In our last lecture, we were discussing the
operated algebra involving the angular momentum
operators. We will continue our discussions
on that. We first said that the operators
J square and let me just consider the z component
of the angular momentum J z. They commute
with each other and since, they commute and
since, they are observables, we can have a
set of simultaneous Eigen vectors of the operator
J square and J z.
So, we wrote down J square ket lambda m is
equal to so much and J z ket lambda m is equal
to m lambda m. Then, we had defined two, we
have defined two operators J plus and J minus.
J plus was J x plus i J y and J minus was
J x minus i J y and now, let me consider the
commutation of J plus, J z. So, this is J
plus J z minus J z J plus. So, this is equal
to J plus is equal to J x plus i J y. So,
this will be J x, J z plus i J y, J z. Now,
you know that J y J z commutator is just i
J x. So, this will be i times i is minus J
x and this is in the wrong order. So, J z,
J x as you would recall from the previous
lecture is i J y. So, J x J z is the minus
of that. So, this is minus i J y.
So, if I take the minus sign, so you will
get J x plus i J y and so, this is equal to
minus J plus. I hope I have done it correctly.
So, the left hand side is J plus J z minus
J z J plus is equal to so much. What we do
is we operate this on ket lambda m, that is
we operate these both sides on ket lambda
m. So, you will get, you had here. So, let
me put a plus sign here, plus sign here and
minus sign here and take this to the other
side. So, we will have, please see this J
z J plus. Let me do this once more.
So, we have J plus, J z the commutator this
was equal to J plus we derive that so we had
J plus J z. Let me leave some space minus
J z J plus let me leave some space. This is
equal to J plus.
What we do is as we have mentioned operate
each term on ket lambda m operate this on
ket lambda m operate this on ket lambda m.
Then, you can see that J z operating. Did
I do it correctly? J plus J z is equal to
this was minus J plus minus. So, this is minus,
I am sorry, there is a minus sign here. So,
J z operating on ket lambda m is m times J
plus ket lambda m minus J z J plus ket lambda
m minus J plus ket lambda m.
So, I multiply all sides by a minus sign.
So, I get a minus sign here, plus sign here
and a plus sign here. Now, I take this term
on the right hand side. So, I get J z operating
on J plus ket lambda m is equal to m plus
1, m plus 1 operating on J plus ket lambda
m. Thus, if ket lambda m is an Eigen ket of
the operator J z belonging to the Eigen value
m, then J plus ket lambda m is also an Eigen
ket of the same operator J z, but now belonging
to the Eigen value m plus 1.
So, let me denote this file, say ket p. So,
J z ket p is equal to m plus 1 ket p. Now,
I do the same thing. Instead of operating
this by ket lambda m, I operate this on ket
p. So, I will obtain J z J plus ket p will
now be equal to m plus 2 J plus ket p of course,
provided ket p is not a null ket. Similarly,
if ket p is not a null ket, then J plus ket
p is also an Eigen ket of the operator J z
belonging now to the Eigen value m plus 2,
provided this is not a null ket.
We had seen that J, therefore we had seen
that J plus ket lambda m. In the last lecture,
we had shown that is also an Eigen ket of
the operator J square belonging to the same
Eigen value J plus ket lambda and J z J plus
ket lambda m is equal to m plus 1 J plus ket
lambda m. Therefore, if ket lambda m of course,
a non-null ket is a simultaneous Eigen vector
of the operator J square and J z. Then, J
plus ket lambda m is also a simultaneous Eigen
ket of J square and J z belonging to the same
Eigen value of J square, but 1 raised Eigen
value of J z.
Therefore, J plus ket lambda m must be something
like some constant times m plus 1 m increases
by 1, but lambda value remains the same. I
can go on doing that. Then, we will have J
square operating on J plus J plus ket lambda
m. That will belong to the same Eigen value
of J square J plus J plus ket lambda m, but
for this, we say will now be m plus 2. That
is why, they are known as ladder operators
provided this is not a null ket. So, the value
lambda remains the same on m increases, but
I have the inequality that lambda must be
greater than equal to m square. So, at some
m value, we must have a maximum value of m
because this process cannot be on indefinitely.
Otherwise, it will violate the inequality
lambda is greater than or equal to m square
and therefore, at some maximum value of m.
So, let there must exist a maximum value of
m and let the maximum value of m be equal
to J. Then, the max, this Eigen ket is a non
null ket, is not a null ket, but J plus ket
lambda J is a null ket because J m value with
the maximum value of m is J. So, you cannot
have m plus 1 and therefore, J plus ket lambda
J must be a null ket.
So, therefore, we must have that J plus ket
lambda J. That what is J. This is a null ket
that J is the maximum value of m. I operate
this by J minus and I had shown that you remember
that J minus J plus. We had evaluated this.
This is J x minus i J y multiplied by J x
plus i J y. So, this is J x square plus J
y square and this is plus i J x J y minus
J y J x. So, that is i J z. So, this is equal
to J square minus J z square minus J z. So,
I operate J minus J plus this thing.
So, you must understand the equation that
J plus ket lambda J is a null ket. This is
not a null ket. Of course, this is not a null
operator, but operating on a ket gives you
a null ket. So, I operate this by J minus.
So, I get J square minus J z square minus
J z operating on lambda J must be a null ket.
This is a non null operator, this is a non
null ket, but this operating on this is a
null ket.
Now, this is an Eigen ket of J square. So,
the J square operating on that is lambda.
The Eigen value of J z is J of J z square.
If J z ket lambda J is equal to J lambda J,
if this is the Eigen value of J z and therefore,
if I operate this again by J z, so J z square
lambda J will be equal to J operating J z
lambda J. So, that will be J square lambda
J. So, this will be lambda minus J square
minus J operating on lambda J must be a null
ket.
Now, here this quantity is a number, this
is an operator. It is a non null operator,
but this is a non null ket by definition.
So, this number must be 0. So, lambda minus
J square minus J. This is really beautiful
is 0 which gives me the beautiful result that
lambda must be equal to J square plus J or
J into J plus 1.
Now, if you have not been able to follow the
algebra, let me do it all over again for the
J minus operator. So, you have the J minus
operator. Let us forget about everything,
but keep this result somewhere. So, the J
minus operator is equal to J x minus i J y.
So, J square commutes with J x, J square commutes
with J y. So, J square commutes with J minus.
So, J square, J minus is equal to J square,
J x minus i times J square, J y. Both are
0, both vanish. So, this is zero.
So, you have J square J minus is equal to
J minus J square. I operate both sides by
lambda m. So, we have this is equal to lambda.
So, this will be lambda J minus ket lambda
m. So, you will have J square J minus. Thus,
if ket lambda m is an Eigen ket of the operator
J square belonging to the Eigen value lambda
as expressed by this equation, then J minus
ket lambda m is also an Eigen ket J square
belonging to the same Eigen value lambda.
Similarly, let me denote this by say, ket
a. So, then, we can write down J square exactly
in a similar way J square J minus ket a will
be equal to lambda J minus ket a. So, if I
denote this by ket B which is J minus ket
A which is equal to J minus J minus ket lambda
m, then we say.
Listen, please listen carefully. If ket lambda
m is an Eigen ket of J square belonging to
the Eigen value lambda, then J minus J minus
is also an Eigen ket of this same operator
J square belonging to the same Eigen value
lambda provided of course, a is not a null
ket and J minus a and b is also not a null
ket. Therefore, once again we must expect
the genre is and this is an Eigen ket, this
is an Eigen ket and this is also an Eigen
ket. Then, we consider the commutation relation
of J z.
So, let me consider the commutation relation
of J z of J minus. This is J z, J x J J minus
is J x minus i J y. So, this is J z, J y.
So, J z, J x is equal to i J y. I J y and
J z J y will be equal to minus. If you look
up the first, we wrote down that J z, J y
was minus i J x. So, minus minus plus and
i square is minus. So, this becomes minus
J x. So, if I take minus sign outside. So,
I get J x minus i J y, sorry i J y. So, this
will be minus J minus. So, I had J z J minus.
Please leave some space minus J z J minus
J z, leave some space is equal to minus J
minus.
Now, what I will do is operate this on lambda
m, operate this on lambda m and operate this
on lambda m and you have here J z ket lambda
m is m lambda m. So, I get this first term
becomes J z J minus ket lambda m and if you
take it to the other side, then it will become
m minus 1 J minus ket lambda m. Therefore,
if ket lambda m is an Eigen ket of the operator
J z, if ket lambda m is an operator is an
Eigen ket of the operator J z belonging to
the Eigen value m, then J minus ket lambda
m is also an Eigen ket of the same operator
J z.
Now, belonging to 1 less Eigen value m minus
1 provided of course, this is not a null ket
and I can continue this again, so J z operating
on J minus J minus ket lambda m. I hope you
can now immediately say that this must be
m minus 2 J minus J minus ket lambda m.
So, therefore, if ket, let me summarize If
ket lambda m is a simultaneous Eigen ket of
J square belonging to the Eigen value lambda
and J z belonging to the Eigen value m, then
J plus ket lambda m is also a simultaneous
Eigen ket of J square belonging to the same
Eigen value lambda and of J s J z. No, I will
put minus sign here, but now this will become
m minus 1 provided this is not a null ket
and then, J minus ket lambda m will also be
a simultaneous Eigen ket of J square belonging
to the same Eigen value lambda, but of J z
belonging to the Eigen value m minus 2 and
so on provided we do not hit a null ket, but
I have. So, this is an arithmetical preparation
m m minus 1 m minus 2 m minus 3. If it goes
on indefinitely, then this quantity will become
much greater than lambda and we have the inequality
that lambda must be greater than m square.
So, there must be a minimum value of m and
let that minimum value be equal to J prime.
That means that ket lambda J prime is not
a null ket. However, J minus ket lambda J
prime is a null ket. Even then J prime is
not a minimum value. So, once again this is
not a null ket, this is not a null operator,
but an operator operating on this will give
you a null ket. So, let me multiply this by
J plus. Then, my left hand side is equal to
left hand side. So, this J plus J minus is
equal to J x plus i J y J x minus i J y. Let
me do it carefully.
So, this is equal to J x square plus J y square
i times i is minus 1 minus minus plus 1 and
then, you will have minus i J x J y minus
J y J x. So, this is i J z. Actually, i h
cross J z. So, this will be i square is minus
1, so plus. So, this will become J square
minus J z square plus J z. So, this is my
J plus J minus, but we are seeing that J plus
J minus operating on lambda J prime. So, this
I operate on ket. This whole thing I operate
this on lambda J prime. So, this will be a
null ket. So, J square operating on this,
this is an Eigen ket of J square. So, this
will become lambda minus J prime square plus
J prime operating on lambda J prime must be
0.
This is not a null ket and this is a number.
Now, so this must be 0, so this must be 0
and therefore, lambda must be equal to J prime
square minus J prime. So, this will be equal
to J prime J prime minus 1.
So, please remember two things, ket lambda
m is a simultaneous Eigen ket of J square
and J z m is the Eigen value of J z and lambda
is the Eigen value of J square. First, we
derive that lambda must be greater than or
equal to m square and we said that what the
maximum value is. So, we said that since this,
because of this inequality the maximum value
of this is J and we found that lambda must
be equal to J into J plus 1.
Then, we use ladder operators J minus operators.
We found that m must have a minimum value
and let that be J prime and we just now found
that lambda must be equal to J prime into
J prime minus 1. So, J prime into J prime
minus 1 is equal to J into J plus 1 and J
prime. This is a quadratic equation in J prime,
but since J J prime has to be less than J,
so the root is J prime is equal to minus J
because if I put minus J here, so you will
get minus J into minus J minus 1. So, that
is J into J plus 1.
So, the solution of this equation is J prime
must be equal to minus J. So, the maximum
value is J. The minimum value is minus J and
now, see that I start with a value J and I
go down in steps of 1 J minus 1 J minus 2
J minus 3 J minus 4 J minus 5 J minus 6 and
then, I must hit minus J. That is only possible
if J itself is equal to 0 half 1 3 by 2 etcetera
not for any other number because we have said
that if m is an Eigen value, then m plus 1
is also an Eigen value. If m is an Eigen value,
m minus 1 is an Eigen value, it can hop in
units of 1 and it has to land up on minus
J. Otherwise, you will never get a null ket
and therefore, as you start with plus J, you
hop in units of 1 and you go to minus J. That
will only be possible if J takes these values.
So, if you have J is equal to 1, then you
start with 1 decrease by 0. They are the m
values. You start with J is equal to 3 by
2, so you have 3 2 half minus half and minus
3 by 2.
Let me say that I take a, let us suppose I
have J is equal to 7 by 3 to 7 by 3 minus
1 is 4 by 3, 4 by 3 minus 1 is 1 by 3, 1 by
3 minus 1 is minus 2 by 3 minus 2 by 3 minus
1 is minus 5 by 3 and then, minus 8 by 3 and
so on. So you have not hit minus 7 by 3 to
7 by 3 is not possible, but if you use 5 by,
so 5 by 2 minus 1 is 3 by 2, 3 by 2 is half
minus, half minus 3 by 2 and minus 5 by 2.
So, that is possible. So, J is equal to 3,
then 2, 1, 0, minus 1, minus 2, minus 3.
So, we get the result that the value of J
can only take 0 to half, 1 and lambda is equal
to J into J plus 1. These are the Eigen values
of J square operator. Now, you see a remarkable
result that we have obtained. Earlier, we
had said we had used differential operator
representation.
We had found that the Eigen functions are
the spherical harmonics and the Eigen values
were l square Y l m theta phi were equal to
l into l plus 1 h cross square Y l m theta
phi. In fact, these were simultaneous Eigen
functions of l square and l z, but here l
is equal to 0, 1, 2, 3, but here I will get
that J square ket J m ket lambda m. This is
equal to J into J plus 1 h cross square ket
J m ket, sorry lambda m and J z ket lambda
m is equal to m h cross, but the difference
between the two is that here J can take half
integral values.
As we know that the electron and the proton
and the neutron are endowed with a intrinsic
angular momentum which is half h cross and
that is predicted by quantum mechanics that
are now derived from quantum mechanics and
just by using commutation relations and nothing
else. So, we finally obtain actually instead
of lambda, the convention is to label the
Eigen kets as ket J m. This is not the Eigen
value. The Eigen value is really lambda, but
the convention is J square ket J m. This is
equal to J into J plus 1 h cross square ket
J m and J z ket J m is equal to m h cross
ket J m I have put. So, these and these values
of J that we can take are 0, half, 1, 3 by
2 etcetera and m can take from minus J to
plus J.
So, I had shown that J prime which is the
minus value of which is the minimum value
of J was equal to minus J. Therefore, these
are the m values for a given value of J. The
m values go from J to minus J. I had now proved
that just by using commutation relation. So,
all the previous results are proved that because
the differential operators also satisfy the
same commutation relations.
So, therefore, I have proved that for a given
value of J, the m values go from J to minus
J. Now, let me do now one more algebra and
that is we have shown that J plus ket J m.
If ket J m is an Eigen ket of J square and
J J z, then J plus ket J m is also an Eigen
ket, but now belonging to the Eigen, same
Eigen value lambda which is J into J plus
1 lambda is equal to J into J plus 1, but
m value is raised. So, let me write this down
and this must be a multiple of J m plus 1.
May be you put a, here. What we must always
remember on the right hand side, I may write
it down that J square ket J m.
Let me write it on a fresh piece of paper.
J square ket J m is equal to J into J plus
1 h cross square is there, but we assume that
in a system of units where J h cross is 1
and J z J m is equal to m h cross is 1. We
are working in a system of units of this.
Now, we say that J plus J m and these are
orthonormal ket. These are simultaneous Eigen
ket, but they are orthonormal, that is J prime,
m prime, J m here. All delta, J J prime delta,
m m prime, they are all orthonormal. Both
of them and this is normalized, that is J
m J m is 1. We assume that because they belong
to operators, so J plus ket J m is therefore
a multiple of this ket J m plus 1.
So, let this be c plus, let this be ket p.
Then, bra p is equal to bra J m J plus bar.
This will be equal to this is just a number,
so c plus star bra J m plus 1. So, I now write
bra p ket p, this is equal to J m J plus bar
is J plus is equal to J x plus i J y. So,
J plus bar is J minus because this will be
J x minus i J y. So, this is J minus J plus
J m is equal to c plus mod square and this
is 1 this times this is 1. So, once again
J minus J plus is equal to J x minus. This
you must remember by now J x plus i J y. So,
this is J x square plus J y square plus i
times J x J y minus J y J z. So, that is i
times J z. So, this is J x square plus J y
square is J square minus J z square minus
J z. You have derived this earlier.
So, we have therefore, bra p ket p is J m
J square J minus J plus is J square minus
J z square minus J z operating on J m is equal
to mod c plus square 2. So, this is operating
on this, this is an Eigen ket of J square
J z square and J z. So, J square that will
be J minus m square minus m J m J m. This
is c plus square. I am sorry this will be
J into J plus 1. I am sorry J square. I am
sorry this is sorry I have to write it again.
J square operating on J into this is J into
J plus 1 minus m square minus m. This is not
the Eigen value. The Eigen value is J into
J plus 1 as we had done here J into J plus
1.
So, this will be J into J plus 1 minus m square
minus m bra J m ket J m which is 1 because
they are orthonormal ket. This is equal to
c plus square. So, this is equal to 1 and
therefore, we obtain c plus is equal to the
under root of that. So, this will be J minus
m into J plus m plus 1. Therefore, J plus
ket lambda m, I am sorry let me do it again.
So, therefore, we will have J plus ket J m
will be equal to c plus and that is square
root of this. You have to remember J minus
m J plus m plus 1 into J m plus 1. These are
ladder operators. It raises the value of m
to m plus 1 and similarly, I leave it as an
exercise exactly in a similar way you show
that the operator J minus m is proportional
to J m minus 1 and this will be equal to J
plus m J plus m J minus m plus 1. These two
relations we had to remember and we see that
when m is equal to J, this is 0. So, J m J
is the maximum value of m.
So, J, J is a non null ket. It is a well defined
ket but, J plus operating on J J is a null
ket. Similarly, J minus J is a non null ket,
but J minus operating on that will be a null
ket. That comes out automatically from here.
Now, let me do an example. Let me consider
the case where this spin half J is equal to
half. This is an extremely important problem.
So, J is equal to half. So, m can be only
half and half minus half, sorry m can be half
and minus half. So, we can only have two states.
State 1 which I denote as half, half. This
is the value of J. So, the Eigen value of
J square is J into J plus 1 half into half
plus 1, that is 3 by 4 h cross square.
So, J square ket 1 will be half into half
plus 1, that is 3 by 4 h cross square ket
1 and J square and the other state will be
ket 2 half ket minus half. Now, so this I
take as the base vectors. I make a representation.
I will try to make a representation of the
operators J square J x J y J z. Now, when
we make a representation, you have to first
tell me what your base vectors are like in
an I want to describe a vector. If I want
to describe a vector and in terms of its component,
I will first ask you tell me your x and y
axis. So, this is the base vectors. So, in
terms of the base vectors 1 and 2, I will
write the operators as a matrix. So, for example,
since this is a two-dimensional space, there
are only two base vectors. We may have 2 by
2 matrices.
If I had J is equal to 1 where m can take
1, 0 and minus 1, so there will be three independent
vectors and it will span a three-dimensional
space. We will have 3 by 3 matrices. So, of
any operator, the matrix representation will
be let us suppose of operator p, there will
be four numbers p 1 1 which is 1 p 1 2, that
is 1 p 2 p 2 2 and p 2 2 1. These are the
four numbers which will give you a representation
of the operator. So, let me take these as
my base vectors.
So, I will take ket 1. So, I consider J is
equal to half. Then, m is equal to plus half
and minus half. So, the base kets are half
half and the second is half minus half. These
are simultaneous Eigen kets of J square and
J x, that is J square ket 1 is equal to half
into half plus 1. That is 3 by 4 h cross square
ket 1. J square ket 2 is also 3 by 4 h cross
square ket 2 and J z ket 1 is equal to this
is the Eigen value of J z half h cross ket
1 and J z ket 2 is equal to minus half h cross
ket 2.
Now, let me write down what is J plus ket
1. J plus ket 1 will be equal to J plus half
half. So, this will be square root of J minus
m. So, this is 0. So, remember that J plus
ket J m was let me write it below. J plus
ket J m was under root of J minus m into J
plus m plus 1 J of m plus 1. So this is 0
and J minus ket 1 will be J minus half half.
So, this will be please see this J minus half
half will be under root of J plus m into J
minus m plus half 1 J, m minus 1. So, J is
half m is half. So, half plus half is 1 square
root of 1 half minus half is 0, 1. So, this
will be half minus half. So, that is ket 2.
Now, let me write down what is J plus ket
2. So, here J is half m is minus half. So,
here J minus m half minus minus half half
plus half 1 half minus half is 0. So, this
is equal to 1 and J minus ket 2 J minus ket
2 means m will go down, but that cannot be
minus 3 by 2. So, this will be a null ket.
Ok, so we all have the relations now.
We can write down the matrices. First, let
me write down for J square. So, we had first
we wrote down J square ket 1 is equal to 3
by 4 h cross square ket 1. J square ket 2
is equal to 3 by 4 h cross square ket 2, so
J square. There will be four elements 1 1,
that is bra 1 J square ket 1, so bra 1. So,
this will be equal to 3 by 4 h cross square.
Then, J square 1 2 will be bra 1 J square
ket 2.
So, this will be bra 1 ket 2. So, that will
be 0. So, J square 2 1 will be 2 J square
ket 1. That will be also 0 and J square 2
2 will be equal to, I leave it as an exercise.
So, 3 by 4 h cross. So, the matrix representation
of the operator J square is 3 by 4 h cross
square 1 0 1. Similarly, I have J z ket 1
is equal to half h cross ket 1 because m value
is half and J z ket 2 is equal to minus half
h cross ket 2. So, again since these are Eigen
kets, so if you write J z 1 1, so this will
be multiplied by bra 1. So, this is half h
cross, but J z 1 2 or 2 1 2 1 will be bra
2 J z. So, this will be multiplied by half
h cross. So, this will be 0 and z z 2 2 will
be minus half h cross.
So, the operator representation, the matrix
representation of J z will be half h cross
1 0 0 minus 1. This is the sigma z, the sigma
z, the Pauli matrix. We have derived that
what are the Eigen values of J square at this
matrix 1 1. Therefore, what are the Eigen
values of J square? That is just 3 by 4 h
cross square. What are the Eigen values of
J z? The Eigen value of these matrix are 1
and minus 1. So, the Eigen values of J z are
half h cross and minus half h cross.
So, I have just given you a hint as to I mean,
rest we will do in our next lecture. I want
now to make a representation of x J x. So,
we have J plus is equal to J x plus i J y
and J minus is equal to J x minus i J y. So,
I add them up and get J x is equal to half
J plus J minus. So, J x bra m ket m, this
is the m, mth matrix element is given by this.
So, this will be half plus J plus ket n plus
half m J minus ket and I leave this as an
exercise for you. We will try to do this next
time, but if you can work this out, this will
be very good and we will, you should find
that the matrix representation of J x will
come out to be half h cross 0 1 1 0 and similarly,
for J y.
So, with this we conclude today’s lecture.
What we have achieved is that we have been
able to obtain the Eigen value spectrum, simultaneous
Eigen kets of the operator, J square and J
z. In our next lecture, we will complete the
analysis for obtaining the matrix representation
of J square, J x, J y and J z. Thank you.
