# Chapter 8 Quantum Theory - PowerPoint

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```					            Chapter 8: Quantum Theory
•   Failures of classical physics
-   Heat capacities
-   Photoelectric effects
-   Diffraction of electrons
-   Atomic spectra
•   Quantum mechanics
-   Schroedinger equation
-   Born interpretation
-   Uncertainty principle
•   Applications of quantum mechanics
-   Translation motion: a particle in a box
-   Rotation motion: a particle on a ring
-   Vibration motion: the harmonic oscillator
•Failures of classical physics

Around the turn of 19th century, 3 principles of classical
physics were challenged by new findings.
1. A particle travels in a trajectory
-location and velocity (momentum) of a particle
are known at all time (Newton’s law)
F  ma
dV   d 2x
    m 2
dx   dt
2. A particle can possess any arbitary energy
-There exists continuous excited states
3. Waves and particles are distinct concepts

Black-body is a body that can emit and absorb all
frequencies of electromagnetic radiations.

Fig 12.3 p 287, Atkins
All substances contain black-bodies.
It was found that radiation emitted or absorbed by
black-body changes with temperature
Iron turns from red to white (red+blue) when heated.

Fig 12.3 p 287, Atkins
Not all frequencies are emitted or absorbed with the
same energy density.
There exists a maximum corresponding to a particular
wavelenght, lmax, in the curve.
This lmax changes with temperature
As temperature is raised, lmax moves toward smaller
wavelenght (from red to blue) called “blue shift”
Wien’s displacement law:
1
Tlmax  c2
5
c2, second radiation constant, 1.44 cm K
Another features of black-body radiation given by Josef Stefan called
Stephan-Boltzmann law is

E = aT4

E  total energy density
or
M = T4

M = emittance i.e. brightness of emission
 = 5.67 x 10-8 wm-2k-4
Rayleigh-Jeans proposed that black-body consists of oscillators
which can emit or absorb light at any frequency
Using classical physics and equipartition principle, they arrived
at Rayleigh-Jeans Law
dE =  dl
where  = 8pkT/ l4
Rayleigh-Jeans law suggested there can be emission of very
short wavelength even at room temperature.
In other word, there should in fact be no darkness. (objects
should glow in the dark.)
Failure of Reyleigh-Jeans law are called “UV-catastrophe”
Planck took Rayleigh-Jeans’s idea but instead of allowing oscillators
to be able to emit or absorb light at any frequency each oscillator
possesses only discrete values of energy
E = nhn

h = Planck’s constant = 6.626 x 10-34 Js

Planck’s hypothesis also implies that “energy is quantized” and is
regarded as “Quantum theory”
By using statistical interpretation Planck proposed that
dE =  dl
where
8phc      1    
  5  hc / lkT 
l e        1 
For very large l
hc / l kT           hc
e            1        ...
l kT
11             l kT
hc / lkt / l kT

e      e hc  1 hc
8p hc lkT
8 hc l kT
 8

l 5 hc
l       hc
  8 kT
8 pkT
                        Rayleigh-Jeans Law
l
l4
4
By integrating l from 0 to , one obtains


E    d l  aT 4
0

4       2p5kk 4
2 5 4
a    ,            5.67 x106 Wm 2 k 4
c      15c 2 h3
-Stefan-Boltzman law

By differentiating   dE
0
dl
hc
T lmax         -Wien’s displacement law
sk
5k
hc
c 1.439cmK
 s 2 =1.439 cm K
k
Example 8.1 Find the surface temperature of sun where
the maximum emission occurs at 490 nm.

Wien’s displacement law:

Tlm ax  2.9 mm  K
2.9 x103 m  K
T                      5.9 x103 K
490x109 m
 6,000K
Example 8.2 Calculate the number of photons emitted by a 100
W yellow lamp in 10.0 s. Take the wavelength of yellow light as
560 nm and assume 100 percent efficiency.

Energy emitted by lamp
Eemitted      = (Power)(time)
= (100 W) (10.0 s)
= 1,000 J

Photon energy
Ephoton = hn = h c/l
=(6.626x10-34 J s)(2.998x108 ms-1)
(560x10-9 m)
= 3.547x10-9 J
Nphoton = Eemitted/Ephoton
= 1,000/3.547x10-9 J = 2.82x1021
• Heat capacity

In 1819, Dulong – Petit proposed that for all monoatomic
solids, heat capacities are about 25 J/Kmol (3R) at all
range of temperature
 Dulong – Petit arrived at this conclusion by assuming
the concept of classical physics
heat capacities of solid is the energy used to oscillate or
vibrate atom at the lattice position
Each atom can vibrate in 3 orthogonal directions
For N atoms solid, their will be 3N motions
From equipartition theory, each motions uses energy of
kT
The total vibrational energy
U = 3NkT
For 1 mole of atoms

U          3NA kT
3 NkT
 3RT
 U m   
Cv , m            
 T     v
 3R        ~25 J/Kmol
Later on with the advance in measurement, it was found that at low
temperature all substances have heat capacities lower than 25 J/Kmol.
Diamond has Cm = 6.1 J/K mol at 25°C
Dulong-Petit
Albert Einstein borrowed Planck’s idea. He proposed that
each atom can possess only a discrete amount of energy
which is an integral multiple of hn
3 N A hn
Um       
e hn kT  1
 U m 
Cv , m            
 T v
3 N A hn                    hn hn
                                    e            kT

e            
2             2
hn kT
1              kT

 hn                     e hn
2                   kT
 3R    2 
 kT            e                   
2
hn kT
1
hn kT        2
 hn   e                     
 3R  2 hn                       
 kT e
kT
 1
Cv , m    3Rf (T )
Debye corrected Einstein’s formula by allowing atom to
oscillate at more than one frequency value
• Photoelectric effects

In late 1800s, it was demonstrated that electrons can be
ejected from surface of certain metals when exposed to
light of at least a certain minimum frequency.

This phenomena is called “photoelectric effect”
In 1905, Albert Einstein using Planck’s quantum concept
was able to explain the “photoelectric effect”

Einstein’s proposal:
1) Light is composed of “light particle” called
“photon”.
2) Each photon has discrete value of energy
E = hn
= h c/l
3) Let F, work function of metal, be the minimum
energy that required to knock electrons out of metal
(binding energy, each metal has different F).
If photon energy is equal to F, electrons will be ejected
from metal when exposed to light.
If photon energy is greater than F, electrons will be ejected
from metal and carry kinetic energy EK when exposed to
light.
EK = hn - F

EK could be measured, F could then be determined
Example 8.3 The work function for metallic caesium is 2.14
eV. Calculate the kinetic energy and the speed of the electron
ejected by light of wavelength 250 nm.

F  2.14 eV  2.14 x1.602 x1019 J
 3.428x1019 J
c
EK  h       F
l
(6.626 x1034 Js )(2.998x108 ms 1 )
                                       3.428x1019 J
(250 x109 m)
 7.946 x1019 J  3.428x1019 J
 4.518x1019
1 2
E K  mv
2
19
2 EK   2 x 4.518x10 kgm2 s  2
v      
m            9.1x1031 kg
 996.5 x103 ms 1
• The diffraction of electrons

In 1925, Davisson and Germer observed the diffraction of
electrons by a crystal
Davisson-Germer’s experiment has been repeated using other
particles such as H2
“particles have wave-like properties”
photoelectric effect suggests that
“waves have particle-like properties”
This phenomena is called
“wave-particle duality”
In 1924, de Broglie suggested that any particle traveling with a
linear momentum, p , should have a wavelength λ according to
λ = h/p
faster object will have smaller wavelength
heavier object will have smaller wavelength
Example Estimate the wavelength of electrons that have
Example 8.4 8.4 Estimate the wavelength of electrons that have been

accelerated from rest through a potential difference 1.00 kV.
been accelerated from rest through a potentialdifference of of 1.00 kV

Ek        e
1 2
      mv
2
p2
p  mv         
2m
p                 2me
h            h
l                    
 2me 
2
p
6.262 x1034 Js

 2 x9.110 x10                                        
1
31                  19            3             2
kg 1.602 x10         C 1.0 x10 V
 3.88 x1011 m
 38.8        pm
• Atomic spectra

Radiation is emitted (or absorbed) at a series of discrete frequencies
This phenomena can be explained if we assume energy of atomic
state is quantized and light omits or absorbs as electron transfer
from lower to higher state or vice versa and releases photon
Quantum Mechanics

Failures of classical physics made scientists realized that new kind
Of mechanics was needed.
This new mechanics should not give accurate trajectories of
particles.
This new mechanics was later called “Quantum mechanics”
Uncertainty principle

In 1927, Heisenberg pointed out that for very small particles
(quantum particles) like electrons it is not possible to measure
accurately their positions (x) and linear momentum (px) at the same
time.
This statement is now known as “the uncertainty principle”.
2sin 


hn
e-

x
For us to find the position of electron, light (hn) must hit electron and
reflect back to the lens of the microscope.
l
The accuracy of measurement can be determined from
2sin 
where l is the wavelength of light and 2sin is the dimension
of the lens
From de Broglie’s relation,
h    h h
l  p  
P
P
p    l c
h/l’
h/l

e-    
mv
e-
x

As photon hits electron, momentum is transferred.
The leaving electron carries momentum of mv while leaving photon
has momentum of h/l’ (or energy hn’)
From law of conservation

incoming momentum = outgoing momentum

Along x-axis

h        h
        cos   mv cos            (1)
l       l'
Along y-axis
h
0           sin   mv sin            (2)
l'
From (1)
h       h
mv cos                       cos 
l       l'
Generally l’ > l but for simplicity assuming l’  l then
h
Px
px      (1  cos  )
l

                    90 -  <  <- 90 + 
e- 
Thus

h                           h
(1  cos(90   )  Pxx  (1  cos(90   )
<p <
l                           l
h                      h
(1  sin  )  Px < (1  sin  )
< px 
l                      l
The accuracy of measurement of momentum is then
h
P
pxx      sin 
l
The accuracy of measurement of position is
l
x
sin 
h          l
xp
xP        sin 
l        sin 
xp h
x P                      “uncertainty principle”
-Schrödinger equation

Wave concept was introduced to explain “quantum” particle.
Schrödinger adapted Maxwell’s equation for electromagnetic radiation for
describing behavior of quantum particle.

Maxwell’s equation

 2W  2W  2W 1  2W
 2  2  2                    W=amplitude of wave
x 2
y  z   c t 2
E
particle has energy     E  hn  n 
h
and momentum                h        h
P l 
p l          P
p
where l is de Broglie wavelength of the particle
velocity of the particle can be given by
n  ln
h E E
 . 
P h P
Replacing wave’s amplitude by wave function  (psi) which gives the
information about positions of electron

 2  2  2 P 2  2
 2  2  2 2
x  2
y  z  E t

(x,y,z,t) is not trajectory, it gives information about position but not
exactly like the amplitude of the wave
For localized or standing wave

Thus
2pnit
 ( x, y, z, t )   ( x, y, z )e    2 n it

  2  2  2  2pnitit P 2      2pnit
 2     2  2 e     2 n
 2e    2 n it
(2n i ) 2
2pnit
 x     y    z              E
  
2      2     2     p2
P2
 4 n2 2
 2  2  2 2 4p22n 
x 2   y    z    hn
 2
  2
 2           p2p22
2 P
( 2  2  2 )  4              
x     y   z             h 2
Kinetic energy T=p2/2m and total = kinetic + potential (E=T+V)

2  2  2      2m4p2 P2
2mn 4 2 p 2
( 2  2  2 )                
x y z           h 2
2m
2

8 2 m
-
 8p2m T
h2
h2
 2  2  ( E  V )
8 m
p2 m
 2  Laplacian operator
  2 22
h2                                       h
or   2   V   E
     V                             
                                        2p
 82mm      
H  hamiltonian operator

or
H  E                           “Schrodinger equation” or SE
SE is differential equation
SE is an eigen-value equation
E is an eigen-value which represents energy of particle
Solution of SE is wave function 

SE can be solved using methods of differential equation
Born Interpretation

In wave theory, square amplitude of wave is interpreted as intensity.

Using the same analogy, Born gave interpretation of wave function as

“The probability of finding a particle in a small region of space dV
is proportional to ll2dV, where  is the value of wave function in the
region.”

In other word, ll2 is the probability density
Applications of quantum mechanics

-Solution to SE
H             E
 2 2
H                 V
2m

For 1-dimensional system
 2 d2
H               2
V
2m dx

> Free electron V = 0
 2d2
H     
2m dx 2
 d 22 
2
 E
2m dx
  Aeikx  Be ikx                             A, B  coefficient 
d
       Aikeikx                  B  ik  e  ikx
dx
d 2
 ik                           ik        Be ikx
2                               2
                    Ae   ikx

dx 2
      
2
-kk
2
 Ae   ikx
    Be  ikx   
2
2  1
ii =-1          
-kk  
2 2

 2 2 2
     -kk  
                          E
2m
 2k 2
 E         
2m
> 1D –Square potential well

 x  L 
            
V    0 0  x  L 
 x  0 
            
Area =1 ≡ prob.
Probability

I         II      III

              0             L                       
xL             0
0 x L          Aeikx  Beikx
x0             0
We can substitute
eix     cos x  i sin x
eix    cos x  i sin x
Aeikx  Beikx      A  B  cos kx       A  B  i sin kx
 C cos kx  D sin kx
At x = 0
I      II

0     C cos k 0  D sin k 0
0     C 1 D 0
C     0
     II  D sin kx
at x  L
II = III
I     II

D sin kx  0
If D = 0, trivial solution is obtained. Then D must not be equal to 0 and
sinkL must be equal to 0.

sinkL = 0       only when kL = nπ

k = nπ/L
npx
nx
  II     D sin
L
Using Born’s interpretation

               2



dx  1

0                   2                 L         2                   2



I           dx           
0
II       dx    
L
II       dx  1

L         2

0                       
0
II       dx  0  1

2
np x
L
D  sin
2
dx                                   1
0
L
From       cos (A+B) = cosAcosB - sinAsinB
cos2x    = cosxcosx - sinxsinx
= cos2x - sin2x
= (1-sin2x) - sin2x
cos2x    = 1- 2sin2x
sin2x   = (1 – cos2x)/2
L     2
np x
L   1  cos 2 np x
L d np x      L

0
sin
L
dx         
0
2                    L np
 1 L np x                         1

 

L

L
 d
1
  cos np x d 2np x
np             L 2          L        L 2

 20           0                    
L    1 np x L L
               1    2np x 
L

L0 
                     d  sin
np                            L0

 2        0
4           
L  np
1  0    sin 2np  sin 0 
1
                                              
np  2                  4                  
L

2
2
np x
L
L
D 2  sin            dx        D2               1
0
L                      2
2
D 
L

2      np x
 n  x        sin
L          L
2 2
k            h2n2
E                 
2m             8mL2
h 2 n 2p 2
                   n  1, 2,3,....
8p m L
2        2
The lowest energy is achieved when n = 1, and equal to h2/8mL2.
The lowest energy is non-zero called “zero-point” energy.
Excitation or transition to adjacent wall could be calculated as

E  En 1  En


h2
8mL2

 n  1  n2
2

h2

8mL2

n 2  2n  1  n 2   
h2
  2n  1
8mL2
>Expanding to 3D                      particle in a box

  x, y, z     x   y   z 
c                        2     np x 2     mp x 2     lp x
   sin        sin        sin
b                            a      a   b      b   c      c
a
8      np x     mp x     lp x
 nml  x, y, z            sin      sin      sin
abc      a        b        c
h2  n2 m2 l 2 
En , m , l        2 2  2
8m  a  b  c 

Particle in a box is a model representing “translation motion”
Particle on a ring

Particle is held on the ring, hence travels
in 2 dimension with V=0
Thus
2 2 2 )
H  (     2
2m x2 y

H  E
2 2 2 ) E
 (    
2m x2 y2
y
r is fixed        To solve this equation, one must
transform from cartesian (x,y) to polar
r                         coordinate (r,)
p                     0
2p         x
 2  2   1 d 2            Aeim
 2  2
x  2
y  r d 2           d
 A(im)eim
2
d 2                 d
             E
2mr d
2     2
d 2
 A(m 2 )eim  m 2
d 2    2 IE               d 2
 2
d  2
2 IE
m  
2
2

I moment of inertia = mr2
m2 2
Em 
2I
2p


0
 * d  1
2p
A2  e  im eim d  1
0

A2 |0 2p  1
2p A2  1
1
A
2p
1 im
 m ( )     e
2p
From boundary condition

 (0)   (2p )
 ( )   (  2p )                  eix  cos x  i sin y
1 im
e 
1 im (  2p )
e
eip  cos p  i sin p  1
2p          2p                   ei 2p m  (1)2 m  1
1 im i 2p m
          e .e                      m  0, 1, 1,...
2p
e i 2p m  1          Thus, 2m must be positive or negative even integer
From classical interpretation

Angular momentum j = pr

2
Linear momentum
p
P2
j 2
E        
2m 2mr 2                  from Schrödinger equation
j 2 m2 2
E       
2I     2I                  From deBroglie’s relation p=h/l

j 2  m2 2  j  m                         hr   h
j    m
l     2p
2p r
l
m
Particle on a ring represents rotation motion
Harmonic oscillator

θ
k

m

x
x0

k
m

x   V = 1/2kx2
x0
 21 22
V 
H        V
kx
2m 2
 2 d2     1 2
H             2
   kx
2m dx     2
H  E
 2 d 2     1 2
      kx   E
2m dx 2       2
d 2     2m       1 2
   2 
E  kx   0               (1)
         
2
dx                2
2m              2m 1
Given   2 E        ;  2 k
2

2
mk


d 2
Then (1) is, 2
dx

  2  
   2x2     0   (2)
Let  ( xi ),    x
d         d  d                d
                    
dx         dx d                 d
d 2       d  d           d d      d 
                              
dx 2       dx  dx          dx d     d 
 d 2 
  2 
 d 
d 2               
Then (2) is,                  2             0   (3)
d 2               
 2
If we assume     u   e               2

d 2u        du      
Then (3) is,           2       1 u  0             (4)
d 2        d      
Hermite ' s equation
d 2u       du
2
 2x     2nu  0
dx         dx
which has solution

H n  x    1 e
n   d n x 2
x2

dx n
e   
called " Hermite polynimials "
Thus
    1
Ev       v   hn        v  0,1, 2,....
    2
The first 5 polynomial are:

H0(x) = 1
H1(x) = 2x
H2(x) = 4x2-2
H3(x) = 8x3-12x
H4(x) = 16x4-48x2+12


If we set      1 = 2n, then (4) is the Hermite’s equation.

Equation (4) has Hermite polynomials for solution. The wave
function for the harmonic oscillator is then given by,

 2
 v    H v   e          2

v
 2                    2
v=    0  0    N 0 H 0   e                   2
 N0e           2

 2                      2
1  1    N1 H1   e                 2
 N1 e          2
…

* N0,N1 are normalizing factor
Take definition of α and β

      2mE
1                       1  2v
        2
mk
m 2p E      1                    1 1        m
 v              E  (v  )h
k h         2                    2 2p       k

From classical physics
1   k
(vibration frequency)      
2p   m

1
Thus         E  ( v  ) hn       v = 0,1,2,…
2
“Harmonic oscillator represents mode of vibration”
ax          ax 2
Example 8.5 Show whether (a) e or (b) e          is eigenfunction of
the operator “d/dx” and find the corresponding eigenvalue

ˆ
f(x) would be an eigenfunction of an operator  only when
ˆ
f ( x)   f ( x)

where  is an eigenvalue

d ax
(a)           (e )  a ( e ax )
dx
d ax 2             ax 2
(b)           (e )  a ( xe )
dx

e ax is an eigenfunction of d/dx with eigenvalue “a”
ax 2
e is not an eigenfunction of d/dx
Example 8.6     Proof that “sinx” and “sin2x” which are
eigenfunction of d2/dx2 are mutually orthogonal.

Orthogonal means

 f ( x) g ( x)dx    0
2p
Thus      sin x sin 2 xdx
0
 ?

From
cos( A  B)  cos A cos B  sin A sin B            (a)
cos( A  B)  cos A cos B  sin A sin B            (b)
(b)  (a)
cos( A  B)  cos( A  B)  2s in A sin B
1
2sin A sin B        cos( A  B)  cos( A  B)
2
2p                       2p

  cos (2x – x)- cos  2 x  x  dx
1
 sin x sin 2 xdx                  A  B 
0
20
2p
1        2p    sin 3x  
    sin x 0            
2                3 0   
 0
“eigenfunctions of the same operator must be orthogonal”
Example 8.7 Find the normalizing constant for ground state and first
excited state of harmonic oscillator

 2                        2
 0    N 0 H 0   e              2
 N 0e              2

 2                          2
 1    N1 H1   e            2
 N1 e               2



  0   0   d
*
             1



e
 2
              N   2
0               d


              N   2
0   e
  x2
d      x   

1

1 3 5  ...  2n  1  p    2
x
2 n  ax 2
for even number                          e         dx                                          
0
2n 1 a n      a

n!
x
2 n 1  ax 2
or odd number                          e           dx                          ,   a0
0
2n 1

    x2 
2           
 N0       e   x dx                  N 0 2  e dx 
2
2
even n=0
                                      
 0        

1
1                      p
         N 2
2       2
 1
2
0

1
1             2              1
2
             1 

p
N   0
2   
1
                               p     2

1
N0                      1
p     4
                                                  

  )1( ) d
*

2  2
 (
0 1   0                          N    1
2
e    d
                                                


 x e
2  2
       N       1
2
d x


  2   x2 
3           
        2 N1 2  x e dx                          even n=1
 
 0          

12
3                         1          p 
1                N     2          2
2              
4           
1

3
1                 2           2
2
N    2
                                            
1

3
2              p   12
p1 2
 2 
N1               12 
p 

```
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