Principles of quantum mechanics
 Quantum theory was able to explain the nature of light in terms of packets (or) bundles known as
Photons (or) Quanta.
 Some phenomena like interference, diffraction, polarization etc… are explained on the basis of wave
nature of light.
 However, phenomena like photoelectric effect, Compton effect, Zeeman effect etc … are explained
the basis of particle.
 Thus the wave- particle properties of light is known as dual nature of light.
Black body radiation:- It is the one which absorbs 100% radiation incident on it.
According to Kirchoff law of radiation a good absorber is also good emitter at high temperature.
AS, black body is a perfect absorber it is a perfect emitter also emitting all possible wave lengths
from 0   .
Planks Quantum theory:- According to Max Planks radiation emitted in the form of Quanta
(packet) of energy.
The energy of each packet is E  h
Where h  6.626 1034 J  sec
 = frequency of light
Matter waves:- In 1924 Louis de- Broglie suggested that wave – particle duality exhibited by
radiation to matter. Therefore particles like electrons, protons , photons, neutrons atoms (or)
molecules will have waves associated with them, known as matter waves (or) pilot waves (or)
guiding wave (or) de- Broglie waves.
de- Broglie proposed that the wave length of matter wave would be related to its momentum by
the same same relationship for photons.
i.e The particle of mass ‘m ’ travelling with speed ‘v’ the wavelength  is give by  
h
h
.

p mv
de- Broglie wavelength:- Let us consider particle with frequency ‘  ’ travelling the velocity of
light ‘C’
The velocity of light C=   -------------------------------> (1)
 is wavelength of light
According to Planks Quantum theory E  h
Where h  6.626  1034 J
According to Einstein mass energy relation E  mc 2
mc 2  h

C

mc 2  h
C

h

 mc  p
mc
h

(Wave behavior)
p
Consider a particle of mass ‘m’ moving with velocity ‘v’ Then the wavelength associated with it

h
h

mv p
P is momentum of the particle.
de- Broglie wavelength in terms of K.E:- Consider a particle of mass ‘m’ moving with velocity ‘v’
Then the K.E of the particle is given
E
p2
1 2 m2v 2
mv 

2
2m
2m
1
The momentum p 
Substitute in  

2mE
h
we get.
p
h
This is the de- Broglie wavelength in terms of K.E
2mE
de- Broglie wavelength for an electron:- Consider a electron of mass ‘m’ and charge’q’ that is
accelerated through a potential difference ‘V’ . Then the K.E of electron is equal to the loss in P.E
1 2
mv  qV
2
Substitute in  
m2v 2  2mqV
p  mv  2mqV
h
h
we get. de- Broglie wavelength of an electron  
------------>(2)
p
2mqV
Substitute the h, q, and m values in eq(2)
We get Wave length of an electron

1.
2.
3.
4.
5.
6.625 1034 Js
2  9.11031 Kg 1.6 1019 C  V

12.26 1010
12.26 0   150 A0
m
A
V
V
V
The above equations give the de-Broglie wavelength of electron.
The wavelength of the matter wave associated with electron in motion and confirmed thedeBroglie hypothesis
Characteristics of matter wave:Lighter is the particle, grate is wavelength associated with it.
Smaller the velocity of particle, grater the wavelength associated with it.
When v=0;    wave length of matter waves becomes infinity and this indicates that matter
waves are to associated with only particles. V   ;   0 Particle associated with matter waves.
Matter wave wavelength does not depend upon charge of the particle.
No single experiment (or) Phenomenon can determine (or) exhibit both wave and particle
properties simultaneously. Matter waves propagate with a velocity greater than that of the particle
and will have a velocity even greater than that of light. From Planks Quantum theory E  h
Where h  6.626  1034 J From Einstein mass energy relation E  mc 2 mc 2  h
Velocity of matter wave   . 

mc 2
h
h mc 2 c 2
.

mv h
V
Where ‘V’ is the velocity and is always less than ‘C’
The above analysis indicates that matter waves travels much faster than the particle and hence are
called Pilot waves (or) guiding waves.
6. The wave nature of matter introduces an uncertainty in the location of the position of the particle
because of wave cannot be said exactly at this point (or) that point.
2
Davisson and Germer Experiment:The first experimental evidence of matter waves was given by two American Physicists, Davisson &
Germer in 1927. They also succeed in measuring the de-Broglie wavelength associated with slow
electrons. They were studying the reflection (diffraction) of electrons from Nickel target.
The Nickel target was subjected to such as heat treatment that the reflection (or) diffraction
becomes anomalous. Now the reflected (or diffracted) intensity showed striking maxima & minima.
Thus they suspected that electrons are diffracted like X- rays i.e they behave like a wave under
certain conditions.
Thus the experimental arrangement is as shown in fig. The apparatus consists of an electron gun
(G) where electrons are produced and obtained in fine pencil of electron beam of known velocity.
The electron gun consists of Tungsten filament (F) heated to dull red so that electrons are emitted
due to thermionic emission. Now the electrons are accelerated in electric field of known potential
difference (P.D) . The beams of electrons are directed to fall on single crystal of Nickel known as
target (T). The electrons acting as the wave it diffracted in different direction. The angular
distributions are measured by an electron detector (Farady cylinder) which connected to
Galvanometer. The Farady cylinder can move graduated scale (S) between 290 to 900 .
The electrons accelerated through a potential 54 Volts hit a Nickel target and scattered electrons
distribution showed maximum at an angle of 500 with incident beam. The incident and diffracted
beams were making an angle of 650 with incident crystal planes at that inclination of the target.
They are rightly concluded that electrons are getting diffracted from Nickel crystal target similar to
X- rays and applying Bragg’s equation for diffraction maxima 2d sin   n .
For different accelerating potentials number of electrons beams collected at different angles. A
graph is plotted between galvanometer current (Intensity of diffracted beam) against angle 
between beam and beam of entering the cylinder. The observations are repeated for different
accelerating potentials. The corresponding curves are as shown in fig.
3
1. With increasing of potential, the bump moves upwards.
2. The becomes most prominent in the curve of 54 Volts electrons at   500
3. At higher potentials the bump gradually disappears.
According to de-Broglie wavelength associated with electron accelerated at a potential of 54 Volts is
 
12.26
12.26

 1.668 A0
V
54
By Bragg’s condition for known‘d’ value of Nickel crystal d=0.91 A0 at an angle   650
Diffracted beam at   500 . The
corresponding angle of incident
relative to the family of Bragg’s
plane.  
1800  500
2
2d sin   n 2  0.91sin 650   n=1   1.65A0
Hence this experiment gives good agreement for Bragg’s law and de- Broglie concept of matter
wave.
Hesigenberg’s uncertainty:We are trying to measure the position of an object with photon of wavelength  . The position can
be measured best to accuracy about  .
The uncertainty in the measurement in the position is x . x 
Assume that momentum of a single photon
h
P

When the photon strikes the object hence the final momentum of the object is uncertain by a
quantity. p 
h

h
The product of the uncertainties in the position and momentum gives xp
It states that product of the uncertainty in the position and momentum of moving particle can never
be less than
2
i.e Where

h
2
If x & px are the uncertainties in position and momentum xpx 
2
or xpx 
h
4
i.e Both position & momentum of the particle cannot be determined simultaneously and accurately.
The other pairs which obeys this uncertainty principle.
yp y 
2
, E t 
2
, z pz 
2
, J  
Where
2
 1.05 1034
Applications:1. The uncertainty principle has confirmed the non existence of electrons in nucleious.
2. It has proved stability of atoms & nuclei.
4
Schrodinger wave equation:Let the wave function associated with a moving particle be denoted as
 (x t)   0e  i ( t  kx)
(1)
From quantum theory, we have E   or  
E
Similarly from de-Broglie concept p  k or k 
p
Substituting  &k values in eq(1)
We get
 (x t)   0e
 (x t)   0e
i
i (
E
t
p
x)
(Et  p x)
(2)
Differentiating (2) with respect to‘t’
i
(Et  p x)  -iE 
 (x t)
  0e


t


 
 (x t) - iE
E (x t) 
 (x t)

 (x t )
i t
t
Differentiating (2) with respect to ‘x’
(3)
i
(Et  p x)  ip 
 (x t)
  0e
 
x
 
 (x t) -ip

 (x t )
x
Again differentiating with respect to ‘x’
i
(Et  p x)  ip  ip 
 2 (x t)


e
0
  
2
 x
  
2
2
  (x t)  p
 2  (x t )
p 2  (x t)  
2
x
As total energy of particle is E = K.E + P.E
E
Multiplying both sides with  (x t)
p 2 (x t)
 V (x t)
2m
E & p2 
Substituting the values of
i.e from eq(3) & eq (4) we get
 2 (x t)
 2
 
 2 x  V (x t)
 (x t) 
i t
2m
2
2

   (x t)
i
 (x t) 
 V (x t)
t
2m  2 x
p2
V
2m
E (x t) 
(5)
5
2
 (x t)
x 2
(4)
The above equation is called Schrodinger time dependent wave equation in 1D (1 dimension)
Eliminating time from the above equation we get Schrodinger time independent wave equation
Splitting  (x t) into two small functions
 (x t )   (x) f(t)   f
d
 2 d 2
f (t) 
f
 V f
dt
2m d 2 x
Dividing throughout by  f we get
i 
1d
 2  1  d 2
i   f (t) 
V
 
2m    d 2 x
 f  dt
This identity is possible when both L.H.S and R.H.S are separately equal to a constant
From R.H.S we get
 2  1  d 2
V  E
2m    dx 2
Multiplying both sides with  we get
 2 d 2
d 2
2m
 V  E

(E  V)  0
2
2m dx 2
dx 2
(6)
eq (6) is Schrodinger time independent wave equation in 1D (Dimension)
 2 (xyz) 
2m
(E  V) (xyz)  0
2
(7)
eq (7) is Schrodinger time independent wave equation in 3D (Dimension)
Where 2 
d2
d2
d2


dx 2 dy 2 dz 2
Physical significance of  (Born interpretation):According to Max Born the wave function ( ) of a material particle is simply
a mathematical function with no physical meaning for itself, whereas square of the wave function
 (x t) denotes the probability of finding the particle at that point (x) and at that instant (t)
As  may be real or complex but as probability is always real it is written as
p   2   *
Where 
* is complex conjugate of 
The probability of finding the particle in a small interval of length (dx) is p  
6
2 dx
As the particle is surely to be found somewhere in the entire space we have
p   2 dx  1
A wave function satisfies this condition is said to be normalized to unity.
Applications of Schrodinger’s wave equation:
Particle in 1 – D box (Infinite Square well potential)
Consider a 1-D box of length (a) extending from x=0 to x=a as shown. A particle is
moving inside an infinite square well as shown in fig. The potential is infinite everywhere except in
the region 0  x  a where it is zero
The potential specifications are V(x) =0 for 0  x  a (Region 2)
V(x) =  for x  0 ; x  a (Region 1 & Region 3)
Schrodinger time independent equation is
d 2
2m

(E  V)  0
2
2
dx
Inside the box we have V(x) = 0
i.e
d 2 2m
 2 E  0
dx 2
Where  2 
d 2
  2  0
2
dx
2mE
2
Solution to eq (1) is of the form
(1)
(2)
 (x)  Asin  x  Bcos  x
Where A, B are constants
7
(3)
The boundary conditions are
At x= 0 ;  =0
At
x=a;  = 0
always
Applying first boundary condition to eq(3) (x = 0 ;  =0)
We get
0  Asin 00  B cos 00 B=0 (always) because (cos 00  0  1)
Solution becomes  (x)  Asin  x
(4)
Applying 2nd boundary condition to eq (4) (X = a;  = 0)
We get 0  Asin  x
As A  0
sin  a  0  a  n

Eq(4) becomes
where n = 1,2,3,……. n  0
n
a
 (x)  Asin
n x
a
(5)
This is the general solution for particle in a box
2
We have  
n 2 2
2mE

2
2
a
The eign values are
En 
n2 h2
8ma 2
1. Particle cannot have zero energy. The minimum energy for particle is E1 
h2
8ma 2
Which is called zero point energy?
2. Particle in a box energy levels are quantized as E1 
8
h2
E2 =4E1 E3 = 9E1
8ma 2
Normalized wave function:As particle is surely to be found somewhere inside the box always
We have
a
 n x 
 A2 sin 2 
 dx  1
0
 a 
a
  dx  1
2
0
a
 n x 
A2  sin 2 
 dx  1
0
 a 
A
2
a 1  cos 2  n x  
 a   dx  1

2

 

Evaluate the above integral we get A2 
 Normalized wave functions are
2
2
 A
a
a
 n (x) 
2
n x
sin
a
a
From above it can be seen that in quantum mechanical picture the particle
1. Can have only certain discrete energies as given by E n 
n2 h2
with n = 1,2,3,4, - - - - - This indicates that the energies
8ma 2
levels are quantized. (Fig. 1)
2. Different regions inside the well have different probabilities of containing the particle, and this probability changes with
particle energy. (Fig.3)
3. When the particle is in lowest energy level i.e. n=1, maximum probability is at center of the well, i.e. at x = a/2 and the
minimum at, x=0 and x= a. For second energy level n=2, maximum probability is x = a/4 and also at x =3a/4. The minimum
probability is at x = 0, x=a/2 and X=a
4. In classical picture, the particle can have any amount of energy and all points inside the well have same probability of
containing the particle at all energies.
9
Extension to three dimensions:The potential V(x y z ) in the box as
V x y z  0
  x y z   X (x) Y(y) Z(z) 0  x  a , 0  y  b , 0  z  c
The time independent Schrodinger equation is
 d 2 d 2 d 2  2mE

 2 
  2  E V  x y z    0
dy 2
dz 2 
 dx
The wave function  can be written as
Reduced 1-D equation
d 2Y
d2X
d 2Z
2
2


x

0
  32 z  0


y

0
1
2
2
2
2
dx
dz
dy
Where  1 ,  2 ,  3 are constants obeying the equation
12   2 2   32 
2mE
2
Proceeding in the similar manner as 1-D box
X (x)  A1sin
n1 x
a
Y(y)  A 2sin
n2 y
n
2  2
b
b
n2 =1,2,3 …………
Z(z)  A 3sin
n3 z
c
n3 = 1,2,3 ………….
   2  3 
2
1
2
2
En1n2 n3
1 
3 
2mE
2
2
n1
a
n3
c
n1 =1,2,3 ……….
 n12 n2 2 n32 
  2  2  2 
b
c 
a
2
 n12 n2 2 n32 

 2  2 

2m  a 2
b
c 
2
The function  is given by  n1n2 n3  A1 A2 A3 sin
According to Normalized condition A1 A2 A3 
nn n 
1 2 3
nz
n1 x
ny
sin 2
sin 3
a
b
c
8
abc
nz
nx
ny
8
sin 1 sin 2 sin 3
abc
a
b
c
10
For Cubic potential a=b=c En1n2 n3 
nn n 
1 2 3
2
2
2ma
2
n
2
1
 n2 2  n32 
nz
nx
ny
8
sin 1 sin 2 sin 3
3
a
a
a
a
Pb) calculate the first three allowed energy levels from electron in 1- D box of length 10 A0 .
En 
Sol:
n2 h2
n2 (6.625 1034 )2

Joul
8ma 2 8  9.11031  (10 1010 ) 2

`
n 1
n 2 (6.625 1034 ) 2
 0.37 n 2 eV
31
10 2
8  9.110  (10 10 )
E1  0.37eV
n  2 E2  0.37  4  1.48eV
`
n3
E3  0.37  9  3.33eV
11