So, in this discussion ah we shall talk about
different representations in quantum mechanics.
And ah its almost obvious to you that the
physical observables would not depend upon
ah the representation that we are using rather
should be independent of the representations.
So, the expectation values of operators which
denote physical observables should remain
same.
However, ah it is possible to go from one
representation to another. ah In particular
we are talking about the coordinate representation
or the position representation and the momentum
representation, and we will see that ah how
they are connected, or rather ah what is the
form of an operator in momentum representation
and coordinate representation. And to begin
with let us talk about the x operator.
So, we start with position representation.
And x coordinate of a particle 
is associated with the operator x. So, we
will write an operator with a hat and so it
is associated with x and where of course,
my x variable ah that goes from ah minus infinity
to plus infinity. So, the gate x, so this
ah denote 
a particle at x such that ah x cap acting
on this ket will give me the position and
will return me back the ket and of course,
your x cap is a Hermitian operator which will
come and for. So, if a particle is in some
state psi, psi is an arbitrary state, the
amplitude of the particle this is important.
The amplitude ah to find the particle find
a particle at x is the complex number 
x psi.
And ah so basically the functional dependence
of this number 
of this number on x is what we call as wave
function psi x. So, let us see so the definition
of psi of x is the overlap with x and the
arbitrary state psi.
So, ah since the basic postulate says that
the total probability of finding the particle
should be one ah between ah minus infinity
to plus infinity. So, we can write this as
minus infinity to plus infinity mod psi x
square is equal to 1 and that is equal to
minus infinity to plus infinity, and ah you
have a dx and then there is a psi x, x psi
and so that is the ah definition.
And now, let us calculate ah x and x cap psi.
So, that tells that this is x cap star and
x psi. Now, ah x is equal to x cap star ah
this is called as the Hermitian operator ah
x has to be a Hermitian operator because x
is related to the measurement of the position
of the particle. So, it is a physical observable
and hence the ah operators should be Hermitian.
So, if this is true then ah this x x cap psi
is same as x cap x psi which is same as x
x psi and nothing, but x psi x. So, this is
the, ah so x cap operator ah being taken between
these states ah x ah which we know that x
acting on ah x cap acting on x will give me
x and we will return the ket x and so that
is now, taken with this with an arbitrary
ah state psi and it gives me ah the position
of the particle x and returns me the wave
function ah which is the function of x.
So, if you want to do the same calculation
in a continuum notation. So, this is in discrete
notation where we have extensively used the
Dirac bra and ket formalism if you want to
do it in. ah So, let us write it in continuum
notation is the same thing excepting that,
we are writing it in ah continuum notation
now. So, it is this and this is minus infinity
to plus infinity dx and x x cap and x prime
and ah so ah this is psi. So, we have simply
introduced states ah completeness of states
and this is nothing, but ah minus infinity
to infinity dx and x x cap ah x prime and
a psi of x prime and this is nothing, but
minus infinity to plus infinity dx x prime
delta x minus x prime and a psi of x prime
which is nothing ah by using the property
of the delta function its nothing, but x psi
x
So, it is a the same thing that we have got
here ah and here ah we have used to different
notations in order to arrive at the same result,
all right. So, we are dealing with the position
representation and let us now, ah talk about
the momentum operator.
So, ah the momentum operator is written with
a p cap ah and the definition ah we can prove
this, but however, we will skip the proof
and take it as a definition. So, definition
is that a p is written as a minus i h cross
del del x. So, ah we may actually define a
p and a psi of x to be a state a momentum
eigenstate let us call it as u p x, where
p this subscript denote the momentum and similarly
as earlier. So, it acts on a ket p and returns
back the momentum of the particle and the
ket itself and so this is nothing, but then
minus i h cross ah del, del x of u p x which
is equal to p, u p x.
And similarly when we want to write ah the
expectation of the momentum operator between
the position ket x and an arbitrary state
psi this is nothing, but minus i h cross del
psi del x. And ah so this if you want to ah
contrast with what we have just obtained in
the last slide this is equal to x psi x.
So, ah as I told you that it is possible to
prove ah the definition of ah the momentum
operator. However, we will skip that rather
what we would do is that we will go ahead
to prove the form of the position operator
in the momentum space. And in order to do
that ah let us the question is ah find ah
x cap in the momentum representation, ok.
And to do that let us ah take an arbitrary.
So, an arbitrary state psi ah and we can use
the ah commutation relation between ah x and
p to be. So, remember ah x and p the commutation
is equal to i h cross, i, where this is an
identity ah operator or you can call it an
identity matrix ah and basically the x and
p which are the conjugate variables here they
do not commute and the commutation give rise
to a finite quantity which is equal to i h
cross. And remember that they both are infinite
dimensional matrices each of x and p and infinite
dimensional matrices do not ah commute. ah
So, now, I can write or rather take this commutation
relation between ah the ket p ah and psi and
this is nothing but equal to ah p ah x cap,
p cap minus, p cap, x cap and a psi. And which
is nothing but equal to if I separate it out
it is x cap p caps I and minus p ah p cap
ah x cap psi that is separating it out. Now,
ah we have p ah x cap p cap, I leave this
first term as it is and apply or rather operate
p cap on the drop p which gives me a p and
so this is equal to a x cap and a psi. So,
ah this is also again taken as momentum is
also a Hermitian operator which means a p
equal to p star hm
Now, if I ah write this as equation 1, ah
if I rearrange, rearrange equation one and
use the commutation relation 
then I can write this as a p ah x ah rather
ah p x psi, equal to a p x p ah psi and a
minus i h cross p psi that is just rearranging
this and using the ah commutation relation
and now, take the first term. So, let us call
this as to take first term in the right RHS
of equation 2 that is this term and this term
can be written as a p ah x cap p caps psi
which ah we use our continuum notation which
is equal to p x cap ah x x ah again use the
completeness of states and this, and this
is nothing but this is equal to a dx x exponential
i p x by h cross and a minus i h cross d dx
of psi x. Now, I leave one step for you to
ah figure out and complete.
So, using integration by parts, please ah
look at calculus book and or rather mathematical
physics book and see what integration by part
ah is and this can be written as. So, this
same term I write it here ah this is equal
to a i h cross a dx, and d dx of x ah exponential
i p x by h cross psi x. So, this can be written
as i h cross d dx of exponential i p x by
h cross psi of x plus i h cross dx hm x i
p over h cross exponential i p x by h cross
psi of x. If you do it carefully you will
get it.
So, this is nothing, but this is equal to
the Fourier transform of psi of x which is
nothing, but psi of p. So, this is psi of
p and this is i h cross p ah and a d dp of
dx exponential i p x by h cross psi of x.
So, this whole thing is written as d dp of
this hm and one can check that that is true
its i h cross psi p plus i h cross p ah d
dp of psi of p, ok.
So, this is equal to ah where of course, we
have used definitions such as um ah let us
ah see the definitions being used ah we have
x p psi ah is equal to minus i h cross d dx
of psi of x which is known and x of psi is
equal to psi of x and p of psi is equal to
psi of p. So, this is ah our ah that first
term on the RHS of 2. If we put it back ah
we will get ah putting it back in into we
will get p. So, we will get up p and ah p
p x sorry p x psi and this is equal to i h
cross psi p plus i h cross p d dp of psi p,
and minus i h cross p psi and this is nothing
but psi p and so there is a plus i h cross
psi p and there is a minus i h cross i p they
would cancel, and I am going to also lose
this p because p is not equal to 0. So, they
can be ah divided by in both sides. And then
finally, I land up with a p x cap and a psi
which is equal to i h cross d dp of psi p.
So, that is the ah definition. So, if we ah
want to get the operator x cap ah in terms
of the momentum. So, we will knock off the
edge end states and we will get x cap equal
to i h cross d dp. So, that is the definition
of ah position operator in the momentum space.
As I told you earlier that you are aware of
and very familiar with the momentum operator
in position space. However, there is an alternate
ah description possible in quantum mechanics
ah where we do not talk about position representation,
but we talk about momentum representation,
and a position operator takes a form ah of
this kind in the momentum space whereas, the
momentum operator in the position space has
a minus sign with a del d dx or del del x
as you wish to write.
So, this is the position ah operator of course,
this p is in the x direction here is a canonical
variable ah in the x direction corresponding
to the x ah motion. So, let us write down
that how would it ah be.
So, any operator can be written down in any
representation. However, the expectation value
which corresponds to a physical quantities
remains conserved. Let us write an operator
ah o ah in momentum space. So, ah o is written
in o p. So, the expectation value of o p is
equal to psi star p o p psi p dp psi p if
psi of p equal to delta of p prime minus p.
Then of course, your o p its equal to a delta
p prime minus p, o p delta p prime minus p
and dp which gives me nothing, but o p prime.
Sometimes it is important to find powers of
the position operator. So, ah let us take
x to the power n. So, ah x to the power n
expectation is written as psi star x ah x
to the power n psi x dx which is equal to
psi star x ah dx and there is a 1 over root
over 2 pi h cross and a psi p x to the power
n exponential i px by h cross dp. And hence
x to the power n ah or rather, ah this one
ah exponential i p x by h can be written as
minus i h cross d dp whole to the power n
exponential i p x by h cross.
See here every time you take a derivative
of this exponential function ah which is oscillatory
ah you would get of x here. So, if you take
it n times you will get x to the power n and
that is what it is. So, your x to the power
n is nothing, but i h cross d d p of n. So,
this is the, so if you raise x n times ah
the operator has to be multiplied that many
times in order to get ah it in the momentum
space. ah Let us give a simple example.
So, Hamiltonian 
of quantum harmonic oscillator. So, this is
equal to ah H equal to p square over 2 m plus
half m omega square and then x square. So,
I can write x square as i h cross d dp square
which is equal to p square over 2 m minus
half m omega square h cross square and d 2,
dp 2. So, this is the form of the Hamiltonian
in the momentum space and we have to solve
a potential or rather Schrodinger equation
ah with this Hamiltonian.
We have always you know solved the quantum
harmonic oscillator problem writing it in
position space that is changing the p to minus
i h cross del del x, and then I have gone
ahead and solved the Schrodinger equation.
However, an alternate thing can also be done
ah in the form of this.
Let us go to another important discussion
which is called as ah density operators and
density matrices. So, the origin of the problem
is like this that usually we deal with a large
number of particles. When we talk about a
particular experiment or even talk about Young's
double slit experiment we do not talk about
a single photon we talk about a large number
of photons. So, there is a statistical ah
analysis or the statistical theory has to
be inbuilt.
Now, the statistical probability or the weightage
that is associated with the statistical ensemble
that has to be taken into account and as well
as the probabilistic interpretation of quantum
mechanics has to be taken into account. So,
the problem ah as in classical statistical
mechanics which was only ah the probabilistic
interpretation or the probability was coming
as a weights of the statistical ah you know
the entities ah is no longer ah true. We also
have a quantum mechanical probabilistic interpretation
which comes and interferes with that. So,
in order to develop the quantum statistical
mechanics ah it is important to ah study these
density operators or the density matrix and
we will give you ah the important ah points
and probably do one or two examples with that.
Most of the time 
or ah ok, I mean which is true as well you
can ah ah say that there are exceptions to
this as well. So, most of the time we worry
about we start from a given state of the system
and then allow the system to evolve. However,
there are exceptions where the initial state
is not ah completely known. ah We will say
it in a little technical words ah we will
say that that is not pure and may contain
admixture of states. ah
So, this an example can be given as a the
polarization state 
of natural light what we mean by is not laser
ah it is basically natural light is not known
with precision, with certainty. Which means
that there are ah different polarizations
that are mixed for sure we know that the amplitude
ah is actually could be in any direction and
this called as a polarization ah being not
in a given direction or its not known. And
then in this particular case that system is
said to be in in an admixture of states, states
with probabilities p 1 p 2 etcetera such that
sum over i p i is equal to 1.
So, what I mean to say is that the probabilities
intervene at different levels the initial
information about the system ballistic. ah
So, it becomes probabilistic and the final
states that is the time evolved states 
also suffer from probabilistic predictions.
So, we will talk about density operators,
but while we talk about these probabilistic
interpretation, I it one thing that comes
to my mind is that sometimes the students
ah get quite confused with the fact that we
write. ah So, this is just a discussion which
is slightly out of the way, but it is important
never the less. The discussion is that that
we write a total state psi and write it in
terms of C n and a phi n, where C ns are they
depend on time whereas, the basis states phi
n s are constants. So, phi n is a basis that
spans the Hilbert space.
So, it is often said that the probability
ah to find the state in phi 1 is associated
with ah C 1 square and phi 2 is associated
with ah C 2 square and so on. But this statement
is not correct because the cut the correct
statement would be ah the probability to find
the eigen function ah or rather to find an
eigen value corresponding to phi 1. So, the
correct statement is the probability 
of finding 
eigen value associated with with phi k is
C k square. Because you have to ah allow for
because if you say that the probability that
the state is in is found in phi 1 is C 1 mod
square phi 2 is C 2 mod square that defeats
the purpose of quantum mechanics because there
has to be necessarily interference between
these states. So, those interference terms
such as a C k star and a C k prime interference
terms should be present.
So, just to reiterate this ah when you write
ah a total wave function as a sum of these
ah amplitudes and the basis states ah there
is a sum over n. ah So, ah it is not true.
ah So, this is a false or rather this is untrue
that ah the probability of finding the particle
in phi 1 is C 1 mod square and phi 2 C 2 mod
square and so on. The rather the correct thing
is that the probability of finding the eigen
value associated with phi 1 is C 1 square
C 1 mod square and C 2 mod square and so on,
ah because then that allows for the these
the cross terms or the interference terms
to be present which are hm very important
for quantum mechanics.
Anyway, we will come back to this density
operators and see that what they ah or density
matrices what they see, no, what they ah actually
ah mean.
So, to study ah statistical mixture of states
we should use, we should rather introduce
a density operator 
which permits a simple description of the
statistical as admixture of states. But also
there is another hm advantage of using the
density operator, that is without using the
state ah vector or the wave function psi ah
everything could be done in terms of the density
operator. So, we shall show that characterizing
a system 
by its state vector 
is completely equivalent to 
to characterizing it 
by a certain 
operator acting on the state space.
So, that underscores the utility of the density
operator that we can actually do away with
using the state space ah and completely ah
work with the density operator. So, let me
ah box this rather important statement which
we are going to prove.
So, recall that a psi equal to sum over n
c n t, u n and for any operator A we can write
it O also previously we have written O, but
it is a same thing. So, ah the matrix elements
between two ah basis states u n is given by
this. So, the expectation value 
at a time t is A t, its psi t A psi t ah this
is equal to n and p. So, this is C n t, C
p p, A n p, where A n p is defined earlier.
Now, so the expectation value A t involves
ah these quantities C n star t C p t which
are the amplitudes the complex amplitudes.
But see that these amplitudes these product
of amplitudes are the matrix elements elements
of an operator psi t and psi t. So, so this
is this is the projector on to the ket psi
t. Let us see how.
So, let us take a an operator say let us call
this as B which is equal to we will of course,
soon change its nomenclature from being B
to rho. ah So, this is equal to ah a psi of
t. So, this tells you the B acting on psi
t its equal to a a psi t and a psi ah p, and
psi ah t, and so on and so this is nothing
but a C n star t ah and C p t and a psi t
and so on. So, this is the matrix element
or rather the eigen value of this operator.
So, this B acting on psi t ah it will give
you ah this.
So, it is now natural to introduce 
density operator rho of t, where rho of t
is exactly what we said about B that is rho
of t is equal to ah this projector psi t and
so on ah. So, in the basis, so psi s are written
in the basis this u n, ah hm u n's are assumed
to be orthonormal basis, but it is not a big
problem if they are not there are always going
from ah non orthonormal basis to orthonormal
basis by doing a transformation called as
a ah this a Gram-Schmidt Orthogonalization.
And so this is a, so the matrix elements of
the ah operator are. So, rho mn t equal to
u m rho t u m which is equal to cm star t
C n t. So, ah now, rho t yields all the physical
observables as was the claim that it will
replace the usage of the state vector psi
of t. So, before we show that this is ah so
your ah sum over n ah C n t mod square it
is equal to rho nn t which is equal to trace
of rho t which is equal to 1. So, this talks
about the conservation of probability.
So, the expectation value of any arbitrary
operator is written as ah u p rho t and a
u n, u n A and a u p and this is nothing but
summation over p ah u p, rho t A and ah u
p ah this and this is nothing but trace of
a rho t A. So, we do not; so we simply have
to calculate ah the density matrix multiply
it by the operator and take the trace that
will give the expectation value of the operator.
So, this is completely equivalent.
And the last thing that we can see is that
the equation of motion 
d dt of rho t its equal to ah d dt of psi
t ah and psi t outside. Now, keeping that
psi ah t 
and ah psi t hm taking the derivative on the
other psi t. ah So, this is equal to minus
i by h cross H t psi t, ah psi t and this
is a plus i by h cross ah psi t ah psi t H
t,and this is nothing but minus i by h cross
H t and rho t. And ah so this tells that the
equation of motion of the density matrix of
the density operator ah is the commutation
of of the of the operator with the Hamiltonian.
So, if these two commute then they have same
state of eigenstates or ah they have the same
basis, they can be diagonalized in the same
basis.
So, for pure states, ah I we have def defined
pure earlier it means that the ah the initial
state with which we begin with is completely
known. So, there is no statistical admixture
at all and so ah for pure states rho dagger
of p equal to rho of t, rho square of t equal
to rho of t and trace of rho square of t equal
to ah 1. And as I told that if there is an
admixture of states some of these ah these
properties do not hold good we will see ah
for a particular example.
