PChemCh7_20110222.pdf

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

P.249

The origins of quantum mechanics7.1 The failures of classical physics and

Energy quantization7.2 Wave-particle duality

The dynamics of microscopic systems7.3 The Schrdinger equation7.4 The Born interpretation of the wavefunction

Quantum mechanical principles7.5 The information in a wavefunction7.6 The uncertainty principle7.7 The postulates of quantum mechanics

Atkins & Paula ,Physical Chemistry, 9th Ed., Oxford, 2010

Objectives Objectives -- From Classic to Quantum MechanicsFrom Classic to Quantum Mechanics

Introduction of Quantum Mechanics Understand the difference of classical theory

and experimental observations of quantum mechanics

((Physical ChemistryPhysical Chemistry, 2, 2ndnd Ed., Ed., Thomas Engel, Philip Reid)

OutlineOutline

1. Why Study Quantum Mechanics?2. Quantum Mechanics Arose Out of the Interplay

of Experiments and Theory3. Blackbody Radiation4. The Photoelectric Effect5. Particles Exhibit Wave-Like Behavior6. Diffraction by a Double Slit7. Atomic Spectra and the Bohr Model of the

Hydrogen Atom

Why Study Quantum Mechanics?Why Study Quantum Mechanics?

Quantum mechanics predicts that atoms and molecules can only have discrete (, )energies.

Quantum mechanical calculations of chemical properties of molecules are reasonably accurate.

Quantum Mechanics Arose Out of the Interplay of Quantum Mechanics Arose Out of the Interplay of Experiments and TheoryExperiments and Theory

Two key properties are used to distinguish classical and quantum physics.

1. Quantization ()- Energy at the atomic level is not a continuous variable, but in discrete packets called quanta () .

2. Wave-particle duality ()- At the atomic level, light waves have particle-like properties, while atoms and subatomic particles have wave-like properties.

Chapter 7. Quantum theory: introduction and principles P.250

=c (7.1) frequencyc - speed of lightWavenumber (unit: cm-1):

(7.2)

1

~

==

c

The origins of quantum mechanics

Fig.7.1 The wavelength, , 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, , is the number of cycles per second that occur at a given point.


Fig.7.2 The electromagnetic spectrum and the classification of the spectral regions

7.1 Energy quantization 7.1 Energy quantization -- Key pointsKey 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, specificallytheir 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 Blackbody Radiation (())

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 () with the solid.


A black body is an object of emitting and absorbing all frequencies of radiation uniformly.

Black-body Radiation (, contd)

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.



Energy density (Jm-3):d(,T) = (,T) d (7.3)Total energy density in a region:

(7.4)

Total energy: E(T) = V (T) (7.5)

( ) ( ) dT= 0 ,T

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).



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.


Energy density (Jm-3):d(,T) = (,T) d (7.3)Rayleigh-Jeans law - density of states (Jm-4) : (,T) =8kT/ 4 (7.6) k Boltzmanns constant:1.381 10-23 JK-1


Fig.7.6 The RayleighJeans law (eqn 7.6) predicts an infinite energy density at short wavelengths. This approach to infinity is called the ultraviolet catastrophe.

Blackbody Radiation Blackbody Radiation ((, cont, contdd))

The limitation of energy to discrete values is called the quantization of energy.

E=nh (n=0,1,2,) (7.7)h- Plancks constant 6.626x10-34Js

Planck distribution:dE= d (7.8)

For short wavelength, hc/kT 1, faster than 5->0

For long wavelength, hc/kT1,

Taylor series: f(x) = f(x0) + f(x0)(x-x0)/1! + f(x0)(x-x0)2/2! +

( ) ( )18

, /5

= kThce

hcT pi

oras 00

kThc

kThc

ekThc

++= 1...11/

kThce /

Blackbody Radiation (contBlackbody Radiation (contd)d)

Max Planck derived the agreement between theory and experiment on radiation energy.

where h = Plancks constantn = 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.

nhvE =


( ))9.7()(15

8

18

(T)densityenergy

3

45

40 /5

hck

awith

aTde

hckThc

pi

pi

=

=

=

Blackbody Radiation Blackbody Radiation ((, cont, contdd))

Fig.7.7 The Planck distribution (eqn 7.8) accounts very well for the experimentally determined distribution of black-body radiation. Planck s quantizationhypothesis essentially quenches the contributions of high frequency, short wavelength oscillators. The distribution coincides with the RayleighJeans distribution at long wavelengths.

Blackbody Radiation (contBlackbody Radiation (contd)d)

Spectral density ())is the energy stored in the electromagnetic field of the blackbody radiator.

Heat CapacityHeat 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=3NAK=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 T0

Einstein assumed that each atom oscillated about its equilibrium position with a single frequency in 1905


Heat capacities (contd)

)10.7(9.243C

:capacityheat olumeconstant vmolar )10.7( 33U

:energy internalmolar

11,

bmolJKRT

U

aRTkTN

V

mmV

Am

==

=

==

khetemperaturEinsteine

e

TTf

TRfT

E

T

TE

E

EmV

E

E

/

)11.7(1

)(

)(3)(C :formulaEinstein 2

/

2/2

,

=

=

=

( )

(7.12b) e )(

:)( )2

(7.12a); 11/1

2/1)(

:)( )1

/-22

/

2/2

22

E TET

TE

E

E

E

EEE

E

Tee

TTf

TestemperaturlowAtT

TT

Tf

TestemperaturhighAt

E

E

=

>

L

L

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.


Heat capacities (contd)

( )khetemperaturDebye

dxe

exTTf

TRfT

DD

T

x

x

DD

DmV

D

/:

)13.7(1

3)(

)(3)(C :formula Debye/

0 2

43

,

=

=

=

By averaging over all the frequencies present from zero up to a maximum value D:

Fig.7.9 Debyes modification of Einsteins calculation (eqn 7.13) gives very good agreement with experiment. For copper, T/D=2 corresponds to about 170 K, so the detection of deviations from Dulong and Petits law had to await advances in low-temperature physics.

The Photoelectric Effect (The Photoelectric Effect ())

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 (contThe Photoelectric Effect (contd)d)

Albert Einstein states that the energy of light,

where = constantv = frequency

From energy conservation the energy of the electron, Ee, is

(7.15)where = 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.

= vEe

vE =


The results of is identical to Plancks constant, h, thus

hvE =



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.


= hvme 221 (eqn 7.15) provides a technique for

the determination of Plancks constant, for the slopes of the lines in Fig.7.13 are all equal to h.


1) Photoejection cannot occur if h < because the photon brings insufficient energy;

2) (Eq.7.15) predicts that the kinetic energy of an ejected electron should increase linearly with frequency;

3) 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.

= hvme

2

21

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).

Example 1Example 1

Light with a wavelength of 300 nm is incident on a potassium surface for which the work function, , is 2.26 eV. Calculate the kinetic energy and speed of the ejected electrons.

SolutionSolution

We write and convert the units of from electron-volts to joules:

Electrons will only be ejected if the photon energy, hv, is greater than . The photon energy is calculated to be

which is sufficient to eject electrons.

( ) == /hchvEe

( )( ) JeVJeV 1919 1062.3/10602.126.2 ==

( )( ) Jhc 199834

1062.610300

10998.210626.6

=

=

Solution (contSolution (contd)d)

We can obtain .

Using , we calculate that

( )sm

Jm

Ev e /1010.8

10109.91099.222 5

31

19

=

==

22/1 mvEe

=

( ) JhcEe 191099.2/ ==

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 h, so the total # of photons needed to produce energy E is E/h. The frequency of the radiation ( = c/=?) and the total energy(?) emitted by the lamp E =?. E = Pt, P - the power (in watts), t - the time interval for which the lamp is turned on.

Answer: The number of photons is

Note: it would take ~40 min to produce 1 mol of these photons.

( ) ( ) ( )( ) ( ) 2010998.210626.6 0.11001060.5 108.2 )(

EN

1834

17

==

=

==

msJssJsm

hctP

chtP

h

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: [51014]

Particles Exhibit WaveParticles Exhibit Wave--Like BehaviorLike 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 pshould have (in some sense) a wavelength given by the

de Broglie relation:

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 waveparticle duality.

ph

=

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 wavefunctionof the particle). A particle with high momentum has a wavefunction with a short wavelength, and vice versa.

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: =h/p (7.16)The linear momentum of the electrons, p=?E=? The energy acquired by an electron accelerated through a potential difference is e , 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=p2/2meLet p2/2me = e p

Carry through the calculation algebraically before substituting the data.

Example 7.3 Estimating the de Broglie wavelength

Answer: p2/2me = e p = (2mee )1/2

de Broglie relation = h/p

1 V C = 1 J and 1 J = 1 kg m2 s2. 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).

( )

( ) ( ) ( ){ }m

VCkgsJ

em

h

e

12

2/141931

34

21

101.6100.410602.110109.92

10626.62

=

=

=

Self Test 7.2

Calculate: (neutron mass: mn=1.67510-27kgPlancks const.: h=6.62610-34Js

Boltzmanns const.: k=1.38110-23J/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 1034 m

Example 4Example 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

( )( )( ) eVm

hm

pE 17or 107.2100.310109.92

10626.622

181031

234

2

22

=

===

Diffraction by a Double Slit (Diffraction by a Double Slit ())

Diffraction of Light

Diffraction is a phenomenon that can occur with any waves, including sound waves, water waves, and electromagnetic (light) waves.

Diffraction by a Double Slit (contDiffraction by a Double Slit (contd)d)

For diffraction of light from a thin slit, b >> a.


Maxima and minima arise as a result of a path difference between the sources of the cylindrical waves and the screen.


The condition that the minima satisfy is

where = wavelength

.....,3,2,1,sin == na

n


For double-slit diffraction experiment,

One slit or the other blocked

both slits open

Physical Chemistry Fundamentals: Figure 7.15

Fig. 7.15 The DavissonGermer 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 AtomAtomic 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, of all lines in an atomic hydrogen spectrum is given by

( ) ( ) 1221

11,

11~ nn

nncmRcmv H >

=

v~/1 =


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).

Atomic spectraBohr frequency condition:

)14.7(E h=


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.


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.

The failures of classical physics - Atomic and molecular spectra

SUMMARY: Energy quantization SUMMARY: Energy quantization -- Key pointsKey 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, specificallytheir 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 () 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 ( TEM), the electron beam passes through the specimen and the image is collected on a screen.

In scanning electron microscopy ( 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 (contd) The wavelength of and the ability to focus the incident beam

electronsgovern 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=2.

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 cross-sections of a specimen and SEM observations must be made on dry samples. Neither technique can be used to study living 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 m long. (Brian Bowes)

Electron microscopy is very useful in studies of the internal structure of cells

Objectives - The SchrThe Schrdinger Equationdinger 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() Wave Equation

3. Waves Are Conveniently Represented as Complex Functions

4. Quantum Mechanical Waves and the Schrdinger Equation

Outline

1. Solving the Schrdinger 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? (contd)

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.

where ni = number of atoms or moleculesi = energy of atoms or molecules

gi = degeneracy () at energy level i (energy level vs. energy state)

[ ] kTeej

i

j

i jiegg

n

n /=

Example 5

Consider a system of 1000 particles that can only have two energies, , with . The difference in the energy between these two values is . Assume that g1=g2=1.

a. Graph the number of particles, n1 and n2, in states as a function of . Explain your

result.b. At what value of do 750 of the particles have

the energy ?

21 and 12 =12 >

21 and

/kT

/kT

1

Solution

We can write down the following two equations:

Solve these two equations for n2 and n1 to obtain

1000 and / 21/

12 =+= nnenn kT

kT

kT

kT

en

e

en

/1

/

/

2

11000

11000

+=

+=

Solution

Part (b) is solved graphically. The parameter n1 is shown as a function of on an expanded scale on the right side of the preceding graphs, which shows that n1=750 for .91.0/ =kT

/kT

Classical Waves and the Nondispersive Wave Equation

Transverse( ), Longitudinal( ), 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 ()

A transverse wave is a moving wave that consists of oscillations occurring perpendicular to the direction of energy transfer.

Transverse plane wave

Longitudinal wave () 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.

Longitudinal harmonic Sound waves:

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 is the angular frequency of the wave. The quantity x/c is the time that the wave takes to travel the distance x.

=

c

xty sint)y(x, 0

Plane pressure wave

Classical Waves and the Nondispersive Wave Equation

The wave amplitude is:

It is convenient to combine constants and variables to write the wave amplitude as

where k = 2/ (wave vector) = 2v (angular frequency)

( )

=

TtxAtx pi2sin,

( ) ( )wtkxAtx = sin,

pi2,

hkp = hh

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

( ) ( )2

2

22,1,

t

tx

vx

tx

=

where v = velocity at which the wave propagates

Example 6

The nondispersive wave equation in one dimension is given by

Show that the traveling wave is a solution of the nondispersive wave equation. How is the velocity of the wave related to k and ?

( ) ( )2

2

22

2,1,

t

tx

vx

tx

=

( ) ( ) += tkxAtx sin,

Solution:

We have

( ) ( )( ) ( )( ) ( )

kv

tkxAvtv

tkxA

tkxAkx

tkxAt

tx

vx

tx

/ gives results two theseEquating

sinsin

sinsin

,1,

2

2

22

2

22

2

2

2

22

2

=

+

=+

+=

+

=

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 b. Express the complex number in the form (a+ib)

2/33 piie

( ) sincos irrei +=

Solution

a. The magnitude of 4+4i is The phase is given by

Therefore, 4+4i can be written

421

cosor 2

124

4cos 1

pi ====

( )( )[ ] 244444 2/1 =+ ii

( )4/24 piie

Solution

b. Using the relation can be written

( ) iii 3032

3sin

23

cos3 ==

+

pipi

( ) 2/33 , sincosexp pi ii eiie +==

The dynamics of microscopic systems

Quantum mechanics acknowledges the waveparticle 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, (psi).

7.3 Quantum Mechanical Waves

and The Schrdinger equation

Key point: Schrdinger equation is a second-order differential() eqn used to calculate the wavefunction of a system

(1926) The time-independent Schrdinger equation for a particle of mass m moving in one dimension with energy E is

Study the stationary states of quantum mechanical systems.

)17.7()()()()(2 2

22

xExxVdx

xdm

=+ h

Quantum Mechanical Waves and the Schrdinger Equation

An analogous quantum mechanical form of time-dependent classical nondispersive wave equation is the time-dependent Schrdinger equation, given as

where V(x,t) = potential energy function This equation relates the temporal and spatial

derivatives of (x,t) and applied in systems where energy changes with time.

),(),(),(2

),(2

22

txtxVx

tx

mt

txi +

=

h

h

Quantum Mechanical Waves and the Schrdinger Equation

For stationary states of a quantum mechanical system, we have

Since , we can show that that wave functions whose energy is independent of time have the form of

),(),( txEt

txi =

h

tEiiEt exextx )/(/ )()(),( hh ==

Using the Schrdinger equation to develop the de Broglie relation

( )

( )

pipi

pi

hhkp

mpmkE

mEVEmk

waveharmonicxkx

VEmdxd

k

k

===

==

=

=

==

=

22

2/2/

22

)()/2cos(cos

2

222

2/1

2

2/1

2

22

2

h

h

hh

h

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 wavefunctionmust satisfy.

Born interpretation of the wavefunction (contd)

If the wavefunction of a particle has the value at some point x, then the probability of finding the particle between x and x + dx is proportional to ||2dx.

||2 is the probability density. The wavefunction itself is called the probability amplitude.

If the wavefunction of a particle has the value at some point r, then the probability of finding the particle in an infinitesimal volume d = dxdydz at that point is proportional to ||2d.


Fig. 7.18 The wavefunction is a probability amplitude in the sense that its square modulus (* or ||2) is a probability density. The probability of finding a particle in the region dx located at x is proportional to ||2dx. We represent the probability density by the density of shading in the superimposed band.


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 d = dxdydz at some location r is proportional to the product of d and the value of ||2 at that location.

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. 7.20 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 and depicted by the density of shading).

Example 7.3 Interpreting a wavefunction

The wavefunction of an electron in the lowest energystate 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 within it and write the probability, P, as proportional to the probability density (2) evaluated at the point of interest multiplied by the volume of interest, V. i.e., P 2V, with 2 exp(2r/a0).

Example 7.3 Interpreting a wavefunction (contd)

Answer In each case V = 1.0 pm3. (a) At the nucleus, r = 0,

(b) At a distance r = a0 in an arbitrary direction,

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.

( ) ( ) ( )330 0.10.10.1P pmpme =( ) ( ) ( )332 0.114.00.1P pmpme =

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 in the region can be ignored.

Normalization of wave-function

)19.7(1)18.7(1N

2/1

2

=

=

dxN

dx

N - normalization constant

dxdydzdwhere

cd

bdxdydz

sDimensionIn

adx

DimensionIn

=

=

=

=

)20.7(1)20.7(1

:3

)20.7(1:1

A normalized wavefunction satisfies:

Spherical polar coordinates for system with spherical symmetry

Fig. 7.21 The spherical polarcoordinates used for discussing systems with spherical symmetry.

x=rsincos, y=rsinsin, z=rcos

r radius [0,) - colatitude(), [0, pi] - azimuth(), [0,2pi]d=r2sindrdd

)20.7(1sin 0 0

2

0

2 ddddrr =

pi pi


Fig. 7.22 The surface of a sphere is covered by allowing to range from 0 to , and then sweeping that arc around a complete circle by allowing to range from 0 to 2.

The Eigenfunctions of a Quantum Mechanical OperatorAre 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 - , the 3rd -.

A useful integral:

where n! denotes a factorial: n! = n(n 1)(n 2)...1.

10

!+

= naxn

a

ndxex

Answer to Ex.7.4 Normalizing a wavefunction:

The integration required is the product of three factors:

and the normalized wavefunction is

Note: a0 is a length, the dimensions of are 1/length3/2 and those of 2 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) 2.2106, ~ 1 chance in about 500,000 inspections of

finding the electron in the test volume; (b) 2.9 107, ~ 1 chance in 3.4 million.

12241

sin

230

30

2

0

2

0

/2

0

22 0

===

=

NaaN

dddrerNd ar

pipi

pi pi pi 2/1

30

1

=

aN

pi

0/2/1

30

1 are

a

=

pi

Example

Normalize the function over the interval

Solution:Volume element in spherical coordinates is

, thus

re

pipi 20 ;0 ;0 r

ddrdr sin2

14

1sin

0

222

0

22

0

2

0

2

=

=

drerN

drerddN

r

r

pi

pipi

Solution

Using the standard integral ,

we obtainThe normalized wave function is

Note that the integration of any function involving r, even if it does not explicitly involve , requires integration over all three variables.

re

re

pi

1

( )integer positive a is ,0 /!0

1naandxex naxn >=

+

pipi

1 that so 1

2!24 3

2== NN

or

Test 7.4 Normalize the wavefunctiongiven in test 7.3. He+ g.s.~ exp(2r/a0).

Correct Answer: N = (8/a03)1/2 7.4 Born interpretation of wavefunction (b) Quantization must be continuous, have a continuous slope (the 2nd

derivative of must be well-defined), be single-valued, and be square-integrable - The wavefunction 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 Schrdinger 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.


Fig. 7.23

The wavefunctionmust 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 () of a free particle with a specific linear momentum corresponds to a uniform probability density ().

(b) The Schrodinger eqn is an eigen value eqn () in which the wavefunction is an eigenfunction () of the Hamiltonian operator().

(c) Observables () are represented by operators (); the value of an observable is an eigenvalue () of the corresponding operator constructed from the operators for position and linear momentum ().

(d) All operators that correspond to observables are hermitian; their eigenvalues are real and their eigenfunctions are mutually orthogonal (). Sets of functions that are normalized and mutually orthogonal are called orthonormal ().

(e) When the system is not described by an eigenfunction of an operator, it may be expressed as a superposition () of such eigenfunctions. The mean value () of a series of observations is given by the expectation value () of the corresponding operator.


QM principles - 7.5 The information in a wavefunction (contd)

Fig.7.24 (a) The square modulus of a wavefunctioncorresponding 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.

)22.7(2

,:

)21.7(2

22

2

22

m

kEBeAeSolutions

Edxd

m

ikxikx h

h

=+=

=

( ) ( ) ( )( )

(7.26) kxcosA4(2Acoskx)(2Acoskx):densityy probabilit

(7.25) 2Acoskx)eA(e :BA(7.24)AAeeAAeAe(7.23) Ae:0B

222

ikxikx

2ikxikxikxikx2

ikx

==

=+==

===

==

node

7.5 The information in a wavefunction

(b) Operators, eigenvalues, and eigenfunctions

)27.7()(2

:dimension)-one(in operator n Hamiltonia

)27.7( :form)operator (in eqn dinger oSchr

2

22

bxVdxd

mH

aEH

+=

=

h

&&

(7.28c) tion)(eigenfunc e)(eigenvalution)(eigenfunc(Operator))28.7( :equation Eigenvalue

(7.28a) function) (same factor)(constant (function)(Operator)

=

==

b

Solving the Schrdinger Equation: Operators, Observables, Eigenfunctions, and Eigenvalues

We would need to use operators, observables, eigenfunctions, and eigenvalues for quantum mechanical wave equation.

The time-independent Schrdinger equation is an eigenvalue equation for the total energy, E

where {} = total energy operator or It can be simplified as

( ) ( ) ( )xExxVxm

nnn =

+

2

22

2h

H)

nnnEH =

Example

Consider the operators . Is the function an eigenfunction of these operators? If so, what are the eigenvalues? Note that A, B, and k are real numbers.

( ) ikxikx BeAex +=22 / and / dxddxd

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:

In this case, the result is not multiplied by a constant, so is not an eigenfunction of the operator d/dx unless either A or B is zero.

( ) ( )ikxikxikxikxikxikx BeAeikikBikAedx

BeAed

==+

( )x( )x

Solution

This equation shows that is an eigenfunctionof the operator with the eigenvalue k2.22 / dxd

( )x

( ) ( ) ( )( )

( )xkBeAek

BikAeikdx

BeAed

ikxikx

ikxikxikxikx

22

222

2

=

+=

+=+

Example 7.5 Identifying an eigenfunction (p.268)

Show that eax is an eigenfunction of the operatord/dx ; find the corresponding eigenvalue. Show that exp(ax2) is not an eigenfunction of d/dx.

Method: Operate on the function with the operatorand check whether the result is a constant factor times the original function.

Answer:

Test 7.5: Is the function cos ax an eigenfunction of (a) d/dx, (b) d2/dx2?

Correct Answer: (a) No, (b) yes

===

===

axaxeedxd

aaeedxd

axax

axax

22

22

(c) The construction of operators (p.268)

[7.29] p , xx :operators ofioon Specificat

)observable of (value)observablean toingcorrespond(Operator (energy)operator)(Energy

:operators momentum andposition following thefrombuilt ,

operators,by drepresente are , s,Observable

x dxd

ih

==

=

=

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?

Method Operate on with the operator corresponding to linear momentum eqn 7.29:

Answer:

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 , where is its angular position. What is the angular momentum

of a particle described by the wavefunction e2i? Correct Answer:

m

kEBeAe ikxikx

2,

22h

=+=

px dx

dih

=

h

hhhhh

h

hhhhh

kp

kBekeikBidx

deidx

di

p

Aakp

kAekikeAidx

deA

idxd

ip

Ba

x

ikxikxikx

x

x

ikxikxikx

x

=

=====

=

+=

=====

=

)(B

:0)(

:0)(

dilz )/( h=

h2=zl

(c) The construction of operators (contd, p.270)

(7.32) 2

EH :operatorn Hamiltonia

(7.31) 22

1E

(7.30) 21V

2

22

k

2

22

k

2

Vdxd

mV

dxd

mdxd

idxd

im

kx

+=+=

=

=

=

h

hhh


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 energyof 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)


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.


Fig.7.27 The wavefunctionof 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:

{ }{ }

{ }

hermitian. is /dxdoperator that theConfirm 7.7

:partsby n integratio

:operator momentumlinear theofy hermiticit The J8.2

[7.33] :yHermiticit

22

ji

jii

ji

Test

dxdxdfgfgdx

dxdgf

dxpdxp

dxdxdx

dd

ixjx

ijj

ij

=

=

==

=

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 that is normalized and is an

eigenfunction of an hermitian operator with eigenvalue ,

hermiticity

* = confirms that is real.

{ }

===

===

dd

ddd

The Eigenfunctions of a Quantum Mechanical OperatorAre Orthogonal

Orthogonality is a concept of vector space. 3-D Cartesian coordinate space is defined by

In function space, the analogous expression that defines orthogonality () is

0=== zyzxyx

J7.4 Orthogonality of wavefunctions (p.272)

Two different functions i and j are orthogonal if the integral (over all space) of their product is zero:

Wavefunctions corresponding to different eigenvalues of an hermitianoperators are orthogonal.

Test 8.8 Confirm that the functions sin x and sin 3x are mutually orthogonal.

(7.34) jifor 0 :ityOrthogonal i = dj

2)b1,(a 02sinsin)cos()cos(sinsin2

sinsincoscos)cos(

)(aconstant )(2)sin(

)(2)sin(

sinsin

2

0

22

===

+=

=

++

+

=

pi

xdxx

bba

xbaba

xbabxdxax

m

=pi2

003sinsin xdxx

Example

Show graphically that sin x and cos 3x are orthogonal functions. Also show graphically that

( )( ) 1for 0sinsin ==

mndxnxmx

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 and

. By similar means, we could show that any two functions of the type sin mxand sin nx or cos mx and cos nx are orthogonal unless n=m. Are the functions cos mx and sin mx(m=n) orthogonal?

( )( ) 03cossin =

dxxmx( )( ) 0sinsin >

dxxx


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 functionand the value of the integralrepeats itself for all replications of the section between 0 and 2, so the integral from to is zero.

7.5 The information in a wavefunction(e) Superpositions and expectation values

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 eikx - definite momentum states.

The total wavefunction is a superposition of more than one wavefunction:

B)A with (eqn.7.22 cos2)A( ==+= kxAee ikxikx

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 k correspond to different momentum states. The functions k

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 k 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, , 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

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

= |c1|21 + |c2|22

= : eigenvalue with ofion eigenfunctan is If

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 , we need the normalized wavefunction (Ex. 7.4):

Test 7.9 Evaluate the root mean square distance, 1/2, of the electron from the nucleus in the hydrogen atom.

Correct Answer: 31/2a0 = 91.6 pm

[7.37] 52.9pma radiusBohr

)4.7.(E 1

0

/2/1

30

0

pi

d

xea

ar

=

=

=

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 operatorgiven 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 (d2/dx2 is large) and the wavefunction itself is large (* 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/2 . 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.


A wavefunctionas a superposition of eigenfunction

Fig. 7.29 The wavefunction for a particle at a well-defined location is a sharply spiked function that has zero amplitude everywhere except at the particle s position.


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 particles 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 m s1. Calculate the minimum uncertainty in its position.

Method Estimate p from mv, 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 s2). 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 one-dimensional region of length 2a0.

Correct Answer: 547 km s1

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 1 and 2 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)

The Heisenberg uncertainty principle in the most general form - For any two pairs of observables, 1 and 2, the uncertainties (to be precise, the root mean square deviations of their values from the mean) in simultaneous determinations are related by:

(7.42) == hh ii

xppx xx

Postulates of Quantum MechanicsPostulates of Quantum Mechanics

Postulate ()1:All information that can be obtained about the state of a mechanical (physical) system is contained in a wave function , 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)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 2Postulate 2

The wave function obeys the time-dependent Schrodinger equation

(1)

where is the Hamiltonian operator of the system.

tiH

= h

H

Postulate 2 (contPostulate 2 (contd)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:

(q, t) = (q)(t) (2)where q stands for all of the coordinates of the particles in the system and where the coordinate wave function 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 Postulate 3 Mathematical Operators

and Mechanical Variablesand 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 ValuesPostulate 4 Expectation Values

(a) If a mechanical variable A is measured without experimental error, the only possibleoutcomes 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

where is the operator corresponding to the variable A, and where =(q, t) is the wave function corresponding to the state of the system at the time of the measurement.

A

A

=

=

Ad

dAA

Postulate 5. Measurements andPostulate 5. Measurements and

the Determination of the State of a Systemthe 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 eigenvalueequal to aj .

A

Postulate 5 (contPostulate 5 (contd)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 for the system, which is a mathematical function found by solving the Schrdinger equation for the system. In one dimension:

The Born interpretation. If the wavefunction of a particle has the value at some point r, then the probability of finding the particle in an infinitesimal volume d = dxdydz at that point is proportional to ||2d.

Acceptable wavefunctions. An acceptable wavefunction must be continuous, have a continuous first derivative, be single-valued, and be square-integrable.

Observables. Observables, , are represented by operators, , built from position and momentum operators of the form

or, more generally, from operators that satisfy the commutation relation

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.

[ ] hipxx

=,

PChemCh7_20110222.pdf

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