28. Quantum Physics28. Quantum Physics28-1. Black-Body Radiation and Plank’s Theory

T

Thermal radiation : The radiation depends on the temperature and properties of objects

Color of a Tungsten filament – Black– Red– Yellow– White

• Classic Point of View

The thermal radiation was considered to be simply due to accelerated charged particles near the surface.⇒ Not right !

• Black-body Radiation

All the light is absorbed. But the radiation depends on the temperature of the inside wall.

KmT ⋅×= −2max 102898.0λ

Wien’s Displacement law

Wavelength

Inte

nsity Experimental

Classical theory

Ultraviolet catastrophe

• Plank --- Explain the black-body radiationwith two assumptions related to the oscillating charges.

1. The radiation energy is Quantized.

nhfEn =

2. The rasonators emit energy, the so-called photon.

hfE =

Plank succeeded in reproducing the black-body radiation curve. But no body including Plank himself did not accept the quantum concept. -- Considered the assumptions unrealistic.

28-2. The Photoelectric Effect

Photoelectric effect (광전효과)

The first discovery by Herz in 1887.

∆Vs : Stopping potential (independent of the radiation intensity)

• Electrons having a kinetic energy K

sVeK ∆=max

• Characteristics in the photoelectric effect

i) Cutoff frequency, fc

⇒ No photoelectronsii) Kmax is independent of the light intensity.iii) iv) Photoelectric effect occurs instantaneously ( ~ 10-15 sec.)

cff <

fK ∝max

• Einstein (1905)

Extend the quantum concept of Plank’sEnergy of the electromagnetic waves⇒ PhotonsEach photon can give its energy to a single electron.

φ−= hfKmax

Work function

φ=chf

Minimum energy bound in the metal (~ 3-6eV)

i) Cutoff frequencyii)iii) iv) The particle theory of light

chf=φφ−= hfKmax

fKmax ∝

Kmax

ffc

φφλ hc

hc

fc

cc ===

/

Cutoff wave length

28-3. The Compton effect

• Einstein hfE =

cE

chfp ==

Energy

Momentum

• Holly Compton and Peter Debye in 1923 carried an experiment to prove Einstein’s point-like particle concept.

cEphfE

==

The photoelectric effect (x-ray scattering): The total momentum of the photon-electron pair

must be conserved.

(a) Classical Model

Doppler effects

(b) Quantum Model

)cos1(0 θλλ −=−′cm

h

e

Compton Shift

Compton Wavelengthnmcm

h

ec 00243.0==λ

( )p,pppp ′<<∆=′−0

Compton’s Experiment

nm071.00 =λ

Example 28.4

°= 45θnm200000.00 =λ

nmcc 000710.0)2/11()cos1( =−⋅=−=∆ λθλλ

nm200710.0=′λ

00354.0/ =∆ EE

28-4. Photons and Electromagnetic Waves

• Light has a dual nature , Wave & Photon.

Low frequency : Long wavelength ⇒ More wave likeHigh frequency : Short wavelength ⇒ More particle like

28-5. The Wave Properties of Particles

Particle also has a dual nature!!

In 1923, Louis Victor de Broglie (Ph. D. dissertation) postulated an electron also has a dual nature.Perhaps all forms of matter have wave as well as particle properties.

• Photon:

The wavelength of photon can be defined by the momentum.

hfE = phh

cEp =⇒== λ

λ

• Electron:

de Broglie wave : de Broglie wavelegnth

frequency of matter

mvp =

mvh

ph ==λ

hEf =

Quantization of Angular Momentum in the Bohr model

de Broglie ; a dual nature of MatterBohr’s theory : Semiclassical theory

A standing wave form

r λ

rn πλ 2= L,3,2,1=n

vmh

ph

e

==λ

rvm

nh

e

π2= hnhnvrm, e =π

=2

Angular momentum

hnvrmL e == Quantization of angular momentum

Standing wave : Discrete frequency

rn πλ 2≠If , no standing wave ⇒ no closed circular orbit.

• De Broglie (1923): All matters have a dual nature. Then an electron must exhibit diffraction and interference effects.

• Davisson-Germer Experiment (1927): Measure the wavelength of electrons.

Crystalized NiO target

Diffraction patterns due to electron beam.

Extended work on many single-crystalline targets

λhp =Conclude

• G. P. Thomson (1928)Electron diffraction patternfrom electrons passing through a gold foil

• Helium atom, Hydrogen atom, Neutron also show the diffraction pattern.⇒ The matter wave is an Universal Nature

28-6. The Double-Slit Experiment

2sin λθ =D

xph

=λ

xDph

D 22sin ==≈

λθθ

Minimum

The number of electrons detected at a certain spot is proportional to the intensity of two interfering matter waves.

How do we understand the wave-character of electrons?

BErr

,Photon ⇒ Electromagnetic Wave

⇒ Interference effects2EI ∝

ψ : Wave function

*2 ψψψ =∝I

21 ψψψ +=

φψψψψ

ψψψ

cos2 212

22

1

22

21

2

++=

+≠=I

• Which slit does the electron pass through? Slit 1 or Slit 2

28-7. The Uncertainty Principle 1927 Werner Heisenberg

• Heisenberg Uncertainty Principle

A measurement of position is made with precision ∆x, anda measurement of momentum is made with precision ∆px.

2h

≥∆⋅∆ xpx

It is fundamentally impossible to make simultaneous measurementsof a particle’s position and momentum with infinite accuracy.

Similarly

2h

≥∆⋅∆ tE ⇒ Life-time of a particle

λhp =

λhpx =∆

Position of electron λ=∆x hxpx =∆⋅∆⇒

hxpx ≥∆⋅∆

28-8. An Interpretation of Quantum Mechanics

de Broglie ; matter waveMax Bohr ; Explain atomic discrete energy levelSchrödinger ; Wave equation

A particle is described with a wave function ψ(x, y, z, t).

The probability density

dxdx)x(P 2ψ=

12 =ψ= ∫∫∞

∞−dxdx)x(P

Expectation valuethe average position x∫

∞

∞−= dxxx 2ψ

xph

=λ

kxAxAx sin2sin)( =

=λπψ

λπ2

=k( )( )ikxAex =ψ

∞=∆⇒=∆ xp 0If k is determined, then

A traveling electron (wave-pocket)

22

222

22

22

2

2

)()(

xx

xxx

xxxx

xxx

−=

+−=

+−=

−=∆

222)( ppp −=∆

28-9. A Particle in a Box

In classical theory

In Quantum Mechanics

from Schrödinger EquationkxAx sin)( =ψ

Lxx== 0

0)(0)(

==

Lx

ψψ

Boundary Conditions

Lnk

nkLπ=

π=

L,3,2,1=n

xLnsinA)x( π

=ψ∴

de Broglie wave

Analogous to the standing wave

2λnL =

nL2

=λ

Lnh

nLhhp

2/2===

λ

22

222

8221 n

mLh

mpmvEn

=== L,3,2,1=n

Energy Quantization!!

The energy of the particle is quantized.

22

2

8n

mLhEn

=

Example. (i) 1g of a ball, L = 1 m m = 10-3 kg , h = 6.63 × 10-34 J·s En = 5.5 × 10-65 n2 J

If v = 10 m/s , J.mvEk 050221 ==

31102.3~ ×n Very large value

(ii) an electron L = 1 nm = 10-9 mm = 9.11 × 10-31 kg

eVEeVEeVE

6.36.14.0

3

2

1

===

eVnJnEn2218 4.0106 ≈×≅ −

JeV 19106.11 −×=

28-10. The Schrödinger Equation

ph=λλ

π= 2kkxAx sin)( =ψ

hppk h == π2

xpsinA)x(h

=ψ

xpApdx

xdhh

cos)(=

ψ

)(sin)( 22

2

2

xpkxApdx

xd ψψ

−=

−=

hh

2

222

dxdp h−=⇒

mpmvEk 22

1 22 ==

)(2

)(2

)( 2

222

xdxd

mhx

mpxEk ψψψ −==

)(2)( 22

2

xEmxdxd

kψψh

−=

A particle in a potential U(x)

UEEk −=

ψψ )(222

2

UEmdxd

−−=h

• Schrödinger equation is originated from a wave equation.

)tkx(iAe)t,x()tkxsin(A)t,x(

ω−=ψ

ω−=ψ

λ

π=

2k kpp

hh=⇒=λ

)t,x(x

)t,x(k)t,x(p ψ∂∂

−=ψ=ψ 2

22222 hh

ω=⋅==π

ωh

2hhfE

),(),(),(),( txt

itxti

txtxE ψψωψψ∂∂

=∂∂

−== hh

h

Um

pUEE k +=+=2

2

),(),(2

),( 2

22

txUtxxm

txt

i ψψψ +∂∂

−=∂∂ h

h

General form

),(2

),( 22

trUm

trt

i rhrh ψψ

+∇−=

∂∂ Time-dependent

Schödinger equation

( )∞→∆tTime-independent ⇒ E is fixed

)(2

)( 22

rUm

rE rhr ψψ

+∇−=

Kinetic energy Potential energyTotal energy

A simple form of the one-dimensional Schödinger equation

)(2

)( 2

22

xUxm

xE ψψ

+

∂∂

−=h

h

mEk 2=

)()(2)( 222

2

xkxEmxdxd ψψψ −≡−=

h

ikxikx eCeCx −+= 21)(ψ

0)0( 21 =+= CCψ

12 CC −=

Boundary condition at x=0

−=⇒

−

ieeiCx

ikxikx

22)( 1ψ

sin kxkxAx sin)( =ψ

Boundary condition at x=L

0sin)( == kLALψ πnkL =⇒

LnmEk π

==h

2

22

22

2

22

82n

mLhn

mLE

==

πh

=

LxnAx πψ sin)(

How to define A

Px =2)(ψ

12

2cos12

sin1)(

2

0

2

0

22

0

2

==

−=

==

∫

∫∫

LA

dxxLnA

dxL

xnAdxx

L

LL

π

πψ0

2/12

=⇒

LA

π

=ψ∴

Lxnsin

L)x(

212

28-11. Tunneling through a Barrier

0 Dd d+L

LeT κ−≅ 2 Transmittance

1<<Tif where

h

)EU(m −=κ

2

1=+ RT

If an electron is in a such potential well, what is the probability that the electron is in each region ?

In region I

)(2

),( 12

22

1 xdxd

mtxE ψψ h

−= kxsinA)x( =ψ⇒ 1

)(2

),( 22

22

2 xUdxd

mtxE ψψ

+−=

h

( ) ( ) ( )

)x(

xEUmxdxd

22

2222

2 2

ψκ=

ψ−=ψh

h

)EU(m −=κ

2

( ) xBex κ−=ψ2

In region II

In region III

)()(2)( 32

3232

2

xkxEmxdxd ψψψ −=−=

hikxikx eCeCx −+= 213 )(ψ

Boundary Conditions

( ) ( )( ) ( )( ) 03

32

21

=ψ+ψ=+ψ

ψ=ψ

LLdLd

dd

Normalized Condition

( ) ( ) ( ) ( )1

23

220

21

2

=

ψ+ψ+ψ=ψ ∫∫∫∫ +

+∞

∞−

D

dL

dL

d

d dxxdxxdxxdxx

Determine the coefficient A, B, C1, and C2