The 
fifth tutorial, so fifth tutorial has interesting
problem which is problem 3. We have been seeing
how to find out the eigenvalues and eigenvectors
in a way which we discussed in the first part
of fifth tutorial, you write the eigenvalue
equation, find out the eigenvalues, then you
find out the eigenvectors and then now let
us try to do some smart work.
Consider a 3-dimensional ket space with simultaneous,
in the question we have 3-dimensional ket
space with vectors or the kets which are orthonormal,
ket 1, 2 and 3 are orthonormal and the form
as a base ket of operator A okay and B okay,
and operator A is represented as a 0 0, 0
–a 0, 0 0 –a, this is operator A. In operator
B what do we have is b 0 0, 0 0 ib, 0 ib 0.
A and B are real. First operator you can see,
you can simply use eigenvalue equation you
need not use eigenvalue equation. So diagonal
matrix. So you can look from here and say
what are the eigenvalues. So eigenvalues will
be a –a –a, so operator A is degenerate
with eigenvalues, 2 eigenvalues as –a okay.
So let us see operator B. Operator B you cannot
do it quickly as we did for A.
So let us write this in some form okay. So
the third A question is to find out the eigenvalues
and eigenvectors of B. A you can just verbally
see it is very simple. So B I can write, let
me define something I, we always represent
I as identity, identity operator right, and
let me call this X as some 0 1 1 0 okay. This
is an identity operator I have swapped these
okay, swapped these operator and I get an
operator X.
So this is an operator, so this looks familiar,
you have seen such operators. They; is one
of the Pauli’s matrix and now I will try
and write operator, these operators B in terms
of identity and X operator. Let us see whether
it is easy to write and recognize the eigenvalues
and eigenvectors.
So operator B, I can rewrite as operator B
can be rewritten as b. This is b times I1
+ ib X2 X3, okay, so this was the operator
X 0 1 1 0, this is the identity operator I
which is 1 0 0 1 okay and you can see from
here that X is a Hermitian operator. If you
take the X dagger to be = X. So X is Hermitian.
So eigenvalues of this Hermitian operator
will be, this is a Pauli matrix. Eigenvalue
of Hermitian operator will be real.
And this operator is also called a swap operator
okay. So eigenvalue of this operator would
be simply 1 and -1. So let me write the eigenvalues
of so eigenvalues of operator X would be +1
-1 and you can write the corresponding eigenvector,
as you can write this corresponding eigenvector
as 1 2 1/root 2 ket. In terms of ket 1 and
2 okay. So for operator X the eigenvalues
are +1 -1 and the corresponding eigenvectors
are given by 1 and ket 2 it is very simple.
Now, we have written down B in terms of X
and I. Let me explain this to you. So B is
the coefficient which is real. I1 correspond
to the first component I11 component and ib
2 3 that means the second so this part is
0 ib ib 0 okay and the rest of the terms are
0. So I can just write in terms of this representation
of the vector, the operator in terms of these
operators. So operator b in terms of identity
operator and the poly operator, poly matrix
okay.
So this we have done using this information
and what we have obtained here can we calculate
the eigenvalues and eigenvectors of B.
What we have here is, that operator B can
be written as I just rewrite this as before
B is bI1 + ib X 23. This can be written as
B as identity + the new operator and now we
have that which forms which is the sum of
the subspace, this will be the sum of the
operators which are orthogonal subspaces.
This will give me one dimensional subspace
which is spanned by the eigen ket 1 and a
2 dimensional subspace this is 2 3. So this
is a 2 dimensional subspace which is spanned
by 2 and 3 ket 2 and 3 okay.
Eigenvectors of operator B is nothing but
1, 1/root 2 ket 2 plus ket 3 1/root 2 ket
3 2 – ket 3 with eigenvalues. okay. From
this we can guess the eigenvalues also b and
ib and ib. Because we have seen ib and –ib,
because we have seen here on the second page
the eigenvalues of this operator is +1 and
-1. This is an identity operator. So it will
have eigenvalue B so and the corresponding
eigenvalue of this operator.
We have written down this and the eigenvalue,
the eigenvector of operator x is this quantity
and eigenvector of identity operator will
be ket 1. So just precisely you can see that
operator B has eigenvectors 1, 2 + 3, ket
2 + ket 3/root 2 ket 2 – ket 3 by root 2
and the corresponding eigenvalues are b, i
and –ib. So you can see that these are nondegenerate
okay; however, the operator A has eigenvalues
a –a –a, so this is a degenerate state
or eigenvalue –a.
So commutator would be multiply matrix A by
B - matrix B by A. So if you get A B is 0
then A and B operator A and B commutes okay.
So this you have to check please check this
okay. It will be a simple exercise. So I will
leave it to you all. Next is to find a new
set of orthogonal ket. Now we have a set of
orthogonal kets 1, 2 and 3 given to us. You
have to find out a new set of orthogonal kets
which are simultaneous ket of both A and B.
So this is possible only when this is true.
So you have to find out the simultaneous eigen
ket of A and B. This is possible when A and
B commutes, then they would form a simultaneous
eigen ket, so you got the hint. So let us
express operator A as, operator A I can express
it as aI1 okay where aI1 is one-dimensional
identity operator - we have aI2 3 again this
is an identity operator because what was A?
Recall, A as a -a –a, 0 0 0 0 0. So this
will give me aI1 – aI2,3.
So this will give me aI1, one dimensional
identity operator and a –a 23 okay. So this
is what we have then what we do is – A,
just note this that - a is double degenerate
okay. So – a, eigenvalue –a is degenerate,
it is double degenerate and what do we have
here is - a is with a 2-dimensional space,
it is double degenerate with 2-dimensional
space. So therefore what do we say that 1
and any pair okay.
What do I mean by any pair, there are 2 kets,
ket 2 and ket 3, so one with any pair of the
orthogonal vector in this span 2 and 3.
So A has eigenvalues a –a –a this is what
we have and in the previous part I said that
ket 1 and any pair of the orthogonal space
vector is in the span of 2 and 3 will form
the basis of A okay. So A has eigenvalues
a –a and –a and we can choose the basis
of vector B to compute the above and it will
be the eigen ket or the basis of A also okay.
This is possible only when A and B commutes.
So if A and B commute and if you choose a
basis for B then automatically it will become
basis for A also, since B is non degenerate
it has 3 eigenvalues b, ib and –ib. So it
has different eigenvalues and the specification
of B, if we specify 
the eigenvalues of B then it can completely
determine the eigen ket. So we will stop here
and we will continue with the next 2 problem
in the next part.
