In the last lecture I gave an introduction
to the linear harmonic oscillator. We showed
that operators called a and a dagger can be
constructed for the linear harmonic oscillator
and these operators raise the energy levels.
So basically I had the Hamiltonian for the
oscillator given by a dagger a plus half h
cross omega. The energy levels of the oscillator
were equally spaced. The notation used, the
ground state was ket 0, the 1st excited state
was ket 1 and so on and that is an infinite
set of energy levels. The energy itself was
given by E 0 is half h cross omega which is
a zero point energy. Then there is E 1 which
is 1 plus half h cross omega and so on. In
general E n is n plus half h cross omega,
n taking values 0, 1, 2, 3.
Basically, we started with the commutation
relation. The commutator of X with P was i
h cross identity, where P which is the linear
momentum conjugate to X and identity are all
operators. And this translated to a a dagger
equals identity, where a itself was defined
in terms of X and p as linear combinations,
a dagger was the Hermitian conjugate of a.
This is the situation regarding the harmonic
oscillator. It turns out that there is a 1
to 1 correspondence between the linear harmonic
oscillator problem and quantum optics. That
is the framework in which one discusses the
quantized electromagnetic field, where we
try to understand, various states of the quantized
electromagnetic field or the radiation field.
Describe and discuss the properties of the
electromagnetic field.
So, basically I will now consider the next
problem, which is the problem of the quantized
electromagnetic field, where one considers
an electromagnetic wave which is quantized.
K is the propagation vector. So, the direction
of propagation will be chosen to be the z
direction. I consider a simple case where
this is the origin of coordinates and this
is a cavity which extends in the z direction
from zero to L.
So the polarisation direction is the X direction,
which basically means, that the electric field
is along the X direction and can in general
be written as e sub X being the unit vector
along the x direction. E 0 now this object
is simply a function of the permittivity,
the frequency of the electromagnetic wave,
the volume of the cavity and so on.
Some function of time because the electromagnetic
field evolves in space time. And sin k z because
of the boundary conditions, because the electromagnetic
field at z is equal to 0 is 0 and at z is
equal to L is also 0, which tells me that
k L is equal to 0 and that quantizes K. So,
K is m pi by L where m can take values 1,
2, 3 etcetera. The frequency omega is related
to k through the standard relation omega is
equal to C k, which means that omega is now
C times pi by L or 2 C pi by L and so on.
For simplicity, we choose m is equal to 1.
This can be done without loss of generality
because I am considering an electromagnetic
wave of frequency omega, which is pi by L
times C.
Now, it turns out that the problem of the
quantized electromagnetic field. This is a
single mode field propagating along the z
axis and so on. This has been chosen for simplicity.
This problem has a one to one correspondence
with the linear harmonic oscillator problem,
and that is the mapping that we will see right
now.
So, in general I have the electromagnetic
field E is e x E 0 q of t sin k z. I have
the Maxwell’s equations del dot E equal
to 0 that is automatically satisfied here,
because e is along the X axis and the function
there is a function of z. Del dot B is equal
to 0, del cross E is minus 1 by C delta B
by delta t and then cross B is 1 by C delta
E by delta t. I am assuming that there are
no free charges or currents. So, these are
the source free Maxwell equations.
Given this electric field, I can write down
the magnetic field. Since the propagation
direction is along the z axis and the electric
field is along the X axis. The magnetic field
is along the y axis. There is B 0 which is
analogous to E 0 and it is clearly a function
of the magnetic permeability epsilon 0 omega,
the volume of the cavity and so on. This equation
here tells me that delta E by delta t is related
to the curl of B. So, the time dependence
of B is going to bring about a q dot of t
that sits here. And then it is clear that
the space dependence of B comes from a del
cross E, which means the derivative of the
sin term will appear and therefore, that gives
me a cos k z. So this is the way I write B.
Now given this E and B, I have the Hamiltonian
for the electromagnetic field. This is integrated
over the entire volume d V. So, this part
is the energy density and integrated over
the entire volume gives me the total energy.
Now I want to look at this term by term. So,
let me look at the 1st term d V is simply
d x d y d z that is a triple integral. So
that is half epsilon zero I can pullout the
q of t and the E naught, which is a constant.
There is a sin k z d z whole square, there
is an E naught square and a q square because
there is an E square. Then of course, there
is a d x d y the z integral goes from 0 to
L.
It is pretty clear that this object can be
written in terms of sin 2 k z and 1 that pulls
out a half. There is an integral d z d X d
y d z which gives me a v and then there is
an integral of sin 2 k z and that is between
limit 0 and l, but we know that k is pi by
L and therefore, it is just cos 2 k z between
0 and L and that quantity vanishes. So this
is all that I have. It turns out I did not
put down the constants there. It turns out
that this E naught is really 2 omega square
by V epsilon 0 square root, where V is the
volume of the cavity and therefore, I can
simplify this.
And I have half epsilon 0 E square has a half
epsilon 0 so that is the quarter epsilon 0
E 0 square, which gives me a 2 omega square
by V epsilon 0. And there is a V by 2 which
I pulled out due to the integration I have
already absorbed the 2 there so that gives
me a V and this object is just half omega
square. But, there was also a q square of
t so I have half omega square q square of
t so that is the 1st term. Then I have the
2nd term which is an integral over b square
d V. 
I can do it the same way that I did the integral
over the 1st term.
So, I need to consider 1 by 2 mu 0 integral
B square d x d y d z. Once more when I substitute
for B, there is a q dot square of t, there
is a cos square k z and can be written in
terms of cos 2 k z. And using the limits of
integration I find that that gives me a V
by 2 can substitute for B naught explicitly,
and once that is done times 2 omega square
by V epsilon 0 there is an epsilon 0 mu 0
out here.
This turns out to be with the q dot square
of t. So the mu zeroes cancel out, the epsilon
zeroes cancel out, the V cancels out. There
is an omega square here which too cancels
out and I will be left with a q dot square
of t. And therefore, the Hamiltonian can be
written as half. The total Hamiltonian is
half omega square q square of t plus q dot
square of t. And this is how one maps the
problem of the electromagnetic field, the
quantized electromagnetic field to the harmonic
oscillator, because I will now re-label q
dot of t as p of t and q of t as x of t.
And I have the Hamiltonian given by half omega
square x square of t plus p square of t. Suppose,
we did not consider the time dependence at
all and decide only to look at that part of
x of t and p of t, which do not have time
dependence. I simply have this object because
I am not interested in the dynamics that is
a simple case that I am taking up for the
moment. And since this is the Hamiltonian
for the electromagnetic field and you can
trace this term back to the electric field
e square and this term to the magnetic field
b square and the electric and the magnetic
field are the observables. The operators in
this case are x and p, not to be confused
with position and linear momentum because
this has to do with the electric field and
this has to do with the magnetic field. The
notation has been used essentially to show
that there is a one to one correspondence
between the quantized electromagnetic field
and the linear harmonic oscillator.
So now, given this I can define an operator
a as I did earlier. This is an operator shows
x and shows p except that this has to do with
the electric field and that with the magnetic
field, and that is an operator which is a
linear combination of objects that are pertaining
to the electric and the magnetic fields. And
a dagger is simply the Hermitian conjugate
of this. So, I have simply renamed things
I have gone from x and p to a and a dagger
and it is easy to check that a a dagger is
the identity operator. Except that, now I
have to give an interpretation for a and a
dagger and that is rather easily done in this
context.
Because, as I pointed out in the case of the
linear harmonic oscillator, a lowers levels
the same algebra of the oscillator is carried
over to the context, that we are discussing
at present 
and therefore, all this is true.
The point is the following: and we know from
the problem of the harmonic oscillator, that
if I started with n is equal to 0 and when
to n is equal to 1 through an a dagger that
increases the energy by 1 h cross omega. I
also know that a quantum of light the photon
carries energy h cross omega. And therefore,
a dagger would be the operator that creates
a photon pumping in energy h cross omega into
the system, a is the photon destruction operator
which destroys photons, n is no longer a label
n is a number of photons. That is the crucial
difference between the harmonic oscillator
problem and the problem that we are discussing
now.
Because, in the case of the linear harmonic
oscillator there was a single linear harmonic
oscillator, n was merely a label and you moved
from ket 0 to ket 1 to ket 2 which are excited
states of a single linear harmonic oscillator.
Here, we have made a leap in our understanding.
We are now discussing a problem where n stands
for the number of photons. Because, if I repeat
a dagger n times if I repeatedly operate a
dagger n times on a state it is going to add
energy n h cross omega and therefore, creates
n photons. So n is no longer a label in this
problem n refers to the number of photons.
So, ket 0 would correspond to the 0 photon
state and this is a crucial difference between
the harmonic oscillator problem and here.
In the quantized electromagnetic field, because
while the algebra is the same and the mathematical
framework is the same the interpretations
are different.
So this is a zero photon state, this is the
n photon state. And as you go from the 0 photon
state to 1 photon state, there is an increase
in energy h cross omega. The more the number
of photons that are added mathematically represented
by repeated application of a dagger on the
original state keep adding more photons and
therefore, more energy is pumped into this
system of photons. So, what is this operator?
This operator would be the photon number operator.
It counts the number of photons in the state.
So it is the state with n photons and that
is what is pulled out here. So, this is the
photon number operator a dagger is the photon
creation operator and a is a photon destruction
operator. It is well worth passing at this
point and getting the identity straight.
In the case of the simple harmonic oscillator
a was the lowering operator which took us
from one energy level to another, a dagger
was the raising operator and n was merely
a label for the states. Now, in the case of
the quantized electromagnetic field, a is
the photon destruction operator, a dagger
is the photon creation operator, n is the
number of photons in the given state. The
Hamiltonian in both cases is simply a dagger
a plus half h cross omega. So this is the
mapping that we are interested in. When I
spoke about the harmonic oscillator, I discussed
the uncertainty principle in that context
and showed that the ground state of the oscillator
was a minimum uncertainty state.
So, let me recapitulate in the case of the
simple harmonic oscillator the ground state
of the oscillator satisfied delta X delta
P is equal to h cross by 2 not only that delta
X was root of h cross by 2 and delta P was
root of h cross by 2. In the case of the oscillator,
if m equals 1 and omega equals 1 it was not
only a minimum uncertainty state, it was a
state where delta X was equal to delta P.
The general uncertainty relationship tells
us that delta X delta P is greater than or
equal to h cross by 2 and certainly for the
excited states of the oscillator the inequality
holds. Delta X delta p will be greater than
h cross by 2.
Now, we say that a state is a squeezed state
in the X quadrature if you wish, if delta
X is less than root of h cross by 2, it is
a squeezed state in the p quadrature. If,
delta P is less than root of h cross by 2.
Now it is clear that since the uncertainty
principle holds if it is squeezed in the X
quadrature, if there is a state which is squeezed
in the X quadrature the corresponding delta
p will be large such that delta X, delta P
is greater than h cross by 2. If it is squeezed
in the P quadrature the corresponding delta
X value will be large so that there is a compensation,
and delta X delta P is greater than h cross
by 2.
You cannot have a state where both delta X
and delta P are less than root of h cross
by 2 because that violates the uncertainty
principle. So, let us look at various states
in the context of optics, quantum optics or
the quantized electromagnetic field. The problem
of the radiation field in other words the
problem of the light quanta. So, specifically
we will consider various superpositions of
photon number states ket n. It will illustrate
things, it will illustrate the property of
squeezing, it will also illustrate a very
important aspect of quantum mechanics and
that is the power of quantum superposition.
So, first of all without much ado we can say
that in the case of quantum optics the zero
photon state it is a minimum uncertainty state.
We have already established this in the case
of the oscillator and all that we have done
is a mapping from that problem to this problem.
So, in the zero photon state certainly delta
X is equal to delta P is equal to root of
h cross by 2 and delta X delta P is equal
to h cross by 2. You could do this by writing
X in terms of a and a dagger, X was root of
h cross by omega a plus a dagger by root 2.
In the case of the harmonic oscillator there
was also a mass sitting there and as I pointed
out earlier root of h cross by m omega has
a dimensions of length. Now there is no m,
but we do have a root of h cross by omega
that gives us X and P is root of omega h cross
a minus a dagger by root 2 i.
So, given this I can find out delta X and
delta p in the 0 photon state, where I already
know that a on ket 0 is equal to 0. And, this
is established in precisely the same manner
as we established that the ground state of
a simple harmonic oscillator is a minimum
uncertainty state. Now, let us look at a superposition
of photon number states. Let us consider this
state. I consider a superposition of the 0
photon state and the 1 photon state. The 1
photon state was got by acting a dagger on
ket 0, a dagger on ket 0 gives me ket 1 with
the root 1 in front as a coefficient. This
is a normalised state because bra psi it has
been constructed such that ket psi bra psi
is a 3 quarter plus 1 quarter and that is
a 1. Let us compute delta X and delta P in
this state psi.
So, first of all delta X square is expectation
value of X square in the state minus expectation
value of X the whole square. So, I start with
X square. Suppose, I set h cross and omega
equal to 1 for convenience. We can always
put it back for the sake of dimensions. This
is essentially a plus a dagger, producted
with itself with a half there. I need to find
psi X square psi. This is expectation value
of X square.
Now this object is half. The half is put here
when I expand out this operator. I get an
a square plus a dagger square. Then I have
an a a dagger plus a dagger a. I have the
commutation relation 
and therefore, a a dagger is 1 plus a dagger
a. So this gives me a 1 plus 2 a dagger a.
There was already an a dagger a and the a
a dagger contributes another because of that
commutation relation. And then on the other
side I have root 3 by 2 ket 0 plus half ket
1. So this is the object that I have to compute.
Let us look at the first term. Now, a square
a acting on ket 0 is 0 so a square simply
destroys this and therefore, the inner product
with this does not contribute does not contribute
with this either. So, there is no contribution
from a square. Look at a dagger square, a
dagger acting on ket 0 gives me ket 1 apart
from some constant number multiplying with
it. Repeat it twice and that takes it to ket
2. Now, ket 2 is orthogonal to both ket 1
and ket 0 and therefore, there is no contribution
from this term.
Similarly, when a dagger square acts on ket
1 takes it through ket 2 to ket 3 and again
by the orthogonality property, it does not
make a contribution. There is a contribution
from here. This is the identity operator and
therefore, from the first term I get 3 by
2 and from the second term I get a quarter
3 by 4 and from the second term I get a quarter.
So this is simply the identity and this is
what comes out of that by way of contribution.
But then there is a contribution from 2 a
dagger a. Now a dagger a picks up an Eigen
value 0 when it acts on this state. So the
overall contribution from the inner product
of ket 0 with bra 0 is 0, but a dagger a here
gives me a 1 and there is already a quarter,
because of these two and that is what I have.
Now, this can be simplified that is a half
that is a 3 quarter plus 1 quarter which is
a 1 and there is a half, so expectation X
square in this state is 3 by 4 that is a 3
quarter.
So, let us look at expectation X the whole
square. I repeat the performance. So, expectation
X would simply give me 1 by root 2 root 3
by 2 bra 0 plus 1 by 2 bra 1 a plus a dagger,
that is expectation X apart from a root 2
root 3 by 2 ket 0 plus half ket 1 and this
is what I need to compute. I work as before,
I have an overall factor 1 by root 2 a destroys
0 ket 0 and therefore, it does not contribute
here. On the other hand, a acting on ket 1
gives me ket 0. So, the inner product of this
state with a acting on ket 1 makes a contribution.
I pick up an overall constant root 3 by 2
from here and a half from there 
and since a acting on ket 1 is ket 0 this
is all I have from this term.
Now look at a dagger, a dagger acting on ket
0 takes it to ket 1, so there is a contribution
from here. There is no other contribution,
because when a dagger acts on ket 1 it takes
it to k 2 which is orthogonal to both of them.
So, this object is twice root 3 by 4 and therefore,
that just gives me a root 3 2 root 2, I need
expectation X the whole square that is 3 by
8 and therefore, delta X the whole square
is expectation X square minus expectation
X the whole square, which is 3 by 8.
What happens in the ground state? Delta X
the whole square when I set h cross equal
to 1. If h cross is equal to 1, which is what
I have been working with for simplicity. In
the ground state or in the zero photon state
delta X is root of 1 by 2 and therefore, the
variance is half. So that is the ground state
and here I have a state where the variance
is less than half. So it is squeezed in the
X quadrature. I would like to check that it
is not squeezed in the p quadrature, because
that would violate the uncertainty principle.
So, let us work out delta p the whole square.
So delta p the whole square is expectation
p square minus expectation of p the whole
square. Now p square is again 1 1 half of
a minus a dagger by i a minus a dagger by
i. That is the minus half a square plus a
dagger square minus of a dagger a plus a a
dagger, as before I use the commutation relation.
Now we will write this as minus half a square
plus a dagger square minus 2 a dagger a plus
1. It is clear, that when I sandwich p square
between the state of relevance here. Expectation
p square is simply going to be root 3 by 2
bra 0 plus half bra 1. I repeat what I did
earlier times half a square plus a dagger
square minus 2 a dagger a plus 1 and then
there is a ket root 3 by 2 ket 0 plus half
ket 1. As before a square and a dagger square
do not contribute, but a dagger a does and
the identity operator there does and we can
simplify this.
So, what is expectation P square in this state
as before a square and a dagger square do
not contribute, and we nearly have expectation
p square the state itself is given here. And,
there is an a dagger a plus 1, there was an
overall half and that comes there. We started
with the a dagger a and therefore, it is just
divided by 2 and the ket is root 3 by 2 ket
0 plus half ket 1,but I know the answer to
this. This has been done earlier by us, that
is 3 quarters. But now look at expectation
P in this state, so the state itself once
more I sandwich the operator in this manner.
I have a 1 by root 2 is an overall factor.
So, essentially one looks at the action of
a minus a dagger on this. Of course, a acting
on ket 0 simply gives me 0 there is no contribution.
They acts on ket 1 brings it down to ket o
so there is a contribution from here which
is a root 3 by 2 from ket 0 from bra 0 here
and a half from there.
Now look at a dagger, a dagger acts on ket
0 takes it to ket 1, so there is a contribution.
But the coefficients are half from here and
a root 3 by 2 from there. There is no contribution
when a dagger acts on ket 1, because it takes
it to ket 2 and I do not have ket 2 on this
side, bra 2 on this side so this 0. So the
crucial thing to note is that unlike the case
of expectation X where I had a plus sign there
and therefore, there was a contribution.
In the case of the expectation P, the individual
contributions cancel out and I get a 0 expectation
P the whole square is 0, that implies that
delta P the whole square is simply expectation
P square which is good news, because delta
P the whole square is greater than half. And
therefore, in this state that I have given
which is the superposition of the 0 photon
state and the 1 photon state that is squeezing
in the quadrature, and there is no squeezing
in the p quadrature as expected. And the product
delta X delta P is greater than root of h
cross by 2. So it is not a minimum uncertainty
state, but certainly there is squeezing.
Now, it is clear that I could have changed
the coefficients and suppose I constructed
a state which is given by half ket 0 plus
root 3 by 2 ket 1 it will be squeezed in the
p quadrature and not squeezed in the X quadrature.
So, this is the power of quantum superposition.
I have constructed a state which shows squeezing
and squeezed states are very important in
optics. The squeezed vacuum is a very important
non classical state. There is no analogue
of squeezing in classical physics and for
that matter there is no analogue of the uncertainty
principle in classical physics either. And
therefore, a squeezed state is an example
of a non classical state. Quantum optics abounds
a non classical states of light. To illustrate
the power of quantum superposition I want
to show another example, not in the context
of squeezing, but in a very different context.
Consider this state the normalised 10 photon
state. So, it is clear this is a 10 photon
state. In my notation it is represented in
this manner, n is equal to 10 and this state
is normalised. So what is the mean photon
number in this state? The mean photon number
is clearly 10. Because, a dagger a acting
on ket 10 is 10 ket 10 and that inner product
becomes 1. So the mean photon number is 10.
Now, let me take this state psi and act on
it by a dagger. In other words, i will add
a photon to the state. So a dagger on ket
psi is root 11 ket 11.
I have to remember that a dagger on a state
ket n is root of n plus 1 ket n plus 1. This
is not a normalised state, and we need to
consider normalised states, because otherwise
the probabilistic interpretation fails the
total probability must be 1. And therefore,
to normalise the state I have the normalised
state should satisfy, if this is a normalisation
factor the bra which is a psi, there is an
n n square where n n is a normalisation factor
this object must be 1. This immediately tells
me that the normalised state is ket 11. I
do not have to do any work for this because
I know that the 11 photon state acts such
as a normalised state.
So, what is the expectation value of a dagger
in the new state? Expectation value is 11,
which is believable because I took the 10
photon state. I added a photon to it by applying
a dagger to it and therefore, the mean value
of a dagger a has moved up to 11 I have added
a photon.
Now instead of this, let me consider this
state psi which is the 0 photon state plus
11 photon state. It has to be normalised so
let me consider this state. This is normalised
I can check it out because bra psi is 1 by
root 2 and therefore, I have the inner product
given by this. This is normalised to 1 and
that is normalised to 1 and therefore, I have
the inner product of psi with itself to be
1. So, this is a normalised state. The total
number of photons in the state is 11, because
this is a 0 photon state and that is a 11
photon state. So, I have a total of 11 photons.
Let me find the mean number of photons in
this state. So, I can calculate the mean number
of photons in this state and I have a certain
mean number of photons. First of all this
gives me a 0, but that gives me 11. That is
the mean number of photons. I have brought
down the mean number of photons from the initial
value by doing a quantum superposition of
the 11 photon state with the 0 photon state.
So, naively one may imagine that since a 0
photon state has no photons it cannot make
a contribution to anything, but it has brought
down the average from 11 to 5.5, this is what
I have for this state. Now, let me pump in
photons into the state.
I wish to do a dagger on ket 0 plus ket 11
by root 2. In other words I am going to add
photons to this state a dagger on ket 0 is
ket 1 and a dagger on ket 11 is root 12 ket
12. And therefore, this is just 1 by root
2 ket 1 plus root of 12 by 2 ket 12. This
is not a normalised state. I need to normalise
this state. So, I need to consider n n square
which is the square of the normalisation with
itself and that should tell me what the normalisation
factor is. Half from here and 6 from there
should be 1. Now, this normalisation is simply
root of 2 by 13 and this is what I have for
the normalisation.
So, I have the new normalised state a dagger
acted on my state psi and I normalised it.
And the new state that I have got is root
of 2 by 13 times ket 1 by root 2 plus root
12 ket 12 by root 2. So this is just 1 by
root of 13 ket 1 plus root of 12 by 13 ket
12. This is what I have when I added a photon
to this state. So, to begin with since I have
added a photon I would normally expect it
to add here or add there, but a dagger acts
on both states and adds a 1 here and makes
it 12 out there. So, this again is the power
of quantum superposition of course, this is
a normalised state, because when I square
it I just get a 1 by 13 plus 12 by 13 and
I am through.
So when you do a quantum superposition the
photon sits here and the photon sits there.
It is not as if the photon gets added to this
state or to that state. So that is another
aspect of quantum superposition which I want
to bring about here. There are 2 aspects to
quantum superposition, 3 in fact which I have
demonstrated in this lecture. The 1st is suitable
superpositions could produce squeezing either
in the X quadrature or in the P quadrature.
The 2nd is this that in a superpose state
the average number of photons can come down
drastically although a photon has been added
that tells us the importance of the zero photon
state, and thirdly when you add a photon it
does not add to just this state or that state,
but adds on the whole. So, these are aspects
of quantum physics which do not see analogues
in classical physics. I will stop here and
take on more interesting quantum superpositions
in the next lecture.
