In today’s lecture I will essentially be
talking about the basic postulates of quantum
mechanics.
Postulates are a set of hypothesis. And then
there is a theory, which is essentially these
postulates together with the mathematical
framework, which makes predictions, which
are supposed to be experimentally verifiable.
The theory is not meant to explain the postulates,
but given the postulates and the mathematical
framework 
you make theoretical predictions based on
these postulates, and compare with experiments,
the consequences of these predictions. Now,
if the experimental results agree with the
theoretical predictions, in a sense it is
an indirect test of the postulates. So, quantum
mechanics too works on a set of basic postulates.
Several of these postulates have been already
told to you by way of examples. The two dimensional
linear vector space illustrates these postulates
quite well as I will explain to you right
away. So, the first postulate 
it says that every physically observable quantity
is represented by a Hermitian operator. If
you wish to give matrix representations to
operators this would be a Hermitian matrix.
The experimentally measured values of the
observable 
can only be 
the Eigen values of the operator corresponding
to that observable.
So, for instance in the familiar example of
the two level atom the Hermitian operators
that we thought up were S squared, and S z.
So, too with the three level atom. These are
Hermitian operators and we had matrix representation
for these operators. In the case of the two
level atom there were two by two Hermitian
matrices or in the case of the three level
atom there were three by three Hermitian matrices,
so spin and the third component of spin if
you thought about it in terms of the spin
doublet which could be the electron or the
proton or the neutron. In the case of the
two and three level atomic systems it would
be just S square and S z which satisfied the
algebra S square is equal to S x square plus
S y square plus S z square and S z satisfied
along with S x and the S y.
The Lie algebra, the experimentally measured
values of these observables can only be the
Eigen values of the operators corresponding
to the observables. And Hermitian operators
are selected because Hermitian matrices have
real Eigen values, and all measurable quantities
will have to be real quantities. So, right
away it is good to see why Hermitian matrices
have real Eigen values. A Hermitian matrix
has this property that H is equal to H dagger,
dagger would just mean interchange rows and
columns and take the complex conjugate of
every element in the matrix.
I give you an Eigen value equation, H psi
is equal to a psi where a is the Eigen value
and psi is the corresponding Eigen vector.
Clearly in the Dirac notation, if I take the
dagger the ket would become the bra and the
number would simply be replaced by its complex
conjugate, but H dagger is equal to H. So
now, I could well find psi H psi h psi is
a state because H is an operator that acted
on psi to produce a new state H psi, and we
are trying to find the inner product of psi
with H psi. This quantity clearly from this
equation is psi a psi, but a is a number and
can be pulled out.
So, you have the first equation psi H psi
is a, inner product of psi with itself, that
is the first equation. Well, you could have
done that with this. You could have started
with psi H and you could have had a psi on
this side that would have given me an a star
psi psi and this is my second equation. But
since both quantities are the same a must
be equal to a star. In other words, the Eigen
value of the Hermitian matrix is real and
that is true for all Eigen values.
So, the set of Eigen values of Hermitian matrices
are real and therefore, the experimentally
measured values of the observable can only
be the Eigen values of the Hermitian operator
corresponding to that observable. We also
assume that the Hermitian operator is a bounded
operator. A concept of a bounded operator
really needs explanation only when we deal
with infinite dimensional spaces. And since,
up to now I have only spoken about finite
dimensional vector spaces. All finite dimensional
vector spaces have this property that the
operators are anyway bounded operators.
The concept of boundedness of an operator
is intimately linked with the concept of continuity
and both of them are best explained in the
framework of infinite dimensional linear vector
spaces, which I will do in a subsequent lecture.
So, for the moment these examples the two
and the three level atomic systems really
have only bounded operators as relevant operators,
having said that in general, there is no need
to imagine that all the Eigen values of the
Hermitian matrix should be a continuous set.
In general there could be discrete Eigen values
and that is precisely why the observable could
be quantized with discrete values, in contrast
to a classical system where the measurement
outcome could take a continuous set of values.
Here, for instance in the familiar example
of the two and three level atom equivalently
the spin doublet.
This was certainly true that S square acting
on psi was S into s plus 1 h cross square
psi, and S z on psi was m h cross psi. In
the case of the two level atom s was half
and m could take values minus half or plus
half, and therefore, I defined for you in
a previous lecture two basis states labelled
by the s and m values and the states were
half half and half minus half. It was on the
basis of this, you will recall that in passing
I mentioned Fermions and Bosons. Fermions
have half integer spin that means s can take
values half, three half and so on and Bosons
have integer spin, which means s can take
values 0 1 2 3 and so on. While Hermitian
operators are certainly the only operators,
that can be used to represent physically measurable
objects. It is also true that other operators
have their own importance in quantum mechanics.
For instance, we have already seen the operators
S plus and S minus. We have seen unitary operators
an example would be e to the i theta S z,
which I spoke about in the last lecture. This
is unitary because that is Hermitian and this
is exponential i times a Hermitian operator.
Talking of Eigen values it is good to digress
at this point and see in general what kind
of Eigen values a unitary operator can have.
So, if u is a unitary operator this is an
example: and if this is the Eigen value equation.
Correspondingly I can take the dagger and
get this equation. A unitary operator satisfies
this and therefore, I can find the inner product
of this bra with this ket. This is a ket and
that is the bra vector corresponding to that
ket vector. But this object is simply mod
a square psi psi because u dagger u is identity,
and if initially psi has been normalised to
unity it is clear that mod a square is equal
to 1.
So the Eigen values of a unitary operator
are complex numbers, whose modulus is unity.
We also have non Hermitian operators in quantum
mechanics which do important jobs. As for
instance, S plus and S minus they were the
raising and lowering operators. S minus was
a lowering operator 
in the context of the atomic system, because
S plus on the ground state of the atom took
it to the excited state and S minus on the
excited state of the atom gave us the ground
state.
We can easily find out the physical significance
of the coefficients here. If you go back to
the two level atom problem you will see that
S plus was defined as h cross e g. Let us
recall that these were the two levels and
S plus acting on g gave me 1 h cross e and
S minus acting on e gave me 1 h cross g, S
minus being the dagger of S plus.
Now in general, because S plus is the raising
operator it acts on the state given by labels
s comma m takes it to the state s comma m
plus 1 with the coefficient, which is s minus
m times s plus m plus 1 h cross. Now, this
is a general relation and we will derive this
in greater detail later. And therefore, S
plus in the case of the spin doublet or the
two level atom would act on the state half
minus half and give us 1 h cross as the coefficient,
and take it to the state half half.
Now similarly, S minus acts on a state given
by the labels s comma m, takes it to the state
s m minus 1 with the coefficient, which is
root of s plus m times s minus m plus 1. And
therefore, in our example S minus acting on
the state half half will take it to this coefficient
which really turns out to be 1 h cross times
s comma m minus 1, which is half minus half.
This is true even for the three level atom
and this is a general expression which can
be derived from the angular momentum algebra,
so much for the first postulate. The second
postulate is the following.
So there is postulate 2. And this postulate
says with every physical system 
there is associated an abstract Hilbert space.
Vectors in this space represent states of
the system, while I have not introduced the
concept of a Hilbert space. A Hilbert space
is a linear vector space on which an inner
product has been defined and which also has
a concept of completeness. Again, completeness
is best described and understood in the context
of infinite dimensional spaces becomes non
trivial there.
But as far as the finite dimensional linear
vector spaces are concerned, the completeness
relation has already been spelt out by me,
and completeness is a concept that needs to
be understood mainly in the case of infinite
dimensional spaces, which I will do subsequently.
The spin system that you have seen is certainly
the concept of completeness is pretty much
there already, so much for the Hilbert space.
In general any state of the system would be
a vector in this space and a general vector
can be written in terms of the basis vectors
as superpositions of the basis vectors.
So, already there is the concept of basis
vectors, I have this space it is spanned by
the basis vectors. So, if it is an n dimensional
space there are n basis vectors, by definition
these are linearly independent vectors and
every vector in the space can be expanded
as the superposition of these basis vectors.
Now, it is very convenient to choose an orthonormal
basis, where the basis vectors are mutually
orthogonal and normalised to unity.
The orthonormalization as I indicated in my
last lecture is done through a procedure called
the Gram-Schmidt orthonormalization procedure.
And before we proceed it is good to understand
this procedure. So, suppose I have a basis
set let me call it phi i, i is equal to 1
to n because it is an n dimensional space.
Let me start with the first of these basis
states I can always normalise the state to
1 so, you consider phi 1 phi one. Suppose
it is not equal to 1 I can always define the
ket phi 1 prime, which is phi 1 by phi 1 phi
1 square root, because then phi 1 prime phi
1 prime is phi 1 phi 1 by phi 1 phi 1, which
is equal to one. So, I have normalised one
of the basis vectors to unity.
So now let me look at this second vector in
this set. The second vector is phi 2 and I
am given that phi 1 prime phi 1 prime is equal
to 1. I need to construct from phi 2 a vector
which is orthogonal to phi 1. In general,
I expand phi 2 as a phi 1 prime plus some
other ket chi. I require that phi 1 prime
is orthogonal to phi 2. So, clearly from the
first term I get this and from the second
term I get this.
So, this should be equal to 0. But, this is
1 because I have already normalised phi 1
prime and therefore, a is equal to the inner
product phi 1 prime chi. In other words, phi
2 should be selected to be minus phi 1 prime
chi this inner product which is in general
a complex number phi 1 prime plus chi. Now
we need to normalise phi 2 to unity, and for
that as before i will divide phi 2 given in
this fashion by the square root of the inner
product of phi 2 with itself. And therefore,
phi 2 prime phi 2 prime inner product is 1.
So, I have normalised the new vector which
is phi 2 prime. I therefore, have two vectors
in the space 
which are normalised to 1 and which are orthogonal
to each other, namely phi 1 prime and phi
2 prime. I have to repeat this procedure with
the 3rd vector phi 3, which I expand as b
1 phi 1 prime plus b 2 phi 2 prime plus some
vector lambda. My requirements are the following;
that phi 1 prime phi 3 is equal to 0 and phi
2 prime phi 3 is also equal to 0. It is evident,
that if I first work with phi 1 prime phi
3 that just gives me b 1 plus phi 1 prime
lambda, because I have already shown that
this inner product is 0, which tells me that
b 1 should be equal to the inner product phi
1 prime lambda with a negative sign.
Similarly, the fact that phi 2 prime phi 3
is equal to 0 implies from here, that b 2
plus phi 2 prime lambda is equal to 0 or b
2 is equal to minus phi 2 prime lambda. Again
I have used the fact that the inner product
of phi 1 prime with phi 2 prime is equal to
0. Therefore, I expand phi 3 as minus phi
1 prime lambda. Phi 1 prime this is a ket
and this is a number minus phi 2 prime lambda
phi 2 prime plus lambda. In this manner I
can proceed and get an orthonormal basis of
n basis vectors. By definition they are linearly
independent and now they are also mutually
orthogonal, and each vector is normalised
to 1, so much for the Gram-Schmidt procedure.
The second postulate clearly said that every
vector in the Hilbert space represents a state
of the system. So, a general vector in the
Hilbert space which I will call psi can be
expanded as a superposition of the phi’s
or the phi primes now in this context, because
these are the orthonormal basis I can without
loss of generality remove the prime, and say
that I have a set of n mutually orthogonal
vectors phi k, which are the basis set and
any vector in that space can be expanded in
terms of this basis set. This is the expansion
postulate; this is quantum superposition because
I can superpose basis states 
to produce all vectors in that state.
Now the question is the following: I have
operators given by my first postulate which
act on states what kind of basis states can
I select? I have already demonstrated in the
context of the two and three level atoms that
you could choose different basis states and
these are unitarily related to each other.
They are related by a unitary transformation
i have also demonstrated that the Lie algebra
is preserved under such a unitary transformation.
Suppose I were making a measurement of some
physical observable. Let us take the familiar
example of S square and if it acts on any
state in that space, quantum mechanics tells
us that the state will collapse to one of
the basis states or Eigen states of S square
in this case with the corresponding Eigen
value. Because the measurement outcome is
simply going to be one of the Eigen values
of S square.
And therefore, the basis state in this context
would be simply the Eigen states of the observable
S square and the measurement of an arbitrary
state will lead post measurement to one of
the Eigen states of S square. With the measurement
outcome which is the Eigen value, I could
well make a measurement of S z as well. Now
if I did that, again the system will collapse
to one of the Eigen states of S z with the
corresponding Eigen value. But the physical
state of the system is the same. And therefore,
the Eigen state that I have finally, a post
measurement of S square and S z must be a
common Eigen state of S square and S z. In
other words, what happens to the system after
measurement is this the system collapses to
a state which is a common Eigen state of the
various observables that are measured simultaneously
giving corresponding Eigen values as the measurement
output.
In this context, one needs to understand that,
if two operators commute with each other,
you can find a complete set of common Eigen
states. First of all one needs to understand
what one means by a complete set of Eigen
vectors of each operator. As I said a bounded
Hermitian operator has a complete set of Eigen
vectors.
Now, to digress a little bit if you take a
Hermitian operator we can show that Eigen
vectors of a Hermitian operator or a Hermitian
matrix of a Hermitian matrix, corresponding
to distinctly different Eigen values are mutually
orthogonal. So, the bounded Hermitian operator
has a complete set of Eigen vectors and if
these Eigen vectors correspond to distinctly
different Eigen values, they are mutually
orthogonal. This can be seen very simply because
if I have a Hermitian matrix and suppose this
is the first Eigen value equation where a
1 is the Eigen value and psi 1 is the Eigen
vector. And I also have another Eigen vector
satisfying this Eigen value equation and the
Eigen values are different from each other.
Then clearly I can do the following thing.
I can find out psi 2 h psi 1. Now it is clear,
that if this equation holds h dagger is the
same as h is a 2 star psi 2. So, if i find
out psi 2 h psi 1 that will be psi 2, but
from my first equation h psi 1 is a 1 psi
1 and a 1 is a number. But, I could have used
the second equation and that just gives me
a 2 star. I know already I have proved that
Hermitian matrices have real Eigen values.
So i can drop this star 
and a 1 is not equal to a 2; implies if this
has to be true psi 2 is orthogonal to psi
1. So that is how you prove that Eigen vector
is of a Hermitian matrix corresponding to
different Eigen values are mutually orthogonal.
So, now the next part is to show that if I
have two such Hermitian operators. Let us
say S square and S z and they commute with
each other. Then I should be able to find
a complete set of common Eigen vectors. This
is an example: In general if A B is equal
to B A that is A and B commute with each other,
and if psi is an Eigen vector of A with Eigen
value a, B A psi is equal to A B psi. But
I know that A psi pulls out an Eigen value
a so this object is simply a b psi. And therefore,
the state B psi is an Eigen state of A the
operator A with Eigen value a. Now two cases
arise, A could be non degenerate. In other
words there is not more than one Eigen vector
corresponding to the Eigen value A, if at
all there is one more that is linearly dependent
on the other Eigen vector. So, i consider
that case that is an easy case to consider.
So, if the Eigen value is non degenerate it
is clear that the new vector B psi is linearly
dependent on the vector psi. So a constant
C 1 B psi plus C 2 psi is equal to 0, if there
is linear dependence C 1 and C 2 not equal
to 0 therefore, B psi can be written as minus
C 2 by C 1 psi. And his is just a number,
which I will call b. And therefore, psi is
also an Eigen state ket psi is an Eigen state
of the operator B with Eigen value b. And
therefore, I have found a common Eigen state
ket psi for the two operators A and B which
commute with each other. Now, in the even
that there is a degeneracy you have to work
a little bit more. Consider linear combinations
of the degenerate Eigen vectors and show once
more that there is a complete set of common
Eigen vectors for the two operators that commute.
Taking the example that we are familiar with,
S square commutes with S z and we know that
they have common Eigen vectors S square e
and S z e is an Eigen value equation. And
this is simply going to be half into half
plus 1 h cross square and this is just going
to be half h cross. Similarly, S square on
the ground state gives me half into half plus
1 h cross square g and S z g is minus half
h cross g. S square and S z commute with each
other. So this state collapses to a common
Eigen state of these observables. The state
itself is a very interesting concept in quantum
physics. Because, in classical physics if
you look at phase space you simply need to
know the value of the generalised coordinate
I call it x, and the generalised momentum
p corresponding to x.
At any instant of time to completely know
the state of the system, then there is the
equation of motion. The equation of motion
will tell you how exactly you could find the
values of x and p at a later time. This state
is completely determined. But here the state
of the system is described by this ket in
an abstract Hilbert space. In order to find
out what is this state of the system, we need
to first of all know what are the objects
that we are measuring. In this context, we
need to know that we are looking out for Eigen
values of S square and S z. So, the dynamical
variable becomes important and then one talks
about what is the state of the system post
measurement.
The 3rd postulate is a very important and
interesting postulate about measurement itself,
and this has to do with Eigen values and expectation
values of operators. So, the postulate basically
tells us this. So, this is postulate 3. It
is clear what the Eigen values are we have
already spoken about the Eigen values being
the measurement outcomes. But since every
state in the linear vector space or the Hilbert
space is a possible state of the physical
system, I could in general have a state psi
which is a superposition of the basis states
in an n dimensional linear vector space, and
it need not be an Eigen state of the operator
a.
This is the physical observable I am interested
in measuring and this is the operator corresponding
to that physical observable. In that case
one does not have an Eigen value equation,
but one talks of the average or expectation
value of A in the state psi. This should be
suitably normalised because psi itself may
not be normalised to unity and this has a
short hand notation, A is identical to this.
So, wherever the state is not an Eigen state
of A, it is a different matter that it will
collapse through the Eigen state after measurement.
The value of the physical observable that
I intend to measure is given by this expectation
value or the average value and that is a symbol
which denotes it. The denominator has been
put in because the state need not in general
be normalised to one, if it is normalised
to one this of course, becomes one and the
numerator will ((Refer Time: 43:10))
So, let us look at this expectation value
in the context of the general state psi. So
I have A psi is summation over k, C k A acting
on phi k where phi k are the basis states
corresponding to A. It should be read off
like this. There are a set of states phi 1
phi 2 up to phi n. So, I have A phi 1 is a
1 phi 1 A phi 2 is equal to a 2 phi 2 and
so on all the way to k is equal to 1 2 3 4
to n. So, in general if i take any one of
these the Eigen value is this and since this
is merely a number, and A acts linearly it
just acts on every term in this expansion
and therefore, I have this. But this object
is clearly given in the following manner.
A on ket psi is summation over k C k and from
these Eigen value equations it is clear that
this is what I have. Now you see I need to
find psi A psi that is the numerator. This
object is simply equal to summation over l
C l star phi l. The ket has become a bra vector
these are numbers these are coefficients,
so when I take the bra each coefficient becomes
is replaced by its complex conjugate, and
since I do not want to confuse indices I have
used l here instead of k. And therefore, I
sum over l here and I sum over k there. I
need to use the fact that this is in an orthonormal
basis. I can write the orthonormality condition
as phi l phi k inner product is delta l k
where delta l k is a Kronecker delta it means
that when l is equal to k is equal to 1 say
the answer is 1. Similarly, when l is equal
to 2 the answer is one and so on, but if l
and k are different the answer is 0. So this
is what I have and these are numbers so I
can well write it in the following manner.
I have psi A psi is summation over l and k
that is a double summation c l star C k a
k. The inner product phi l with phi k and
that was a delta l k, which means you can
get rid of one of the summations and I have
C k star C k a k. I can well write this as
modulus of C k square a k therefore, the expectation
value of A is summation over k mod C k square
a k by this.
This object is easy to determine in the event
that psi is not normalised to unity. And since
psi is expandable in terms of the basis vectors
in that fashion, I can find out the inner
product psi psi. Use the fact that there is
a delta l k out here, remove one of the summations
is a number, and therefore I just have summation
over k modulus of C k square.
And therefore, the expectation value of a,
it is a average value of A as a measurement
outcome is simply given by summation over
k modulus of C k square a k by summation over
k modulus of C k square. Now, the event that
psi is normalised to one it is clear that
if this is 1 modulus C k square summation
over k is 1. And then, it is clear that expectation
value of a simply has the numerator 
because the denominator is one. This has to
be properly interpreted.
It means the following. Suppose, I conduct
a number of trials to experimentally determine
A and I find the average value of these trials
the measurement outcome would be one of the
Eigen values of a, which is a k could be a
1, could be a 2, could be a 3 so one of the
Eigen values with probability modulus of C
k square. And therefore, the Eigen value a
1 will occur with probability in fact with
probability modulus of C 1 square by summation
over k modulus of C k square, but I am assuming
that psi is normalised to 1 and then it makes
it simpler to explain because a 1 will occur
with probability mod c 1 square, a 2 with
probability mod c 2 square and so on.
So, the measurement outcome really occurs
with a certain probability. Each of these
outcomes is possible with a certain probability
and therefore, this is a weighted average
that we have here. This is the sum and substance
of the 3rd postulate which tells us about
expectation values as oppose to Eigen values
and measurement outcomes. I will continue
to describe this and go on to the other postulates
in my next lecture.
