In my last lecture, I spoke to you about linear
vector spaces and introduced the properties
of a linear vector space. In particular, we
discussed a two dimensional linear vector
space. We shall continue our discussion on
linear vector spaces today, using a specific
example the two level atom.
So topic is linear vector spaces. This is
the 2nd of the lectures on linear vector spaces,
with particular reference to the two level
atom. Let me, quickly recapitulate some of
the salient features of what I said last time.
We had a 2 dimensional linear vector space.
The basis vectors were chosen to be orthonormal.
Each of them was normalized to unity and they
were orthogonal to each other. If we used
column vector representation for these basis
vectors, we could have well represented e
x by 1 0 and e y by 0 1. We could use the
Dirac notation, and perhaps refer to these
basis vectors as ket 0 and ket 1. I want to
emphasize that the 0 here does not stand for
the null vector, this does not mean the null
vector.
Earlier on when we discussed the properties
of a linear vector space, I said that there
is a null vector in the space, and on an addition
with any other vector it leaves the vector
unchanged. Also introduced, the scalar, the
number 0 which multiplied with any vector
in the space gives me 0, this ket does not
represent the null vector, nor does it represent
zero it is merely a label. It is a label for
a basis vector and this is another label for
this basis vector, in the case of the two
level atom.
I have two basis vectors, I could refer to
them as ket 0 and ket 1, or in the paralons
of atomic physics they are usually referred
to as ket g and ket e. So, g and e are merely
some symbols, denoting that this is the lowest
energy state of the atom or the ground state.
And, this is the 1st excited state of the
atom and that is why the symbol e. The point
that I am trying to make is that, if energy
is the observable in the system, which is
indeed the way we are going to work with it.
Once a measurement is made on the energy,
the system collapses either to this state
with an energy Eigenvalue E g, or to this
state with an energy Eigenvalue is E sub e.
Denoting by H, the observable energy I have
used H because H is the Hamiltonian, and from
classical physics you would know that the
Hamiltonian operator would represent the total
energy of the system. Ket e and ket g are
Eigenvectors of h with Eigenvalues E sub e
and E sub g. A general state of the atom can
be expanded in terms of these two basis vectors.
So, the general atomic state can be written
in this manner and if indeed the state is
normalized.
So, if the states psi were normalized to 1,
then it automatically follows that mod a square
plus mod b squared, equals 1 where a and b
are complex numbers. Of course, here I have
selected orthonormal basis states. This is
1, this too is 1 and ket e and g are orthogonal
to each other. This obviously, follows from
this fact because this is obtained by simply
taking the complex conjugate of this object.
Now, I could give matrix representations in
that case I would write this as: 0 1, 1 0
and that is 0. I could give matrix representations
for all these and obtain for you for instance
in this case that 0 1 row with 0 1 column
vector is 1.
So, I have this orthonormal basis and these
basis vectors are chosen in the context of
this physical situation to be the energy Eigenstates
of the atom. Now, having said that what are
the various operators that I can make with
these two basis vectors?
Here, are the list of operators, that is an
operator. In matrix notation, this is simply
going to be 
and this is going to be the matrix. Similarly,
I can write the other matrices, this is one
of the operators that I can form with ket
g. It is evident that these two are Hermitian
operators, because when I take the dagger,
that means interchange the rows and the columns
and take the complex conjugate of all the
entries that simply gives me the same operator.
Similarly, this is another Hermitian operator.
Of course, I can have combinations of these
operators, I can for instance think of the
operator 
you will notice, that I use operators and
observables interchangeably and I could give
matrix representations to the operators. Similarly,
I can have an operator of this form, these
are evidently Hermitian operators. There are
more operators that I can think of:
I can have the non-Hermitian operator e g
and g e, this object is clearly the Hermitian
conjugate of the dagger of that operator.
I can think of combinations of these and so
on. So, I can form a set of operators given
the basis vectors. We now see, how the operators
are formed using the basis vectors. Choose
the basis, the basis should be physically
relevant to that system. In the atomic system
and experimentally measurable quantities say,
energy of the atom, here is a two level atom.
So choose the energy Eigenstates, ket e and
ket g as the basis states. This is just a
convenient choice, you may have many more
sets of basis vectors that can be chosen.
And I will discuss various possibilities later
on, in subsequent lectures.
Given those basis vectors I can form all these
operators and combinations of these operators.
Some of them are Hermitian, and some of them
are not but non Hermitian operators such as
these also have a physical significance. They
perform a certain operation on the atomic
system. And we shall see pretty soon, what
kind of operations are performed by these
operators?
In this context, I wish to recall the algebra
of the spin matrices because that is particularly
relevant in the problem of the two atomic
two level atomic system.
The spin matrices, were defined in this way,
S y is h cross by 2 sigma y and S z is h cross
by 2 sigma z. The matrix representation for
these let me recall are as follows: these
are the matrix representations. There was
an algebra of the spin matrices, and I pointed
out that that is the general angular momentum
algebra. In the sense, the same algebra will
be satisfied in the context of orbital angular
momentum.
For the moment, I will refer to these as the
algebra of the spin operators S x, S y and
S z. And to remind you of the algebra, there
is a cyclic relation like this the commutator
of S x with S y is i h cross S z, S z with
S x is i h cross S y and S y with S z is i
h cross S x. I could equivalently define S
plus and S minus as, S x plus or minus i S
y and the algebra can be written in terms
of: S plus S minus and S z in the following
manner. What has that got to do with the atomic
system? We will see that right away. Consider
for instance the action of, let us say e g
this operator on this state on either ket
e or ket g.
So, let me consider the action of this operator
for which I can give a matrix representation.
Recall, that ket e was represented by the
column vector 1 0 and bra g was represented
by the row vector, 0 1. The action of this
operator on ket g which is the same as the
action of this matrix, and this matrix I write
right here on 0 1, is simply ket e because
the action of this matrix on that column vector
is 1 0 which is identical to ket e.
I could have easily found this, without writing
out matrices by realizing that this object
can be expanded, using the fact that g is
normalized to unity, ket g is normalized to
unity and therefore I will have ket e. If
we realize that this is simply the scalar
product, that leaves behind the state ket
e and that is what it is. Similarly, if I
take the Hermitian conjugate operator, g e
and if it acts on ket g, it is clear by the
same kind of argument that because ket g and
ket e are orthogonal to each other, this is
0.
So in my picture, I have the two levels of
the atom and there is this operator, which
I will denote by S plus. You will realize
that I am doing this from hindsight because
I will show soon, that this S plus is the
same as the S plus that I got here. So, the
operator e g takes the ground state to the
1st excited state. The operator g e, brings
down the 1st excited state to the ground state.
And to complete the story, when the operator
S plus acts on the excited state. It simply
destroys it, it annihilates it and when it’s
Hermitian conjugate acts on the ground state,
it simply destroys it. So, I have this two
level atomic system, where I have formed two
operators non Hermitian operators but which
are Hermitian conjugates of each other. So
S minus is the Hermitian conjugate of S plus
and this is the role of these operators, to
raise a state or to lower the state. If you
look at this .and you put in the matrices
for S plus and S y and find out the matrix
representation of S plus.
You will find that, the matrix S plus is precisely
this 0 1, 0 0. You can do that right away
and check out that that is indeed true apart
from a factor h cross, if you use matrix representations
for S plus and S minus.
You find that S x plus i S y is, let me put
in the h cross as well 
and that simply gives me which is precisely,
what I have here apart from the h cross. And
therefore, I can do by a similar procedure,
I can find out that S x plus i S y, which
is S plus is simply h cross e g.
So, I just have to write h cross e g is S
plus and h cross g e, is S minus. S minus
is h cross g e. We have therefore, discussed
the job that these two operators do.
We now know what exactly is the role played
by e g and what is the role played by g e.
Let us look at an operator like e, e. I could
do this by writing matrices explicitly, as
in the earlier case, but I think we will do
well to use the Dirac notation.
And find out happens when, this acts on states
it is clear that because e is a, normalized
vector normalized to 1. You just retrieve
ket e and when e e acts on g that is 0. Similarly,
this operator, acting on g gives me the same
state and this operator acting on the 1st
excited state gives me 0. Now, from this,
it is evident that when the operator e e acts
on a generic ket, which represents a vector
in the linear vector space, it simply gives
me a ket e.
Similarly, if I replace all the e’s by g’s,
I find that the operator g g here would simply
project out b, which is the component of the
state in the direction of the basis vector,
g. In other words, this is a projection operator
And I have identified e e and g g as, projection
operators 
which project a generic state in the direction
of ket e or ket g, depending on which operator
is used. So, these are the projection operators.
So Now, we know what exactly these two do
needless to say,
That if I now consider the operator e e plus
g g, that is, simply going to play the role
of the identity operator. It is obvious because
while one of them gives us the component along
e the other gives us the component along g,
and therefore we can assemble both together
and make the full vector. To go back to our
matrix notation, just to keep correspondence
with it. This object is simply going to be
and this is going to be represented by this.
You can easily check that this is the identity
operator, that is the identity operator. That
is a 1 0 out there.
So, you see the sum over operators formed
like this from the basis states, that sum
adds up to the identity. In other words, in
general this is an example of a statement
which says that let phi i, i is equal to 1
2 as many as there are basis states, be a
complete 
basis set. Then, summation over i, phi i,
phi i is equal to the identity operator. And
here, in our problem i simply takes values
1 and 2 phi 1 being e and phi 2 being g. And
therefore, this relationship is evident, this
is called the completeness relation and what
I have demonstrated to you is the completeness
relation in the context of the atomic system.
Now, let us proceed and see what the other
operators mean.
So we know what this operator is, this operator
is merely the identity operator. Now, let
us consider this operator:
So, I now start with the operator minus g
g and I wish to know what it is. Already,
I know what S plus and S minus are, suppose
I find the commutator of S plus and S minus
in this language. That is the same as trying
to find the commutator of e g, apart from
an h cross with g e. I pulled out h cross
squared because there was an h cross with
this operator, and there was an h cross multiplying
this. The way to do this, is to make it work
on some basis state and see if the algebra
that we have is indeed state independent.
So, let us consider the commutator of S plus
with S minus acting shall we say on the state
e.
So this is the same as h cross squared. Suppose,
I expand the commutator since, the commutator
is a b minus b a, where this is the operator
a and that is the operator b and this is supposed
to act on ket e. So, we take a look at this,
this is 1, this is 1 and I can write it as
e e minus g g acting on ket e. And low on
behold that is what I was trying to find but
I do know that this commutator S plus S minus
is 2 h cross S z and therefore, I understand
that h cross by 2, e e minus g g is equal
to S z.
So, I have found out what exactly this is
h cross by 2 e e minus g g is S z, e e plus
g g is the identity, h cross e g is equal
to S plus, h cross g e is equal to S minus.
And of course, I can form e g minus g e and
so on. So, returning now to the construction
of further operators. Let me consider S squared.
I define S squared as S x squared plus S y
squared plus S z squared. The simplest way
to find out what S squared is, is to realize
that this is h cross squared by 4 sigma x
squared, plus sigma y squared, plus sigma
z squared. But, this is each of them is the
identity operator. You can explicitly check
this, by putting in the matrix representations
for sigma x, sigma y and sigma z. Therefore,
S squared is 3 h cross squared by 4 times
the identity operator. It is obvious that
S squared commutes with S x, S y, S z, S plus,
S minus and so on, because it is essentially
the identity operator.
Let me look at Eigenstates of S squared and
S z or S squared and S x or S squared and
S y but it is conventional to look at the
Eigenstates of S square and S z and pretty
soon we will understand what the purpose of
this is.
It is clear that when S squared acts on ket
e, it simply gives me 3 quarter h cross squared
ket e and S squared acting on ket g, gives
me the same thing. I would like to, write
this as half into half plus 1, like this.
I would like to write this in general as S
squared, acting on an Eigenstate, gives me
S into S plus one h cross squared Eigenstate.
And in this case, this state could be either
ket e or ket g and S is equal to half.
Now, let me look at S z acting on ket e. It
is obvious, that this is the same as h cross
by 2, e e minus g g, acting on ket e, and
that is just h cross by 2 ket e. There is
no contribution from this term, because this
inner product is 0 whereas, from this term
one realizes that, that inner product is 1
leaving behind ket e. I can do S z acting
on ket g and that is just going to give me
minus h cross by 2 ket g. I therefore, realize
that the two basis states in question ket
e and ket g are common Eigenstates of S squared
and S z.
So, what are the Eigen values?
The Eigen values are the following: s is equal
to half and m, where m is defined this way,
takes values plus or minus half. This becomes
very important particularly in the context
of spin doublets like the electron. A model
which I will talk about a little later. Returning
to the two level atom, I find that there are
two commuting observables: S squared and S
z. By commuting observables, I mean that this
commutator is 0, it is also true that S squared
commutes with S x and with S y. But, since
S x, S y and S z do not commute with each
other. The only commuting observables that
I can talk about are S squared and S z or
S squared and S x or S squared and S y. It
is conventional to use S z as the operator
along with S squared.
What do we see from this? If a measurement
is made, to find out the value of s and m,
that means the Eigenvalues of S squared and
S z, for clever experiment is made on the
atomic system. On the state of the atomic
system to, find out these Eigenvalues after
the measurement simultaneous measurement of
S squared and S z. We will get the values
of s and m accurately.
Of course, you have to allow for human errors
and for instrumentation errors but having
said that you can get simultaneous values
for s and m. The Eigenvalues of S squared
and S z accurately, quantum mechanics tells
you, that you can take a complete set of commuting
observables and make a measurement simultaneously
of the Eigenvalues, corresponding to all those
observables, or the actual values, numerical
values of these observables, which are outcomes
of the experiment that is performed. And you
will find, that you can measure all these
with accuracy, subject to human errors and
instrumentation errors.
On the other hand, if you make a measurement
of the Eigenvalues corresponding to S x and
S y. Since, this is not a pair of commuting
variables, if you make simultaneous measurement,
you will find that you cannot measure both
of them accurately, simultaneously. This is
in fact, the preamble, to the uncertainty
principle the Heisenberg uncertainty principle
which I will talk about in more detail in
a subsequent lecture. In this problem, we
have these two commuting observables S squared
and S z and therefore, it is possible to label
the state of the atom in the following manner,
by two quantum numbers, which I get as the
outcome of suitable experiments.
There is this state e: which has a value of
s equal to half and a value of m equal to
half and the state g: which has s equal to
half and m equal to minus half in units of
h cross. So, I can do better than simply use
abstract notation e and g for my states, I
can start finding attributes, I understand
properties of that state, I understand that
these are states with values of quantum numbers
given like this. I also understand that these
are the only two quantum numbers in this problem
that can be measured simultaneously, accurately.
Quantum mechanics, does not forbid us from
making measurements on observables simultaneously
on any set of observables. The question is
how accurate are your answers? And if you
want accuracy, then there should be a set
of commuting observables.
I would now consider another example. This
does not pertain to the two level atom, it
is a different example. It is about the spin
doublet. Now, the reason why I wish to discuss
it here, is because the spin doublet can also
be explained using the same mathematical framework,
that is a two dimensional linear vector space.
The spin doublet, in particular I can talk
about the electron. An elementary particle
which has got charge minus e and rest mass
approximately 0.51 million electron volts
by c squared, where c is the velocity of light
and vacuum. Even as the electrons attribute
is minus e, for the value of charge and the
rest mass with this value, it can exist in
two spin states. Spin is an experimentally
measurable quantity, you could call it the
up state of the electron, which is the half,
half state that I spoke about here.
And the down state of the electron, which
is the half, minus half state. Both these
states are spin states of the electron, before
measurement of the spin of the electron, the
spin state of the electron would in this notation
be represented in this manner. And after measurement,
the system would collapse either to the up
state, or to the down state. We will show
later, that there are other particles, with
spin s value given by 1, 3 by 2, 2 and so
on, in units of h cross.
In fact in the three dimensional world that
we live in, elementary particles are classified
as fermions and bosons, fermions have S equal
to half 3 by 2, 5 by 2 etcetera. Half integers
spin values, in units of h cross and bosons
have S equal to 0 1 2 etcetera in units of
h cross. What I have spoken about, in the
context of the two dimensional linear vector
space, is the physical system. A physical
system, with S equal to half, which could
live in the up state, which means m is equal
to half, that is the Eigenvalue corresponding
to S z, or in the down state where m is equal
to minus half, which is the Eigenvalue corresponding
to S z in the down state of the electron.
There are other particles which have spin
equal to half for instance the proton, the
neutron and many others. Now, examples of
bosons and example of a boson is the photon.
The photon, is an example that is light quanta,
it is an example of a boson. And, it has spin
1. So clearly, the photon will not be described
in the purview of this two dimensional linear
vector space. In general, it is true that
for a given value of spin, the 3rd component
can take values minus s to plus s in steps
of 1. This is a very important result in angular
momentum algebra. And I will derive it in
a subsequent lecture in detail, when I talk
about angular momentum.
It would be pretty clear by now, that the
Dirac notation is extremely powerful and it
would be very cumbersome to use matrix notation
all the time. In the sense of writing explicitly
column vectors and matrices n by n matrices,
as the dimension of the linear vector space
increases and do an explicit matrix multiplication.
The power of the Dirac notation, was what
I was trying to bring about, in this context
by trying to find out, what all these objects
mean through their effect on the basis states.
By using the Dirac notation instead of the
matrix notation.
A good exercise would be to consider the 3
level atom, you could denote the energy Eigenstates
as g, e 1 and e 2. Both in the case of the
two level atom and in the 3 level atom, a
question arises, if the only commuting observables
are s squared and s z. How do you manage to
find out the value of the energy? These are
supposed to be energy Eigenstates and if energy
does not commute with s squares and s z. How
can we fix the energy Eigenvalues corresponding
to the ground state and the excited state
of the atom? The fact of the matter is that,
the Hamiltonian or the energy operator in
this problem can be written in terms of s
z.
As also in this problem, look at it this way.
Suppose I wrote the Hamiltonian as 
omega S z, where omega is a suitable constant,
then the action of the Hamiltonian in the
case of the two level atom on e, is going
to be 
h cross omega by 2 e, and when it acts on
the state g it is minus h cross by omega by
2 g. We would not like to have negative energies
as the Eigenvalues as experimental outcomes.
So it is easier for us to say, that we can
choose a Hamiltonian, which perhaps is of
this form, omega S z plus h cross omega, times
the identity operator. In that case, it is
clear that I pickup an extra h cross omega
from here, and therefore, I will have a net
energy value 3 h cross omega by 2 e for the
excited state. And for the ground state, I
have h cross omega by 2 with the minus sign
there, plus h cross omega. And that is the
same as h cross omega by 2 g which is a positive
quantity. So this is how I would construct
the Hamiltonian for the two level atom. I
would attempt to do a similar kind of operation.
In the case of the 3 level atom the operators
are easy to identify.
I have a set of operators as before e e, e
1, e 1 to show that this is a 1st excited
state, g g and e 2, e 2 and non Hermitian
combinations of this kind, e 1 g etcetera.
The completeness relation, in this case would
simply mean that: e 1, e 1 plus e 2 e 2 plus
g g is really the identity operator. I could
choose for e 2; the column vector 1 0 0, for
e 1; the column vector 0 1 0 and for g; the
column vector 0 0 1. This is a possible representation
for the basis vectors, and in terms of these
I can write out various operators for this
system. I should be able to construct raising
operators, that is operators which act on
ket g and take it to ket e 1, act on ket e
1 take it to ket e 2 and lowering operators
which will bring down the atomic levels, we
should bring down this level here and this
level there.
How exactly is this done? Well, normally in
order to raise the system from a lower energy
state to a higher energy state you pump in
energy by getting the atom, to interact with
light, shine light of appropriate frequency
on the atomic system. So, that energy is provided
to the atomic system which perhaps is in the
ground state enough energy for it to go into
this excited state or that excited state and
so on.
Similarly, when the atom gives energy out
in the form of light quanta, that is photons
it jumps from a higher excited state to a
lower energy state, it could come from here
right there or from here, here and so on.
So, there are very many possibilities and
clearly these experiments are done, using
light atom interactions. Many times, ideal
laser light is used to make ket interact with
the atoms in order to, take care of these
operations and we just saw how exactly the
operators, in the linear vectors space do
the job of raising and lowering of the atomic
levels. It would be a very good exercise,
to try to interpret the various operators
that one can form in the case of the three
level atomic system, and I would urge you
to do it, so that you get a complete understanding
of what exactly happens in this system.
It is evident that, there are three basis
vectors in this case and therefore, it is
no longer the two dimensional linear vector
space that we spoke about. The linear vector
space is larger. It is good to check out all
the properties of the linear vector space
in this context. With this, I conclude my
discussion on the two level atoms. And the
next time, we will continue our discussion
of the spin matrices with further properties
of S x S y and S z and of course, the Pauli
matrices.
