Quantum Physics - Lecture Notes-ZQ One in One 19 Jan 2011

FE1002 Ph i IIFE1002 Physics II

An introduction toAn introduction to

Quantum Physics Q yAssociate Professor Zhang Qingg g

Microelectronics CentreS h l f EEESchool of EEE

Nanyang Technological Univeristy

FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 1

Prescribed Text Book

“Physics for Scientists and Engineers with Modern Physics” (6th Edition)

R A S & J W J tt JR. A. Serway & J. W. Jewett, Jr. (QC23.S492P 2004)

Main ReferencesMain References

“University Physics” by Hugh D.Young & Roger A Freedman (12th

Edition) (QC21.2.Y72U 2008)

“Fundamentals of Physics” (Extended) (7th Edition) by D. Halliday, R. Resnick & J. Walker

(QC21.2H188 2005)

“Physics: principles with applications” (5th Edition)


by Douglas C. C. Giancoli (QC23G433 2005)

Topic One: IntroductionpIsaac Newtonian (1642-1727) developed the basic concepts and laws of

17th

1600

the basic concepts and laws of mechanics and proposed the law of universal gravitation, etc.

cs

18th

1700

ical

phy

sic

19thJames Clark Maxwell (1831 1879)

Dynamic characteristics of an object: position, velocity, momentum, etc

1700

clas

si 19thJames Clark Maxwell (1831-1879) developed the electromagnetic theory of light and kinetic theory of gases, etc.

1800

20th

Characteristics of electromagnetic waves: wavelength, velocity, amplitude, phase angle, etc.


1900

20th

1900A major revolution in physics took place at the beginning of 20th century.

phys

ics

In 1900, Max Plank (1858-1974) introduced the concept of quantum

21st

mod

ern

p

2000

action.

In 1905 Albert Einstein (1879 1955)In 1905, Albert Einstein (1879-1955) published his special theory of relativity.


1.1 BackgroundTopic One: Introduction

g

A few experimental observations could not be i d i h f k f l i h i

Blackbody radiation

interpreted in the framework of classic physics.

Topic 2

Photoelectric effect Emission of sharp spectrum lines from atoms

Topic 3

Topic 5lines from atoms

New concepts and theories must be introduced.

p


1.2 Wave & particle behaviorTopic One: Introduction

In quantum physics, wave and particle are not contradictorycontradictory.

Electromagnetic wave does show particle characteristics, i.e.,

Electrons and other particles are of

particle characteristics, i.e., momentum etc.

Topic 4

wave properties, i.e., wave package and wave length etc. Topic 6

The Schrödinger equation with certain boundary conditions links the wave and particle properties

Topics 7 & 8


the wave and particle properties together.

& 8

1.3 Applications of quantum physics

Topic One: Introduction

1.3 Applications of quantum physics

Classically, a particle will never overcome aClassically, a particle will never overcome a potential barrier whose barrier height is larger than the total energy of the particle. What about in quantum physics?

Topic 9

about in quantum physics?

Why do we have the periodic table for different elements? Topic 10different elements?

What is the difference between laser light and ordinary light? How is laser light Topic 11and ordinary light? How is laser light generated?

Topic 11


Topic Two: Blackbody Radiationp yAll objects at temperature T > 0 K emit thermal radiation.

The thermal radiation consists of a continuous distribution of wavelength from the infrared, visible and ultraviolet portions of the spectrumvisible and ultraviolet portions of the spectrum.

nsity

Mono

Spec

tral

Inte

n

Mono-chromator

Detector

As the temperature of an object increases, the major emitting portion of the spectrum shifts to a shorter

SFE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 8

emitting portion of the spectrum shifts to a shorter wavelength.

Topic Two: Blackbody Radiation

2.1 Spectral radiancySpectral radiancy or spectral intensity R(,T) tells us the radiation energy emitted by per unit area of an

p y

object, per unit time, per unit wavelength at a given temperature T and wavelength . (unit: Js-1m-3)

Intensity R(,T)d gives the radiated intensity or power per unit area for the wavelength from to +d. (unit: Js-1m-2))

ensi

ty R (,T)

pect

ral i

nte

I(T) +d

R(,T)dFigure 2-1: Spectral intensity of an emitter radiation versus wavelength at a temperature.


Sp

( )

+d

I t it f λ t λ b d t i d b i t ti


Intensity from λ1 to λ2 can be determined by integrating in the spectral radiancy with respect to λ. (unit: Js-1m-2)

2

1 21

,I T R T d

(2.1a)

Radiant intensity or power density I(T) is the rediation energy emitted from per unit surface area of an object in

dTRTI

, (2.1)

gy p jper unit time for entire wavelength. (unit: Js-1m-2)

dTRTI 0 , ( )


2.2 Blackbody


2.2 Blackbody

A blackbody is defined as an object that absorbs all the i id t it

Good approximation of a blackbody is a hole leading to the inside of a hollow object.

energy incident on it.

A blackbody is also a perfect emitter since an object in thermal equilibrium emits as much energy as it absorbs.

Figure 2-2 : The opening to the cavity inside a hollow object is a good approximation of a blackbody.


blackbody.

The radiation of a blackbody depends only on the


The radiation of a blackbody depends only on the temperature of the blackbody and not on the material of which the blackbody is made.The radiation in the visible region is only a small portion of the entire

Fi 2 3 S t l i t it f th

small portion of the entire radiation.

inte

nsity

Figure 2-3: Spectral intensity of the blackbody radiation versus wavelength (spectral radiancy) at three temperatures. Note that the 3500 KSp

ectra

l i

pamount of radiation emitted (area under the curve) increases with increasing temperature.


visible region

2.3 The Stefan-Boltzmann lawTopic Two: Blackbody Radiation

The Stefan-Boltzmann Law: total radiated intensity or power per unit surface area of a blackbody isor power per unit surface area of a blackbody is only determined by the temperature T of the blackbody as

4TTI (2.2)

where =5.670 10-8 Wm-2K-4 is the Stefan-Boltzmann constantBoltzmann constant.



Ordinary objects radiate less efficiently than a blackbody and the corresponding radiant intensity becomesbecomes

4TTI (2.3)

where 0< <1 is the emmisivity of the surface of the ordinary objects.

For a blackbody, = 1.


2.4 Wien’s displacement law


2.4 Wien s displacement lawmax means the wavelength where the intensity or radiancy is maximum

• Wien’s Displacement Law: For a blackbody, the product of max and temperature T is a universal

radiancy is maximum.

p max pconstant, i.e,

Km 10898.2 3max T (2.4)

max

y

Figure 2-4: Wavelength max of the

ctra

l int

ensi

ty

peak of the curve shifts to shorter wavelengths at higher temperatures 3500 KSp

ecFE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 15

2.5 Two classical models


Rayleigh & Jeans Model: assuming that atoms are h i ill t th t it l t ti t

2.5 Two classical models

harmonic oscillators that emit electromagnetic waves at all wavelengths, Lord Rayleigh and James Jeans developed the following formula in 1900;

42,

TckTR BRJ (2.5)

where kB= 1.38×10-23 J/K is Boltzmann’s constant and cis the speed of light.

Eq.(2.5) fits the blackbody’s radiation curves well at long wavelengths (> 50 m), but completely fails at short wavelengths, because of RRJ(,T) as 0.


g , RJ( , )

Wien Model: Wilheim Wien also gave an expression


Tb

W eaTR

5, (2.6)

based on a conjecture (“guess”)

W 5, ( )

where a and b are empirical constants chosen to give the best fit to the experimental data.

Wein’s formula fits the curves well at short wavelengths but

best fit to the experimental data.

gdeparts noticeably at longer wavelengths.

Figure 2-5: Comparison of the Wein and the Rayleigh-Jeans theories to that of Planck, which closely follows the experimental data


the experimental data.

2.6 Planck’s modelTopic Two: Blackbody Radiation

In 1900, Max Planck, seeking to reconcile these two laws, made an inspired interpolation which fits the data at all a elengthsall wavelengths.

12, 5

2

hhcTR (2 7)

1, 5

Tkhc

Be (2.7)

h = ?

Tk

hc

B is dimensionless quantity. TkB

1

BB k Tk T mJh Jsc c ms



Plank adjusted the constant h so that his simulated curve could perfectly match the observed blackbody’s radiation curves. y

h is Planck’s constant: h = 6.626 10-34 J·s

h l f h i i d d f h bl kb d ’The value of h is independent of the blackbody’s materials and temperatures.


2 7 Planck’s two assumptions


Assumption 1: Oscillating atoms/molecules can have

2.7 Planck s two assumptions

only discrete values of energy En, given by

nhfEn (2.8). . . ,3,2,1,0n

where n is called a quantum number and f is the natural frequency of oscillation of the atoms/molecules.Radiation energy is quantized.



Each value of n represents a specific quantum state.

molecule energy

Figure 2-6: Allowed energy levels for a molecule that oscillates at its natural frequency f.


natural frequency f.

Assumption 2: The atoms/molecules emit or absorb


Assumption 2: The atoms/molecules emit or absorb energy in discrete packets (called photons) by jumping from one quantum state to another. D d t iti it

Energy of one photon is the product of h and f:Upward transition absorbs energy.Downward transition emits energy.

hchfE (2.9)

Energy of one photon is the product of h and f:

where c is the speed of light in vacuum.

Figure 2-7: A representation of photons. Each photon has a discrete energy hf


has a discrete energy hf.

Example 2-1*Topic Two: Blackbody Radiation

Find the peak wavelength of the blackbody radiation emitted by each of the following objects (as blackbodies):blackbodies):

(a) The human body when the skin temperature is 35oC(b) The tungsten filament of a light bulb operating at

2000 K(c) The Sun, which has a surface temperature of about

5800 K Solution: From Wien’s Displacement Law, Eq.(2.4), we can easily determine max as follows

For human body,

μm4.9Km10898.2 3

max


μm4.9K35273max

This radiation of 9.4 m is in the infrared region and


ginvisible to the human eye.

For tungsten filament,

μm4.12000K

mK10898.2 3

max

It is also in the infrared region so that most of emitted energy is not visible to us. For the sun,

50mK10898.2 3

μm5.05800Kmax

This is near the center of the visible spectrum.


p

2.8 Max Planck (1858 –1947)


Max Karl Ernst Ludwig Planck (April 23, 1858 October 4 1947) was a German

2.8 Max Planck (1858 1947)

1858 – October 4, 1947) was a German physicist. At first Planck considered that thequantisation was only as “a purely formal assumption ... actually I did not think much about it ”; nowadays this assumption incompatibleabout it… ; nowadays this assumption, incompatible with classical physics, is regarded as the birth of quantum physics and the greatest intellectual

li h t f Pl k'Planck was awarded the Nobel prize for physics in 1918.accomplishment of Planck's career.


Data Source: http://en.wikipedia.org/wiki/Max_Planck

Topic Three: The Photoelectric

The photoelectric effect: light incident on certain

Effectp g

metallic surfaces can cause electrons to be ejected from the surfaces. Th j d l ll d h lThe ejected electrons are called photoelectrons.

The induced current is called photoelectric current.

hf photoelectronsFigure 3-1: Photoelectrons are generated from the

metal

are generated from the surface of a metal under impact of an incident light beam.


Topic Three: The Photoelectric Effect

metallic target Electron collection electrode

li d i l photoelectric currentapplied potential difference

Figure 3-2: An apparatus used to study the photoelectric effect. When a beam of monochromatic light with appropriate wavelength shines on the metallic target T photoelectrons will be collected by electrode C and the


metallic target T, photoelectrons will be collected by electrode C and the photoelectric current i is detected by the ammeter.


3.1 Stopping potential For a large positive V, the current reaches a maximum value.

pp g p

The current increases as incident light intensity increases.

Saturated photo-electric current

Figure 3-3: Photoelectric current versus applied potential differences–


versus applied potential differences for two light intensities.

When a negative potential is applied to electrode C


When a negative potential is applied to electrode C, photoelectrons are repelled, leading to a decrease in photocurrent. Stopping potential, VS: Any negative bias V < -|VS|, no photocurrent is detectable.VS is independent of radiation intensity, but dependent ofVS is independent of radiation intensity, but dependent of the frequency of the incident light. The maximum kinetic energy of the photoelectrons is

max | |sK q V (3.1)

where q=1 610-19 C is the elementary charge

Classically Kmax should depend on intensity.

where q 1.610 C is the elementary charge.


max

3.2 Cutoff frequencyTopic Three: The Photoelectric Effect

|Vs| linearly increases with fCutoff frequency, f0: any frequency of the incident light b l f ill h t l t j tibelow f0 will cause no photoelectron ejection. f0 is characteristic of the target materials, independent of radiation intensity.

Classically, effect should occur at

f

sodium target

any frequency provided intensity is high enough.

Figure 3-4: The stopping potential Vstop as a function of the frequency f of the

g g


q y fincident light.


Kmax increases with increasing light frequency.

Cl i ll K h ld b i d d t f f

Electrons ejected from surface almost instantaneously, l li h i i i

Classically Kmax should be independent of frequency.

even at low light intensities.

Classically, electrons were expected to require some time to absorb radiationtime to absorb radiation.



Simulation & Illustration

Photoelectric Effect

Key points: This cartoon illustrates the influences of light intensity, frequency and bi l h l ibias voltage, etc on photoelectric current.


3.3 Einstein’s modelTopic Three: The Photoelectric Effect

In 1905, Einstein assumed that light of frequency f can be considered as a stream of photons with energy Ebe considered as a stream of photons with energy Egiven by E=hf [i,e, Eq.(2.9)].Photon is so localized that it gives all its energy to a single electron in the metal target.

The maximum kinetic energy of photoelectrons can be determined by the difference between photon energy and the work function (or the minimum energy with which an electron is bound to the metal), see Fig.3-5. ), g

hfKmax (3.2)


Einstein’s model explains the previously mentioned


Einstein s model explains the previously mentioned features which the classical theory cannot explain.(1) The effect is not observed below f0 .

(2) K is independent of the incident light intensity.

® The photon energy of incoming photon must be .

(2) Kmax is independent of the incident light intensity.® Increasing intensity only increases the number of

photons.(3) Kmax increases with increasing f. ® Kinetic energy of photoelectrons depends only on f.(4) Th h t l t j t d i t t l(4) The photoelectrons are ejected instantaneously.® Incident energy appears in small packets and there

is a one-to-one interaction between photons and


is a one to one interaction between photons andelectrons.


Electronic energy states in a metal.Case 1: hf1<, no photoelectrons are ejected.Case 2: hf2=, the critical case.Case 3: hf3>, photoelectronsCase 3: hf3 , photoelectrons with certain kinetic energy are generated. Kmax

E=0

The vacuum level

hf3

hfEF

level

Work function: the minimum energy

hf1hf2

The Fermi energy level: the highest

energy level

minimum energy required to move an electron from

the metal to


occupied by electrons at 0 K

vacuum. Figure 3-5: Electronic energy states in a metal.

I i l h l i i iExample 3-1*


In a particular photoelectric experiment, a stopping potential of 2.1 V is measured when ultraviolet light with a wavelength of 290 nm is incident on a metal. Using the g gsame setup and metal, determine the stopping potential if blue light with a wavelength of 440 nm is used, instead of the ultraviolet lightthe ultraviolet light.

Solution

34 8

1 19 9

6.66 10 3.0 10 2.1 4.3 2.1 2.2 (eV)1 6 10 290 10

S

S

hf qV

hf qV

34 8

2 19 9

1.6 10 290 106.66 10 3.0 10 2.2 2.8 2.2 0.6 (eV)1.6 10 440 10SqV hf


2So, 0.6 (V)SV

Example 3-2*Topic Three: The Photoelectric Effect

The human eye can respond to as little as 10-18 J of light energy. Determine the number of photons that will lead to an observable flash at a wavelength of 550 nmto an observable flash at a wavelength of 550 nm.

Solution:

18 9

34 8

10 550 10 36 6 10 3 10

E ENhf hc

6.6 10 3 10hf hc


3.4 Albert Einstein (1879 –1955)Topic Three: The Photoelectric Effect

Albert Einstein (March 14, 1879 – April 18, 1955) was a German-born theoretical physicist, one of the greatest physicists of all timephysicists of all time. He played a leading role in formulating the special and general theories of relativity, quantum theory and statistical mechanics. He was awarded the 1921 Nobel Prize for Physics for his e o ys cs o sexplanation of the photoelectric effect in 1905 (his "wonderful year" or "miraculous year") andyear or miraculous year ) and "for his services to Theoretical Physics".


Data Source: http://en.wikipedia.org/wiki/Albert_Einstein

3.5 X-ray photoelectron Topic Three: The Photoelectric Effect

X Ph t l t S t (XPS) i

spectroscopy*X-ray Photoelectron Spectroscopy (XPS) is a quantitative spectroscopic technique that measures the empirical formula, chemical state and electronic state of the elements that exist within a material.

XPS spectra are obtained by irradiating a testing material with a beam of x-rays while simultaneously measuring the kinetic energy and number of electrons that escape from the top 1~10 nm of the material beingthat escape from the top 1 10 nm of the material being analyzed.



Figure 3-6: Schematic of an X-ray photoelectron spectroscopy.

Figure 3-7: An XPS spectrum for a testing sample which contains Si O C F N Sn etc


Si, O, C, F, N, Sn, etc.

3.6 Photomultiplier tubes*Topic Three: The Photoelectric Effect

Photomultiplier tubes (photomultipliers or PMTs) are extremely sensitive detectors of light in the ultraviolet, visible and near infrared regionsvisible and near infrared regions.

0V

500V500V

1500V

2500V

1000V2000V

Figure 3-8: Schematic of the structure of a typical PMT

4500V3500V3000V

4000V

5000V


structure of a typical PMT and a photo of PMTs.


These detectors multiply the signal (number of electrons) produced by incident light by many times, from which a single photon could be resolvedfrom which a single photon could be resolved.

The combination of high gain, low noise, high frequency response and large area of collectionfrequency response and large area of collection makes PMTs suitable for applications in nuclear and particle physics, astronomy, medical imaging and motion picture film scanning etcand motion picture film scanning, etc.


Topic Four: The Compton EffectIn 1923, A. H. Compton did an experiment that confirmed the particle-like aspect of electromagnetic

p p

p p gradiation.The Compton effect: the scattering phenomenon associated with incident x ray colliding with electronsassociated with incident x-ray colliding with electrons.

Figure 4-1: Compton’s apparatus. A beam of x-rays of wavelength = 71 1 i di t d t71.1 pm is directed onto a carbon target. The rotating crystal plays a role of a spectrometer.


p

4.1 Scattering of x-raysTopic Four: The Compton Effect

g yThe scattered x-ray only y ypeak at 0 and ’ which is a function of

graphite

0o

90o

function of the angle.

Classically, the

45o

graphite

135o

ywavelength of the scattered x-ray is a broad distribution incident

x-ray 0

Figure 4-2: Scattering of x-ray as the function of

135obroad distribution at a given angle.


Figure 4-2: Scattering of x-ray as the function of the angle with respect to the incident direction.

Topic Four: The Compton Effect

4.2 Scattering of electronsThe electron is scattered through an angle with respect to the incident photon as if this was a billiard-ball t pe collisionball type collision.

Figure 4-3: Illustration of the scattered x-ray and electron in the Compton effect.

Classically, the electron would be pushed along the incident direction due to radiation pressure and set into oscillatory


direction due to radiation pressure and set into oscillatory motion driven by the wave’s oscillating electric field.

4.3 Compton’s modelTopic Four: The Compton Effect

To fully explain the effect, Compton and his co-workers had to treat photons, not as waves, but rather as point-

p

had to treat photons, not as waves, but rather as pointlike particles assuming that the energy and momentum of any colliding photon-electron pair are conserved, respectivelyrespectively.

The energy of photons E = hf [Eq.(2.9)] and the t

(4.1)hhfp

momentum

(4.1)c

p


Using these ass mptions and the la s of energ


Using these assumptions and the laws of energy conservation and momentum conservation in both x and y directions, he derived the Compton Shift Equation;

cos1' c0 (4.2)

where ’ and 0 are the scattered and incident light wavelength, respectively, as indicated in Fig.4-3. = h/m c=0 00243 nm is called the Comptonc= h/mec=0.00243 nm is called the Compton wavelength, me=9.10910-31 kg is the rest mass for electron and c is light speed in vacuum.



Compton’s measurements were in excellent agreement with his predictions and these really convinced physicists of the fundamental validity of

Photon transferred energy to the electrons during

convinced physicists of the fundamental validity of the quantum theory.

collision and hence the scattered x-ray has a longer wavelength.The unshifted peak (0) is due to scattering from theThe unshifted peak (0) is due to scattering from the atoms’ tightly bound inner electrons which do not gain significant energy from incident photons.


Example 4-1*Topic Four: The Compton Effect

pA 1.0×10-12 m wavelength photon collides with a free electron initially at rest. After the collision, the photon recoils directly backward. (i) Determine the wavelength, momentum and energy of the scattered photon and (ii) determine the momentum and kinetic energy of the de e e e o e u a d e c e e gy o escattered electron.Solution:

θ =180o

λ f λ′ f ′

Figure 4 4: Illustration of momentum of the scattered photon and

λ0, f0 λ′, f ′


Figure 4-4: Illustration of momentum of the scattered photon and electron in the Compton effect.


Solution: From Eq.(4.2), we have the wavelength of the scattered photon

0 c

3412 o

' 1 cos

6.626 10 J s1.0 10 1 cos180

31 8 1

12 12 12

1.0 10 1 cos1809.11 10 kg 3.00 10 m s

1.0 10 4.8 10 =5.8 10 (m)

The momentum and energy of the scattered photon:34

226.6 10 1 1 10 (k / )h

22 8 141 1 10 3 0 10 3 3 10 (J)hcE p c

2212

6.6 0 1.1 10 (kgm/s)5.8 10ph

hp


1.1 10 3.0 10 3.3 10 (J)ph phE p c


From momentum and energy conservation, we have:h h

0e

h h p

epheph EEEE 00

Therefore, the momentum and kinetic energy of the

and

3412 12

0

1 16.6 101.0 10 5.8 10e

h hp

, gyscattered electron are:

022

1.0 10 5.8 10

7.7 10 (kgm/s)

0 01 1

k h hE E E E E hc

0 00

34 8 1312 12

1 16.6 10 3.0 10 1.6 10 (J)1 0 10 5 8 10

k e e ph phE E E E E hc

12 121.0 10 5.8 10

Note: according to classic physics

22 227 7 10p


13classic 31

7.7 103.3 10 (J)

2 2 9.1 10e

ke

pE

m

Wrong!


4.4 A. H. Compton (1892-1962) p ( )Arthur Holly Compton (September 10th, 1892 - March 15th, 1962) discovered the Compton effect, which clearly illustrates the particle concept of electromagnetic radiation, wasof electromagnetic radiation, was afterwards substantiated by C. T. R. Wilson who, in his cloud chamber, could show the presence of the trackscould show the presence of the tracks of the recoil electrons. For this discovery, Compton was awarded the Nobel y pPrize in Physics for 1927 (sharing this with C. T. R. Wilson).D t S


Data Source:http://nobelprize.org/nobel_prizes/physics/laureates/1927/compton-bio.html

Topic Five: Atomic SpectraLow-pressure gas subject to electric discharge emits discrete line spectrum (emission spectra)

p p

discrete line spectrum (emission spectra)

Mono-chromator

Detector


Figure 5-1: Emission line spectra for hydrogen, mercury and neon.

Absorption spectrum is obtained by passing light from

Topic Five: Atomic Spectra

Absorption spectrum is obtained by passing light from a continuous source through the gas or dilute solution.Absorption spectra of low pressure gases show a series of discrete dark lines superimposed on a continuous spectrum of light source.

Mono-chromator

Detector White light

Figure 5-2: Absorption spectrum for hydrogen gas.



5.1 Hydrogen emission line series In 1884, Johann Balmer realized that the wavelengths of the first 4 lines in the visible spectrum of hydrogen

y g

p y gwere related by

543111

R ...5,4,3, 1

211

22H

n

nR

(5.1)

h R 1 097 107 1 i ll d h R dbwhere RH= 1.097107 m-1, is called the Rydberg constant.

1/λ is called wavenumber.


Other series of lines in hydrogen were soon found and


Other series of lines in hydrogen were soon found and Balmer’s equation was generalized.

111 . . . 2, 1, ,111

22121

22

H

nnn

nnR

(5.2)

B l ’ i f 2 ( i ibl )Balmer’s series for n2 = 2 (visible)Lyman’s series for n2 = 1 (UV)Paschen’s series for n2 = 3 (IR)

All of these equations were purely empirical. No

Brackett’s series for n2 = 4 (IR)

theoretical basis existed for them. Classic electromagnetic theory could not explain these.


these.


5.2 The Bohr atomIn 1913, Niels Bohr proposed an atomic theory that accounted for the spectra lines of hydrogen. The model contains: (1) The electron moves in circular orbits around the proton under the influence of the Coulomb force ofproton under the influence of the Coulomb force of attraction.

Figure 5-3: Diagram represent-g g ping Bohr’s model of a hydrogen atom, in which the orbiting electron is allowed to be only in specific orbits


in specific orbits.


(2) Only certain orbits are stable i e electrons in these(2) Only certain orbits are stable, i.e, electrons in these orbits do not emit or radiate energy.

In classical physics, orbiting electrons should In classical physics, orbiting electrons should continuously radiate, losing energy and causing it to spiral into the nucleus.

(3) Radiation is emitted when electrons jump from a more energetic initial orbit to a lower energy orbit. Frequency of the emitted radiation depends on theFrequency of the emitted radiation depends on the difference between the energy of the initial state, Ei, and the energy of the final state, Ef and Ei > Ef.

hfEE fi (5.3)


(4) The allowed orbits are those for which the electron’s bi l l b h l i


orbital angular momentum about the nucleus is an integral multiple of ħ = h/2.

nvrm (5 4)nvrme (5.4)

Using these four assumptions, the allowed energy levels f h d b l l dof a hydrogen atom can be calculated:

32112

nekE e (5 5)...3,2, 1, ,

2 20

nna

En (5.5)

where ke=1/(40)=8.988109 Nm2/C2 is the Coulomb’s constant and a0= h2/[meke(2e)2]= 0.0529 nm is the Bohr radius, which corresponds to the orbit with the smallest radius.


radius.

R di f ll d bit i


02anrn (5.6)

Radius of any allowed orbit is

Fi 5 4 Th fi hFigure 5-4: The first three circular orbits predicted by the Bohr model of the hydrogen atom.


y g


With Eq.(5.5), we have the energy of hydrogen:

eV 606.132n

En (5.7)...3, 2, 1,n

Only energies satisfying Eq. (5.7) are permitted.n = 1 is the ground state with E1 = -13.606 eV.g 1 Vn = 2 is the first excited state with E2 = -3.401 eV.n = is the free electron state with E = 0 eV.

The energy needed to completely remove an electron from the atom in its ground electronic state is called the ionization energythe ionization energy.The ionization energy for hydrogen is E-E1=13.6 eV


Frequency of the photon emitted when electron jumps


Frequency of the photon emitted when electron jumps from an outer orbit i to an inner orbit f can be determined from Eqs.(5.3) and (5.5):

22

0

2 112 if

efi

nnhaek

hEE

f (5.8) f

The wavenumber or 1/ is:

2 111 ekf

22

0

112

1if

e

nnhcaek

cf

(5.9)

RH

Comparing Eq.(5.9) with Eq.(5.2), the Rydberg constant is

1-72

H m10097.1 ekR e (5 10)


0H 2 hca (5.10)

The theoretical


The theoretical Rydberg constant is accurate to within 1% of the experimentally determined value.The spectra series forThe spectra series for hydrogen can be interpreted as

IR

transitions between the energy levels.

visible

Figure 5-5: An energy level di f h d


diagram for hydrogen

UV

5.3 Niels Bohr (1885-1962)


( )Niels (Henrik David) Bohr (October 7, 1885 – November 18, 1962) was a , )Danish chemist who made fundamental contributions to understanding atomic structure andunderstanding atomic structure and quantum mechanics. Bohr is widely considered one of the greatest h i i f h i hphysicists of the twentieth century

(even though he self-identified as a chemist).)He received the Nobel Prize for Physics for his atomic theory in 1922.


Data Source: http://en.wikipedia.org/wiki/Niels_Bohr

Topic Six: Wave-Particle DualityTopic Six: Wave-Particle DualityBlackbody experiments, photoelectric effect, Compton effect and atomic spectra offer ironclad evidence that when light and matter interact, they behave like particles.particles.

A photon has a speed of c, energy of hf=hc/ and momentum of hf/c=h/ and no mass.f

On the other hand, light and other electromagnetic waves exhibit interference and diffraction effects which could be interpreted only with wave characteristics.

Electromagnetic wave has its amplitude, frequency and


phase angle.

Q Which model (particle or wave) is more appropriate?

Topic Six: Wave-Particle Duality

A. Depends on the phenomenon observed – some experiments are better or solely explained using

Q. Which model (particle or wave) is more appropriate?

experiments are better or solely explained using particle model while others are better or solely explained using wave model.

Light has a dual nature. It exhibits both wave and particle characteristicsand particle characteristics.



To understand why photons are compatible with electro-magnetic waves, let us consider a radio wave and x-ray.

Consider a 2.5 MHz radio wave (=120 m).

Its energy E =hf 10–8 eV – too small to be detected.

Requires a lot of photons to produce a detectable signal

At higher frequencies energy of photon is higher and

q p p g– graininess is lost.

At higher frequencies, energy of photon is higher and graininess can be easily detected.For very high frequencies such as x-ray and -ray, the photons are easily detected as a single event, but wave effects are difficult to observe.



6 1 The Wave properties of particles6.1 The Wave properties of particlesIn 1923, L. V. de Broglie, in his doctoral dissertation, postulated that because photons have wave andpostulated that because photons have wave and particle characteristics, perhaps all forms of matter have wave as well as particle properties.

Electrons have particle-wave nature electrons in motion exhibit wave properties.motion exhibit wave properties.


From the momentum wavelength relationship of p=h/


From the momentum-wavelength relationship of p=h/for photons, de Broglie applied this relation to other particles. de Broglie wavelength: de Broglie suggested that material particles of momentum p are associated with their characteristic wavelength :

mvh

ph (6.1)

their characteristic wavelength :

The frequencies of the matter waves obey the Planck relationship

mvp

relationship.

hEf (6.2)


h


Example 6-1*If a proton and an electron have the same de Broglie wavelength, will they have the same kinetic energy? Account for your answer If not determine the ratio of theAccount for your answer. If not, determine the ratio of the proton’s kinetic energy to that of the electron.

S l i Th h diff ki i i22 1

hpE ek

Solution: They have different kinetic energies

22

mmk

31

2

101921

h

mE 427

31

2 105.5107.1101.9

21

2

p

e

e

p

electronk

protonk

mm

hm

mEE


2 em

6.2 Louis de Broglie (1892-1987)


6.2 Louis de Broglie (1892 1987)Louis-Victor-Pierre-Raymond, 7th duc de Broglie, generally knownduc de Broglie, generally known as Louis de Broglie (August 15, 1892–March 19, 1987), was a French ph sicistFrench physicist. He introduced his theory of electron waves. This included theelectron waves. This included the wave-particle duality theory of matter. For this he won the Nobel Prize in Physics in 1929Prize in Physics in 1929

Data Source: http://en.wikipedia.org/wiki/Louis, 7th duc de Broglie


Data Source: http://en.wikipedia.org/wiki/Louis,_7th_duc_de_Broglie

6.3 Double-slit experimentTopic Six: Wave-Particle Duality

pIf de Broglie’s hypothesis is right, we should observe the interference pattern caused by a beam of electrons incident to double slits.

The experimental requirements: the slit width<< the p qseparation between the two slits; the detector is far behind the slits.

The experimental procedures: (1) Close one slit and only leave the other open for a given time and observe the pattern. (2) Open the both slits at theobserve the pattern. (2) Open the both slits at the same time for the given time.



Figure 6-1: The two blue curves in the middle represent the patterns of individual slits with the upper or lower slit closed.



Figure 6-2: The single blue curve on the right represents the accumulated from the two blue curves. The brown curve represents the interference pattern with both slits open at the same time.


p



Particles through double slits

Key points: This simulation shows interference patterns for a beam of particles

i th h d bl lit Th i flpassing through double slits. The influences of slit width and particle mass on the patterns can be seen.


Example 6-2*Topic Six: Wave-Particle Duality

A particle of charge q and mass m has been accelerated from rest through a potential difference V. Show its d B li l th h/(2 V)1/2de Broglie wavelength =h/(2mqV)1/2.

Solution: when a charged particle is accelerated from rest through a potential difference V the gain in kineticthrough a potential difference V, the gain in kinetic energy mv2/2 must equal the loss in potential energy qV

p1 22

V

mpmvVq22

1 2

where the momentum p=mv=(2mq V)1/2

v=0

vwhere the momentum p=mv=(2mq V)1/2, so that

Vhh

2

Shown!


Vmqp 2


6.4 Determination of the wavelength

I 1927 C J D i d L H G d d i

6.4 Determination of the wavelength of electron

In 1927, C. J. Davisson and L. H. Germer succeeded in measuring the wavelength of electrons.

I th i tIn their apparatus, electrons are accelerated from a

hot filament

heated filament F by an adjustable potential difference V.

Figure 6-3: The apparatus used in the Davisson-Germer

difference V.


experiment.

The electron beam with kinetic energy of eV, is i id i k l l C


incident to a nickel crystal C.A detector at D at an angle reads the current I of electrons entering the detector for various potentialelectrons entering the detector for various potential difference V.

There is a strong diffracted beam at = 50 and V = 54

Figure 6-4: The results obtained by Davisson and Germer for five different accelerating voltages, shown as polar plots of current I as a function of the angle .


There is a strong diffracted beam at 50 and V 54 V but not otherwise.


Figure 6-5: A simplified representation of a nickelrepresentation of a nickel crystal.

Dsin

The crystal surface acts like a diffraction grating with i Dspacing D.

Because of the low energy of the electrons, they cannot penetrate very far into the crystal and diffraction takes


penetrate very far into the crystal and diffraction takes place in the plane of atoms on the surface.

This situation is very similar to light reflected from


This situation is very similar to light reflected from diffraction grating where the maxima must satisfy the following equation.

...3, 2,1, sin mDm (6.3)

For nickel crystal, D = 2.15 Å and m = 1, we have = D sin50o =1.65Å.

Since the kinetic energy of the electrons is gained from electric potential acceleration Eke=p2/2m=eV, we can have the classic momentum p=(2eVm)1/2 and the de Brogliethe classic momentum, p (2eVm) and the de Broglie wavelength is =h/p=h (2eVm)-1/2=1.6710-12 m=1.67 Å.



In 1928, G. P. Thomson of Scotland also observed electron diffraction patterns by passing electrons through

thi ld f ilvery thin gold foils. These results show conclusively the wave nature of electrons and confirmed de Broglie’s hypothesis.electrons and confirmed de Broglie s hypothesis.

Q Would the wavelengths of 10 eV photon, electron and t b th ?proton be the same?



6.5 Principle of complementarityp p yThe dual nature of matter and radiation is conceptually difficult to understand because the two models seem to contradict each other. Neils Bohr’s principle of complementarity states that th d ti l d l f ith tt

Q How do they complement each other?

the wave and particle models of either matter or radiation complement each other.

Q. How do they complement each other? Position of particle can be localized in both space and time but a wave cannot, being spread out in both oftime but a wave cannot, being spread out in both of these dimensions.


6.6 Uncertainty principleTopic Six: Wave-Particle Duality

Localizing a wave in space and considering wave at arbitrary instant t = 0.

0

2cosxby

2b 02b

0 20 0

2 4, 0, 0k k

Δx =

→0

Δk =0b

Figure 6-6: (a) A harmonic wave viewed at t = 0. (b) The distribution f b h l t f th lit d f th h i


of wave numbers, shown as a plot of the amplitude of the harmonic component as a function of its wave number.

The wave can be expresses as:


The wave can be expresses as:

0

2cos, xbtx t

(6.4) 0

where b is a constant.

This wave has a sharply defined wavelength and a corresponding sharply defined wave number k0= 2/0kor k=0.

If this wave is to represent a particle, the uncertainty xi th i i i fi it (i h l th i )in the x axis is infinite (i.e. anywhere along the x axis).


Add sine or cosine waves with properly chosen wave numbers amplitudes and phases see Fig 6 7(b) and


numbers, amplitudes and phases, see Fig.6-7(b), and sum up all the waves over, we have a wave packet only over a certain region x and zero everywhere else, Fig.6-7(a).

Δx ≠

2b

0 20 0

2 4, 0, 0k k

Δx ≠

b

Figure 6-7: (a) A wave packet of length x, viewed at t = 0. (b) The


relative amplitudes of the various harmonic components that combine to make up the packet. The central peak has a width k.



Length of waves x & width of wave number k

Key points: Through the Fourier transform, this illustration shows clearly that wave number (k=2/) distribution becomes broad when the length of the wave reduces.wave reduces.



Wave is now localized, but purity of the wave is sacrificed since the packet contains a spread of wave numbers (k) centered about k0. u be s ( k) ce te ed about k0.

Hence the sharper the wave packet (or smaller x), the broader the range of wave numbers (k) we must use to

In general, as x increases, k decreases. The f ll i l ti hi h ld

build up the wave packet and vice versa.

following relationship holds

1 xk (6.5)



Form de Broglie wavelength Eq.(6.1), we have

x ppk

222 (6.6)xp

hhk

(6.6)

A l i E (6 5) thApplying Eq.(6.5), then

12 xp

hxk x

(6.7)h

2hxp (6 8)

or

2xpx (6.8)


Taking into account that momentum is a vector and for id l i


non-ideal instruments,

4hxpx

4

4

h

hypy

x

(6.9)

4hzpz

These are the mathematical formulations of the Heisenberg Uncertainty Principle for position-momentum relation.


Though an individual measurement of momentum of a


Though an individual measurement of momentum of a particle can yield an arbitrarily precise value, that value can be anywhere in a range px about the “true” px.

Repeated measurements on identically prepared systems generate results clustered about px with statistical di ib i f id h

It is not possible to determine both the position and th t f ti l ith li it d i i

distribution of width px.

the momentum of a particle with unlimited precision.


Observing a wave at x = 0.


g

00

2cos,T

tbtx x (6.10)

where T0 is the period and the angular frequency isdefined as 0 (= 2/T0) and b’ is a constant.

If this wave has a sharp angular frequency of 0= 2f0, it must be a continuous harmonic for infinite time t, see Fi 6 8Fig. 6-8.



t

2b’

T0Δt =

→0

= 0b’

→0

0000

b

Figure 6-8: (a) A harmonic wave viewed at x = 0. (b) The distribution of angular frequencies, shown as a plot of the amplitude of the harmonic component as a function of its angular frequencies.


component as a function of its angular frequencies.

Similarly, add sine or cosine waves with properly


chosen angular frequency , amplitudes and phases and sum up all the waves over, we have an impulse only over a certain duration t and zero beyond it.over a certain duration t and zero beyond it.

If the duration t will decrease, the spread of angular frequencies will increase. Thus we have:

1 t (6.11)


As the energy of the particle is E = hf the uncertainty in


As the energy of the particle is E hf , the uncertainty in frequency is related to the uncertainty in energy.

Ef (6 12)

hf (6.12)

Substitute Eq.(6.12) into Eq.(6.11) with = 2f= 2E/h,2E/h,

2htE (6.13)

The following

is the Energy-Time Uncertainty Relationship

g

4htE (6.14)

It is not possible to determine both the energy and the time coordinate of the particle with unlimited precision

is the Energy-Time Uncertainty Relationship.


time coordinate of the particle with unlimited precision.

ll i i h i


All energy measurements carry in inherent uncertainty. The lowest state of an atom (ground state) has well defined energy because the atom normally exists gy yindefinitely in that state.

All other states at higher energies are less precise because the atom (sooner or later) will move spontaneously to a lower energy state.



Example 6-3*pThe speed of an electron is measured to be 5.00×103 m/s to an accuracy of 0.00300%. Find the minimum uncertainty in determining the position of this electron.

Solution: The momentum of the electron is

m/skg1056.41000.51011.9 27331 mvpx

The uncertainty in px is 0.00300% of this value, so px=0.0000300×4.56×10-27 kg·m/s. From Eq.(6.9),

38000038010626.6 34

h mm38.0m00038.010368.114.34

10626.64 31

xphx


6.7 Werner Heisenberg (1901 –1976)Topic Six: Wave-Particle Duality

Werner Karl Heisenberg (December 5, 1901 – February 1,

g ( )

1976) was a celebrated German physicist. He invented matrix mechanics theHe invented matrix mechanics, the first formalization of quantum mechanics in 1925. His

i i i l d l d iuncertainty principle, developed in 1927, states that the simultaneous determination of two pairedp

He received the Nobel Prize in physics in 1932quantities, has an unavoidable uncertainty.


Data Source: http://en.wikipedia.org/wiki/Heisenberg

Topic Seven: A Particle in a 1-DBox

7.1 Wave function and probabilityWave function is a mathematical expression (usually a complex quantity) to describe the waves

i i l

p y

representing particles.If we know the wave function (x, y, z, t) for every point in space and every instant of time we know all about the

Probability density is determined by | |2 (probability

in space and every instant of time, we know all about the behavior of the particle.

per unit volume, or area, or length), defined as the probability of experimentally finding the particle described by the wave function at x, y, z at time t.


y , y,

Probability that the particle will be somewhere must be

Topic Seven: A Particle in a 1-D Box

Probability that the particle will be somewhere must be equal to unity (100% chance of finding it) in the space.

12 dV (7 1)(normalization condition)1 dV (7.1)(normalization condition)

To normalize a wave function is to multiply it by a constant such that Eq (7 1) is satisfied

For a one-dimension system, where the particle must be located along the x axis and its wave function is only a

constant such that Eq.(7.1) is satisfied.

located along the x-axis and its wave function is only a function of x, we replace dV with dx and

2

12 dxx (7.2)



Figure 7-1: The probability of a particle being in the interval a≤x≤ b is the area under the curve from a to b.

Probability of finding the particle in the interval a xbis

dxPb

aab 2(7.3)


Expectation value of a particle’s position x (for 1 D


Expectation value of a particle s position x (for 1-D case) is determined by

dxxx

2 (7 4)dxxx (7.4)

The expectation value a particle’s position x is the p p paverage value of x. Once the wave function is known, it is possible to calculate the average position x or the expectation value of the particle

In general, the expectation value of a function f(x) is

expectation value of the particle.

dxxfxf

2 (7.5)


7.2 A particle in a 1-D boxTopic Seven: A Particle in a 1-D Box

Classically, if a particle is confined to moving along the x-axis and bouncing back and forth between two impenetrable walls with a speed of v, then its momentum (mv) and its kinetic energy are constants.

Figure 7-2: A particle of mass m and velocity v confined tom and velocity v confined to moving parallel to the x axis and bouncing between two impenetrable walls.


7.2.1 The wave function of the particle


Quantum mechanics’ approach to this problem is very different and requires that the appropriate wave function consistent with the conditions of the situation be foundconsistent with the conditions of the situation be found.If a string of length L is fixed at each end, the standing waves set up in the string must have nodes at two ends.

y(x)=0

y(x)=0

Figure 7-3: Standing waves set up in a stretched string of length L.

y(x)=0


g g

x=0 x=L

R hi d l h l th i i t l


Resonance achieved only when length is some integral multiple of half-wavelengths.

2orn LL n (7 6)n = 1 2 3 or 2

nnL n

n (7.6)n = 1, 2, 3, . . .

The standing waves are only the form of stable waves whose wavelengths are quantizedwhose wavelengths are quantized.

2 x For the standing waves, it can be shown that

2sin sinnn

xy x A k x A

(7.7)

where A is the amplitude. Substitute Eq.(7.6), we have

LxnAxy sin (7.8)


For a particle in a 1 D box of length L the de Broglie


For a particle in a 1-D box of length L, the de Broglie waves of the particle, in analogy with standing waves on a string, must form standing wave.

xnAx sin (7.9)n = 1 2 3

The allowed wave functions for the particle is

Allowed de Broglie wavelengths are those of standing

L

Ax sin ( )n = 1, 2, 3, . . .

waves n=2L/n.



( ) 0

(x)=0|(x)|2=0

(x)=0

(x)=0

|(x)|2=0

|(x)|2=0

|(x)|2=0

Figure 7-4: The first three allowed states for a particle confined to a one-dimensional box. (a) The wave functions for n = 1, 2, and 3. (b) The probability distributions for n = 1 2 and 3


probability distributions for n 1, 2, and 3.

7.2.2 The probability density of the particle


The probability density: 2

can be either positive or negative, but 2 is always p g ypositive.2 is zero at the boundaries, in other words, it is i ibl t fi d th ti l t th i timpossible to find the particle at these points. Where2 is the maximum and zero depends on n.

Th b bili d i di ib i i h iThe probability density distribution is shown in

Fig.7-4(b).


7.2.3 The energy of the particle


The kinetic energies of the allowed (stable) states are

2 222 Lh

gy p

2

222

2

222

hm

Lnhm

pmvEn (7.10)

3,.... 2, 1,n 8

22

n

mLh

E i ti d ith thEnergy is quantized with the lowest energy corresponding to n = 1 (zero-point energy).

Figure 7-5: Energy level diagram for a particle confined to a one-dimensional


particle confined to a one dimensional box of width L. E1=h2/8mL2.

Cl i ll th i i f th ti l


Classically, the minimum energy of the particle can be zero.

When a particle drops from E3 to E2, it emits a photon of energy hf = E3 – E2.It l b b h t d j f E t E ifIt can also absorb a photon and jumps from E1 to E2 if the incident photon has energy hf = E2 - E1.


Topic Eight: Schrödinger EquationTopic Eight: Schrödinger Equation

8 1 1 D tiGeneral form of the wave equation for waves traveling along the x axis

8.1 1-D wave equation

along the x axis

t

txvx

tx

,1 2

22

2(8.1)

tvx where and v are the amplitude and propagation speed of the wave. In quantum mechanics, particles’ behaviors can described using de Broglie waves, which should satisfy Eq.(8.1).


8.2 A system with constant energy Topic Eight: Schrödinger Equation

We consider an isolated bound system and the paticles’ interaction inside it is ignorable, the total energy E = hfg gy fof the particle will remain constant.The wave function of the de Broglie waves is

l t t di t i b

txΨtx cos (8.2)

analogous to standing waves on a string can be expressed as

txtx cos ( )

Substituting into Eq.(8.1),

tΨxΨdt

22

coscos

xΨd

txΨvxd

t

22

2 cos cos

(8 3)


xΨvxd

xΨd

2

(8.3)or

Recall that = 2f = 2 v/ and p = h/, then

Topic Eight: Schrödinger Equation

2

22

2

22

2

2 42

pphv

(8.4)

E th t t l f ki ti (K) d

where ħ=h/2π is the reduced Planck constant.

Express the total energy as sum of kinetic (K) and potential energy (U).

2pE K U U

2

2 2

pE K U Um

p m E U

(8.5)

(8.6) 2

2

2

2 2

2v

p m E U

and


v

Substituting Eq (8.6) into Eq (8.3), we have


(8.7) xEΨxΨxUxd

xΨdm

2

22

2

kinetic energy

potential total energy

This is 1-D time-independent Schrödinger Equation.For 3-D time-independent Schrödinger Equation;

energy energy

z

zyxΨy

zyxΨx

zyxΨ ,,,,,,2

2

2

2

2

2

zyxΨzyxUEmy

,,,,22

(8.8)

In principle, if U(x,y,z) and two appropriate boundary conditions are known, we can solve the Schrödinger equations to obtain the wave functions and energies


equations to obtain the wave functions and energies for the allowed states.

Example 8-1*Topic Eight: Schrödinger Equation

A free electron has a wave function ψ(x)=Aexp(ikx), where x is the position and k is a constant. (i) Show that its non relativistic kinetic energy may be expressed asits non-relativistic kinetic energy may be expressed as

mkE

2

2

em2

where ħ is the reduced Planck constant and me is the rest mass of electron.Solution:For free electron, U=0. The time-independent Schrödinger equation:equation:

22

22d Ψ x

EΨ xm d x


2m d x

d ik Since


22

2

ikdx

d kdx

Since

dx

Substituting the above into the time time-independent Schrödinger equation, we have

2

2

2k x E x

m

g q ,

2mSo

2kh

2 e

Em

shown


8.3 Erwin Schrödinger (1887-1961)


8.3 Erwin Schrödinger (1887 1961)Erwin Rudolf Josef Alexander Schrödinger (August 12, 1887 –Schrödinger (August 12, 1887 January 4, 1961), an Austrian physicist, achieved fame for his contributions to quantumcontributions to quantum mechanics, especially the Schrödinger equation.He received the Nobel Prize in 1933 for his great contributions.

Data Source: http://en wikipedia org/wiki/Schrodinger


Data Source: http://en.wikipedia.org/wiki/Schrodinger

8.4 Determination of


8.4 Determination of The solutions to the Schrödinger Equation should satisfy the following conditions:satisfy the following conditions: Since it is second order differentiation equation, one must need two boundary conditions to determine the wave function. At the boundary of two contiguous regions, the wave function must join smoothly at the boundary betweenfunction must join smoothly at the boundary between regions. For 1-D at the boundary x=x0,

xxΨxxΨ (8 9) xxdΨxxdΨ

xxΨxxΨ

00

00 (8.9)

(8.10)


dxdx

(x) must obey normalization condition :


(x) must obey normalization condition :

12

dxxΨ (8.11)

0xΨ (8.12)

(x) must be single-valued and must also be

Note : Steps leading to Eq (8.7) do not represent a

(x) must be single-valued and must also be continuous for finite values of U(x).

derivation of the Schrödinger Equation. Rather, the procedure represents a plausibility argument based upon an analogy with other wave phenomena that are alreadyan analogy with other wave phenomena that are already familiar to us.


8.5 An example-a particle in a 1-D


8.5 An example a particle in a 1 D box revisited

Figure 8-1: Diagram of a one-dimensional box of width L and

infinitely high walls.

The1-D impenetrable box is a 1-D

infinite infinitely high walls.

Potential energy distribution:U(x) = for x 0 and x L

potential well

U(x) for x 0 and x LU(x) = 0 for 0 < x <L (8.13)

The particle does not feel any external forces exerted to it.

Thus, the particle can never escape from the box and

(x) = 0 for x 0 and x L (8.14)


(x) 0 for x 0 and x L (8.14)

I id th b 0 < < L th S h ödi E ti


Inside the box 0 < x < L, the Schrödinger Equation Eq.(8.7) becomes

mEΨd 22

(8.15)

mE

ΨkΨmEdxΨd

2

2 222

(8 16)h

mEk 2 (8.16)where

The general solution to Eq.(8.15) isg q ( )

kxBkxAx cossin (8.17)

h A d B t i t ti t t t bwhere A and B are two integration constants to be determined as follows


Using the boundary condition, Eq.(8.14),(0)=0, we


Applying the other boundary condition, Eq.(8.14),

Using the boundary condition, Eq.(8.14),(0) 0, we have B=0

0sin kLA (8.18)

(L)=0, we have

We have to require sin(kL)=0 This requirement leads to

We can not require A=0. Otherwise, =0 everywhere.

We have to require sin(kL) 0. This requirement leads to

nkL n=1, 2,3…. (8.19)


Substituting k in Eq (8 19) into Eq (8 17) we have the


Substituting k in Eq.(8.19) into Eq.(8.17), we have the final wave functions, see Fig.8-2:

xnAxn sin n=1, 2, 3…. (8.20)

Ln , , ( )

Applying the normalization condition Eq.(8.11): 2

1sin0

2

dx

LxnA

L

1

(8.21)2

1

2

LA

Th b bili d i Fi 8 2(b) iThe probability density, see Fig.8-2(b), is

Lxn

Lxn

22 sin2 n=1, 2,3…. (8.22)


LL

With Eq.(8.16) and Eq.(8.19), we have obtained the


(8 23)nLmEkL 2 n = 1 2 3

energy of the particle, see Fig.8-3:

(8 24)22hE

(8.23)nLkL

n = 1, 2, 3, . . .

(8.24)228

nmL

En

These results agree with those obtained in the previousThese results agree with those obtained in the previous section.



(x)=0|(x)|2=0

(x)=0

(x)=0|(x)|2=0

Fi 8 2 Th fi t th ll d t ti t t

|(x)|2=0

Figure 8-2: The first three allowed stationary states for a particle confined to a one-dimensional box. (a) The wave functions for n = 1, 2, and 3. (b) the probability densities 2 for n = 1, 2, and 3. p y , ,

Figure 8-3: Energy level diagram for a particle confined to a 1-D box of width L. The lowest


allowed energy is E = h2/8mL.


pA 1.00 mg object is confined to moving between two rigid walls separated by 1.00 cm. Calculate the

i i d f h bjminimum speed of the object.Solution: The minimum speed corresponds to the state for which n=1 From Eq (8 24)for which n 1. From Eq.(8.24),

J1049.501.01018

1063.68

5826

234

2

2

1

mLhE

01.010188mLThis energy corresponds to kinetic energy of the object,

/103131049.52/2 262/158

E m/s1031.3

101/2 26

61

mEv

The speed is so small that it can be considered to be at


rest, which is what one would expect for the minimum speed of a macroscopic object!


A particle confined to a 1-D infinite potential well from x=0 to x=+L has a wave function given by

2 sin n xL L

determine the probability that it will be found between(1) x=0 and x=L/4, (2) x=L/4 and x=L/2, (3) x=0 and x=L/2

22 1 2sin 1 cosn x n xdx dxL L L L

Solution:Since

1 2sin2

L L L L

x n xL L


2L n L

Yielding:


2 / 4/ 4

0 / 4 00

2 1 2sin sin2

LL

Ln x x n xP dx

L L L n L 0

1 1 sin4 2 2

nn

2 / 2/ 2

/ 4 / 2 / 4

2 1 2sin sin2

LL

L L L

n x x n xP dxL L L n L / 42

1 1 sin4 2 2

LL L L n L

nn

21

2/4/4/02/0 LLLL PPP


2


8.6 Characteristics of the particle in the box

Classically the particle would be expected to stay stillClassically, the particle would be expected to stay still with zero kinetic energy at very low temperatures. In quantum mechanics, the particle cannot be at rest in the box.The stable wave functions (standing waves) are caused by multiple reflections of the particle’s de y p pBroglie wave at the two walls of the box. Zero-point energy is the lowest energy (ground state)

d 1 dcorresponds to n = 1 and

E h1

2

28 (8.25)


mL1 28

Depending on the quantum state n, the electron spends


p g q , pmore time at certain position of the box than in others.In the ground state, the electron is more likely to be f d th t th it ll Thi fi difound near the center than near its walls. This finding contradicts classical theory. As n increases, the distribution of electron probability , p ydensity becomes more uniform and approaches classical theory.



8.7 Finite potential wellV0= p0.5 n=3

0.3

0.4

V0=0.3 eV

0.2

E/e

V

n=2

0.1

n=1

5-50.0

x/nm x/nmFi 8 4 T i l ll f ( ) i fi i d h d (b) fi i d h (0 3


Figure 8-4: Two potential wells of (a) infinite depth and (b) finite depth (0.3 eV), respectively, are compared. The wells have the same width of 10 nm.

The following characteristics can be found when we deal


gwith the more realistic case of a finite quantum well:1) Note that0 at the two walls of the well. It

should satisfy boundary conditions Eqs. (8.9) and (8.10). As a result, there is a spilling over of the exponential tail of the probability curve.p p y

2) The energy levels En are lower than the corresponding ones in the infinite deep quantum

ll ith th ll idth3) If the particle energy is higher than the barrier height

V0, it will not be trapped. If lower, it may be trapped

well with the same well width.

0, pp , y ppwith a probability of escaping from the trap. It is also possible to find the particle beyond the well.




Quantum well of finite height

Key points: Influences of the well height, width and particle energy on the wave p gyfunctions, energy of the states, etc are illustrated.


Topic Nine: Tunneling p g9.1 Tunneling through a rectangular

b ibarrier Consider a particle of energy E incident on a rectangular barrier of finite height U and width L where E<Ubarrier of finite height U and width L, where E<U.

Classically, the particle must be reflected as it does not havereflected as it does not have sufficient energy to overcome the barrier. U

EIn quantum mechanics, the amplitude of the de Broglie wave associated with the particle is L

E


pnonzero everywhere.

Tunneling or barrier penetration: the possibility of

Topic Nine: Tunneling

finding the particle on the far right side of the barrier.

Figure 9-1: for a particle incident from the left on the b i f h i ht U ibarrier of height U is sinusoidal in regions I and III, but exponentially decaying in region II. y g g

The probability of tunneling can be described with a transmission coefficient T and a reflection coefficient R.

1RT


1 RT (9.1)

9.2 Transmission coefficient of Topic Nine: Tunneling

tunnelingT i i ffi i t f t li T i thTransmission coefficient of tunneling T is the probability that the particle penetrates the energy barrier.

An approximate expression for T (when T <<1) is

EUmL 24

hEUmL

T24

exp

(9.2)

h h i Pl k’ t t d E th dwhere h is Planck’s constant, m and E are the mass and energy of the incident particle, L and U are the width and barrier potential height, respectively.




Quantum Tunneling

Key points: We pay our attention to the waveforms of incident, penetrated and reflected de Broglie waves and the influences of U, L and E on them.


9.3 Low contact resistance th h t li *


through tunneling*For a metal in contact with highly doped semicon-ductor ND1019 cm-3, the barrier width becomes very narrow, the tunneling current becomes dominant and contact resistance Rand contact resistance RCdecreases rapidly with increased doping.

Figure 9-2: Calculated and measured values of specific contact resistance. Upper insert shows the tunneling

L i h h i i


process. Lower insert shows thermionic emission over the low barrier.

9.4 Scanning tunneling microscope* Topic Nine: Tunneling

g g pAn electrically conducting probe with a very sharp tip p obe w t a ve y s a p t pis brought near the surface to be studied.

The empty space between the tip and the surface represents the “barrier”.

Figure 9-3: Schematic view of a scanning tunneling microscope (STM). The tip is mounted on a piezoelectric xyz p p yscanner. A scan of the tip over the sample can reveal contours of the surface down to the atomic level. An STM image is

d f i f di l d


composed of a series of scans displaced laterally from one another.

The tip and the surface are two walls of the “potential


well”.The STM allows highly detailed images of surfaces with resolutions comparable to the size of a single atom.gAn STM image of the surface of graphite is shown in Fig 9-4shown in Fig.9 4.A conductive surface is needed.

Figure 9-4: The surface of graphite as “viewed” with an STM. This type of image is capable of a lateral resolution of about 0.2 nm and a vertical resolution of 0 001 nm


vertical resolution of 0.001 nm.

Example 9-1*Topic Nine: Tunneling

A 30-eV electron is incident on a barrier whose cross-session is a rectangle of height 40 eV. What is the probability that the electron will tunnel through the p ob b y e e ec o w u e oug ebarrier if its thickness is 1.0 nm? and 0.1 nm?Solution: In this situation, U-E=40-30=10 eV=1.6×10-18 J. F E (9 2)From Eq.(9.2),

24exp

hEUmL

T

1831 106.11011.9214.34exp L

h

34106.6exp

When L=1 nm T8 510-15 If L=0 1 nm T0 039


When L 1 nm, T8.510 . If L 0.1 nm, T0.039.

Topic Ten: Quantum States ofTopic Ten: Quantum States of an Atom

In a hydrogen atom a single electron is bound to a

10.1 Hydrogen atom revisitedIn a hydrogen atom, a single electron is bound to a single proton by the attractive Coulomb’s force.Atomic system can be viewed as an electron trapper in

Electron can exist only in a discrete set of quantum t t h ith t i

y ppwhich an electron is confined to a region of space.

states each with a certain energy.


Potential energy of the hydrogen atom is

Topic Ten: Quantum States of an Atom

rekrU e

2

(10.1)

h k 8 99 109 N 2/C2 i h C l bwhere ke=8.99 109 N·m2/C2 is the Coulomb constant and r is the radial distance between the proton and electron.Substitute U(r) into a 3-D Schrödinger equation and find the appropriate solution.Solving the equation and the energies of the allowed states for the hydrogen atom are,

2 eV 6.1312 22

0

2

nnaekrE e

(10.2)n = 1, 2, 3, . . .


ll d d d h b


Allowed energy states depend on the quantum number n.

This result is exactly the same as that obtained in the B h thBohr theory. For circular electron orbit, one quantum number n is sufficient to characterize a stationary statesufficient to characterize a stationary state. n is called the principal quantum number, an integer, ranging from 1 to .


10.2 Quantum numbers of atomic Topic Ten: Quantum States of an Atom

For a more general electronic orbit, three quantum system

numbers are needed to define it.They correspond to three independent degrees of freedom.

Quantum Number

Name Allowed Values Number of Allowed States

n Principal quantum 1, 2, 3, . . . Any positive number integer

l Orbital quantum number

0, 1, 2, . . . , n - 1 n

mlOrbital magnetic quantum number

- l, - l + 1, . . . , 0, . . . , l - 1, l

2l + 1


Any orbits that violate the above rules cannot exist.

For historical reasons all orbits having the same


For historical reasons, all orbits having the same principal quantum number are said to form a shell. Those orbits having the same value of n and l are said to

n Shell Symbol l Subshell symbol

gform a subshell.

y y

1 K 0 s

2 L 1 p

3 M 2 d

4 N 3 f

5 O 4 g5 O 4 g

6 P 5 h

. . . . . .


10 3 Spin magnetic quantum number


10.3 Spin magnetic quantum number The spectra of certain gases, such as sodium vapor, show two very closely spaced lines called a doubletshow two very closely spaced lines called a doublet.To explain this, Goudsmidt and Uhlenbeck proposed a new quantum number mS, called the p p q S,spin magnetic quantum number. Convenient (but incorrect) to think of mS as describing

l t i i it i it bit than electron spinning on its axis as it orbits the nucleus.


There are only two values mS = ½, corresponding to


two directions, i,e, the electron can “spin up” (mS = ½) or “spin down” (mS = -½).

I t l tiIn external magnetic field, the energy of the electrons differs slightly for the two spin directions. This accounts for the doublet.for the doublet. Electron is a point particle, without

Figure 10-1: The spin of an electron can be either (a) up or (b) down relative to an

spatial extent.


( ) p ( )external magnetic field.

O h bi l i h “ i ” ( ½)On each orbit, an electron can either “spin up” (mS = ½) or “spin down” (mS = -½).

A f b d fi d b fA quantum state of an atom must be defined by a set of four quantum states, i.e., n, l, ml and ms.

Q H ibl ldQ. How many possible quantum states could an atom have?

A Determined by a combination of all possible values ofA. Determined by a combination of all possible values of n, l, ml and ms.



Example 10-1*For a hydrogen atom, determine the quantum numbers associated with the possible states that correspond to h i i l bthe principal quantum number n=2.

n l ml ms subshell shell Number of states in subshell

1/2112

-1/2002

1/20022s L 2

1/2012

-1/2112

1/2112

1/2-112

-1/2012

1/2012 2p L 6


-1/2-112

10.4 The exclusion principleTopic Ten: Quantum States of an Atom

p p

It turns out that n, l, ml, and ms can be used to describe all the electronic states of an atom regardless of theall the electronic states of an atom regardless of the number of electrons in its structure.

Q. How many electrons can have a particular set of quantum numbers?

orHow many electrons can a particular set of quantum numbers accommodate? qua tu u be s acco odate?


Pauli, in 1925, answered this question in the


Pauli, in 1925, answered this question in the following statement:

No two electrons in the same atom can ever be inNo two electrons in the same atom can ever be in the same quantum state. Therefore no two electrons can have the same set of quantum numbersof quantum numbers. If this principle were not valid, an atom could radiate energy until every electron in the atom is in the lowestenergy until every electron in the atom is in the lowest possible energy state and the chemical behavior of the elements would be grossly modified. Nature as we know it would not existknow it would not exist.


10.5 Wolfgang Pauli (1900-1958)Topic Ten: Quantum States of an Atom

g g ( )Wolfgang Ernst Pauli (April 25, 1900 – December 15, 1958) was

i h i i d fan Austrian physicist noted for his work on the theory of spin, and in particular the discovery of p ythe exclusion principle, which underpins the whole of chemistrychemistry.In 1945, he received the Nobel Prize in Physics for his “decisive contribution through his discovery in 1925 of g ya new law of Nature, the exclusion principle or Pauli principle.” He had been nominated for the prize by Einstein


Einstein.Data Source: http://en.wikipedia.org/wiki/Wolfgang_Pauli

Q. How are the quantum states for an atom filled with l ?


electrons?Electrons intend to fill up the state (subshell) with the lowest energy state firstlowest energy state first.

To a first approximation, energy depends only on the principal quantum number n. It increases with

Once a subshell is filled, the next electron goes to the t l t t b h ll

p p qincreasing n.

next lowest energy vacant subshell.If an atom is not at its ground state (or the lowest energy state available), it will radiate energy until itenergy state available), it will radiate energy until it reaches this state.


10.6 The periodic tableTopic Ten: Quantum States of an Atom

pAn orbit for an electron in an atom refers to the quantum state of an electron characterized by the quantum

The exclusion principle suggests that there can be only t l t i bit l ith +½ () d

y qnumbers n, l and ml.

two electrons in any orbital with ms = +½ () and ms=-½ ().

Allowed Quantum States for an Atom having n = 3

210100l

3M

2L

1K

n

Allowed Quantum States for an Atom having n 3.

-2-1012-1010-10100ml

2d

1p

0s

1p

0s

0s

l


ms

2 8 18 Total=28

n = 1 K shell can only accommodate 2 electrons


n = 1 K shell can only accommodate 2 electrons. n = 2 L shell has 2 subshells and 4 orbitals and is capable of accepting a total of 8 electrons. p gn = 3 M shell has 3 subshells and 9 orbitals and is capable of accepting a total of 18 electrons.

In general, each shell can have up to 2n2 electrons.


Hydrogen has 1 electron with the ground state


Hydrogen has 1 electron with the ground state described by 2 sets of quantum numbers : 1, 0, 0, ½ or 1, 0, 0, -½. The electronic configuration is written as 1 11s1.

Electronic configuration: 1s1

numbers of electronson the subshell

n l2p2s1s

or

Helium has 2 electrons with ground state quantum numbers of 1, 0, 0, ½ and 1, 0, 0, - ½. The electronic configuration is written as 1s2 - K shell filled.

2p2s1s


Lithium has 3 electrons with 2 in the 1s subshell and


1 in the 2s subshell (2s subshell has slightly lower energy than 2p subshell). The electronic configuration is written as 1s22s1is written as 1s 2s .

2p2s1s

or

Beryllium has electronic configuration of 1s22s2.

2p2s1s


Boron has electronic configuration of 1s22s22p1. The


2p electron may be described by 6 sets of quantum numbers corresponding to 6 states of equal energy.

221 2p2s1s

b h l d h l h hCarbon has 6 electrons and the 2p electrons are such that they occupy different orbits with unpaired spins ().

2p2s1s

2p2s1s

or


H d’ R l Wh t h bit f l


Hund’s Rule: When an atom has orbits of equal energy, the order in which they are filled by electrons is such that a maximum number of electrons have unpaired spins. Exceptions to the rule occur in an atom having subshells close to being filled or half filledclose to being filled or half-filled.


The Exclusion Principle can be illustrated as follows :


3

Atomic number

K L

3

4

5

6

Figure 10-2: The filling of electronic states

7

8must obey both the exclusion principle and the H d’ l

9

10


Hund’s rule. 10

Atomic number Z

Symbol Ground state configuration

1 H (Hydrogen) 1s1

K1 H (Hydrogen) 1s1

2 He(Helium) [inert] 1s2

3 Li (Lithium) 1s2 2s1

4 Be (Beryllium) 1s2 2s2

L

4 Be (Beryllium) 1s 2s

5 B (Boron) 1s2 2s22p1

6 C (Carbon) 1s2 2s22p2

7 N (Nitrogen) 1s2 2s22p3

8 O (Oxygen) 1s2 2s22p4

9 F (Fluorine) 1s2 2s22p5

10 Ne (Neon) [inert] 1s2 2s22p6 M11 Na (Sodium) 1s2 2s22p6 3s1

12 Mg (Magnesium) 1s2 2s22p6 3s2

13 Al (Aluminum) 1s2 2s22p6 3s23p1

14 Si (Sili ) 1 2 2 22 6 3 23 214 Si (Silicon) 1s2 2s22p6 3s23p2

15 P (Phosphorus) 1s2 2s22p6 3s23p3

16 S (Sulfur) 1s2 2s22p6 3s23p4

17 Cl (Chlorine) 1s2 2s22p6 3s23p5


17 Cl (Chlorine) 1s 2s 2p6 3s 3p5

18 Ar (Argon) [inert] 1s2 2s22p6 3s23p6

19 K (Potassium) 1s2 2s22p6 3s23p6 4s1

20 Ca (Calcium) 1s2 2s22p6 3s23p6 4s2

K L M

M

N

( ) p p

21 Sc (Scandium) 1s2 2s22p6 3s23p6 4s2 3d1

22 Ti (Titanium) 1s2 2s22p6 3s23p6 4s2 3d2

23 V (Vanadium) 1s2 2s22p6 3s23p6 4s2 3d3

24 Cr (Chromium) 1s2 2s22p6 3s23p6 4s1 3d5

25 Mn (Manganese) 1s2 2s22p6 3s23p6 4s2 3d5

26 Fe (Iron) 1s2 2s22p6 3s23p6 4s2 3d6

27 Co (Cobalt) 1s2 2s22p6 3s23p6 4s2 3d7

28 Ni (Nickel) 1s2 2s22p6 3s23p6 4s2 3d8

29 Cu (Copper) 1s2 2s22p6 3s23p6 4s1 3d10

30 Z (Zi ) 1 2 2 22 6 3 23 6 4 2 3d10 N30 Zn (Zinc) 1s2 2s22p6 3s23p6 4s2 3d10

31 Ga (Gallium) 1s2 2s22p6 3s23p6 4s2 3d10 4p1

32 Ge (Germanium) 1s2 2s22p6 3s23p6 4s2 3d10 4p2

33 As (Arsenic) 1s2 2s22p6 3s23p6 4s2 3d10 4p3

N

33 As (Arsenic) 1s 2s 2p 3s 3p 4s 3d 4p

34 Se (Selenium) 1s2 2s22p6 3s23p6 4s2 3d10 4p4

35 Br (Bromine) 1s2 2s22p6 3s23p6 4s2 3d10 4p5

36 Kr (Krypton) [inert] 1s2 2s22p6 3s23p6 4s2 3d10 4p6 O


( yp ) [ ] p p p

37 Rb (Rubidium) 1s2 2s22p6 3s23p6 4s2 3d10 4p6 5s1

38 Sr (Strontium) 1s2 2s22p6 3s23p6 4s2 3d10 4p6 5s2 (4d5p6s4f5d)

s1 s2 p1 p2 p3 p4 p5 s2p6

K (n=1)

L ( 2)L (n=2)

M (n=3) d3d2d1 d10 d10

N (n=4)?

O (n=5)?

P (n=6)?( )



Noble gases formed when either shell or subshell are

Elements in periodic table are arranged so that those in a column have similar chemical properties. Noble gases formed when either shell or subshell are filled or there is a large gap in energy before the next possible level is encountered.

He (helium, Z=2): 1s2 first (K) shell filled

Ne (neon, Z=10): 1s22s22p6 second (L) shell filled( , ) p

Ar (argon, Z=18): 1s22s22p63s23p6 p subshell in (M) shell filled

Kr (krypton Z=36):1s22s22p63s23p63d104s24p6 p subshell in (N) Kr (krypton, Z=36):1s22s22p63s23p63d104s24p6shell filled

Xe (xenon, Z=54): 1s22s22p63s23p63d104s24p64d105s25p6


p subshell in (O) shall filled

Topic Eleven: Lasers andTopic Eleven: Lasers and Laser Light

LASER means Light Amplification for Simulated Emission of Radiation.

Laser light is an intense, concentrated, and highly parallel beam of coherent light.

L i h h f MASER i il d iLaser is the outgrowth of MASER, a similar device used in microwaves instead of visible light.

In 1960 the first laser was built by T H Maiman of theIn 1960, the first laser was built by T. H. Maiman of the Hughes Aircraft Company Laboratories.


11.1 Absorption

Topic Eleven: Lasers and Laser Light

To understand the operation of a laser, we must be familiar with the processes describing the emission and

11.1 Absorption

absorption of radiation by atoms. Consider atomic system with twosystem with two lower states, of energies E1 and E2.

Figure 11-1: Stimulated absorption of a photon. The

dots represent electrons. One electron is transferred from the ground state to the excited state

when the atom absorbs a


when the atom absorbs a photon of energy hf = E2 – E1.

Stimulated absorption: when a photon of frequency f


Stimulated absorption: when a photon of frequency f(such that the photon energy hf = E2 - E1) is incident and acting with the atom, the photon vanishes and the atomic s stem is e cited to make the p ard transitionsystem is excited to make the upward transition.

Excited states: atoms are raised to allowed higher energy levelsenergy levels.

Excited states are metastable. Instability of the excited states depends on the excited atoms or systemsstates depends on the excited atoms or systems.


11.2 Spontaneous emissionTopic Eleven: Lasers and Laser Light

Spontaneous emission: in the instable excited states (typically ~10-8 s), the atom will jump back to a lower

l l d it h t i th d denergy level and emit a photon in the downward transition process.

Figure 11-2: Spontaneous emission of a photon by an p yatom that is initially in the excited state E2. When the

atom relaxes it to the ground state it emits a photon ofstate, it emits a photon of

energy hf=E2 – E1.


In spontaneous emission process, a photon of energy hf


(= E2 - E1) is emitted under no external influence.

Spontaneous emission happens naturally. Emitted h d l diff f h h iphotons are randomly different from each other in

propagation direction, phase angle, etc.

Phosphorescent materials glow because of a similar process, but the excited atoms may remain in an excited state for periods ranging from a few seconds to several hours. Because of this reason phosphorescent materials emitBecause of this reason, phosphorescent materials emit light after being placed in the dark.


11.3 Stimulated emissionTopic Eleven: Lasers and Laser Light

Consider an atomic system in its excited state in the presence of a radiation field of frequency f, such that hf E Ehf = E2 - E1.

Figure 11-3: Stimulated emission of a photon by anemission of a photon by an

incoming photon of energy hf. Initially, the atom is in the

excited state. The incoming photon stimulates the atom to

emit a second photon of energy hf = E2 – E1.


Sti l t d i i th h t i t t ith th


Stimulated emission: the photon interacts with the excited system to drive it downward transition to its lower energy by emitting an additional photon. The emitted photon is identical with the ‘triggering’ or ‘stimulating’ photon - same energy, direction, phase, and state of polarizationThese two photons can cause other stimulated emissions, leading to a chain reaction of similar

state of polarization.

gprocesses - amplification.


11.4 Conditions for laser action


11.4 Conditions for laser action1. Population inversion must be realized.

For an atomic system at thermal equilibrium, the number of atoms occupying a state at an energy Eis determined by the exponential factor of s de e ed by e e po e a ac o oexp(E/kBT), governed by the Maxwell-Boltzmann distribution.

The ratio of the number of atoms in the upper excited level E2 to the number in the lower level E1isis

TkEE

EnEn

B

12

1

2 exp (11.1)


TkEn B1


Since E2 > E1, the ratio of n(E2)/n(E1) will always be less than unity, meaning that there are fewer atoms in the higher energy state than in the lower one in thermalthe higher energy state than in the lower one in thermal equilibrium.

5

6

7

TFigure 11-4: The

2

3

4

k BT

(E2-

E1)

/kB

gMaxwell-Boltzmann

distribution. 0

1

0 0.2 0.4 0.6 0.8 1 1.2


exp(-E /k BT )n(E2)/n(E1)e-1=0.368e-3=0.050


Example 11-1*Estimate relative populations at room temperature (300 K) of two energy levels such that a transition from the hi h t th l l l i i ibl di ti fhigher to the lower levels gives a visible radiation of 550 nm.

Energy difference between the two levels,

105501031063.6

9

834

12

hcEE

gy ,

From Eq.(11.1), eV 25.2J106.3

1055019

q ( ),

192 37

231

3.6 10exp exp 87 101.38 10 300

n En E


1

Population inversion: there are more atoms in a higher


Population inversion: there are more atoms in a higher energy state than in a lower state at non-thermal equilibrium. P i i th f ti l ti

This can be done optically by creating an intense

Pumping is the process of creating a population inversion. This can be done optically by creating an intense, continuous light source around the lasing material, or electrically by gas discharge.

2. The excited state of the system must be a metastable state so that when population inversion is met, stimulated emission occurs before spontaneous emission.


3. The emitted photons must be confined in the system


p ylong enough to stimulate further emission from other excited atoms.

Fi 11 5 S h i di f l d i Th b i hFigure 11-5: Schematic diagram of a laser design. The tube contains the atoms that are the active medium. An external source of energy “pumps” the atoms to the excited state. The parallel end mirrors confine the photons to the tube. One mirror is made totally reflecting and the other is slightly transparent to


y g g y pallow part of the laser beam to escape.


Spontaneous emission, stimulated emission, laser light

Key points: This illustration shows clearly spontaneous emission, stimulated emission, importance of an optical resonator formed by two mirrors for lasing, etc.lasing, etc.


11.5 Helium-neon lasers*


The three principal elements of a laser are (1) an energy pump (2) an optical gain medium and (3) an

11.5 Helium neon lasers

energy pump, (2) an optical gain medium, and (3) an optical resonator.

(1)Energy pump:A 1400 V high voltage, DC power supply maintains a glowmaintains a glow discharge or plasma in a glass tube containing

ti l i tan optimal mixture (typically 5:1 to 7:1) of helium and neon gas.

Fig re 11 6: A photo of a t pical HeNe laser


Figure 11-6: A photo of a typical HeNe laser.laser=632.8 nm


Figure 11-7: Diagram of optical and electrical components in a typical HeNe


g g p p yplaser.

(2) Optical gain medium: To achieve laser action, it is


necessary to realize population inversion.

20.61 eV 20.66 eVHe* Ne*632 8nm

18.70 eV

Spontaneous

Through the collision, Ne atoms are excited into a

632.8nm

Figure 11-8: Simplified atomic energy level diagram showing li i

emissions deplete the lower level to maintain the population

excited into a metastable state.

diagram showing excited states of atomic He and Ne relevant to the

ti f th

Helium is “pumped” up to excited

population inversion.

operation of the laser at 632.8 nm.

excited states by electrical discharge


Helium Neon

(a) An energetic electron collisionally excites a He atom


to the state He* with an energy 20.61eV. (b) The excited He* atom collides with an unexcited Ne

atom and the atoms exchange their internal energyatom and the atoms exchange their internal energy, causing the unexcited Ne atom to be excited, i.e., Ne* with 20.66 eV. This energy exchange process occurs

ith hi h b bilit l b f th id t lwith high probability only because of the accidental near equality of the two excitation energies of the two levels in these atoms.

(c) The Ne* is metastable and it deexcites to a lower excited energy level of 18.70eV by emitting a photon of

l th 6328 Åwavelength 6328 Å.


(d) The excited neon rapidly deexcites to its ground state by


emitting additional photons or by collisions with the plasma tube walls.

(d) Because of the extreme quickness of the deexcitation(d) Because of the extreme quickness of the deexcitation process, at any moment in the HeNe plasma, there are more Ne atoms in Ne* state (20.66 eV) than in the l it d t t (18 70 V) d l tilower exited state (18.70 eV) and a population inversion is said to be established between these two levels.

(3) Optical resonator or cavity is formed by two highly reflecting mirrors along the axis of the discharge. Thus, the photons of 632 8 nm along the axis of the cavitythe photons of 632.8 nm along the axis of the cavity can be reflected hundreds of times. These reflecting photons can interact with other excited Ne* atoms and

h i 632 8 li h i k


cause them to emit 632.8 nm light in a process known as stimulated emission.

11.6 Laser Applications*Topic Eleven: Lasers and Laser Light

ppOptical CommunicationHolographyLaser cutting Laser marking Laser weldingLaser welding

Figure 11-9: Components of an optoelectronican optoelectronic

communication system and a bundle of optic febres .


Quantum Physics & Classic

Quantum theory must agree with classical theory in the

Physics Quantum theory must agree with classical theory in the limit in which classical theory is known to agree with experiments.I h d h i h l i lIn other words, quantum theory must agree with classical theory in the limit of large quantum numbers.

This is because En-En-10 when n . The energy becomes continuous, not discrete. *For a particle in a1-D box (or infinite deep quantum p ( p qwell), the probability density becomes more uniform when n .


For electromagnetic waves, when their hf is muchFor electromagnetic waves, when their hf is much larger than kBT (classical measure of the mean translation energy of a particle at temperature T), a single photon can cause a detectable measurement

For a particle which is trapped to a small space whose

single photon can cause a detectable measurement. Thus, the waves could show particle properties.

For a particle which is trapped to a small space whose dimension is comparable to the particle’s de Broglie wavelength, the wave character of the particle becomes

ti l d th b ti dessential and the energy becomes quantized. Electron behaviors in nanoelectronic devices could only be interpreted through quantum theorybe interpreted through quantum theory.


Acknowledgments

I thank Professor Tjin Swee Chuan (EEE) for hi t h l ith i th this great help with preparing the notes.


Quantum Physics - Lecture Notes-ZQ One in One 19 Jan 2011

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Transcript of Quantum Physics - Lecture Notes-ZQ One in One 19 Jan 2011