Quantum mechanics

1. Recapitulate
 Particle in a rigid one dimensional box.
 Found the energy states.
 Discussed the wave functions.
 We calculated various expected values.

2. Example: Particle in a Rigid Box
0
V 
V  
V  
x=0 x=L
( ) 0 for 0
for x 0 or
V x x L
x L
  
   

3. 2 2 2
2
2
n
E
mL


n
2
(x)= sin for 0
0 elsewhere
n x
x L
L L

  


4. Wave function and Probability
0
V 
x=0 x=L
0
V 
x=0 x=L
( )
n x
 ( )
P x

5. 2
2
2 2
2
2
2
3
1
3 2
0
2
x
x n
L
x
L
x
n
p
n
p mE
L



 
 
 
 

 
 
 
 

6. 2 2
6
12
0.57 for n=1
=1.67 for n=2
x
n
x p
 
  


7. Orthonormality of Wave function
0
* ( ) ( )
L
n m nm
x x dx
  


We can show that
Here the Kronecker-Delta function is
defined as follows.
1 for
0 for
n m
nm n m
 



8. Expansion
Any function f(x) that is defined between
x=0 and x=L and zero everywhere else can
be expressed as follows.
1
0
( ) ( ) where
* ( ) ( )
n n
n
L
n n
f x c x
c x f x dx









9. The interpretation of the
coefficients
( ,0) ( )
n n
n
x c x
 
 
We had earlier said that the following is a
valid wave function at time t=0.

10. We now give the physical interpretation of
the coefficients. Starting with normalized
ψ(x,0), implying
is equal to the probability of finding the
particle energy to be En.
2
1
=1
n
n
c



2
n
c

11. Example
1
4
2
(x)= sin
2 4
(x)= sin
x
L L
x
L L




Following are the two normalized
wave function corresponding to n=1
and n=4 states.

12. Following wave function is not a
normalized wave function.
2 2 4
( ,0)= sin + sin
x x
x
L L L L
 

The above can be seen by evaluating
0
* ( ,0) ( ,0)
L
x x dx
 


13. 2
1
=2
n
n
c



Also, we can see that
The normalized wave function would thus
be
1 2 1 2 4
( ,0)= sin + sin
2 2
1 1 4
sin + sin
x x
x
L L L L
x x
L L L L
 

 


14. The physical interpretation implies
imagining a large number of boxes where
the wave function of the particle is given
by above. If a measurement of energy is
done, in half of them we shall find the
particle to be in n=1 and in another half in
n=4 state.

15. Question 1: Is the following wave
function a normalized one?
2 8 4
( ,0)= sin + sin
5 5
x x
x
L L L L
 


16. 1 4
1 2
( ,0)= (x)+ (x)
5 5
x
  
We can write the above as follows.
In this case in 20% of the boxes will give an
energy corresponding to n=1 and 80%
corresponding to n=4.

17. Question 2: What would be the expected
value of energy in such a case?
2
1 4
1
0.2 0.8
n n
n
E c E E E


  


18. Question 3: If no measurement was done
what would be the wave function at a time t.
1 4
- -
2 8 4
( , )= sin e + sin e
5 5
iE t iE t
x x
x t
L L L L
 

We can check that this would not be a
stationary state and the probability of finding
the particle at a location would be a function
of time.

19. Question 4: If a measurement is done in
one of the boxes at t=0 and the energy is
found to be E4, what would be the wave
function at a later time t.
The wave function now collapses and the
time dependence would be given by.
4
-
2 4
( , )= sin e
iE t
x
x t
L L



20. Question 5: What would the
measurement of energy yield on this box
at a later time?
The particle is now in stationary state.
Hence the measurement would lead to
E4.

21. SOME POSTULATES OF
QM

22. 1. System Description and Time
Evolution
 A particle under a potential V(x) is
described by a wave function ψ(x), which
contains the information about all the
physical properties of the particle.
 The time evolution of ψ(x) is governed by
the time dependent Schrödinger
Equation.

23.  The wave function ψ(x) is single valued,
finite and a continuous function of x.
The position derivative is also
continuous , unless V(x) shows infinite
jump.
d
dx


24. Compare
From Krane (Modern Physics):
When an object moves across the
boundary between two regions in which
it is subjected to different [forces,
potential energies], the basic behavior of
the object is found by solving [Newton’s
second law, the Schrodinger equation].

25. The [position, wave function] of the object
is always continuous across the boundary,
and the [velocity, derivative dψ/dx] is also
continuous as long as the [force, change
in potential energy] remains finite.

26. 2. Operators
 Each dynamical variable that relates to
the motion of the particle can be
represented by an operator, satisfying
certain criteria.
 The only possible result of a
measurement of the dynamical variable
represented by an operator is one or the
other Eigen values of the operator.
ˆ
n n n
G g
 


27.  The Eigen values are real numbers for
the operators representing dynamical
variables.

28. Hamiltonian Operator
It is defined as follows
Eigen Value Equation of Hamiltonian
Operator is thus
2
ˆ
ˆ
2
p
H V
m
 
2
ˆ
2
n n n
p
V E
m
 
 
 
 
 

29. Replacing by their operators we get
2
2
2
2
n
n n n
V E
m x

 

  


30. 3. Completeness
 The Eigen States of an operator
representing a dynamical variable are
complete.
 Any admissible wave function can
always be expressed in the following
way in terms of Eigen functions of any
operator.
( ) n n
n
x c
 
 

31.  These Eigen functions form the basis.
 We had discussed this aspect in detail
with Eigen function of the Hamiltonian
operator.

32. 4. Probability
 The probability that an Eigen value gn
would be observed as a result of
measurement is proportional to the
square of the magnitude of the
coefficient cn in the expansion of ψ .
 The proportionality become equality if we
have normalized wave function.
2
( )
n n
P g c


33. 5. Collapse of Wave function
 If the measurement gives a particular
value of Eigen value gn, the wave
function discontinuously collapses to .
n


34. Particle in a Finite Potential Well
( ) 0 for 0
for x 0 or
o
V x x L
V x L
  
  
0
V 
o
V V

x=0 x=L
o
V V

RI RII RIII

35. We take E<Vo for the bound state problem.
 
2
2 2
( ) 2
( ) 0
I
o I
d x m
E V x
dx


  
 
2
2 2
( ) 2
( ) 0
II
II
d x m
E x
dx


 
 
2
2 2
( ) 2
( ) 0
III
o III
d x m
E V x
dx


  
Note the sign of quantity in square bracket for
different regions.
RI
RI
I
RIII

36. General Solutions
   
2
2
( ) sin cos ;
2
II x C kx D kx
mE
k
  

 
2
2
2
( ) ; o
x x
I
m V E
x Ae Be
 
 
 
  
 
2
2
2
( ) ; o
x x
III
m V E
x Ee Fe
 
 
 
  
R
I
RII
RII
I

37. Well-behaved Wave Functions
   
0
0
( )
( ) sin cos
( )
x
I
II
x
III
B
E
x Ae
x C kx D kx
x Fe




 



 


38. Boundary Conditions Applied
RI-
RII
   
   
0
0
(0) (0)
sin 0 cos 0
(0) (0)
cos 0 sin 0
I II
I II
Ae C D
d d
dx dx
A D
A Ck
A e Ck Dk
 
 



 

 



39. RII-
RIII
   
   
sin cos
cos si
( ) ( )
( ) ( )
n
II III
II III
L
L
L L
d d
L L
dx
C kL D kL Fe
Ck kL Dk kL F e
dx


 
 





 
  

40.    
   
sin cos
cos sin
L
L
A D
A Ck
C kL D kL Fe
Ck kL Dk kL F e








 
  

41. Solving the Equations
Express all constants in terms of A.
   
   
sin cos
cos sin
L
L
D A
C A
k
A kL A kL Fe
k
Ak kL Ak kL F e
k










 
  

42. Energy Eigen Values
Divide the last two equations.
   
   
 
cos sin
;
sin cos
2
2
o
kL k kL
kL k kL k
m V E
k mE
 



 

 

The above equation governs the allowed
energy levels.

43. Classically Forbidden Region
 There is a non zero probability of finding
the particle in the classically forbidden
region.
 This probability decreases exponentially.

44.  However the uncertainty principle
prohibits one to localize the particle in
that region and measure its kinetic
energy.
 From wave nature we can understand
the finite penetration.

45. Normalization
2
0
2 2
0
2
2 2 2
sin cos
sin cos ) 1
L
x
L x
L
A e dx kx kx dx
k
A e kL kL e dx
k

 





 
 
  
 
 
 
 
 
 
 
 
 
 


46. Free State Problems
x=0
V=0 V=Vo
E
A particle approaching from left on a Step
Potential
RI RII

47. 1 1
2 2
2
1 2
2
2 2
2
( ) ;
2
( ) ;
ik x ik x
I
ik x ik x o
II
mE
x Ae Be k
m E V
x Ce De k




  

  
General Solution
R
I
RI
I

48. Physical interpretation of different
components yields
Boundary Conditions yield
0
D 
1 2
( )
A B C
ik A B ik C
 
 