Welcome back to this. We will continue fromthe
last part on the eigenvalue eigenvector for
a simple 2 by 2 matrix. You know we will have
some examples to actually work them out in
detail.
And also probably slightly more involved examples
in this session ok. So, first let us find
out the eigenvalue for thisfamous matrix known
as the Pauli Spin Matrix x component and so
the eigenvalue equation is this matrix multiplied
by a column x y is equal, to lambda times
x y hm therefore, if you recall that it was
a 1 1, minus lambda. So, you can immediately
write the determinant that goes to 0 as minus
lambda 1 1, minus lambda is equal to 0, please
recall that
You see a 1 1, minus lambda, a 1 2, a 2 1,
a 2 2, minus lambda, a 1 1 and a 2 are zero's
therefore, you have simply, minus lambda and
these are one's. So, you have that which gives
you, it is very simple is, lambda square minus
1 is 0 therefore, lambda is equal to plus
or minus one, two solutions. Let us look,
at the eigen vectors, so if it is 0 1 1, 0,
x y is equal to lambda is 1 and therefore,
this is x y now, if you write this as an equation
you see that, it is y is equal to x and here
x is equal to y, which is the same equation
, but we want x square plus y square to be
1 therefore, if y is equal to x then, what
we have is y square plus y square. So, I am
using the fact that, x is equal to y, y square
plus y square is equal to 1 or y is equal
to 1 by root two ok.
We will take the positive root for this, therefore
x is equal to 1 by root 2, y is equal to 1
by root 2 ok. So, the eigenvalue lambda is
equal to 1 for this matrix 0 1 1 0, the eigenvector
x y is equal to 1 by root 2, 1 by root 2,
now for the eigenvalue lambda is equal to
minus 1, you have to solve this equation 0
1 1 0, x prime, y prime is equal to minus
x prime, y prime and that this equation is
nothing, but x prime is equal, to minus y
prime ok the first equation is actually y
prime is equal to minus x prime.
And the second equation is; obviously, x prime
is equal to minus y prime hm, but again x
prime square, plus y prime square, we will
keep that as 1. Therefore since it is square,
you know that it is going to be y prime square
plus y prime square, is equal to 1, which
means y prime is equal to 1 by root 2 ok,
but the solution, is if y prime is 1 by root
2 then, the x prime is minus 1 by root 2 because
it is minus y prime. Therefore, lambda is
equal to minus 1, the solution x prime, y
prime is 1 by root 2, minus 1 by root 2 and
1 by root 2 ok.
Now this minus, plus the sign is arbitrary.
Because if, I choose y prime square is equal
to 1 by 2 and y prime, I can also choose minus
2 by root 2, the overall sign of the eigenvector,
is not important sign of the eigenvector is
not important, it is not significant, we do
not need to worry about that.
That is ,if you write a 1 1, a 1 2, a 2 1,
a 2 2. If you had, x y is equal to lambda
times x y then this equation is the same as
the other equation a 1 2, a 2 1, a 2 2, minus
x, minus y is lambda times minus x, minus
y. Therefore, it does not matter, but the
relative signs between x and y are important,
because that is, how the eigenvectors are
defined, you recall the previous one.
In the first, eigenvector the elements are
different 1 by root 2 1 by root 2, in the
second eigenvector the elements have an opposite
sign between them that is important, but if
instead of writing this minus 1 by root 2
plus 1 by root 2, if I, write this as plus
1 by root 2, then this must be minus 1 by
root 2, in order for this equation to valid.
So if, I multiply this whole vector by a minus
sign it does not matter, it is irrelevant.
So, eigenvectors are important and eigenvectors
are defined to within your sign technically
to within your face factor, the face factor
is e to the i delta and the such things will
come in when the eigenvectors are complex
and then the normalization conditions will
be slightly different for that ok. We will
see it in the nextexample of the Paulin spin
matrix anyway ok.
So, therefore, now we have two eigenvectors
namely for this, matrix 1 0 0 1, sorry for
this for this matrix 0 1 1 0, lambda 1, lambda
2, 1 by root 2, 1 by root 2 and 1 by root
2, minus 1 by root 2, now if we want to write
this in a very specific form namely, if we
want to write this as 0 1 1 0, incorporate
both the eigenvectors together, if we write
this 1 by root 2, 1 by root 2 and then you
write this as, sorry this is minus 1 by root
2, ok minus 1 by root 2, 1 by root 2, if you
want to write that then this equation, you
can verify by a simplealgebra that, this is
what itis minus 1 by root 2, 1 by root 2,
1 by root 2, 1 0 0 minus 1.
So, now you call this as an orthogonal matrix.
These are real eigenvalues therefore, you
see that this matrix is also hermitian, this
is a hermitian matrix. So, now you call this
as an orthogonal matrix. These are real eigenvalues
therefore, you see that this matrix is also
hermitian, this is a hermitian matrix, or
symmetric matrix essentially, because the
elements are real. For a symmetric matrix,
you see that the symmetric matrix is now A,
an orthogonal matrix O, is equal to the orthogonal
matrix O, times the eigenvalue matrix lambda,
where the eigenvalues or the eigenvalues along
or the matrix elements along the diagonals,
ok this is a very standard form, if you have
to worry about only one eigenvalue, then this
is not a matrix, but this is a column vector,
we had the two previous equations A of x y
is equal to lambda 1 times x y. So, those
are two independent equations this is taking
care of both equations in a single form.
But what is important is if now you multiply
by O of T on the left hand side of this matrix
and you have this matrix A and O, then you
have O of T on the right hand side also to
be multiplied on the left and O times lambda
matrix and you know, this is for an orthogonal
matrix this is the identity matrix 2 by 2
times lambda and therefore, what you see is
O of T, A, O, is now the lambda matrix.
So, what you have done is to take the matrix
A 0 1 1 0, which is off diagonal 
and you have through the process of this O
of T is going to be 1 by root 2, minus 1 by
root 2, 1 by root 2, 1 by root 2. Now you
put this matrixin between the 2orthogonal
matrices 1 by root 2, 1 by root 2, minus 1
by root 2, 1 by root 2, you see that, what
you have done is you have got the eigen value
0 1 0 0 1 minus 1 ok.
Therefore you have diagonalized, this matrix
diagonalize this matrix so, you have performed
a transformation on the matrix, that you were
given, the transformation is if this is called
a similarity transformation matrix. When you
multiply a matrix with a matrix on the left
hand side and it is inverse or it is our in
this case the orthogonal is the inverse.
Therefore, the similarity transformation isfor
any matrix, if you have B, the similarity
transformation if you multiply by A A inverse,
and this is C, then B and C are similarity
transforms of each other.
In the case of orthogonal matrix, what you
have is you have used the orthogonal the inverse
us the O of T, the sorry this is A is O of
T and A inverse the inverse of O of T is O.
Therefore, you have done that so, diagonalization
is also an immediate consequence if you find
all the eigenvectors, if you find all the
eigenvectors and you put them as columns in
your matrix and you take the inverse of that
eigenvectors on the left hand side and you
multiply your original matrix, you have actually
diagonalize therefore, the eigenvalue process
is the process of diagonalizing and obtaining
what are called the fundamentallycharacteristic
values associated with that matrix, the word
eigen means characteristic and we have found
the characteristic values associated with
the matrix 0 1 1 0.
Now, the eigenvalue matrix is 1 0 0, minus
1. Two things are immediately obvious, determinant
of sigma x, this is sigma x, if you recall.
Determinant of sigma x is minus 1, determinant
of lambda is minus 1, the determinants of
the original matrix and the determinants of
the eigenvalue matrix, you see are the same
and the determinant of the eigenvalue matrix
is the product of eigenvalues.
Therefore, you see that if you take any matrix
A determinant of A is nothing, but the product
of the eigenvalues of A; A.
Then, there is a quantity called a trace.
The trace for any matrix, if you have a, i,
j as it is all it is elements are denoted
in this way, then the trace of the matrix
is that, sum of all it is diagonal elements
sum over i, a, i, i it is called a trace of
the matrix A. So, in this case 0 1 1 0 the
diagonal elements are 0 therefore, the trace
of this matrix is 0 the trace of the lambda
matrix, which is 1 0 0, minus 1 is also 0
therefore, you can see that trace of A is
the sum of it is eigenvalues.
By the way, I mean I am not proving any of
these things, in the first instance let us
worry about what the properties are and how
do we make use of them, what do we understand
from them and so on. In the second stage of
quantum mechanics and the later, you can worry
about, what are the proofs and other I mean
how do I how do I verify that these things
which were told to me are actually true, the
proof is yeah small amount of mathematics
you can definitely do that, but let us understand
that the eigenvalues, that is a reason why
they are called the characteristic values
that for any given matrix the property, determinant
is a property is a number associated with
the matrix and that number is nothing, but
the product of the eigenvalues.
So, finding out the eigenvalues is also one
method. The sum of the i are the diagonal
elements is the is a trace of the matrix that
is a sum of the eigenvalues. So, these are
all what are called the invariants associated
with the matrix ok, now this is for the real
problem. Let us do a complex or what is called
an imaginary matrix, like what we have sigma
y, let us do that, 0 minus i, i 0.
Now, let us find out this eigenvalues 0 minus
i, i 0 times x y, is equal to lambda times
x y ok. Therefore, lambda isthis whole thing
isdeterminant, you have to write as minus
lambda, minus i, i minus lambda, that is equal
to 0 and this is nothing, but lambda square.
Please remember minus i into minus i, therefore
this is minus 1, is equal to 0 lambda is still
plus and minus 1, 2 eigenvalues ok Now, therefore,
let us find out the first eigenvalue 0 minus
i, i 0, x y is equal to x y lambda is equal
to plus 1, ok then you can see right away
that this is minus i times y, the first line
is that is equal to x. What is the second
line? obviously, it should be the same as
the first line, but just verify, it i times
x, is equal to y, ok if you multiply this
by minus i the second line, i x you get minus
i y and of course, minus i into i is 1 therefore,
you get x is equal to minus i y . So, it is
the same equation ok.
It is a linear system the determinant is 0
ok that is, how we got the eigenvalues. Therefore,
this is the solution, so if i y, minus i y
is equal to x, essentially y is x by minus
i, which I would write it as, i x because
1 by minus i is plus i.
So, we have an equation which is y is equal
to i x, if we do y square plus x square to
get the normalization factor something funny
happens, you see y square plus x square is
actually, x square minus x square , which
is equal to 0 this cannot be normalized, the
way we have done it, but when complex entities
and imaginary entities are involved the normalization
is not done this way, please remember the
hermitian adjoint matrix.
Hermitian adjoints, please remember the transformation
from [cimmet]- orthogonal to unitary. You
recall, when we did the orthogonality, we
proved that the rows are orthogonal to the
col[umn]- the other rows, the rows are orthogonal
to each other, the columns are orthogonal
to each other, but in a unitary matrix we
were very careful in defining that one row
and the complex conjugate of another row or
the orthogonal things, I mean you take the
product of those that is, what will become
0. Therefore, here if you write y is equal
to i x, you must take absolute y square plus
absolute x square is equal to 1, this is required
for complex eigenvectors and this is the definition
and therefore, you will see now if the absolute
value of i x is of course x.
So, y square plus x square is equal to 1,
is nothing other than absolute x square plus
absolute x square is equal to 1 or x is equal
to 1 by root 2, ok and now we can say y is
equal to i by root 2. Therefore, the eigenvector
x y is 1 by root 2, i by root 2. The second
eigenvector is obtained by solving with the
eigenvalue lambda is equal to minus 1.
So, you have 0 minus i, i 0, x prime, y prime
and that is equal to minus x prime, y prime.
So, you have minus i y prime is equal to minus
x prime or i y prime is equal to x prime the
second equation is the same as that.
Therefore y prime is equal to x prime by i
or which is minus i x prime, again absolute
y prime square plus absolute x prime square
is equal to 1, implies that the, i disappears
because minus i into plus i, when you take
the absolutesquare, essentially it is the
number into its complex conjugate. Therefore,
minus i x into its complex conjugate, which
is plus i x that power that becomes one x
square. So, this is equal to 1 implies again
x is equal to 1 by root 2 and therefore, your
eigenvalues x prime, eigenvectors experiment
y prime or now since we wrotey prime is equal
to minus i x prime and if we write x prime
as 1 by root 2, this will be minus i by root
2.
So, the second eigenvector has a sign change,
except that that is also a complex number,
ok it is a imaginary number therefore, what
you have is 0 minus i, i 0 times, the first
eigenvector is 1 by root 2, i by root 2 the
second eigenvector is 1 by root 2, minus i
by root 2. So, if you put both the eigenvectors
together in the same way that we did from
the previous problem, then the matrix is that
this eigenvector matrix comes first i by root
2, minus i by root 2 and then you have the
eigenvalue matrix 1 0 0 minus 1 ok.
Now, another important thing this is a unitary
matrix. This is not an orthogonal matrix,
ok it is very easy to check that because the
the square of the columns are rows, absolute
square of the columns are rows will give you
a total of one.
And if you take column one, 1 by root 2 and
i by root 2 and if you multiply this with
the in the sense, element by element if you
multiply this with the complex conjugate of
the column two, please remember 1 by root
2, does not change, this becomes plus i by
root 2, multiply element by element, then
what you get is 1 by root 2 into 1 by root
2 plus i by root 2 into i by root 2 and that
is minus 1 therefore, you see this is 0.
So, this is a unitary matrix the columns and
rows are orthogonal to each other in the sense
of the complex quantities involved therefore,
the this being a unitary matrix what is it
is inverse? u inverse, u inverse is, u dagger
which is the hermitian adjoint. So, if it
is a hermitian adjoint, this is nothing other
than, u transpose star. So u is 1 by root
2, i by root 2, 1 by root 2, minus i by root
2. So, what is u T star u transpose is the
transpose. So, let us take the first column,
it becomes the first row, i by root 2, but
its complex conjugate that you have to take
therefore, this is minus i by root 2 and the
second column, if you take that this becomes
1 by root 2, minus i by root 2, but it is
a complex conjugate that before this becomes
plus. So, this is u T star, this is u dagger
ok.
Now, you can easily see that if you write,
u dagger 0 minus i, i 0 u, if you write that
which is this matrix is now, u dagger because
you are multiplied it on the left with u dagger
and this u and times lambda and the definition
of the unitary matrixes that, this is equal
to identity therefore, this is lambda. So,
eigenvalue, eigenvalues are still real eigenvectors
or now complex, eigenvalues of a hermitian
matrix will always be real, there is a very
easy proof to do that I would not do it now
maybe, I will give it as an assignment or
a problem therefore, eigenvalues of these
are all hermitian matrices 0 minus i, i 0
is a hermitian matrix.
Therefore, the unitary the hermitian matrix
is diagonalized, by a unitary matrix. A symmetric
matrix is diagonalized by an orthogonal matrix,
the connection between these two things is
very important that is why this was mentioned
in the previous lecture also, now to see that
more and more such examples and some other
realproblems as we go along will reinforce
this concept again and again and I am sure
you will get familiar with that. So, eigenvalues
and eigenvectors are extremelyI mean they
are they are very resourceful now, what about
the eigenvalue for the z component sigma z
1 0 minus 1.
It is a diagonal matrix already therefore,
the lambda is already, 1 0 0 minus 1. Therefore
what is the unitary matrix, which will diagonalized
this or what is the orthogonal matrix, which
will diagonalized this identity matrix because
it is already diagonal therefore, you can
only write 1 0 0 1, multiplied by 1 0 0 minus
1 it is adjoint or whatever it is orthogonal
is still the same thing and that gives you
1 0 0 minus 1 trivial silly to write this
ok, since the matrix is already in diagonal,
it is eigenvalues eigenvector it is just the
identity matrix ok.
It is very important to understand these things
in detail. So, let me leavethis lecture with
a few specific problems of the eigenvectors
ok, now we will do one; one more simple problem
of a 3 by 3 matrix, before we close this lecture
ok and that is ait is illustrative of slightly
moredetailed algebra and very often, when
we do these lecturesyou will find out that
the simplest things are explained very well
and the slightly more difficult things are
left to you as problems and you always wonder
does the teacher really know the more difficult
problems and. So on.
Now, yes we do know, but it is important for
you to work this out a 3 by 3 matrix is a
slightly more involved example. So, let us
take this particular matrix A is equal to
5, 0, 2 0 1 0, 2 0 2 ok, and let us find outthis
of course, I have picked it up from one of
the most famous mathematical physics textbooks
or can the book on mathematical methods for
physics physicists of course, you can find
enumerable such matrices anyway, physicist
ok.
Now, let us work this out. So, we have to
write 5, 0 2, 0 1 0, 2 0 2, x y z, now the
eigenvector has three entries, because it
is a 3 by 3 matrix now for there are three
very threecomponents and if you want to write
this as lambda x y z ok, and find out lambda,
now look at this matrix this is symmetric
anyway therefore, I already told you that
it is easy to prove that symmetric and hermitian
matrices have real eigenvalues. So, our lambda
cannot be anything other than real let us
find out what they are, it is 5 minus lambda
0 2 0, 1 minus lambda 0, 2 0 2 minus lambda,
this determinant should be 0 hm.
So, let me not do this algebra, but I will
tell you that after factorization, the result
turns out to be 1 minus lambda, into lambda
square, minus 7 lambda, plus 6 is equal to
0. This is easy for you to verify, just expand
the determinant along a row or a column, you
will see a cubic equation and this cubic equation
is factorizable into this ok and quite; obviously,
the solutions are lambda is equal to 1, sorry
1 and lambda square, minus 7 lambda, plus
6 is basically, lambda minus 6 into lambda,
minus 1 is equal to 0.
So, now you have lambda is equal to 1, lambda
is equal to 1, lambda is equal to 6. I chose
this example for the specific reason that,
we have two eigenvalues, which are identical
to each other, this is called a degenerate
system.
Degenerate eigenvalue system, and the degeneracy
here is 2 degeneracy of the eigenvalue lambda
is equal to 1 is 2 therefore, the eigenvectors
will have a little bit of a tricky thing to
do ok. Let us look at the first non degenerate
eigenvalue, lambda is equal to 6. We have
the matrix equation as 5 0 2 0 1 0 2 0 2 times
x y z is equal to 6 times x y z ok, andthe
equations are it is very clear it is a 5 x,
minus 6 x therefore, it is minus x, plus 2
z is equal to 0 ok therefore, x is equal to
2 z ok.
Now, we need only one more equation because
this isthe determinant of this has to be 0
anyway as a homogeneous equation. So, the
next equation you see is 0 times x, plus 1
times y, because 0 times z, is equal to 6
y or y minus 6 y, which is equal to 5 y is
equal to 0, which gives you y is equal to
0 ok, and the last equation will tell you
2 minus 2, x plus 0 y, plus 2 z minus 6 z,
is equal to 0. Which gives you 4 2 x, which
gives you 2 x minus, 4 z is equal to 0. That
is the same thing as this equation namely
x is equal to 2 z ok.
So, the solution is that x is equal to 2 z
and y is equal to 0, this is the eigenvalue
eigenvector system for eigenvalue lambda is
equal to 6 ok. Therefore now we have a challenge
namely, x square plus, y square plus, z square.
If this is equal to 1, then what we have is
essentially x square is 4 z square, plus z
square, is equal to 1. Which means z is equal
to 1 by root 5 ok.
If z is 1 by root 5, then x is 2 z, which
is 2 by root 5 and then y is already 0, therefore
the eigenvector that we have x y z is 2 by
root 5 0, 1 by root 5.
Associated with the eigenvalue lambda is equal
to 6. Now, what about the eigenvalue lambda
is equal to 1? remember this is doubly degenerate,
that is this matrix, which I wrote down namely
the matrix 5 0 2, 0 1, 0 2, 0 2, has now an
eigenvector x prime, y prime, z prime is equal
to lambda is equal to 1 therefore, it is x
prime, y prime, z prime. You see immediately,
when you expand this to the system of equations,
you see this 5 x prime, plus 0 y prime, but
it is 2 z prime is equal to x prime. Which
gives you, 4 x prime, plus 2 z prime, is equal
to 0 ok.
5 x prime, plus 2 z prime, gives you that
is equal to x prime therefore, yeah. So, it
is 4 x prime plus 2 z prime is 0. The next
one is again, y prime is this is y prime,
is equal to y prime, that is all, it says
it does not, say anything else and the third
equation is 2 x prime, plus 2 z prime, is
equal to z prime. Which gives you 2 x prime,
plus z prime, is equal to 0, which is the
same as these these two are identical equations
ok, multiplied by 2 therefore, we have two
equations, in which now you have two quantities,
which are not determined between x and z,
we have to determine one of them of course,
we can do that by having thethe squares of
x prime, plus the square is of y prime, plus
the square of z prime, is equal to 1.
But we do not know what the y prime is the
equation, is y prime is equal to y prime.
So, what are the possible values, can you
take any value, can you take this will always
be is the case, when you have degeneracies,
if the degeneracy is two, that will be at
least one variable. We will have two choices,
we have to make two choices, one choice is
y prime is 0, if y prime is 0, that gives
you one eigenvector, the other choice is y
prime is equal to y prime. Does not tell you
that y prime is 0 therefore, y prime can be
anything.
So, the other choice is that. So, how do we
choose this let us do, the first one namely
choose y prime is equal to 0 and see what
happens then you have 4 x prime, plus 2 z
prime, is equal to 0. Which means x prime,
is equal to minus 1 by 2 z prime.
So, if you write x prime square, plus y prime
square, plus z prime square, is equal to 1,
then you get again x prime square, is 1 by
4 z prime square, plus y prime square is 0,
plus z prime square, is simply z prime square,
this is equal to 1. Which gives you 5 by 4
z prime square, is equal to 1 or z prime is
equal to 2 by root 5 ok. So, if that is the
case then x prime is minus 1 by 2, which is
minus 1 by root 5 ok therefore, we have one
column vector namely x prime is minus 1 by
root 5, y prime is 0 and the last one is 2
by root 5.
This is lambda is equal to 1 eigenvector,
eigenvalue, and lambda is equal to 6 eigenvalue,
we have already chosen that and we remember
that to be 2 by root 5 0, 1 by root 5 ok.
So, there is the third eigenvector, if we
choose y prime to be any number, what number
should it be? Now comes the property of the
eigenvector matrix for the eigenvalues, which
are real and we know that for a real eigenvalueah
system. The transformation matrix or the matrix
of the eigenvectors will be orthogonal. So,
if it is orthogonal then, what it means is
the following.
So, we have one column vector which is given
by 2 by root five. We have another column
vector, 0 which is given by 1 by root 5 this
is the first eigenvector and the second vector
we have is minus 1 by root 5 0, 2 by root
5 and the third eigenvector also, if you put
lambda is equal to 1, the question is only
about choosing y prime, the question is not
about choosing x prime and z prime.
Therefore, you remember x prime and z prime
are going to be in that ratio. So, no matter
what you choose the x prime is minus one half.
So, if you write to this as 
a this will be, minus 1 by 2 a, and this will
be sum constant c, this has to be an orthogonal
matrix. Yeah now please remember, this column
is orthogonal to this column as well as this
column now, if c is arbitrary you will see
that a into 2 by root 5 and then you have
minus a into 1 by root 5, will not give you
orthogonality it will not be 0 a into minus
1 by root 5 and then minus one half a, here
take the orthogonality of these two vectors.
If you do that then this gives you minus a
by root 5, multiplying this element with this
element and multiplying this by this, we will
also get minus a by root 5 is 0. Which tells
you a is 0, therefore, if this is 0 and if
this is 0 then you do not have a choice c
has to be 1 ok therefore, you see when you
have degenerate eigenvalues, it is important
to go one step at your time calculate the
first eigenvector and then find out, what
are all the variables that you can fix; fix
them one at your time and ensure that that
eigenvector is orthogonal to the existing
the pre determined, already determined eigenvector.
In this case it is a 3 by 3 matrix therefore,
we are very easily,we are able to say that
c has to be 1 and these two have to be 0.
Therefore, the eigenvector matrix now, is
2 by root 5 0 1 by root 5, then you have minus
1 by root 5, 0 2 by root 5 , and then you
have 0 1 0, as the eigenvector, what I did
not tell you is that this answer would have
been obvious to me and obvious if you knew
a little bit more matrix algebra from the
structure of the matrix that we were trying
to diagonalize you recall the structure of
the matrix is five 2, 0 1 0, 2 0 2, ok.
Now, in taking the determinant you recall
that this row and this column which has one
in the middle is not connected to any other
row or any other column you see that all the
other elements of this row are 0 all the other
elements of this column are 0. When you see
such a matrix you can write down immediately
that to be the eigenvalue itself 1 eigenvalue
and given the fact that it is 0 everywhere
and 0 everywhere the eigenvector will also
be the nonzero, if this is second it is second
this is nonzero 1 this will be 0 this will
be 0 that will be the eigenvalue lambda is
equal to 1, and the eigenvector lambda is
equal to eigenvector is is this one is easy
to see ok.
So, it is important when you look at the matrix,
you should also see matrices given for example,
if I give you a matrix 0 1 0 1 0 0 0 0 1 if
I give you a matrix like this same answer
is applicable, you see this has two eigenvalues
you remember this element 1 has 0 everywhere
on the row and also 0 everywhere in the column,
this is not connected to any other therefore,
remember the structure of the matrix is very
very important the off diagonal elements tell
you which elements are connected to which
through the row column vectors. So, matrix
is not something that you think is arbitrary
or a mathematicians invention or whatever
it has real values which elements are connected
to which and you can see that one is not connected
to any other therefore, this has a standalone
eigenvector 0 0 1 ok I can easily verify that.
And you recall that since this is standalone
with respect to the eigenvectors of the other
two this will not contribute therefore, this
will be 0. So, you will have 1 by root 2,
1 by root 2 as the other eigen vector and
third eigenvector has 1 by root 2, minus 1
by root 2, 0 as you have seen in this case,
you see the 0 in the middle, 0 in the middle
that is because, this element is not connected
to this element or this element through the
off diagonal links. You have to go through
a row you have to go through a column to see
that structure. So, you can see that since
it is not connected, you can see that the
eigenvector for this is independent of the
eigenvector the way it is and in a similar
way, if you do this matrix you will see that
the eigenvector here isgiven by these three
and you can immediately also see that these
three or the the rows the columns of an orthogonal
matrix and they are orthogonal to each other
ok.
So, when degenerate eigenvalues are there,
you have to be a little bit more careful and
it is important we will see degeneracy in
quantum mechanics, the hydrogen atom hasenergy
levels for it is atomic orbitals, which are
degenerate and they are n square degeneracy
as n increases from the principal quantum
number value 1 2 many and we have the degeneracies,
there in other systems when they have a particle
in the two dimensional box, we have degeneracies
for the energy levels therefore, matrix representations
and matrix algebra associated with some of
these hamiltonians, will have this characteristic
feature. So, this lecture was only to introduce
you to some of the more elementary, but more
important conceptual details. We will continue
nowwe will stop the the matrix description
at this point of time and maybe I will give
a supplementary lecture at the end of the
course or some time to add some additional
properties and things which have been left
out of this description, but the next lecture
said this the other sets will continue with
respect to matrices, this course will contain
only this much ok, we will continue with other
topics until then.
Thank you very much .
