operators,Ω̂ , built from
position and momentum operators of the form
•
• or, more generally, from operators that satisfy the commutation relation
[xˆ, pˆ x ] = ih
• The Heisenberg uncertainty principle. It is impossible to specify simultaneously,
with arbitrary precision, both the momentum and the position of a particle and,
more generally, any pair of observable with operators that do not commute.