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

and By.

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

dimensional constants.

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

a =1.

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.

So

we have

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

mechanics.

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.