Today let me start off on a slightly new topic because I would like to spend sometime giving
an application of quantum mechanics specifically in quantum optics and photon statistics. And
this is one of the applications. There is a large number of such applications but i
thought this would be useful and a good take off from were we have stopped. We studied
a little bit about the harmonic oscillator and its properties as an exercise in quantum
mechanics and we have also done a little bit about spin 1/2. I would to combine these and
talk about 2 level atoms and models of atom radiation interaction. To set the stage for
that, I would like to introduce you to where the harmonic oscillator plays a role in understanding
the behavior of the electromagnetic field.
We saw that you could start with particle mechanics and then do quantum mechanics by
converting Poisson brackets into commutators. The natural question that arises is what happens
if you start with the classical field, like electromagnetic field and would it have a
corresponding quantum version or not. This is really the purview of quantum field theory
which is not part of this course but i would like to point out that there is a simple way
of understanding how the harmonic oscillator plays a fundamental role in quantum field
theory. So I am going to do the quantization of the electromagnetic field but I am going
to introduce the results in a very simple physically appealing way. And then at some
stage, we will make certain assumptions which will tell you how the quantization actually
occurs and after that we will study the consequences of it. So what we intend to do is to understand
how photons behave in the quantum mechanical way. So let me start by writing down Maxwellís
equations and then imposing certain boundary conditions on it to get you familiar with
the idea of quantized modes.
Classically, if I look at the electromagnetic field in the absence of sources, then the
equations as you know are del dot E is 0, no sources and a region without any sources
or currents. Del dot B is 0, del cross E is - delta B over delta t and del cross B is
mu 0 epsilon 0 delta E over delta t. these are the source free Maxwellís equations in
free space. And as you know, itís easy to show from here that both E and B obey the
wave equation and the speed of these waves is given by 1 over square root of mu 0 epsilon
0. Thatís the speed of light in vacuum.
Now suppose I consider propagation of these waves along the z direction, between 2 conducting
plates placed in the xy plane, and lets suppose that at z = 0 and z = l, i have 2 parallel
plates which are conducting and look at what happens to standing electromagnetic waves
in the region in between these 2 plates. The electric field must be 0 at the conducting
plates they are earthed and now i am going to assume that the radiation is polarized
along the x direction.
So I am going to assume that E, the electric field as a function of r and t is just ex
times the unit vector in the x direction. So the electric field is along this direction
and since its radiation, the magnetic field could be along the y direction and the direction
of propagation is z. as you know E B and K where K is a propagation direction for electromagnetic
waves, they form a right handed triangle. The unit vector in the E and B directions,
we take the cross product, you get the unit vector in the direction of propagation.
So this is Ex and that is a function of z and t alone and similarly the magnetic field
B ( r, t) is Ey By and thatís a function of z and t alone. So I am looking at plane
polarized light. The plane of polarization is the xy plane and the direction of polarization
is the x direction for the electric field. now these fields have to obey these equations
here and its easy to see that suppose i put this whole thing in a cavity of volume V,
I havenít told you what the boundary conditions on these sides are but we will assume its
very large container of some kind with some volume V then, Ex ( z, t) is of the form sin
kz, where k is the wave number of this propagation and its going to be such that sin k sin kz
is 0 at z = 0 and 0 at z = l, those are the boundary conditions times the normalization,
so let me write down the solution. If the angular frequency is omega and the wave number
is k then, the solution is 2 omega squared by V epsilon 0 to the power 1/2, thatís the
normalization factor and then some time dependent factor which i called q (t).
So these are some constant factors, omega is the angular frequency of this. Mod k is
the wave number and the 2 are related as you know by omega = ck. V is the volume of this
container, epsilon 0 is the constant that appears in these Maxwell equations. q (t)
has the physical dimensions of a length. Thatís how I chose my normalization constant and
you can check that this quantity has the right dimensions for the electric field and this
is dimensions of length here. Then if you put that in into this equation and solve,
then delta e over delta t will have a q dot of t and thatís del cross B and B we already
know is in the y direction. So if you work it out, the del cross B is going to bring
out a factor k.
So itís By of z and t turns out to be mu 0 epsilon 0 divided by k. thatís because
the del operator is going to produce a factor k that comes down in the denominator times
2 omega squared over V epsilon 0 to the power 1/2 q dot of t and then this is a cos kz because
it involves a spatial derivative and it becomes a cosine. Now itís a simple matter to check
that these solutions actually satisfy the Maxwell equations here. Now what are the possible
values of k? Itís clear that k must be such that sin kl vanishes at this end. So k is
= n pi over L, n =1 2 3 etc. Itís like waves on a string clamped at 2 ends.
We just have a classical solution of Maxwellís equations but if you put this into the energy
of the system; the Hamiltonian, now whatís the energy density of the electromagnetic
field? Whatís the energy per unit volume? Itís 1/2 epsilon 0 E squared + over mu 0
B squared. so the total energy is =1/2 integral over the volume the energy density and the
energy density is epsilon 0 E squared +1 over mu 0 B squared. And you put this in and perform
the volume integration. We have conveniently chosen a thing so that you get1 over B to
the 1/2 here and when you square it you get a 1 over V which cancels the volume integral
there. And then when you do this, you end up with 1/2 omega squared q squared + q dot
squared. That reminds you of the harmonic oscillator.
In fact, this problem is identical to that of the harmonic oscillator. Omega is any1
of these numbers. It is ck where k is n pi over l. so you have many modes possible and
letís look at any one mode. Some allowed value of k, and then the energy in that mode
in that mode is given by just a harmonic oscillator. In fact itís a harmonic oscillator with mass
=1 m =1 effectively. So instead of q dot, let me just write this as p because m q dot
is p. so 1/2 p square + omega q squared. So you can see where the harmonic oscillator
is appearing by the backdoor in the problem of the electromagnetic field. E squared +
B squared is essentially like q squared + p squared. This is the way the field is quantized.
I havenít quantized the field i havenít started with canonical commutation relations
and quantized it. But I am just motivating the fact that the oscillator would appear
naturally in this fashion. It really appears from the fact that E and B, for a given mode
of the electromagnetic field, act like the position and momentum respectively. So letís
accept that thatís the way field quantization appears and now what I am going to do is to
take a big jump and say that I am going to take p and q to be operators as we do for
the harmonic oscillator and impose the canonical commutation relation. So from now p q and
p are operators which satisfy ih cross thatís the unit operator. What happens then? It implies
that the electric field and the magnetic field are operators. Ex and By are themselves operators.
We could ask whatís the commutation relation between Ex and By for example. That would
be related to the commutation relation between q and p you can write down explicitly what
it is. Now we solve the harmonic oscillator by going over to operators a and a dagger
which was essentially q + ip and q ñ ip. Then we obtained a number operator, a dagger
a, we found the eigenstates of a dagger a, they were just natural numbers and we associated
a number operator. Exactly the same thing will happen here except that this number operator
now has a physical meaning. It is the number operator corresponding to the number of field
quanta or the number of photons. so straight away from the oscillator, we go over to a
picture where the electromagnetic field is pictured as a collection of oscillators. One
oscillator for each mode of the electromagnetic field and then the number operator corresponding
to that oscillator is just the number of quanta of the field. Number of photons with that
particular frequency and wave number and state of polarization. We havenít talked about
the polarization. I will come to that in a second.
So if you go ahead and say, at this stage let a = m has been set =1, so its omega q
+ ip over root 2 h cross omega and a dagger = omega q - ip over 2 h cross omega, then
this goes over into h cross omega into a dagger a + Ω exactly as it happened in the harmonic
oscillator. Together with, of course we have adjusted those constant such that a with a
dagger = the unit operator. Now the rest is a just a copy from the harmonic oscillator.
All we have to do is to play around with these operators a and a dagger and we have a complete
set of eigenstates of this and so on. Now please notice that this whole thing is for
one mode of the electromagnetic field.
a mode thatís propagating in the z direction with the particular wave number k which is
1 of these numbers and a fixed state of polarization. In reality, if you wanted to describe the
electromagnetic field, you have to be a little more careful than this. And thatís not the
total Hamiltonian. We could write down a better Hamiltonian than this. And whatís that going
to be? Well first of all, that even in this direction of propagation, there could be 2
states of polarization. You can have 2 independent states of circular polarization for the electromagnetic
wave. One of them, if you resolve it into plane polarization states, you have an Ex
So the electric field is along that direction and the magnetic field is coming out of the
plane of the board. There is another linearly independent state of polarization which would
correspond to Ey. so the electric field points along this direction and the magnetic field
along ñ Bx such that E cross B still gives you the direction k.
these are independent states of polarization. You can combine these 2 and form left circularly
polarized and right circularly polarized corresponding to the electric field vector making a circle
either in the positive sense or in the negative sense. So if I denote these 2 states of polarization
and i pointed out earlier that the 2 states of polarization correspond to 2 helicity states
for individual photons, then what would the Hamiltonian actually look like?
Well, the total Hamiltonian really would be a summation over the states of polarization
which could be either left circularly polarized or right circularly polarized, let me use
a symbol little s for the state of the polarization, and then the summation or an integration over
all allowed values of the wave number k and thatís a summation over all allowed values
of n. but then we have assumed the radiation is propagating in the z direction. It need
not. It could propagate in any direction. And therefore you must really sum over all
possible k vectors as well and for each of these, you would have an h cross omega subscript k because for a given
k omega is ck, multiplied by a dagger k vector s a k vector s comma s + Ω. For each mode,
each wave number k, you would have an a a and for each of the s values you would have
an a dagger and a. so those operators would have 2 labels and this would be the total
Hamiltonian of the electromagnetic field. What would this commutation relations look
like? These are independent modes completely.
So itís clear that these commutation relations would go over into ak s with a dagger k prime
s prime is, independent modes these operators would completely commute. But if they both
correspond to the same mode and the same s, then the answer should be1. So itís clear
that this would be = a Kronecker delta, a 3 dimensional delta by the way of k k prime
delta of s s prime times the unit operator. So this is the start of field quantization.
So you start by postulating these commutation relations and then look at what the consequences
would be. I am not going to do because thatís really the task of quantum field theory. We
are not going to talk about that at all. We are going to come back here and i ask what
consequences can be deduce from just this thing here for a single mode, what does it
look like what happens if you put this in equilibrium with the thermal heat bath at
some temperature which will take us to black body radiation. But the crucial physical ingredient
is that the electromagnetic field is also quantized. We are not talking about it in
interaction with anything at the moment. The free electromagnetic field acts like a collection
of a harmonic oscillators in the quantum mechanical sense.
In this picture, in any state of this field, what would this correspond to? <ak dagger
ak s>. What would this expectation value correspond to? What would be the physical meaning in
any state of this field, what would be the physical meaning of this? It would give you
the number of photons. the average number of photons, itís a number operator of wave
vector k and state of polarization s. and it would therefore, if you would now integrate
this over some range in k, if its continuous then you get the intensity of this electromagnetic
field in that range. But thatís proportional to the number as you can see. Now that we
have this, letís focus on one mode and ask what happens. So letís write down for example
what E does. Notice that we already have Ex (z, t) = 2 omega squared over V epsilon 0
to the power 1/2 q ( t). Now what is q (t)?
We already have these equations. So this is 2 omega q over squared, root of 2 h cross
omega = a + a dagger and this is a root 2 omega there over h cross. So this multiplied
by h cross by 2 omega is q. so q (t) is a + a dagger root h cross over 2 omega. so lets
put that in. um yes root h cross over 2 omega a + a dagger sin az which is = some amplitude,
let me call it just E 0, the operator part i want to retain sin kz. This factor came
from q of t. the t dependence of classical field; I put in this factor q of t. so thatís
true even here they must be t dependence. So this means we are working in the Heisenberg
picture. These operators are time dependent. So we must be careful. Let me write that down
properly. So this is a (p) + a dagger of t and this is some constant whatís the value
of this constant. E 0 is, it has dimensions of electric field so
h cross omega over V epsilon 0 to the power Ω, just for reference. and similarly the
magnetic field By of z this is the y direction field so By of zt is = there was a mu 0 epsilon
0 over k and then there was this 2 omega squared over V epsilon 0 to the power 1/2 and then
q dot or p. so what was p? 2 ip over root of 2 h cross omega was = a - a dagger. So
the 2 becomes root 2 here and over i root h cross omega over 2.
So if I put that in for p here, this was a - a dagger over i and then root of h cross
omega over 2 cosine kz. And if I put this in this becomes omega cubed and the 2 cancels,
I can put this h cross inside here so that goes away and thatís what it is. so let me
call this whole thing B 0 and write this as a of t - a dagger of t over i cosine kz and
call B 0 this whole amplitude mu 0 epsilon 0 over k h cross epsilon 0 omega cubed over V to the
power 1/2 and this goes away. Thatís B (t). So now we have our quantized fields. Ex is
a function of z and t. for this mod of propagation is a + a dagger and B is a - a dagger. As
you can see they play the role of a position and momentum so to speak, apart from some
We can compute the commutator of E with B. thatís like the commutator of x with p and
you can see the commutator is a with a dagger is going to give you1 and a dagger with a
is going to give -1. The question is whatís the time dependence like? Whatís the solution
like? Thatís not hard to find because there are many ways of finding this time dependence
and in this problem, itís completely trivial.
The Heisenberg picture operator, da over dt is = [ A, H] over ih cross at any instant
of time. Thatís trivial to find because this is =1 over i or rather omega over i because
H is h cross omega a dagger a. so this is the commutator of a with a dagger a. the 1/2
part doesnít contribute to anything. And that = - i omega a because a with a dagger
is unity and there is an a which comes out. Whatís the solution to this? This implies
a(t) = e to the - i omega t a of 0. Similarly, a dagger of t can also be computed but itís
just the Hermitian conjugate of this. So itís a dagger of 0 e to the i omega t. this is
just a number. So it doesnít matter where I put it but just by shear force of habit,
the moment I take a Hermitian conjugate, I invert the order here. So all you have to
do is to put that in here and you have the exact the solutions where a and a dagger are
Heisenberg picture operators. By the way, you can solve in another way. You know the
formal solution to this equation. Now the question is: what is the expectation value?
I am going to say a dagger a on n is n | n> and n is 0, 1, 2, etc. the eigenvalue n I
identify with the number of photons in the number state n. The state of this radiation
field in which the number of photons is given, is that an eigenstate of that electric field?
So, the question being asked is the Hamiltonian is H cross omega a dagger a + 1/2 .a dagger
a is the number operator. These are the eigen states of this number operator. So the question
asked is, in a state in which the number of photon is sharp, is that also a state in which
is electric field is known precisely. No, because a and a dagger donít commute with
a dagger a. similarly the magnetic field is not known precisely. In fact, you can ask:
whatís the uncertainty in a state in which the number of photons is fixed?
Well, all you have to do is to compute the expectation value of Ex in whatever state
you want. This = . Whatís the expectation value of a or a dagger in a given number state
n? So this is some constant. this is proportional to n a + a dagger on n. now this is 0 because
a acts on n and lowers it to n -1. a dagger acts on n and raises it to n +1. So a and
a dagger now can be called photon creation and photo destruction operators. Because a dagger acts on the number state
n and makes it n + 1.
In other words, it creates a photon and a destroys a photon. It reduces the number of
photons by 1. Since these have no diagonal elements at all, this is identically 0. So
the average value of the electric field is certainly 0. But the mean square value is
not 0. Whatís the mean square value? Ex squared = E 0 squared sin squared kz times a + a dagger
squared. So n a + a dagger squared on n now what is this equal to? So you should be able
to compute these expectation values fast. Itís got to become proportional n + 1/2 because
this term here is a squared + aa dagger + a dagger a + a dagger squared on n. a dagger
squared has no diagonal elements. Thatís 0. a squared has no diagonal elements and
that is 0 as well. aa dagger is a dagger a +1. So this can be written as a dagger a +1.
But a dagger a on n is just n times n. therefore this works out be = E 0 squared sin squared
kz times 2 n +1.
Therefore, in this number state, delta Ex in the number state n is the standard deviation.
So this is = E 0 modulus sine kz = the square root times root of 2 n +1. Whatís By going
to be? Well itís clear that once again the square of these is going to give a 0 and letís
quickly compute what the expectation value of delta By is.
Delta By whole squared in this number state n = B 0 squared - cos square kz then expectation
value of aa dagger and a dagger a both with ñ sign. So this - sign goes away and thatís
going to give me 2 n +1. So delta By in this number state n = B 0 cos kz root 2 n +1. and
the important thing to see is that even in the ground state, the uncertainty in the eclectic
field and magnetic field is not 0. So there are vacuum fluctuations these are called vacuum
fluctuations. Because in a quantum field it is very subtle. Even in the state where there
are no quanta present and you have a vacuum. There is still a fluctuation of the electric
and magnetic fields and they have physical consequences. This has a physical effect which
is measurable. So note delta Ex at 0 and delta By at 0 is not identically = 0. They are still
vacuum fluctuations. One of the physical effects is the following. if you took 2 parallel plates
as we done here at z = 0 and z = l conducting plates, then even in the absence of an electromagnetic
field in the region between them, there would still be a force between the 2. This is called
the Casimir effect and it happens because of the 0 point energy. Notice that there is
0 point energy in this problem and itís formally infinite which is completely crazy but then
this is an artifact of quantum field theory.
The ground state energy = when n is 0, we know the energy is h cross omega into n +
Ω. So itís actually h cross omega over 2 per mode. Every mode contributes an energy
h cross omega over 2. So every wave number contributes an energy H cross omega over 2.
2 polarizations, the factor 2 goes away because there are 2 such contributions. Then, you
have to sum over all possible omegas and thatís infinite because the number of kís is actually
infinite. So the 0 point energy of the electromagnetic field in a vacuum is actually infinite. But
you could ask for this force between 2 plates, how do I compute it?
Well, I put the whole thing in a box. So I take the side walls of the box to be very
large and the distance between the 2 plates to be small compared to the linear dimensions
of the side walls and I ask what is the energy it takes to bring the second plate from infinity
up to a distance l. The difference in these two 0 point energies will be the work done
in bringing this plate out here and from that if you take a gradient, you end with the force.
And you can actually compute this force and it was experimentally measured in the 1950s
and there is such a force. Itís a delicate experiment because you have to make sure that
you got rid of all the other effects and the conditions of the experiment have to satisfy
the assumptions made deriving this expressions. But this has been done. The 0 point energy
leads to a real force called Casimir effect. It has been experimentally seen. There are
other quantum electrodynamic effects.
One of them is the famous Lamb shift which occurs in this spectral lines of atoms including
that of hydrogen and that too has been experimentally verified to high degree of accuracy. So we
are confident that the quantization prescription is actually a correct1 and this is the way
the electromagnetic field behaves. Now I leave you to figure out what the uncertainty products
are for n, Ex Ey and so on and so forth. Notice that you must definitely have, in any state
delta Ex delta By is greater = 1/2 the magnitude of the commutator of Ex with By of the expectation
value the commutator in thatís state. Thatís easily verified
in this case.
Now one thing is very clear that, when you make a measurement on any physical observable,
the postulate of quantum mechanic says the answer you get is one of the possible eigenvalues
of this operator. The expectation value we have is in the number operator state, we havenít
said anything at all about the actual eigenvalues of a and a dagger or eigenstates of a and
a dagger. But we know the classical harmonic oscillator that the spectrum of p is actually
continuous. So these operators, Ex and By actually have continuous spectra. You make
a measurement; you are going to get one of those answers. What we are now saying is if
you prepare the system in a state of a definite number of photons, then you measure what the
uncertainty in the electric field is, even in the ground state. Even if there are no
photons, the answer is nonzero.
Now of course, you could ask, in a real black body cavity, what kind of radiation do you
have? It is certainly not standing waves operating in 1 direction. There is a whole admixture.
So letís do that. Now there are
several other questions which I will come back to here, which we need to answer. One
of them is, in the case of a classical electromagnetic wave, I have an amplitude as well as a phase
and then interference of it happens because of this phase. Whatís the quantum analog
of the phase? This is a very important question and we will come and talk about it a minute.
The second question is, Ex and By in some sense are like the position and the momentum
and therefore, they are conjugate variables. The commutator of Ex and By is like the commutator
of x with p and gives you ih cross times 1 and so on. What about a dagger a? Whatís
the conjugate variable to a dagger a? In other words, what operator should it be such that
if you take a dagger a, commutator with that will give you unity? Now the square root of
a dagger a is taken to be the amplitude. The classical intuition tells us that there is
the amplitude and the phase. Therefore, there must be some phase operator somewhere and
maybe this is the conjugate of a dagger a. Let me show thatís not as simple as it seems.
We will then do black body radiation, comeback and talk about the phase of this operator
because I would like to introduce the idea of coherent states and go on with quantum
optics. So letís try and see what happens if you try to define a phase. You run into
trouble as follows.
a dagger a is the number operator N. so just as I take a complex number z and write this as r e to the i theta,
where r is really the square root of zz star. in exactly the same way, instead of z i have
a, instead of z star i have a dagger. And i have this combination mod z squared which
is a dagger a and that is a Hermitian combination. Let me see what the standard notation is for
this. a = e to the i phi square root of N. Let me say I do a polar decomposition of a
just like I do a polar decomposition of complex number. I do a polar decomposition of this
operator a by writing it in this form. And then whatís a dagger? Phi is supposed to
an operator and N is an operator. Now this becomes square root of N e to the - i phi.
So letís see if this works or not. And the commutation relation is aa dagger - a dagger
This implies that e to the i phi N e to the - i phi - a dagger a which is N. because a
dagger is root N e to the - i phi e to the i phi root N and e to the - i phi and e to
the i phi commute with each other and give you a 1. And this must be =1. Let me call
phi as a phase operator. I put a question mark to see if this is really sustainable
or not. This was Diracís idea originally. If you do this polar decomposition and see
if you can consistently define a phase operator of this kind. This commutation relation between
a and a dagger which came from the original xp = ih cross goes over into this.
e to the i phi is1 + i phi - phi squared over 2 factorial Ö with N and 1 - i phi - phi
squared over 2 factorial Ö - N = the unit operator. i(phi N - N phi) =1. It looks like
this is sustainable if i define [N, phi] = i. so it looks like we are almost there. Because
just as I had a with a dagger =1, i have N with phi = i times the unit operator. You
can check that if i have this commutation relation, then this relation itself is satisfied.
So it looks like we are in business. you can write a in this form, a dagger in the other
polar form and then N and phi are complimentary to each other, in the sense that the commutation
relation between the number operator and the phase operator is in fact just i.
Now if you want these to be physical operators, you want N and phi to be Hermitian. in other
words, you want e to the i phi to be a unitary operator. Otherwise it doesnít make any sense.
so you certainly like this to be a unitary operator. If itís a unitary operator, it
means U U dagger is =1 and U dagger U =1. look at what happens. e to the i phi = a N
to the - Ω. if i operate with N to the - 1/2 on the right hand side, i get e to the i phi
is a into the - Ω. And if I do the same thing here, e to the - i phi = N to the - 1/2 a
dagger. If this is an operator U, a unitary operator, i expect this to be U dagger or
U inverse. This is U inverse and we want this to be U dagger for it to be unitary.
So U U dagger = e to the i phi into e to the ñi phi. This gives me a N inverse a dagger
and thatís not identically equal to the unit operator by any means. On the other hand,
if i do you U dagger U, you get N to the - 1/2 a dagger a N to the - Ω. But this is N to
the - 1/2 N N to the - 1/2 = the unit operator. So hereís our first indication that everything
is not all right because these are infinite dimensional Hilbert spaces. There is no guarantee
that if there is a left inverse, there is also right inverse. If something is unitary,
you require not only that U U dagger be = i, but also U dagger U be equal i. but with this
set of operators, you end up with U U dagger is not 1 but U dagger U is the unit operator.
So you begin to see there is already a problem this is an infinite dimensional Hilbert space
and you have to be very careful here. There is another way of seeing this. Its 1 over
N doesnít have any meaning. 1 over an operator doesnít have any meaning. By that you mean
N inverse, the inverse of the operator.
said N phi - phi N and i take this between the state n and a state n prime and thatís
= n prime the unit operator n and there is a factor i. But this quantity i know is i
delta n prime n because these are orthonormal states. On the other hand, when this N acts
on this n prime, you get an n n prime. So there is an n prime from here. When phi N
acts on that, you get an n. then the matrix element is this. If this commutation relation
is true, then that is true. but what happens if n is = n prime, (n-n prime) is 0 and <n
prime |phi| n> = i. so you immediately see this commutation relation is not sustainable
in this Hilbert space spanned by the states n. so although it looked as if you had no
difficulty in writing a phase operator in this fashion, really this is not sustainable.
Phi cannot be a physical Hermitian operator because if it were, this is unitary operator
but we see explicitly itís not a unitary operator. Besides, the commutation relation
between N and phi is not maintainable. This leads to an inconsistency. It says 0 = i,
to start with. So there is a serious difficulty in defining the phase operator in quantum
Classically, there is no difficulty at all. And this has arisen for a variety of reasons.
Firstly, it is an infinite dimensional space. Secondly, itís not very clear how to define
a physical phase operator here. There are several resolutions that have been proposed
and we will talk about just 1 of those things to get over this mathematical difficulty.
It brings out the fact that these operators in this infinite dimensional space sometimes
could have a left inverse but not a right inverse. So we wonít spend too much time
on this. The reason itís important is because you do need to know whatís the phase of the
electromagnetic field. The first thing I do next time is to talk about what happens if
you have a radiation field inside, like black body radiation and we will talk about the
photon number distributional in a black body radiation and then go over to the interaction
of atoms with radiation with matter. we will look at some simple models for it which will
help us make a connection with the spin 1/2 problem. So let me stop here today.