Transcript 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 P rojectionoperatorassociatedwith normalisedket V
Any ket  V parallelto V is an eigenket with eigenvalue1
PV V  V V V   V V V   V
Any ket V perpendicular to V is also an eigenket with
eigen value 0
PV V  V V V  0 V
 
Eigenvalues of R  iˆ 
2 
R1  1
Kets thatare neitherand of theform V  V
PV ( V  V )  V
 1 is also an eigenket with eigen value 1
We will shortlysee thateigenketscan only be determined
upto a multiplicativeconstant.
Any other vector whichis not parallelto 1 will get rotated
and hence will not be an eigenketof R
राघव वर्ाा
The characteristic Equation & the solution
to the Eigenvalue problem
V  V
  I  V
 0
Mult iplying by   I  on bot h sides we get
1
V    I 
1
0
Any finit eoperat oract ingon a null vect orcannotproduce
a finit e vect or.
T heequat ion cannotbe t rue.T heonly wayit will not be t rue
is if   I 
1
does not exist .
M 1  Cofactor M T / det M
If det M  0 M -1 does not exist
Comingback t oour earlier equat ion then t heinversedoes not
exist if and onlyif det  - I   0
T o det ermine one st art swit h t heeigen value equat ion
 - I  V
 0
T hisis a vect orequat ion.T o solvefor  we need t o choose
basis.
राघव वर्ाा
Solving the eigen value equation

Choosing the basis and taking projection along the i axis.
  I  V
0
T akingt heproject ionalongi axis
i   I  V  0
Havingchosen t heaxis we can now writ e V also in t erms
of it s component s
i   I v j j  0

ij
 ij v j  0
Setting the det erminant equal t o zero we get
c
m  0  charact erist ic quat ion
m
P n ( )  cm m charact erist ic polynomial
राघव वर्ाा
Solve the eigen value problem of R(/2 i)
1 0 0 
  ˆ
R i   0 0  1
2 
0 1 0 
Charact erist ic Equation
1- 
0
0
det (R - I)  0
-
-1
0
1
-
T hecharacterist ic equat ionis (1   )( 2  1)  0
  1,  i
  1 corresponds 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 thefreedomin scale to normaliseto write the
1
eigenketas 0
0
Here please notethatone can multiply his
t eigenket with
a number of modulus unity without changinganything.More
on thislater!!!!!
राघव वर्ाा
The other two eigen values of R
1 0 0   x1   x1   ix1 
 x1  0
1
0 0  1  x    x   ix  
 i 

 2   3   2 

2
0 1 0   x3   x2  ix3 
 1 
 x1  0
1 
  i 
i 

2
 1 
Label thekets with their eigen values
T hephenomenonof single eigen value representing more than
one eigenvector is called degeneracyand corresponds to
repeatedrootsof thecharacteristic polynomial
T heeigenvalues of a HermitianOperatorare real
  
    
- - - - (1)
T aketheadjoint
 T       - - - - (2)
But  is Hermitianand therefore   T
Subtract the secondfromfirst
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 linearlyindependent basis in V n
T henanotherbasis is generatedby theactionof  i.e.
 V1 ,  V2   Vn is also a LI basis
T o see thislets assume to thecontrary hat
t is thereexists
a relationof theform   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 thena set of linearlyindependent vectors
i
thenany vectorV can be writtenin termsof them
V     i  Vi
राघव वर्ाा
i
Hermitian Operators (Contd…)
In termsof V    iVi we see thatevery V in V
i
may be writtenas V   V where V is unique
Every V thereforearises froma unique V under action
of 
Definean operator whose actionon any vector V is
to takeit back toits unique source V
If thesource of V was not unique say we had two vectors
V1 & V2 thatare mappedinto V we could not define
 for actingon V it would not know whetherto give V1
V2 . T heactionof  is then
 V  V where V   V
We may identify as inverseof 
राघव वर्ाा
Operator R
राघव वर्ाा