Download EP-307 Introduction to Quantum Mechanics

Eigen Value Problem

Consider some linear operator  acting on a ket |V. It will cause a non  V  V 
trivial change to the ket.

If you remember the Stern-Gerlach Experiment. There a non-uniform
magnetic field say in z-direction caused the vector |S to become |Sz+,
|Sz-

If you remember once we had applied the magnetic field in the zdirection repeated application of SG apparatus with magnetic field in the
same direction did not do anything to my vector except for rescaling it.

In general if for an operator we have the following relation then we say
that |V is an eigen ket of the the operator 

In Quantum Mechanics one has to determine the eigenkets and
eigenvalues. A liitle later we will try to show you where does physics
enter here. Or why calculation of eigenket and eigenvalues is made.
However, when we start talking of serious quantum mechanics it will be
eigen-eigen all the way…
राघव वर्ाा
 S  S Z
 S Z 
 
SZ
2
V  V
Eigenkets of some familiar operators
The only eigenvalue of I is 1. All vectors are its eigen vectors with
eigen value 1
What about the projection operator
IV  V
  Pv Projection operator associated with normalised ket V
Any ket  V parallel to V is an eigenket w ith eigenvalue 1
PV V  V V V   V V V   V
Any ket V perpendicu lar to V is also an eigenket w ith
eigen valu e 0
PV V  V V V  0 V
 
Eigenvalue s of R  iˆ 
2 
R1  1
Kets that are neither and of the form V  V
PV ( V  V )  V
 1 is also an eigenket w ith eigen valu e 1
We will shortly see that eigenkets can only be determined
upto a multiplica tive constant.
Any other vect or which is not parallel to 1 will get rotated
and hence will not be an eigenket of R
राघव वर्ाा
The characteristic Equation & the solution
to the Eigenvalue problem
V  V
  I  V
 0
Multiplyin g by   I  on both sides we get
1
V    I 
1
0
Any finite operator acting on a null vector cannot produce
a finite vector.
The equation cannot be true. The only way it will not be true
is if   I 
1
does not exist.
M 1  Cofactor M T / det M
If det M  0 M -1 does not exist
Coming back to our earlier equation t hen the inverse does not
exist if and only if det  - I   0
To determine  one starts with the eigen valu e equation
 - I  V
 0
This is a vector equation. To solve for  we need to choose
basis.
राघव वर्ाा
Solving the eigen value equation

Choosing the basis and taking projection along the i axis.
  I  V
0
Taking the projection along i axis
i   I  V  0
Having chosen the axis we can now write V also in terms
of its components
i   I v j j  0

ij
  ij v j  0
Setting the determinan t equal to zero we get
c
m  0  characteri stic quation
m
P n ( )  cm m characteri stic polynomial
राघव वर्ाा
Solve the eigen value problem of R(/2 i)
1
  ˆ
R i   0
2 
0
Characteri stic
0 0
0  1
1 0 
Equation
1-  0
0
det (R - I)  0 -  - 1
0
1 -
The characteri stic equation is (1   )( 2  1)  0
  1,  i
  1 correspond s to 1
x1  x1
1 0 0   x1   x1 
0 0  1  x    x    x  x  x  x  0
3
2
2
3

 2   2 
0 1 0   x3   x3 
x2  x3
One uses the freedom in scale to normalise to write the
1
eigenket as 0
0
Here please note that one can multiply t his eigenket w ith
a number of modulus unity with out changing anything. More
on this later! !!!!
राघव वर्ाा
The other two eigen values of R
1 0 0   x1   x1   ix1 
0 0  1  x    x   ix   1

 2   3   2 
2
0 1 0   x3   x2  ix3 
 x1  0
1 
  i 
i 

2
 1 
Label the kets with thei r eigen valu es
 x1  0
 i 


 1 
The phenomenon of single eigen valu e representi ng more than
one eigenvecto r is called degeneracy and correspond s to
repeated roots of the characteri stic polynomial
The eigenvalue s of a Hermitian Operator are real
  
    
- - - - (1)
Take the adjoint
 T       - - - - (2)
But  is Hermitian and therefore    T
Subtract t he second from first
0  (    )  
  
राघव वर्ाा
Of the three simultaneous equation the
First is not an equation. In general there
Will only be n-1 independent equations
Hermitian Operators

To every Hermitian operator , there exists (ATLEAST) one basis
consisting of orthogonal eigenvectors, It is diagonal in this eigen basis &
has eigen values as its diagonal entries.

Before we prove that lets prove another theorem

If |V = 0 implies |V = 0 then 
-1
exists
Let V1 , V2  Vn be linearly independen t basis in V n
Then another basis is generated by the action of  i.e.
 V1 ,  V2   Vn is also a LI basis
To see this lets assume to the contrary t hat is there exists
a relation of the form   i  Vi  0 with not all  i  0
i
Because  is linear    i Vi  0
i
Because  is not zero, hence

i
Vi  0 with not all
i
 i  0 which is not true.
If   i Vi are then a set of linearly independen t vectors
i
then any vector V can be written in terms of them
V     i  Vi
राघव वर्ाा
i
Hermitian Operators (Contd…)
In terms of V    iVi we see that every V in V
i
may be written as V   V where V is unique
Every V therefore arises from a unique V under action
of 
Define an operator  whose action on any vector V is
to take it back to its unique source V
If the source of V was not unique say we had two vectors
V1 & V2 that are mapped into V we could not define
 for acting on V it would not know whether to give V1
V2 . The action of  is then
 V  V where V   V
We may identify  as inverse of 
राघव वर्ाा
Operator R
राघव वर्ाा