In the last couple of lectures, I have worked
out a series of exercises for you pertaining
to operator algebras and uncertainty relations
and so on. Today, I will continue to work
out some more exercised problems. These will
be problems of direct relevance to various
physical situations and therefore, I will
explain the importance of the problems that
I will attempt to work out as I go along.
So, again we have exercises. The 1st exercise
pertains to the harmonic oscillator. The linear
harmonic oscillator for which if I set m equals
omega equals h cross equals 1 for simplicity
I can always put them back later, once I do
the calculation. Then X would simply be a
plus a dagger by root 2 and p would be a minus
a dagger by root 2 i and of course, the commutator
X P is i times the identity operator. I have
set h cross equals 1. This would also correspond
to a a dagger equals 1 the identity operator
again.
So, I will work with this example where m
is equal to omega equals h cross equals 1
and what I will attempt to find out are the
following: Expectation value of x in any state
n of the oscillator, expectation value of
p in any state n of the oscillator. Recall
that n takes value 0,1,2 and so on n equals
0 is the ground state of the oscillator then
we can build up the excited states from the
ground state. We also want to know what is
the expectation value of X P plus P x in the
state n. So that would be the 1st problem
that I will attempt I wish to calculate these
3 objects.
So, let us start with the expectation X. This
is identical to n X n and that is equals 1
by root 2 n a plus a dagger n that simply
n a n plus the expectation value of a dagger
in the state n, a n takes this apart from
some coefficient to the state n minus 1 and
a dagger takes this apart from a coefficient
to the state n plus 1. But the states are
orthonormal to each other they are orthogonal
to each other and therefore, this would simply
be n n minus 1 apart from this coefficient
and therefore, this is 0. Similarly, this
would be n n plus 1 and that is also 0 and
therefore, the expectation value of X in any
state of the harmonic oscillator. By state
I mean one of the natural basis states in
that linear vector space or the fock states.
In quantum optics language you would refer
to them as the photon number states and X
as a quadrature variable. So expectation X
is 0.
Now suppose I wish to find out expectation
P in any of these states. I would repeat this
substituting P is a minus a dagger by root
2 i.
So, I wish to find 
this is identical to expectation P and that
is given by 1 by root 2 a minus a dagger there
is also an i n. Once more I use the same argument
a acts on n to lower it to n minus 1, but
ket n is orthogonal to ket n minus 1. Similarly,
a dagger raises this to n plus 1 and the 2nd
term also is 0. I use the same arguments as
I used earlier and therefore, expectation
value of P is 0.
This can be explained well when one goes to
the possession representation and the momentum
representation, which I will subsequently.
It has to do with the fact that all the energy
Eigen states when written as functions of
X, when written as wave functions that is
functions of X they have a definite parity.
It is rather early in the day to discuss that,
but I will come back to this problem when
I discuss the harmonic oscillator in the position
representation later on. The next thing I
wish to prove is that this object is also
equal to 0. So, let me work that out.
So we start with n x p plus p x n, but we
know that the commutator x with p is i therefore,
x p is i plus p x. Actually it is i h cross
times the identity operator out there, but
we would set h cross equals 1. So, x p is
simply i plus p x and therefore, I can write
this as i plus 2 p x, but I want the expectation
value in the state n. The 1st term simply
gives me i, the 2nd term is really this. I
now expand p and x in terms of a’s and a
daggers. p x is a minus a dagger by root 2
i times a plus a dagger by root 2. This object
is half i a square minus a dagger square plus
a a dagger minus a dagger a. I use this and
I find out expectation x p plus p x.
So, I get expectation x p plus p x in the
state n is i plus 1 by 2 i expectation a square
that is a minus a dagger square plus a a dagger
minus a dagger a in the state n. If you look
at the 1st when a square acts on ket n, when
the 1st term around a reduces n to n minus
1 then the 2nd time it reduces n minus 1 to
n minus 2. So, an a square acts on ket n I
get ket n minus 2 and again ket n is orthogonal
to ket n minus 2. So, this term does not contribute.
So this expectation value is 0. Similarly,
a dagger square’s expectation value is 0.
What I would be left with would be objects
like this. So this is i minus i by 2 expectation
value of n a a dagger minus a dagger a n,
but I can always use the fact that a a dagger
is 1 plus a dagger a.
I have x p plus p x and this gave me a 2 p
x. So there was already a two outside and
therefore, I need to put in x p plus p x.
I need to put a 2 there. So the twos cancel
and I am left with i minus i because the inner
product of n with n, bra n with ket n is 1
and therefore, the answer is 0.So, you see
the expectation value in any state n of the
oscillator of the object x p plus p, x that
operator x p plus p x is 0.
So, what we had established in the 1st exercise
is that expectation x expectation p and expectation
x p plus p x all vanish in any state of the
oscillator. I now move on to next exercise
to find out delta x and delta p in any state
n of the oscillator. We have already shown
that delta x, delta p take on minimum values
in the ground state of the oscillator and
if I set h cross equals 1, delta x is 1 by
root 2 and delta p is also 1 by root 2 and
therefore, the product delta x delta p takes
its minimum value which is half.
So now in any state of the oscillator, I wish
to find out delta x delta p. Recall that delta
X square is expectation X square minus expectation
X the whole square, that we have just shown
that expectation X is 0 in any state of the
oscillator. So, I need to calculate expectation
X square in units where m omega and h cross
are 1 is simply a plus a dagger the whole
square by 2, which is the half a square plus
a dagger a plus a a dagger plus a dagger square.
So, if I need to find the expectation value
of x square in the state n. I need to calculate
the expectation value of these four objects
and then sandwiched between ket n and bra
n. Look at the 1st term, as I argued before
the last term will not make a contribution.
So I have half a dagger a plus a a dagger.
The expectation value of just this operator
once more, I can write a a dagger as 1 plus
a dagger a and I need to compute this. I have
used a fact that a a dagger is 1 plus a dagger
a and that is why I have got that. The 1st
term just gives me a half the 2nd term gives
me there was a 2 from here and a 2 from there
so that cancels out it gives me n and therefore,
I get n plus half.
So, the delta X the whole square is n plus
half in the state n of the oscillator. And
that is why in the delta X in the ground state
is just the positive square root of n plus
half and that is why in the ground state of
the oscillator 
delta X was 1 by root 2. I have set h cross
equals 1 you could put it back and then you
will have root of h cross by 2. In any other
state delta X is root of n plus half. So,
in the 1st excited state delta X is equal
to root of 3 by 2, again in units of h cross
equals 1 and therefore, if I put back the
h cross it is root of 3 h cross by 2 and so
on. So, you have the ground state which is
the minimum uncertainty and delta X keeps
increasing as the state label increases.
Now, let us look at delta P once more I use
the same procedure. Delta P the whole square
is expectation P square minus expectation
P the whole square. We have just shown that
this object is 0 at any state n of the oscillator
and therefore, I need to compute expectation
P square P square is a minus a dagger by i
root 2, a minus a dagger by i root 2. And
this object is a minus half a minus a dagger
times a minus a dagger. So, I can expand it
and write it in this manner. I need to find
n P square n and that is minus half expectation
a square minus expectation a dagger a minus
expectation a a dagger plus expectation a
dagger square all expectation values obtained
in the state n. As before this is 0 and so
is this I need to only worry about the contributions
from a dagger a and a a dagger.
So n P square n is half a dagger a plus a
a dagger, but this is precisely what we computed
earlier and this will give me an answer n
plus half. Because once again I will write
a a dagger as 1 plus a dagger a. Therefore,
delta P the whole square is n plus half of
course, if I put in the h cross I get this
and therefore, delta P is square root of n
plus half the positive square root. While
in the ground state, I have the minimum value
in the 1st excited state it is 3 h cross by
2 the square root of 3 h cross by 2 and so
on and we find that delta X is equal to delta
P in any state ket n of the oscillator.
So this is what the 1st exercise was about
and we have established that while the ground
state is a minimum uncertainty state, in all
other state delta X and delta P do not vanish.
Because, expectation X square and expectation
P square are non zero, but we have shown explicitly
that delta X equals delta P and the uncertainty
product gets larger and larger as we go to
higher values of n.
My next exercise concerns the qubit. I now
wish to look at the state psi which is ket
0 plus e to the i theta ket 1.The theta is
e to the i theta is just a phase it is normalised.
I can check that out it’s normalised to
unity, because this object is 1 and so as
this. And therefore, I take this normalised
state ket psi, it is a qubit. It is a superposition
of ket 0 and ket 1 and I wish to find expectation
X and expectation P in the state psi.
So, expectation X in the state psi implies
computing this. I have set m equals omega
equals h cross equals 1, but that object is
the same as 1 by root 2 bra 0 plus e to the
minus i theta bra 1 by root 2 a plus a dagger
ket 0 plus e to the i theta ket 1 by root
2 and that gives me 1 by 2 root 2 and I have
to compute this. So, this is what I have.
This involves the following terms: 1st I have
0 a 0 then I have 0 a e to the i theta ket
1, that is what I have worked with 1st, then
I have 0 a dagger 0 plus e to the i theta
0 a dagger ket 1.
Now I move to the 2nd term here, so that gives
me plus e to the minus i theta 1 a 0. There
is an e to the minus i theta times e to the
i theta so I just have 1 a 1. I work with
the dagger now, so that gives me an e to minus
i theta 1 a dagger 0 and the last term which
is simply 1 a dagger 1. This is what I need
to compute. It is clear that this is 0 because
a and ket 0 is 0 so this will make a contribution
because a and ket 1 is ket 0 and therefore,
this object simply turns out to be e to the
i theta a dagger on ket 0 is ket 1 and by
the fact that ket 0 and ket 1 are orthogonal,
that is 0, this is ket 2 and I have a 0 here
so this is also 0. This annihilates the vacuum
or the ground state and therefore, it is 0
this brings it down to the ground state and
because ket 1 and ket 0 are orthogonal, this
is 0. This contributes because a dagger on
ket 0 is root 1 ket 1 so this whole term is
simply e to the minus i theta. This is 0,
because this takes it to k 2 and ket 1 and
ket 2 are orthogonal to each other.
So I have psi x psi where psi is the qubit
that I have selected. This object is 1 by
2 root 2 e to the i theta plus e to the minus
i theta and that is what I have. There was
an e to the i theta here and there was an
e to the minus i theta there. The whole thing
is multiplied by 1 by 2 root 2.
So, this is here cos theta by root 2. What
about psi P psi? I can repeat this argument
and expand putting in the appropriate values
the appropriate operators for P. I leave that
as an exercise that is again non 0. So, here
is a state where expectation X is not 0 and
expectation P is not 0. Again as an exercise,
one can compute delta X and delta P in this
state; they are not minimum uncertainty superpositions
of 0 and 1. The uncertainty product is higher
than what you would expect for the ground
state of the oscillator. Calculations like
this, exercises like this would become relevant
in quantum computation where qubits are used
and expectation values of various operators
in a qubit state become important.
My next exercise is to compute the expectation
value of X. In a very interesting state I
define that state as a dagger acting on ket
alpha. I have to normalise this where ket
alpha is the standard coherent state and as
you know this is an Eigen state of a with
Eigen value alpha, alpha being a complex number.
Now, first of all let us normalise that state.
So, consider this object I can always write
this in terms of a dagger a. The idea is to
get a to this side, because then I know that
I can use this property and therefore, this
property. So, a 
a dagger is 1 plus a dagger a and I need to
find this and this is something I know. The
first term is simply one because alpha is
a normalised state, normalised to 1 plus mod
alpha square, that comes from the expectation
value of a dagger a in the state alpha.
So, this object is just 1 plus mod alpha square
and therefore, the state psi which is 1 by
root of 1 plus mod alpha square a dagger on
ket alpha. This state you can easily check
is a normalised state. Now, in the parlance
of quantum optics what is it that I have done.
Taken these standard coherent state of lights
acted ones with the photon creation operator
so this is a photon added coherent state suitably
normalised for the single photon added state,
coherence state.
Now, I want to calculate expectation X in
this state. This is what I wish to find but,
X itself is a plus a dagger by root 2 and
therefore, the expectation value of X in this
state is 1 by root 2. I have just taken to
the bra psi of course, that is going to be
a normalisation and that is 1 plus mod alpha
square. So, a 1 by root 2 that came from X
and a 1 plus mod alpha square that comes from
the normalisation of the state, a plus a dagger
a dagger on ket alpha so this is what I have.
Now, that is just 1 by root 2 mod 1 plus mod
alpha square. I have two terms here. The 1st
is the expectation value of a square a dagger
in the state alpha. The next is the expectation
value of a a dagger square in that state.
Now, it is clear that this operator here its
dagger is here and therefore, if I know this
expectation value I can always take the complex
conjugate and get the other expectation value.
So, I have 1 by root 2 1 plus mod alpha square.
Look at this once more it is good to shift
a to that side. So, I have a times a a dagger
which is 1 plus a dagger a. So, I have just
written this operator out here plus this operator.
So, this number is the complex conjugate of
that number and what do I get here? That is
1 by root 2 1 plus mod alpha square.
The 1st term is just the expectation value
of a in this state alpha and that just pulls
out an alpha, because of this property. The
next is expectation value of a a dagger this
a acting on alpha pulls out an alpha so that
is what I have. By this I mean just take down
this term and put it there. Again I write
a a dagger as 1 plus a dagger a and now I
know what is happening.
This expectation value therefore, becomes
1 by root 2 1 plus mod alpha square. The 1st
term gave me an alpha and here there is already
an alpha. So, I just have to look at this
operator and what it gives me. So the 1st
term gives me an alpha times a 1 plus mod
alpha square and so I have alpha from the
1st term plus alpha times 1 plus mod alpha
square and then of course, its complex conjugate
that came from the 2nd term. So, basically
I can now simplify and write this as 2 alpha
plus alpha mod alpha square and from there
2 alpha star plus alpha star mod alpha square.
This can be further simplified. I leave it
to you to put this in a better form and realise
that a dagger on ket alpha the single photon
added coherent state is a very important state,
because 1st of all they are produced in the
laboratory about 5 years ago and it is not
a perfect coherent state, but it departs from
coherence quantifiably added 1 photon to the
coherent state and to that extent created
a departure from the coherent state.
Now, I could have a in general created an
m photon added coherent state in principle
and that is got by repeated application of
a dagger m times on ket alpha suitably normalised.
This is the normalisation constant. And I
would call this state the m photon added coherent
state. Such states become important in various
context. I now move on to a nonlinear Hermitian
operator which is nonlinear in a and a dagger.
Consider the operator a dagger square a square.
I have deliberately used H there because if
I put in some overall constants outside. I
find that this Hamiltonian is the effective
Hamiltonian for nonlinear optical medium specifically,
the Kerr medium. This is the effective Hamiltonian
for a Kerr medium, for a nonlinear optical
medium. I wish to find out the Eigen spectrum
of this Hamiltonian so let me proceed to work
out that exercise. I consider a dagger square
a square. I can write this. The idea is to
write it a dagger a a dagger a so I write
a dagger a in terms of a a dagger. Recall
that a a dagger is 1 and therefore, a a dagger
is 1 plus a dagger a so I can write a dagger
a as a a dagger minus 1 and there is an a
there. So, this object is simply a dagger
a, a dagger a minus a dagger a.
In other words, in terms of the photon number
operator n this is n square minus n or n into
n minus 1, where n is the photon number operator.
I therefore, need to find out the Eigen value
of Eigen values and Eigen functions of this
Hamiltonian h cross chi n times n minus 1.
It is evident that the basis set ket n, the
n photon state for instance is an Eigen state
of this Hamiltonian. Because, h acting on
ket n is h cross chi n into n times n minus
1 ket n. This just pulls out a number. Recall
that the number operator acting on ket n is
simply n times ket n and therefore, this is
n minus 1 and again the number operator acts
on ket n pulls out an n. So I can just write
this as h cross chi n times n minus 1 ket
n.
So, all the photon number states are Eigen
states of this Hamiltonian and this is the
Eigen value n times n minus 1 essentially.
So, let us look at what happens when n is
0 the Eigen value is 0 similarly, when n is
1 the Eigen value is 0. Therefore, 0 is a
doubly degenerate Eigen value and there are
2 states corresponding to this degeneracy
that is ket 0 and ket 1. So, both the 0 photon
state and the 1 photon state are Eigen states
of this Hamiltonian the Kerr Hamiltonian,
because they are Eigen states corresponding
to a degenerate Eigen value 0.
All other Eigen values are non degenerate.
For instance, if you look at n is equal to
2 that just gives me 2, then n equals 3 gives
me 3 times 2 which is the 6 and so on. So,
none of the other Eigen values are degenerate,
but the ground state is the 0 is degenerate.
We have now looked at a large class of states
of relevance to qubits quantum computation
of relevance to quantum optics. The single
photon added coherence states of the relevance
to harmonic oscillator, because we looked
at fock states and we computed delta X delta
P, expectation values of other objects like
X P X P plus P X and so on in these states.
I would now like to move on and work out a
problem in angular momentum and this would
be of relevance to particle physics. Suppose,
I have a Hamiltonian which commutes with the
angular momentum operator J square, in particular
J is spin so; it is not orbital angular of
momentum. So let me look at the spin angular
of momentum so h a square suppose I have a
Hamiltonian which commutes with this and the
Hamiltonian commutes with S x S y S z and
therefore, with S plus.
The spin Eigen states are denoted by s m of
course, S z s m is m h cross s, m these are
the spin labels. In this picture I want to
show that an object with a given spin has
the same mass independent of whether it exists
in the state m, or the 3rd component having
a value m or a value m plus 1 or m plus 2
or m minus 1. Independent of the value of
m provided, the spin of the object is given
to me I want to show that the object has the
same mass whether it exists in the state s,
m or s, m plus 1 or s, m minus 1 it is independent
of m. Provided, the Hamiltonian commutes with
S plus in other words, it commutes with S
x S y and of course, S z.
The way I go about it is as follows. I know
that this is true, because H commutes with
S plus that is given to me. So, I expand this
commutator out 
so this is what I have. But, I know the action
of S plus on s, m it raises the value of m
by 1. So, let me put that in.
You will recall that S plus acts on s, m to
give me root of s minus m into s plus m plus
1 h cross s comma m plus 1, m itself takes
2 s plus 1 values ranging from minus s to
plus s in steps of 1. Similarly, S minus 1
s, m gives me the following. I use this in
my problem therefore, I started with this
object 
and this implied 
H S plus 1 s, m gives me H s, m plus 1. Pulling
out this coefficient root of s minus m into
s plus m plus 1 h cross, h cross is outside
the square root. That is the 1st term and
this quantity is equal to s m plus 1 S plus
H s m, that if S minus acting on s, m gives
me this its dagger is this and that is a real
quantity therefore, I have the same coefficient.
So, I use that here.
So, S plus with an s, m on this side of it
except that I have m plus 1 instead of m.
So, I use this and I have, it reduces m plus
1 to m. Notice that this coefficient is a
same as that coefficient that scores off and
I just have the same coefficients. Therefore,
root of s minus m times s plus m plus 1 h
cross s n plus 1 H s m plus 1. The expectation
value of the Hamiltonian in the state s comma
m plus 1 minus the expectation value of the
Hamiltonian in the state s, m is 0, m is not
equal to s and therefore, this coefficient
does not vanish.
So it tells me a very important property that
the expectation value of the Hamiltonian in
the state s, plus 1 is a same as the expectation
value of the Hamiltonian in the state s, m.
So, let me look at this spin doublet for instance
what is this means for the spin doublet.
In the case of the spin doublet, I have s
is half and m is half and s is half with m
minus half. So this boils down to the statement
that half half h half half is equal to half
comma minus half h half comma minus half that
is what it means. In other words, look at
the Hamiltonian. The Hamiltonian has in general
a kinetic energy plus a potential energy.
For particles of a certain mass, this would
mean the kinetic energy of the particle and
the rest mass energy of the particle. Let
us go with the rest frame of the particle.
So, let us go to the rest frame of the electron
which is a spin half of the object for instance
or a proton or the neutron. The statement
is this: in the rest frame this expectation
value would correspond to the rest mass of
the particle in question. So, the rest mass
of the electron in the spin up state equals
the rest mass of the electron in the spin
down state. Similarly, for protons and neutrons
the rest mass of the proton in the spin up
state is the same as the rest mass of the
proton in the spin down state. The manner
in which I take care of this is by starting
off with a Hamiltonian which commutes with
S plus.
Once this is given to me it automatically
follows that objects in a certain state of
spin in the sense that s is fixed, but m has
a definite value. Objects in a state m s,
m would have the same rest mass as the state
with s, m plus 1 or m minus 1. In other words,
in a spin multiplet the same holds for a spin
triplet. For instance, in a spin triplet the
rest mass of the object in the state 1, 1
in the state 1, 0 and the state 1, minus 1
would all be the same. So, within a spin multiplet
the rest mass is the same.
We discuss a certain particle in a certain
state of spin therefore, but it could exist
in different states of m whatever may be the
state of m the particle has the same rest
mass, that is guaranteed by the fact that
H commutes with S plus. So, I have worked
out certain exercises for you using commutator
algebras. Essentially, the algebra involved
in the case of the harmonic oscillator X P
commutator is i h cross equivalently a a dagger
commutator is 1 and the spin algebra, where
I have used the fact that if I have a Hamiltonian,
which commutes with S plus certain consequences
follow.
