Transcript Chap 15 Quantum Physics Physics

Chapter 15
QUANTUM PHYSICS
Chapter Index
Physics
15-0 Basic Requirements
15-1 Blackbody Radiation, Planck
Hypothesis
15-2 Photoelectric Effect, Wave-particle
Duality of Light
15-3 Compton Effect
15-4 Bohr’s Theory of Hydrogen Atom
*15-5 Franck-Hertz Experiment
15-6 de Broglie Matter Wave, Wave-particle
Duality of Particles
Chap 15
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Chapter Index
Physics
15-7 Uncertainty Principle
15-8 Introduction to Quantum Mechanics
15-9 Introduction to Quantum Mechanics of
Hydrogen Atom
*15-10
Electron Distributions of Multi-electron
Atoms
*15-11 Laser
*15-12 Semiconductor
*15-13
Superconductivity
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15-0 Basic Requirements
Physics
1. Understand experimental laws of
thermal radiation：Stephan-Boltzmann
law and Wein displacement law, and
difficulties of classical physics theory in
explanation of energy-frequency
distribution of the thermal radiation.
Understand Planck quantum hypothesis
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15-0 Basic Requirements
Physics
2. Understand difficulties of classic physics
theory in explanation of experimental
discoveries of photoelectronic effect.
Understand Einstein photon hypothesis, grasp
Einstein equation
3. Understand experimental laws of
Compton effect, and its explanation by photon.
Understand wave-particle duality of light.
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15-0 Basic Requirements
Physics
4. Understand experimental results of
Hydrogin atom spectra, and Bohr’s theory
5. Understand de Broglie hypothesis and
electron diffraction experiment and waveparticle duality of particles; Understand the
relation between physical quantities (wavelength, frequency) describing wave property
and ones (energy, momentum) describing
particle property.
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15-0 Basic Requirements
Physics
6. Understand 1-dimension coordinate
momentum uncertainty principle
7. Understand wave function and its
statistical explanation. Understand 1dimension stationary Schrodinger equation,
and the quantum mechanical method deal
with 1 dimensional infinity potential well
etc.
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Physics
The foundations of quantum mechanics
were established during the first half of
the twentieth century by Niels Bohr,
Werner Heisenberg, Max Planck,
Louis de Broglie, Albert Einstein,
Erwin Schrödinger, Max Born, John
von Neumann, Paul Dirac, Wolfgang
Pauli, David Hilbert, and others.
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Physics
In the mid-1920s, developments in quantum
mechanics led to its becoming the standard
formulation for atomic physics.
In the summer of 1925, Bohr and Heisenberg
published results that closed the "Old Quantum
Theory". Light quanta came to be called photons
(1926).
Quantum physics emerged, its wider acceptance
was at the Fifth Solvay Conference in 1927.
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Physics
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Physics
The study of electromagnetic waves such as
light was the other exemplar that led to
quantum mechanics
M. Planck, in 1900, found that the energy of
waves could be described as consisting of
small packets or quanta, A. Einstein further
developed this idea to show that an EM wave
could be described as a particle - the photon with a discrete quanta of energy that was
dependent on its frequency
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Physics
1. Thermal Radiation
(1) Fundamental concepts and
basic laws
(1a) Monochromatic radiant
emittance: the power of electromagnetic radiation whose
M (T ) W  m -2  Hz -1
frequency around  (or
M  (T ) W  m -3
wavelength  ) per unit area and
unit time radiated by a surface.
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Physics
(2) Radiation emittance
power emitted from a surface per unit time
and unit area

M (T )   M (T )d
0

M (T )   M  (T )d
0
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Physics
SUN M(T )/(10 8 W  m 2  Hz1 )
Ti M(T )/(10 9 W  m- 2  Hz1 )
Monochromatic radiation
emittance of
Sun and Ti
12
T  5 800 K
SUN
visible
10
8
6
4
2
0
Ti
 / 1014 Hz
2
4
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8
10 12
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Physics
(3) Monochromatic absorption ratio and
reflection ratio
 monochromatic absorption ratio  (T) ：
The ratio of absorbed energy to the incident
energy between wavelength  and   d
absorption
Incident
Reflection
Chap 15
transmission
Quantum Physics
15
Physics
 monochromatic reflection ratio r(T ):
the ratio of reflected energy to the incident
energy between wavelength  and   d
For opaque object (T ) + r(T )=1
absorption
Incident
Reflection
Chap 15
transmission
Quantum Physics
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Physics
(4) Black body
An idealized physics object whose absorption
ratio equals 1, i.e., it absorbs all incident EM
radiation, regardless
of its frequency
Blackbody is an
idealized model
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Physics
(5) Kirchhoff’s Law
For a body of any arbitrary material, emitting
and absorbing thermal EM radiation in
thermodynamic equilibrium, the ratio of its
emissive power to its dimensionless coefficient
of absorption is equal to a universal function
only of radiative wavelength and temperature,
the perfect black-body emissive power
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Physics
In other words, for a body of any arbitrary
material, emitting and absorbing thermal EM
radiation in thermodynamic equilibrium, the
ratio of M(T ) to (T) equals to MB( ,T)
under the same temperature T
M  (T )
 M B ( , T )
  (T )
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Physics
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Physics
2. Experimental Observations of
Blackbody Radiation
M  (T ) /(1014 W  m 3 )
1.0
Visible
Region
Exp.
Curve
0.5
6 000 K
3 000 K
0
Chap 15
m1 000
 / nm
2 000
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Physics
1. Stephan-Boltzmann Law
M  (T ) /(1014 W  m 3 )
Total Radiation Emittance

M (T )   M  (T )d  T 4
Visible
region
1.0
0
0.5
where
  5.670 108 W  m 2  K 4
Stephan-Boltzmann
const.
Chap 15
6 000 K
3 000 K
0
1m000
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 / nm
2 000
22
Physics
2. Wien’s Displacement Law
M  (T ) /(1014 W  m 3 )
mT  b
Visible
1.0
region
Peak wave length
6 000 K
Const. b  2.898 10 3 m  K
3 000 K
0
Chap 15
1m000
Quantum Physics
 / nm
2 000
23
Physics
E.g-1 (1) Suppose a blackbody with
temperature T= 20  C , what is the wavelength of its monochromatic peak？(2) the
monochromatic emittance peak wave
length m  483 nm , estimate the surface
temperature of the sun; (3) 上what is the
ratio of above two？
Solu: (1) From Wien’s displacement law
3
b 2.898 10
m  
nm  9 890 nm
T1
293
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Physics
（2） From Wien displacement law
3
2.898 10
T2 

K  6 000 K
9
m
483 10
b
（3）From Stephan-Boltzmann law
M (T2 ) M (T1 )  (T2 T1 ) 4  1.76  10 5
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Physics
3. Rayleigh-Jeans formula
Failures of classical physics
M (T ) /(10 9 W  m -2  Hz -1 )
Rayleigh-Jeans
6
*
*
4
*
*
2
*
0
1
**
Rayleigh-Jeans
* Exp. Curve
*
* T = 2 000 K
*
*
*
* *
*
2
3
Chap 15
2 π 2
M (T )  2 kT
c
Violet Catastrophy
 / 1014 Hz
Quantum Physics
26
Physics
M. Planck (1858 - 1947)
German theoretical physicist
and the founder of quantum
mechanics and one of the most
important physicists of the 20th
century.
His talk under the title “On the
Law of Distribution of Energy in
the Normal Spectrum” *in 1900,
was regarded as the “birthday of
quantum theory” (by M. Laue)
* M. Planck, On the Law of Distribution of Energy in the Normal Spectrum, Ann.
der Physik, Vol. 4, 1901, p. 553 ff.
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Physics
4. Planck’s hypothesis and blackbody
radiation formula
(1) Planck’s blackbody radiation formula
2 π h  d
M (T )d  2 hν / kT
c e
1
3
Planck’s Constant
h  6.63 10 34 J  s
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Physics
M (T ) /(10 9 W  m-2  Hz1 )
Reighley - Jeans
Exp. Data
vs. Planck
theoretical
curve
6
*
4
**
*
*
Planck’s formula
*
*
*
*
* Exp. Data
* T = 2 000 K *
* *
*
*
 / 1014 Hz
0
1
2
3
2
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Physics
2. Planck’s quantum hypothesis
The vibration modes of molecules and atoms in
blackbody can be viewed as harmonic oscillators
(HO). The energy states of these HOs are discrete,
their energies are integer of a minimum energy,
i.e.,  , 2 , 3, … n,  is called energy quanta,
n is quantum number ε  nh (n  1,2,3,)
Planck quantum hypothesis is the milestone
of quantum mechanics
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Physics
E.g-2 Suppose a tuning fork mass m = 0.05 kg ，
frequency   480 Hz , amplitude A  1.0 mm
(1) quantum .number of vibration；
(2) when quantum number increases from n
to n  1，how much does the amplitude
change？
Solu: (1)
1
1
2 2
2 2
E  m A  m(2 π ) A  0.227 J
2
2
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Physics
E  nh
energy
E
29
n
 7.13  10
h
h  3.18 10
31
J
（2） E  nh
E
nh
A 

2
2
2π m
2π 2 m
2
h
2 AdA 
dn
2
2π m
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Physics
n A
A 
n 2
n  1
A  7.0110
34
m
Macroscopically, the effect of energy
quantization is not obvious, namely, the
energy of macroscopic object is continuous
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Physics
1. Photoelectric Effect and Phenomenon
(1) Experimental Setup and Phenomenon
V
Chap 15
Quantum Physics
A
34
Physics
(2) Discoveries
(2a) Current linearly proportional to the
intensity.
i
im2
im1
I2
I1
I 2  I1
U0
o
U
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Physics
(2b) threshold frequency  0
For a given metal, electrons only emitted
if frequency of incident light exceeds a
threshold0. 0 is called threshold
frequency
Threshold frequency depends on type of
metal, but not on intensity
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Physics
(2c) Stopping Voltage U 0
Applied reverse voltage that
makes zero current is socalled stopping voltage U 0 ,
different metal has different
U0
O
U0
C s Z n Pt
0

Stopping voltage linearly related incident light frequency
(2d) Current appears with no delay
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Physics
(3) Failures of Classical Theory
Threshold frequency
Electrons should be emitted whatever the
frequency ν of the light, so long as electric
field E is sufficiently large
No time delay
For very low intensities, expect a time lag
between light exposure and emission, while
electrons absorb enough energy to escape
from material
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Physics
2. “Photon”, Einstein Equation
(1) “light quanta” hypothesis
Light comes in chunks (composed of
particle-like “photon”), each light quanta
has energy ε  h
(2) Einstein photoelectric equation
1 2
h  mv  W
2
Chap 15
Escape work
depends on material
Quantum Physics
39
Physics
Approximate escape work value of
different metals (in eV)
Na
Al
Zn
Cu
Ag
Pt
2.46 4.08
4.31
4.70
4.73 6.35
Theoretical Explanation:
the greater the intensity, the more photons,
the more photo-electrons, and hence the larger
current (   0)
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Physics
Stopping voltage
Applied reverse
stopping voltage U 0
stops electrons
V
A
1 2
eU 0  mv
2
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Quantum Physics
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Physics
Threshold frequency: 
W  h 0
0
0  W h
thrshold
frequency
No lag: photon energy h (   0 ) is
absorbed by a electron and the electron
then emits without time delay
Einstein’s theory successfully explained the
photoelectric effect and won 1921 Nobel prize
of physics (not for relativity)!
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Physics
(3) Measurement of Planck const.
1 2
h  mv  W
2
Stopping voltage
vs. frequency
U0
h  eU 0  W
h
W
U0   
e
e
U 0   h e
Chap 15
O
0

U 0
h
e

Quantum Physics
43
Physics
E.g.-1. Consider a thin circular plane with
radius 1.0  10 3 m , 1.0 m far from an 1W
power light source. The light source emits
monochromatic light with wave length 589
nm. Suppose the energy goes off all directions
equally. Calculate the number of photons on
the plate per unit time.
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44
Physics
Solution:
3
6
S  π  (1.0 10 m)  π 10 m
2
2
S
7
1
EP
 2.5 10 J  s
2
4πr
E E
11 1
N

 7.4 10 s
h hc
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Quantum Physics
45
Physics
3. Applications in Modern Technology
Photo-relay circuit, Automatic
counter, measuring device etc.
Demo. of photo-relay
light
Amplifier
Controller
Chap 15
Photomultiplier
Quantum Physics
46
Physics
Chap 15
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47
Physics
4. Wave-particle Duality of Light
(1) wave:diffraction and interference
(2) particle: E  h , photo-electric effect
etc.
Relativistic energymomentum relation
photon
Chap 15
E  p c E
2
E0  0 ,
2 2
2
0
E  pc
Quantum Physics
48
Physics
photon
E0  0 ,
E  pc
E h
h
p


c
c

Particle
character
E  h
h
p

Chap 15
Wave
character
Quantum Physics
49
Physics
Compton (1923) measured
intensity of scattered Xrays from solid target, as
function of wave- length
for different angles. He
found that peak in
scattered radiation ()
shifts to longer wavelength than source (0), i.e.,
 > 0. Amount depends on
θ, but not on the target
material.
Chap 15
A.H. Compton, Phys. Rev. 22 (1923) 409
Quantum Physics
50
Physics
1. Experimental Asparatus
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Quantum Physics
51
Physics
2. Experimental
Results
(1) shift in wave
length     0
depends on 
(2)  is indep.
I
Relative Intensity
  0
0
  45

  90
 
of targets
0
Chap 15
Quantum Physics
 135

52
Physics
3. Difficulties of Classical Theory
According to the classical picture:
oscillating electromagnetic field causes
oscillations in positions of charged
particles, which re-radiate in all directions
at same frequency and wavelength as
incident radiation.
Change in wavelength of scattered light
is completely unexpected classically!
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53
Physics
4. Quantum Explanations
(1) Physical model
Photon  0
y
y
Electron
v0  0
x
Photon



x
Electron
Incident photon (X-ray or  -ray) with
higher energy
4
5
10
~
10
eV
E  h
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54
Physics
electron energy of thermal motion  h , so we can
treat electron as at-rest approximately
phton 
y
0
electron
v0  0
photon
y
x



x
electron
electrons near the surface of solids with weak
binding, quasi-free
electron with large bouncing velocity, use
relativistic mechanics
Chap 15
Quantum Physics
55
Physics
(2) Qualitative Analysis
(1) “billiard ball” collides between particles of light
(X-ray photons) and weak-binding electrons in the
material, part of energy is transported to electron,
leads to the energy decrease of scattered photon,
hence the frequency, wavelength increases
(2) photon collides with tight-binding electron,
without significant lost of energy, results in the
same wave-length in scattered light
Chap 15
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56
Physics
(3) Quantitative Calculation
Energy conservation
hv0  m0 c  h  mc
2
2
Momentum conservation
h 
y
e
h 0   c
e0 e

c

e0

x

h 0  h 

mv
e0 
e  mv
c
c
2 2
2
2 2
h 0 h 
h  0
2 2
m v  2  2  2 2 cos 
c
c
c
Chap 15
Quantum Physics
57
Physics
h
h  0
h
m v  2  2  2 2 cos 
c
c
c
2
2
2
2
0
2
2
2
v
2 4
2
2
m c (1  2 )  m0 c  2h  0 (1  cos )  2m0c h( 0  )
c
2
2 4
2 1/ 2
m  m0 (1  v / c )
2
c
c
h
 
(1  cos  )    0  
  0 m0 c
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58
Physics
h
2h
2
 
(1  cos ) 
sin
m0 c
m0 c
2
Compton
Wavelength
h
12
C 
 2.43  10 m
m0 c
Compton Formula
h
 
(1  cos )  C (1  cos )
m0 c
Chap 15
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59
Physics
(4) Conclusions
Scattered light wave length change 
depends only on 
  0,   0
  π, ( ) max  2C
scattered photon
energy decrease
h


y
e
h 0   c
e0 e

c

e0
  0 ,   0
Chap 15
Quantum Physics

x

mv
60
Physics
-10


1.00

10
m
Eg-1. X-ray with wavelength 0
elastically collides with a electron at rest,

90
observing along the direction with respect
to scattering angle,
(1) Change of the scattered wavelength  ?
(2) Kinetic energy bouncing electron gets?
(3) Energy that photon loses during collision?
Chap 15
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61
Physics
Solution
(1)   C (1  cos )  C (1  cos 90  )  C
 2.43 10 12 m
(2) bouncing electron kinetic energy
0
hc hc hc
Ek  mc  m0c    (1  )  295 eV
0  0

2
2
(3) Energy photon loses＝ Ek
Chap 15
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62
Physics
1. Review of Modern View of Atomic
Hydrogen Structure
(1) Experimental discoveries of atomic
hydrogen spectrum
Chap 15
Quantum Physics
63
Physics
Light Bulb
Hydrogen Lamp
Quantized, not
continuous
Chap 15
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64
Physics
(1) Experimental discoveries of atomic
hydrogen spectrum
Joseph Balmer (1885) first noticed that the
frequency of visible lines in the H atom
spectrum could be reproduced by:
n2
  365.46 2
nm ,
2
n 2
Chap 15
n  3,4,5,
Quantum Physics
65
Physics
Johann Rydberg (1890) extends the Balmer
model by finding more emission lines outside
the visible region of the spectrum:
1
1
1
 R( 2  2 )
wave number  

n f ni
n f  1,2,3,4,, ni  n f  1, n f  2, n f  3,
Rydberg const. R  1.097  10 7 m 1
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66
Physics
Ultraviolet
Lyman
1
1 1
   R( 2  2 ) , n  2,3,

1 n
Visible
Balmer
1
1 1
   R( 2  2 ) , n  3,4,

2 n
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67
Physics
Infrared
1
1
1
Paschen    R( 2  2 ) , n  4,5,

3
n
1
1 1
Brackett    R( 2  2 ) , n  5,6,

4 n
1
1 1
   R( 2  2 ) , n  6,7,
Pfund

5 n
1
1 1
Humphrey   R( 2  2 ) , n  7,8,

6 n
Chap 15
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68
Physics
Balmer spectrum of H atom
364.6 nm
410.2 nm
434.1 nm
486.1 nm
656.3 nm
Chap 15
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69
Physics
Hydrogen atom spectra
Visible lines in H atom spectrum
are called the BALMER series.
6
5
4
Energy
3
2
1
Ultra Violet
Lyman
Chap 15
Visible
Balmer
Quantum Physics
Infrared
Paschen
n
Physics
(2) Rutherford’s model of atomic structure
1897, J. J. Thomson discovered electron
1904, J. J. Thomson proposed“plum
pudding model” of atomic structure
the atom is composed of
electrons surrounded by a soup
of positive charge to balance
the electrons' negative charges,
like negatively-charged
"plums" surrounded by
positively-charged "pudding".
Chap 15
Quantum Physics
71
Physics
Ernest Rutherford (1871 – 1937)
New Zealand-born British chemist
and physicist who became known as
the father of nuclear physics. He
discovered the concept of radioactive
half-life, differentiated and named α,
β radiation.
He was awarded Nobel prize of Chemistry in 1908 "for
his investigations into the disintegration of the elements,
and the chemistry of radioactive substances"
Chap 15
Quantum Physics
72
Physics
In 1911, he proposed the Rutherford model of the
atom, through his gold foil experiment. He discovered
and named the proton. This led to the first
experiment to split the nucleus in a fully controlled
manner.
He was honoured by being interred with the greatest
scientists of the United Kingdom, near Sir Isaac
Newton’s tomb in Westminster Abbey.
The chemical element rutherfordium (element 104)
was named after him in 1997.
Chap 15
Quantum Physics
73
Physics
Rutherford atomic model (Planetary model)
the atom is made up of a central charge (this is
the modern atomic nucleus, though Rutherford
did not use the term "nucleus" in his paper)
surrounded by a cloud of orbiting electrons.
Chap 15
Quantum Physics
74
Physics
2. Bohr’s Theory of Atomic Hydrogen
(1) Failures of Classical Atomic Models
According to the classical electromagnetic theory, electrons rotate
around atomic nucleus, accelerated
electrons radiate electro-magnetic
wave and hence lose energy
Chap 15
Quantum Physics
75
Physics
electrons orbiting a nucleus – the laws of classical
mechanics, predict that the electron will release
electromagnetic radiation while orbiting a nucleus.
Hence would lose energy, it would gradually spiral
inwards, collapsing into the nucleus.
e
e
Chap 15
Quantum Physics
76
Physics
As the electron spirals inward,
the emission would gradually
increase in frequency as the orbit
got smaller and faster. This would
produce a continuous smear, in
frequency, of electromagnetic
radiation, one should observe
continuous light spectra
Chap 15
Quantum Physics

 v
e F
r
e
77
Physics
Niels Bohr (1885 - 1962)
Danish theoretical physicist,
one of the founding fathers of
quantum mechanics.
He uses the emission spectrum
of hydrogen to develop a
quantum model for H atom
and explains H atom spectrum
1922 Nobel Prize in physics
Chap 15
Quantum Physics
Physics
(2) Bohr’s Theory of H Atom
In 1913, N. Bohr uses the emission spectrum
of hydrogen to propose a quantum model for
H atom, with the following three assumptions
(a) Stationary hypothesis
(b) Frequency condition
(c) Quantization condition
Chap 15
Quantum Physics
79
Physics
(a) Stationary hypothesis
Electrons do not radiate EM wave if they
are on some specific circular trajectories,
they can keep staying on those stable
states, i.e., so-called stationary states
Energies corresponding
to stationary states are
E1, E2… , E1 < E2< E3
Chap 15
E1 +
E3
Quantum Physics
80
Physics
(b) Frequency condition
h  Ei  E f
Ei
emmision
absorption
Ef
(c) Quantization condition
h
L  mvr  n
2π
n  1,2,3, Principal quantum number
Chap 15
Quantum Physics
81
Physics
(3) Calculate H-atom energy and orbital radii
(a) Orbital radii
2
v
n
Classical

m
mechanics 4π  0 rn2
rn
h
Quantization
mvn rn  n
condition
2π
e2
rn 
 0h
2
π me
2
+
rn
n 2  r1n 2 (n  1,2,3,)
Chap 15
Quantum Physics
82
Physics
rn 
 0h
2
π me
n  r1n (n  1,2,3,)
2
2
2
 0h
2
n  1 , Bohr radius r1  π me2  5.29 10
11
m
(b) Energy
The nth orbital electron’s energy：
2
1
e
En  mvn2 
2
4π  0 rn
Chap 15
Quantum Physics
83
Physics
me4
1
E1
En   2 2  2  2
8 0 h n
n
ground state
energy (n  1)
4
me
(Ionized
E1  
2 2  13.6 eV
8 0 h
energy)
Excited state
energy (n  1)
En  E1 n
Chap 15
2
Quantum Physics
84
Physics
Energy level transition and
Spectrum of H-atom
n=
n=5
n=4
0
-0.54 eV
Brackett
Paschen
n=3
-0.85 eV
-1.51 eV
Balmer
-3.40 eV
n=2
Lyman
n=1
-13.6 eV
Chap 15
Quantum Physics
85
Physics
(4) Explanations of Bohr’s Theory on
H-atom Spectrum
4
me
1
h  Ei  E f
En   2 2  2
8 0 h n
1

4
me
1
1
    2 3 ( 2  2 ),
 c 8 0 h c n f ni
me4
7
1

1
.
097

10
m
8 02 h 3c
Chap 15
ni  n f
 R (Rydberg const.)
Quantum Physics
86
Physics
3. The Successes and Failures of Bohr
Theory
(1) Successes
(a) Correctly predicted the existence of atom
energy level and energy quantization
(b) Correctly proposed the concepts of stationary
state and angular momentum quantization.
(c) Correctly explained H-atom and H-like-atom
spectrum
Chap 15
Quantum Physics
87
Physics
(2) Failures
(a) Does not work for multi-electron atoms
(b) Microscopic particles do not have certain
trajectory
(c) Can not deal with the widths, intensity etc. of
spectrum.
(d) Half classical half quantum theory: on one
hand microscopic particles have classical
properties, on the other hand, quantum nature
Chap 15
Quantum Physics
88
Physics
1. de Broglie Hypothesis
In 1923, de Broglie postulated that ordinary
matter can have wave-like properties, with
the wavelength λ related to momentum p in
the same way as for light
 E  mc 2  h
Particle nature 
Wave nature
P  mv  h / 
Chap 15
Quantum Physics
89
Physics
L. de Broglie (1892 – 1987)
French physicist and a Nobel
laureate in 1929. His 1924
Recherches sur la théorie des
quanta (“Research on the Theory
of the Quanta”), introduced his
theory of electron waves, thus set
the basis of wave mechanics,
uniting the physics of energy
(wave) and matter (particle).
Chap 15
Quantum Physics
90
Physics
h
h
de Broglie relation   
p mv
E mc 2
 
h
h
de Broglie wave or Matter wave
Note
(1)if v  c then m  m0
if v  c then m  m0
Chap 15
Quantum Physics
91
Physics
2. de Broglie wavelength of a macroscopic
object is too tiny to be measured, this is why a
macroscopic object behaves particle-like nature
E.g.-1 In a beam of electron, the kinetic
energy of electron is 200eV , Calculate its de
Broglie wavelength.
1
Solution: v  c, Ek  m0 v 2 v 
2
Chap 15
Quantum Physics
2 Ek
m0
92
Physics
2  200 1.6 10
v
9.110 31
 v  c
19
1
m  s  8.4 10 m  s
6
-1
34
h
6.63 10
 

nm
31
6
m0 v 9.110  8.4 10
  8.67 10 2 nm
Roughly the order of X-ray wavelength
Chap 15
Quantum Physics
93
Physics
E.g.-2 Derive quantization condition of angular
momentum in Bohr’s theory of hydrogen atom
Solution: Consider a string
with two ends fixed, if its
length equals wave-length
then a stable standing wave
can form
to form a circle 2π r  
2π r  n
n  1,2,3,4,
Chap 15
Quantum Physics
94
Physics
Electron’s de Broglie wavelength
h

mv
2π rmv  nh
We get quantization condition of angular
momentum
h
L  mvr  n
2π
Chap 15
Quantum Physics
95
Physics
2. Experimental confirmation of de Broglie
matter wave
Quantum Corral: 48 iron atoms form a circular
quantum corral (radius 7.13nm) on the Cu (111)
surface
Chap 15
Quantum Physics
96
Physics
ELECTRON DIFFRACTION
The Davisson-Germer experiment (1927)
θi
θi
The Davisson-Germer
Davisson G.P. Thomson
experiment: scattering a
beam of electrons from a Ni
crystal. Davisson got the
1937 Nobel prize.
At fixed accelerating voltage (fixed
electron energy) find a pattern of
sharp reflected beams from the
crystal
At fixed angle, find sharp peaks in
intensity as a function of electron energy
C. J. Davisson,
"Are Electrons
Waves?," Franklin
Institute Journal
205, 597 (1928)
G.P. Thomson performed similar
interference experiments with thin-film
samples
Chap 15
Quantum Physics
Physics
2. Experimental Confirmation of
de Broglie Wave
(1) Davisson-Germer Diffraction Exp.
U
K
Electron
gun
  50 
I
检测器
Electron 
beam
Scatter
ing
M
G
35
54
75
U /V
beam
Ni-crystal diffraction
Chap 15
Current vs. Acceleration
voltage,   50 
Quantum Physics
98
Physics
The exp. results of single crystal diffraction by
electron beam agree with “Bragg’s law” in Xray diffraction
Interference condition:
d
 2
 2
2d sin

2
cos
. . . . . . . .
. . . . . . . . d sin   k
. . . . . . . .

2
 k
 2
d sin
2
Chap 15
k  1,   50
Quantum Physics
99
Physics
For Ni crystal
d  2.15  10 10 m
  d sin   1.65 1010 m
Wavelength of electron wave
h
h


 1.67 10 10 m
me v
2me Ek
1
d sin   kh
2emU
Chap 15
Quantum Physics
100
Physics
1
d sin   kh
2emU
kh
1
sin  
d 2emU
sin   0.777k
when k  1
  arcsin 0.777  51 , agree well
with experimental results.
Chap 15
Quantum Physics
101
Physics
(2) G. P. Thomson electron diffraction exp.
Electron beam from polycrystalline foil
generates diffraction fringe similar to the
X-ray diffraction fringe
Diffraction of electron beam from
polycrystalline foil
K
U
D
M
Chap 15
P
Quantum Physics
102
Physics
3. Applications
Scanning Tunneling Microscopy (STM)
Developed by Gerd Binnig and Heinrich Rohrer
at the IBM Zurich Research Laboratory in 1982.
Binnig
Rohrer
The two shared half of the 1986 Nobel Prize in physics
for developing STM.
Chap 15
Quantum Physics
103
Physics
4. Statistical interpretation of de Broglie
wave
Classical particle undividable unity, with
certain momentum and trajectory
Classical wave periodic spatial distribution
of some physical quantity, with property of
interference
Wave-particle Duality United wave and
particle natures within one unity
Chap 15
Quantum Physics
104
Physics
(1) Explanation by particle nature
Single particle randomly appears, but large number of
particles show a statistical regularity. The probability
that a particle appear at different position is different
Electron
beam

slit
single-slit diffraction
Chap 15
Quantum Physics
105
Physics
(2) Explanation by wave nature
The more intense the electrons at some
place, the higher intensity of wave; or
vice versa.
Electron
beam

slit
single-slit diffraction
Chap 15
Quantum Physics
106
Physics
(3) Statistical Interpretation
At some place the intensity of de Broglie
wave proportioned to the probability that
the particle appears around that place
M. Born (1926) pointed out , de Broglie
wave is probability wave.
Chap 15
Quantum Physics
107
Physics
1. Heisenberg Uncertainty principle
of Coordinate and Momentum
x
Electron diffraction
b ph 
Position uncertainty
of the electronx  b
the 1st order min.
diffraction angle
sin    b
Chap 15

ph 
y
o
Electron Single-slit
Diffraction Exp.
Quantum Physics
108
Physics
W. Heisenberg (1901 – 1976)
German theoretical physicist, who
made foundational contributions to
quantum mechanics and proposed the
uncertainty principle (1927). He also
made important contributions to
nuclear physics, quantum field theory,
and particle physics.
Awarded the 1932 Nobel Prize in Physics for the creation
of quantum mechanics, and its application especially to the
discovery of the allotropic forms of hydrogen
Chap 15
Quantum Physics
109
Physics
x-direction momentum uncertainty after
passing the slit
x
sin    b
p x  p sin   p
h

p
h
p x 
b

b ph 
b

ph 
y
o
xpx  h
Chap 15
Quantum Physics
110
Physics
the 2nd order diffraction xp x  h
Heisenberg proposed uncertainty principle in 1927
Microscopic particles can not be described
by simultaneous coordinate and momentum
Uncertainty
Relation
Chap 15
xp x  h
yp y  h
zp z  h
Quantum Physics
111
Physics
Implications
(1) a fundamental limit on the accuracy with which
certain pairs of physical properties of a particle, such as
position and momentum, can be simultaneously known
(2) this uncertainty deeply roots in the wave-particle
duality, which is the fundamental property of particles
(3) for macroscopic particles, since h is extremely
small, xpx  0 , hence in macroscopic limit, the
momentum and position can be simultaneously
determined
Chap 15
Quantum Physics
112
Physics
For microscopic particles, h can not be
ignored and x px can not be simultaneously
determined. To describe their motion one has
to borrow the concept of probability. In
quantum mechanics, wave function is used to
describe particle’s states.
The uncertainty principle is one of the
foundational postulates of quantum
mechanics.
Chap 15
Quantum Physics
113
Physics
E.g.-1. The mass of a bullet is 10 g, speed
1
200 m  s . Momentum uncertainty is 0.01%
of its momentum (this is good enough in
macroscopic world), What is the position
uncertainty of the bullet?
Solution: Bullet’s momentum
p  mv  2 kg  m  s
Uncertainty of momentum
1
4
p  0.01%  p  2 10 kg  m  s
Chap 15
Quantum Physics
1
114
Physics
4
p  0.01%  p  2 10 kg  m  s
1
Uncertain range of the position
h 6.63 1034
30
x 

m  3.3 10 m
4
p
2 10
E.g.-2. An electron’s speed is 200 m  s -1 . The
degree of momentum uncertainty is 0.01%
of the momentum, what is the uncertainty
of position of the electron?
Chap 15
Quantum Physics
115
Physics
Solution: electron’s momentum
p  mv  9.11031  200 kg  m  s 1
28
p  1.8 10
kg  m  s
1
Uncertain range of the momentum
p  0.01%  p  1.8 1032 kg  m  s 1
Uncertain range of the position
h 6.63  1034
2
x 

m  3.7  10 m
32
p 1.8  10
Chap 15
Quantum Physics
116
Physics
1. Wave Function and Its Statistical
Explanation
(1) Wave Function
Due to the wave-particle duality of
microscopic particle, one can not determine
its position and momentum spontaneously,
the classical way of description of its states
breaks down, we use wave function
Chap 15
Quantum Physics
117
Physics
(1a) Classical wave and wave function
x
mechanical wave y ( x, t )  A cos 2π (t  )

E ( x, t )  E0 cos 2π (t 
em wave
H ( x, t )  H 0 cos 2π (t 
x

)

)
x
classical wave is a real function
y ( x, t )  Re[ Ae
Chap 15
Quantum Physics
x
i 2 π (t  )

]
118
Physics
(1b) QM wave function (complex function )
Wave function that descibe the Ψ( x, y, z, t )
motion of the microscopic particle
Wave-particle duality of   E ,   h
h
p
microscopic particles
The energy and momentum of free particle are of
certain values, its de Broglie wave length and
frequency are invariant, so it is plane wave with
infinity wave train, the x-position of the particle is
fully uncertain due to the uncertainty principle
Chap 15
Quantum Physics
119
Physics
Free particle plane wave function
Ψ ( x, t )   0 e
i
2π
( Et  px )
h
(2) The statistical interpretation of wave
function
Probability Density: the probability that the
particle appears in unit (spatial) volume
Ψ   *
2
Chap 15
Positive Real number
Quantum Physics
120
Physics
Probability that the particle appears at some
moment in a volume element dV
2
Ψ dV  ΨΨ dV
*
Hence de Broglie wave (or matter wave) is a
probability wave, it is very different with
electromagnetic wave
Chap 15
Quantum Physics
121
Physics
At some moment the probability one finds the
particle in entire space is
Normalization
Condition
Ψ
2
dV  1
(Bound State)
Standard Condition
Wave function is single-valued, real, finite
function
Chap 15
Quantum Physics
122
Physics
Erwin Schrodinger，1887 - 1961
Austrian theoretical physicist
Proposed the famous wave
equation with his name,
founded wave mechanics, and
its approximation methods.
1933 Nobel Prize for Physics
(with P. Dirac)
Chap 15
Quantum Physics
123
Physics
2. Schrodinger Equation
(1) free particle Schrodinger equation
Free particle plane wave function
Ψ ( x, t )   0 e
i
2π
( Et  px )
h
taking 2nd order partial derivative with
respect to x and 1st order partial derivative
with respect to t
Chap 15
Quantum Physics
124
Physics
One gets
Ψ
4π p

Ψ
2
2
x
h
2
2
Free particle
2
Ψ
i 2π

EΨ
t
h
(v  c) E  Ek
p  2mEk
2
1-dimension free particle time-dependent
Schrodinger equation
h Ψ
h Ψ
 2
i
2
8π m x
2π t
2
2
Chap 15
Quantum Physics
125
Physics
(2) Particle in potential field with potential
energy V p : E  Ek  Vp
1-dimensional time-dependent Schrodinger
equation
h 2  2Ψ
h Ψ
 2
 Vp ( x, t )Ψ  i
2
8π m x
2π t
(3) particle in stationary potential
2
p
Vp ( x ) time-indep.
E   Vp
2m
Chap 15
Quantum Physics
126
Physics
Ψ ( x, t )   0 e
 i 2 π ( Et  px ) / h
  0e
e
i 2 πpx / h
  ( x) (t )  ( x)   0 e
i 2 πpx / h i 2 πEt / h
1-dimensional stationary Schrodinger equation in
any potential field
d
8π m

( E  Vp ) ( x)  0
2
2
dx
h
2
2
Chap 15
Quantum Physics
127
Physics
Stationary Schrodinger equation in 3dimensional potential field
      8π m
 2  2  2 ( E  Vp )  0
2
x
y
z
h
2
2
Lapalce operator
2
2
2
2
2



2  2  2  2
x
y
z
2
8
π
m
2
   2 ( E  Vp )  0
h
Stationary wave function
Chap 15
 ( x, y , z )
Quantum Physics
128
Physics
e.g., stationary Schrodinger equation for
hydrogen atom
2
2
2
e
8
π
m
e
2
Vp  

  2 (E 
)  0
2
2
4πε0 r
h
4πε0 r
Properties of stationary
wave function
(1)E is time-independent
2
(2)  is time-independent
Chap 15
Quantum Physics
129
Physics
wave function single-valued, finite, continuous
(1)
 x, y , z 
2
dxdydz  1 normalization
  
,
,
(2)  和
continuous
x y z
(3)  ( x, y, z ) is finite, single-valued
Chap 15
Quantum Physics
130
Physics
3. 1-dim. Potential Well
Particle potential energy V p satisfies
boundary condition
0, 0  x  a
Vp 
Vp  , x  0, x  a
(1)Simplified model for free electron gas
model of metal in solid physics
(2)Demonstrate QM basic concepts and
principles with simple math
Chap 15
Quantum Physics
131
Physics
 Ep  , x  0, x  a
 Ep 
  0, ( x  0, x  a)
Ep  0,
0 xa
d 2 8π 2 mE

 0
2
2
dx
h
o
a
x
8 π 2 mE
k
2
h
d 2
2

k
 0
2
dx
Chap 15
Quantum Physics
132
Physics
d
2
k  0
2
dx
 Ep 
2
 ( x)  A sin kx  B cos kx
o
a
x
Wave function single-value, finite, and
continuous
 x  0,   0,  B  0
 ( x)  A sin kx
Chap 15
Quantum Physics
133
Physics
 x  a,  A sin ka  0
 Ep 
sin ka  0
 sin ka  0,  ka  nπ
nπ
k  , n  1,2,3,
a
2
8π mE
k
2
h
o
a
x
quantum number
2
h
E  n2
8ma 2
Chap 15
Quantum Physics
134
Physics
 ( x)  A sin kx
 Ep 
nπ
k  , n  1,2,3,
a
nπ
 ( x)  A sin
x
a
Normalization
2 a
2
A sin
0


 
nπ
xdx  1
a
Chap 15
o
2
a
a
x
dx    dx  1
*
0
2
A
a
Quantum Physics
135
Physics
hence
 Ep 
nπ
2
k
A
a
a
 ( x)  A sin kx
o
a
x
2
nπ
 ( x) 
sin
x , (0  x  a)
a
a
d  8 π mE

 0
2
2
dx
h
2
wave equation
Chap 15
2
Quantum Physics
136
Physics
 Ep 
wave function
 (x) 
0 , ( x  0, x  a)
2 nπ
sin x , (0  x  a)
a
a
o
a
x
2 2 nπ
Prob. density  ( x)  sin
x
a
a
2
Energy
Chap 15
2
h
2
En  n
8ma 2
Quantum Physics
137
Physics
Discussions：
 Ep 
1. energy quantization
Energy
2
h
En  n
2
8ma
2
o
h
E1 
, ( n  1)
2
8ma
2
g.s. Energy
a
x
2
h
2
excited state En  n 2

n
E1 , (n  2,3,)
2
8ma
the particle’s energy in 1-dim. infinity square well is
quantized.
Chap 15
Quantum Physics
138
Physics
(2) the prob. density that particle appears
in the well is different
Wave function  ( x) 
2
nπ
sin
x
a
a
2 2 nπ
Prob. density  ( x)  sin ( x)
a
a
2
e.g., when n =1, the maximum probability is
at the place x = a /2
Chap 15
Quantum Physics
139
Physics
(3) wave function is standing wave, the nodes
locate at the wall, the No. of valley equals
quantum number n
 ( x)  A sin
nπ
x
a
 ( x) 2 
n
2
nπ
sin 2
x
a
a
n
2
n4
16E1
n3
9E1
n2
n 1
x0
a
Chap 15
x0
Quantum Physics
4E1
E1
a
Ep  0
140
Physics
4. 1-dim. Square Well, Tunneling Effect
1-dim. Square Well
Vp ( x) 
0, x  0, x  a
Vp0 , 0  x  a
Particle’s
Energy
Vp ( x )
Vp 0
E  Vp 0
Chap 15
Quantum Physics
o
a
x
141
Physics
Tunneling Effect
Wave functions
in different
regions
 (x)
2
1
o
a
3
x
When particle’s energy E < Vp0 , the region
x > a is classically forbidden, however in
quantum mechanics, particle can penitrate
in the region with a non-zero probability
Chap 15
Quantum Physics
142
Physics
Applications
 STM (1981)
Scanning Tunneling
Microscopy
AFM (1986) Atom
Force Microscopy
Xenon on Nickel
Single atom lithography
Chap 15
Quantum Physics
143
Physics
Quantum Corrals
Iron on Copper
Chap 15
Imaging the standing wave
created by interaction of
species
Quantum Physics
Physics
1. Schrodinger Equation of Hydrogen Atom
Potential energy of electron in H-atom
2
e
Vp  
4πε0 r
Stationary Schrodinger equation:
2
2
8
π
m
e
2
   2 (E 
)  0
h
4πε0 r
Chap 15
Quantum Physics
145
Physics
Spherical Coordinates
Transform to spherical polar
coordinates because of the
radial symmetry
x  r sin  cos 
y  r sin  sin 
z  r cos 
r  x2  y 2  z 2
z
r
y
  tan 1
x
  cos 1
 Polar angle 
 Azimuthal angle 
Chap 15
Quantum Physics
Physics
In Spherical coordinates:
1  2 
1


1
 2
(r
) 2
(sin
) 2 2
2
r r
r
r sin  

r sin   2
8π m
e
 2 (E 
)  0
h
4πε0 r
2
2
Separable solution, let
 (r , , )  R(r )Θ( )Φ( )
Chap 15
Quantum Physics
147
Physics
We get
d 2Φ
2
 ml Φ  0
2
d
2
ml
1
d
dΘ

(sin 
)  l (l  1)
2
d
sin  Θ sin  d
1 d 2 dR 8π 2 mr 2
e2
(r
)
(E 
)  l (l  1)
2
R dr
dr
h
4πε0 r
Chap 15
Quantum Physics
148
Physics
2. Quantization condition and quantum
number
Solve Schrodinger equation we get the following
quantum number and quantization properties:
(1) Energy quantization and principal
quantum number
1
En  2 E1 n =1,2,3,... Principal quantum
n 4
number
me
E1   2 2  13.6 (eV)
8 0 h
Chap 15
Quantum Physics
149
Physics
(2) Angular momentum quantization and
angular quantum number
h
Angular momentum： L  l (l  1)
2π
l  0，
1，
2，
，
(n  1)
Orbital angular
quantum number
E.g.，n =2，l = 0,1 corresponds to
h
L0 L 2
2π
Chap 15
Quantum Physics
150
Physics
(3) Angular momentum spatial quantization
and magnetic quantum number
In applied magnetic field, angular momentum L
can only take some specific directions,
projection of L along magnetic field satisfies
h
Lz  ml
 ml 
2π
ml  0,1,2,,l magnetic quantum
number
  h / 2 π reduced Planck const.
Chap 15
Quantum Physics
151
Physics
h
h
e.g., when l  1 L  l (l  1)  2  2
2π
2π
magnetic quantum number
ml =0, 1 and Lz  0, h , h
2π
z
z
LZ
2π
L
ħ
o
L
2
ħ
Chap 15
Quantum Physics
152
Physics
(4) Spin and spin quantum number
Spin angular momentum
S  s( s  1)
1
3
where spin quantum number s 
S

2
2
Spin angular momentum takes only two
components along applied magnetic field:
1
S z  ms  ms  
2
ms spin magnetic
quantum number
Chap 15
Quantum Physics
153
Physics
1
ms  
2
S z   / 2
Spin angular momentum and spin magnetic
quantum number of electron
z
Sz
S
Sz
1

2
o
1
 
2
Chap 15
Quantum Physics
1
ms 
2
S 
3

2
1
ms  
2
154
Physics
(5) Summary
The states of electron in hydrogen atom can
be represented by 4 quantum numbers (qn.),
(n, l ,ml , ms)
Principal qn. n determines energy
Angular qn. l determines orbital angular
momentum
Magnetic qn. ml determines direction of
orbital angular momentum
Spin qn. ms determines direction of spin
angular momentum
Chap 15
Quantum Physics
155
Physics
3. Ground state radial wave function and
distribution probability
(1) Ground state energy
Ground state
n=1
l=0
Radial wave function equation:
1 d 2 dR 8π mr
e
(r
)
(E 
)0
2
R dr
dr
h
4πε0 r
2
2
2
solution R  Ce  r / r1
Chap 15
Quantum Physics
156
Physics
where
2
r1
 h /(8π mE )
2
2
 8π 2 me2 2 



r 0
2

r1 
 4πε0 h
Substitute into
ε0 h
r1 
 0.052 9 nm
2
πme
2
get
2
h
E   2 2  13.6 eV
8π mr1
Chap 15
Quantum Physics
157
Physics
(2) Ground state radial wave function
R  Ce
 r / r1
the probability that electron appears in volume
element dV:
Ψ dV  R Θ Φ r sin drdd
2
2
2
2
2
let the prob. density along radial vector p, the
prob. that the electron appears in (r , r+dr)
2
pdr  R r dr
2

Chap 15
π
0
2
Θ sin d
Quantum Physics

2π
0
2
Φ d
158
Physics
2 2
from normalization pdr  R r dr


2
0 pdr  0 R r dr  1  R  Ce


0
2
C e
2
 r / r1
1/ 2
 2 r / r1
r dr  1
2
4
 C   3 
 r1 
1/ 2
 4   r / r1
g.s. radial wave function is R (r )   3  e
 r1 
Chap 15
Quantum Physics
159
Physics
(3) Probability Density Distribution of Electron
p(r)
p(r )  r 2 2
o
r1
r
Chap 15
Quantum Physics
160
Physics
Light
Amplification by
Stimulated
Emission of
Radiation
Chap 15
Quantum Physics
161
Physics
1. Spontaneous and stimulated radiations
(1) Spontaneous radiation
the process by which an atom in an excited state
with higher energy E2 undergoes a (spontaneous)
transition to a state with a lower energy E1 , e.g., the
ground state, and emits a photon, the frequency of
the radiation is determined by
E2  E1

h
Chap 15
Quantum Physics
162
Physics
Spontaneous Radiation
E2
.
E2
。
h
.
E1
E1
Before Radiation
After Radiation
E2  E1

h
Chap 15
Quantum Physics
163
Physics
(2) Absorption of light
the process by which an atom in a state with lower
energy E1 , e.g., the ground state, absorb a photon
energy h , spontaneously transit to a state with a
higher energy E 2 , and E2  E1  h
E2
Excited
Absorption
E1
.
E2
h
E1
Before Absorption
Chap 15
.
。
After Absorption
Quantum Physics
164
Physics
(3) Stimulated radiation
the process by which an atomic electron at energy
level E2 , interacting with an electromagnetic wave
of a certain frequency may drop to a lower energy
level E1 , transferring its energy to that field. A
photon created in this manner has the same phase,
frequency, polarization, and direction of travel as the
photons of the incident wave, and satisfies h  E2  E1
Chap 15
Quantum Physics
165
Physics
Stimulated Radiation
E2
.
h
E1
Before
E2
。
E1
.
After
h
h
Amplification of
stimulated radiation
when a population inversion is present, the rate of
stimulated emission exceeds that of absorption,
results in a coherent amplification  laser
Chap 15
Quantum Physics
166
Physics
2. The principle of laser
(1) Normal and inverse distribution of
population
N i  Ce
 Ei / kT
N1 / N 2  e
N1  E1
( E1  E2 ) / kT
N 2  E2
known E2  E1
N1  N 2 shows that the electron population
at lower energy level greater than that at
higher level, this is normal distribution
Chap 15
Quantum Physics
167
Physics
N 2  N1 is instead inverse distribution of
population, or simply population inversion
Population normal distribution and inversion
E2
E1
N2
E2
.............
..
。
。
。
。
。
。
。
。
。N1
。
。
。
。
E1
。
。。
。。 N1
.. .. .
E2  E1
Normal
Chap 15
E2  E1
N2
Inversion
Quantum Physics
168
Physics
T. H. Maiman (U.S. physicist) made the first
functional ruby laser in sept., 1960
E3
。
Excited state
.
E
Metastable state
2
.
。
Ground
E1
state
Energy level of ion Cr in Ruby laser
Chap 15
Quantum Physics
169
Physics
(2) Optical resonant cavity
Formation of laser light
Light confined in the cavity reflect multiple times
producing standing waves for certain resonance
frequencies. When the standing wave condition is
satisfied the light is amplified, one obtains laser

standing wave condition l  k
2
Chap 15
Quantum Physics
170
Physics
Optical resonator
.
Laser beam
l
HRM
PTM
Demonstration of O.R.
Chap 15
Quantum Physics
171
Physics
3. Laser
(1) Helium-Neon Gas Laser
PTM: partially transmissive mirror
HRM: highly reflectance mirror
A
K
PTM
HRM
He-Ne Laser
Chap 15
He 1
2
Metastable
Ne
632.8 nm
3
Ground state
Energy levels of He and Ne
Quantum Physics
172
Physics
HELIUM-NEON GAS LASER
Chap 15
Quantum Physics
173
Physics
(2) Ruby (CrAlO3) laser
Its active medium is ruby crystal rod, generates
pulse laser with wavelength 694.3 nm.
Pulse
High
reflectance
mirror
。。
U
0
Ruby rod
Partialy
transmissiv
e mirror
。
U。
Demo. of Ruby Laser
Chap 15
Quantum Physics
174
Physics
NEODYMIUM YAG LASER
Rear Mirror
Adjustment Knobs
Safety Shutter Polarizer Assembly (optional)
Coolant
Beam
Tube
Adjustment
Knob
Output
Mirror
Beam
Q-switch
(optional)
Beam Tube
Nd:YAG
Laser Rod
Flashlamps
Pump
Cavity
Laser Cavity
Harmonic
Generator (optional)
Courtesy of Los Alamos National Laboratory
Chap 15
Quantum Physics
175
Physics
4. Characteristics and Applications
of Laser
(1) highly-directional, a laser collimator can reach
accuracy of 16 nm/2.5 km.
(2) highly-monochromatic, 1010 better than ordinary
light
(3) focusing, laser light focuses 100 times better
than ordinary light
(4) coherent, ordinary light source generates
incoherent light, while laser light is highly coherent
Chap 15
Quantum Physics
176
Physics
Incandescent vs. Laser Light
1. Many wavelengths
1. Monochromatic
2. Multidirectional
2. Directional
3. Incoherent
3. Coherent
Chap 15
Quantum Physics
177
Physics
1. Energy Gap of Solids
Fully Separated Energy Levels of Two H-atom
2p
2s
2p
2s
1s
1s
e
A
+
e
Chap 15
e
+
B
e
Quantum Physics
178
Physics
Six closed H-atom’s
energy level split
Two closed H-atom’s
energy level split
E
E
O
2p
2s
2s
1s
1s
r
r
O
E
Energy Band
of Solids
2s
r
O
Chap 15
Quantum Physics
179
Physics
2(2l  1) quantum states
per energy level
Electron distribution
of different energy
bands in Na
3p
2(2l  1) electrons per
N
3s
6N
2p
2N
2s
2N
1s
energy level
2(2l  1) N electrons per
energy band
Chap 15
Quantum Physics
180
Physics
Experiments show that:
The interval between the highest and the lowest
energy level in a energy band is less than the
2
10
eV , the number of N atoms is of
order of
19
3
order 10 mm , hence the distance of the
neighboring energy levels is about
10 2 eV/1019  10 17 eV
Chap 15
Quantum Physics
181
Physics
Energy band of crystals
E
Conduction
Forbi band
Empty
band
Eg
CondForbi uction
-dden band
band
Eg
Valence
band (not
full)
-dden
band
Valence
band
(full)
Chap 15
Quantum Physics
182
Physics
Comparison between Conductor,
Semi-conductor and Insulator
Conductor Semiconductor Insulator
Resistance 10 8 ~ 10 4
(Ω m)
Temp.
Coeff.
Pos. +
F-band
V-band
Not full
Chap 15
10 4 ~ 108
108 ~ 10 20
Neg. -
Neg. -
Small
Large
Full
Quantum Physics
Full
183
Physics
Typical Semiconductors
Silicon
GaAs
Diamond Cubic Structure
ZnS (Zinc Blende) Structure
4 atoms at (0,0,0)+ FCC translations
4 atoms at (¼,¼,¼)+FCC translations
Bonding: covalent
4 Ga atoms at (0,0,0)+ FCC translations
4 As atoms at (¼,¼,¼)+FCC translations
Bonding: covalent, partially ionic
Chap 15
Quantum Physics
Physics
2. Intrinsic and Extrinsic semi-conductor
(1) Intrinsic: pure, no dopants
Normal Bond in Ge
electron
e
Eg
C-band
Ge
Ge
Ge
Ge
Ge
Ge
Ge
Ge
Ge
F-band Electrons are excited, Holes appear
e
Full
band
Ge
Ge
 eG
Ge
Ge
e
e
Ge
Ge
Ge
Ge
hole
Chap 15
Quantum Physics
185
Physics
(2) Extrinsic semiconductor)
Electron type (n-type)
Phosphorus atom are dopant

Si atoms are hosts，As
Si
Si
Si
Si
i
Si
As
Si
C-band
Si
Si
Si
Si
Si
Si
eS
Si
Si
Si

Si
Si
Si
Donor Level
Si
Donor level
Si
Si
Si
Chap 15
V-band
Si
Quantum Physics
186
Physics
p-type semiconductor
Boron atom doping into Ge
atom lattice
Acceptor level
Hole
C band
Ge
Ge
Ge
Ge
Ge
B
Ge

Ge
Acceptor level
Ge
Ge
Chap 15
Ge
Quantum Physics
V band
187
Physics
3. PN Junction
Current-Volt Characteristics of pn Junction
p n
I
U
I


U
p n
U
I


Chap 15
Quantum Physics
188
Physics
p
e
e
Hole
n
p
e
e
-- --- ---
n
+
+
++
++
++
++
x0
Electron
Voltage variation between p-layer and n-layer
U0
x
x0
Chap 15
Quantum Physics
189
Physics
4. Photovoltaic effect
e
e
n


e
P
e
Light
γ  e  e
Photovoltaic effect is the creation of voltage or
electric current in pn upon exposure to light
Chap 15
Quantum Physics
190
Physics
1. The transition temperature of
superconductor
R ( )
around
T=4.20K
risistance
is ZERO
0.150
**
*
0.100
0.050
Tc
: the critical
temperature
0.000
4.00
Chap 15
4.20
4.40
Quantum Physics
T /K
191
Physics
2. Major Properties of Superconductors
(1) Null resistance
When T  Tc , I  I c (critical electric flow)
resistance   0 conductance   
(2) Critical magnetic field
The critical point of applied magnetic
fields that breaks the superconducting
states
Chap 15
Quantum Physics
192
Physics

T 2
H c  H 0 1  ( ) 
Tc 

H
H c (T )
T  0 K , Hc  H0
Normal
Super-
conductor
(3) Meissner effect

o
TC
T
 
 
dΦ
d( B  S )
E  dl  

dt
dt
Chap 15
Quantum Physics
193
Physics
in superconductor
E 0
when H applied  H c
dB / dt  0
H in  0

H

H
H in  0
S
N
I
Chap 15
Quantum Physics
194
Physics
3. BCS Theory of Superconductivity
BCS Theory: proposed by Bardeen, Cooper,
and Schrieffer (BCS) in 1957, is the first
microscopic theory of superconductivity
since its discovery in 1911. Interestingly, this
theory is also used in nuclear physics to
describe the pairing interaction between
nucleons in an atomic nucleus.
Chap 15
Quantum Physics
195
Physics
BCS=Bardeen, Cooper, Schrieffer
Chap 15
Quantum Physics
196
Physics
An electron moving through a conductor will attract
nearby positive charges in the lattice. This deformation
of the lattice causes another electron, with opposite
"spin", to move into the region of higher positive charge
density and to be correlated. A lot of such electron pairs
overlap very strongly, forming a highly collective
"condensate"
deformation of local area
Normal
location of
Lattice
e
Chap 15
deformation
of lattice
Quantum Physics
197
Physics
Chap 15
Quantum Physics
198
Physics
Phonon: a collective excitation in a periodic lattice
of atoms, such as solids. It represents an excited
state in the quantum mechanical quantization of
the modes of vibrations of elastic structures of
interacting particles.
Cooper Pair: two electrons couple by exchanging
phonon, and form the coupled electron pair called
Copper pair
The distance between two electrons is about 10 6 m
their spins and momenta are opposite, the total
momenta is zero.
Chap 15
Quantum Physics
199
Physics
4. The Perspectives of Superconductor
(1) Create strong magnetic field
(2) Energy & power industry, e.g., power
storage etc.
(3) Magnetic levitated high-speed train
(4) Medical applications, e.g., nuclear
magnetic resonance imaging
Chap 15
Quantum Physics
200