In the 
past few lectures 2 lectures, we have discussed
the basic phenomenon behind the development
of quantum mechanics namely the wave particle
duality; and then we continued our discussions
to understand the Dirac delta function. We
will continue on discussions on the Dirac
delta function, and we will introduce Fourier
transform of a function.
However, before we do that, I thought I will
give you a list of references. There are a
large number of books written on quantum mechanics,
but some of the extremely good ones are listed
below, are listed on this slide. The first
one is of course, Professor Dirac book it
is a classic on principles of quantum mechanics
published by oxford university press. And
then we have the famous Feynman lectures on
physics volume 3; the first few chapters discuss
the wave particle duality and the uncertainty
principle in a very beautiful way. I would
advise all students, all students to go through
the first few chapters of the book.
And then there is a book, a classic a very
nicely written text book by Libof, an introductory
quantum mechanics and then, there is book
by Townsend a modern approach to quantum mechanics
published by McGraw hill. And for the history
of the development of quantum mechanics, there
is a very nice book written by Max Jammer
in title the conceptual development of quantum
mechanics.
I will be following my own our own book which
is in title quantum mechanics theory and applications,
which is currently in its 5th edition and
it has been published by Macmillan in India,
New Delhi and also the 5th edition has been
reprinted by Kluwer academic publishers in
Dordrecht.
There is another book that I have written
a very thin small smaller thin book which
is in title basic quantum mechanics and which
includes a CD. We have developed a software
which will demonstrate in the next few lectures
and it has also been published by Macmillan
India and the paperback edition now contains
the CD.
So as I mentioned earlier, we will continue
our discussion on the Dirac delta function,
and we will introduce Fourier transforms,
which plays an extremely important role in
the development of quantum mechanics.
We had given one of the representation of
the Dirac delta function in terms of a rectangle
function. Here the rectangle function is such
that its value is value is 1 over 2 sigma
and then 0 beyond 2 minus sigma and 2 plus
sigma, so that as the value of sigma becomes
smaller and smaller, the width of the rectangle
become smaller and smaller, the height becomes
larger. But the area under the rectangle always
remains unity.
This is perhaps the most straightforward the
simplest definition representation of the
Dirac delta function and we define and we
write this is one of the representation, delta
of x minus a as the limit of the rectangle
function in the limit of sigma tending to
0. When sigma tends to 0 the width of the
rectangle becomes extremely small the height
becomes extremely large but the area under
the curve remains unity.
And because of this property the delta function
is such that, if I integrate the delta function
with multiplied by an arbitrary but well behaved
function f of x. Then it picks up its value
only at the point x is equal to a and we define
the Dirac delta function through these two
equations delta of x minus a is 0 for x not
equal to a at x equal to a. It is not defined
actually it is it has an infinite value but
we represent the function by a spike of unit
height. This is how a Dirac delta function
is represented and it is defined by these
two equations.
We next consider the Gaussian function and
the Gaussian function is, we define is such
that G sigma of x is equal to 1 over sigma
under root of 2 pi e to the power of minus
x square by 2 sigma square. Now this integral
we will be we had derived last time and we
will be using this very frequently e to the
power of minus alpha x square plus beta x
dx is equal to pi by alpha e to the power
of beta square by 4 alpha the limits all limits
are from minus infinity to plus infinity.
So if I substitute if I now right down the
integral of G sigma of x G sigma of x, the
Gaussian function G sigma of x dx, then it
will be I will have 1 over sigma by root 2
pi outside and then minus infinity to plus
infinity e to the power of minus x square
by 2 sigma square dx. If I compare the integrant,
this integrant with this integrant we find
that alpha is equal to 1 over 2 sigma square
and beta is 0.
So therefore, this will become the integral
will be 1 over sigma under root of 2 pi under
root of pi alpha that is pi divided by alpha.
So 1 over alpha is 1 over 1 over 2 sigma square
so, this is 2 sigma square and then e to the
power of beta square beta is 0. So this factor
becomes 1, so as you can see the 2 pi cancels
out under root of sigma square is sigma so
this is equal to 1.
So therefore, this integral G sigma of x dx
is equal to 1 is unity the area under the
curve is unity for all values of sigma. So,
let me plot this function let me plot this
function this G sigma of x that I have written
down and of course, sigma is greater than
0 sigma is greater than 0.
So if you plot this you will find that if
I take sigma is equal to 1 then it will have
a width something like this. This is a Gaussian
function and this width this width is of the
order of sigma this is something like the
full width at half maximum if this value is
1, if you take this the peak value as 1, then
the full width at half maximum is about sigma
may be square root of 2 sigma or 1.5 sigma
or something like that.
Now, as I make the area under the curve I
have just now shown to be unity. So as I make
sigma smaller because of the factor of one
over sigma the value of the function at x
is equal to 0 becomes smaller and smaller,
larger and larger and the width becomes smaller
and smaller. But the area under the integral
area under the curve remains unity. So therefore,
it has all the properties of the delta function
and therefore, we can write that limit of
sigma tending to 0 the G sigma of x that is
1 over sigma under root of 2 pi e to the power
of minus x square by 2 sigma square, this
is a yet another representation of the Dirac
delta function.
So this is a delta function centered at x
is equal to 0. If I want to displace the curve
if I want to shift the origin by a. So therefore,
we can write delta of x minus a is equal to
limit of sigma tending to 0 1 over sigma under
root of 2 pi e to the power of minus x minus
a whole square by 2 sigma square. This is
known as the Gaussian representation of the
Dirac delta function.
So here on this slide we are shown that the
function G sigma of x I define this as 1 over
sigma under root of 2 pi and plot. In the
figure below we have plotted this function
for sigma is equal to 0.1, 0.2 and 0.4. So
the width of the function is of the order
of sigma and as the value of sigma becomes
smaller and smaller, the width becomes smaller
and smaller, the localization of the wave
function becomes smaller and smaller and the
height becomes larger and larger but the area
under the curve is always unity.
So we have what is known as the in Gaussian
representation of the Dirac delta function.
So this is an important representation of
the Dirac delta function. But there are many
representations of the Dirac delta function,
this is yet one more representation. Now,
we will consider yet another representation
of the Dirac delta function and that is known
as the integral representation of the delta
function.
So let me do it carefully there is a function,
there is an integral definite integral which
is equal to minus infinity to plus infinity
sin g x divided by x the whole thing divided
by x. So this is this sink function something
like that and integral this is equal to pi
for any value of g we assume g to be greater
than 0. For any value of g the evaluation
of this is extremely straightforward using
Laplace transform technique and is given in
almost all books of mathematical physics.
But the but the thing that I would like to
emphasis is sin g x by x dx is equal to pi.
So I write this as so I write this as minus
infinity to plus infinity sin g x by pi x
dx is equal to 1. If I plot this function
you will find you will find that sin g x over
pi x in the limit of extending to 0 in the
limit of extending to 0 the numerator becomes
g x and the denominator becomes pi x so this
becomes g by pi g by pi.
And as you can see the first 0 of this function
occurs at g x is equal to pi. In fact the
0s of the function appears occurs as g x is
equal to pi 2 pi 3 pi 4 pi minus pi minus
2 pi etcetera. So if you plot this function
you will find that the value of the function.
Let us suppose let us suppose g is 1 so g
is 1 so 1 over pi is 0.3 something like this
and the first 0 will occur if g is 1 at x
is equal to pi something like this. So this
is a damp sin curve damp sin curve and it
will be symmetric on the other side so, it
will be something like this.
If the value of g is increased, if the value
of g is increased then that let us suppose
g becomes 10. So this value increases by a
factor of 10 I cannot show you this but it
beco[mes]- and then the first 0 occurs at
x is equal to pi by 10 if g is 10. So, therefore,
it becomes very shapely peaked and it becomes
like this but the area under the curve remains
unity.
Therefore, therefore, we have yet another
representation of the Dirac delta function,
that limit as g tends to infinity sin g x
by pi x has therefore, all the properties
of the Dirac delta function and I can write
this as delta x. So I will use this representation.
Now let me write down what is sin g x by pi
x, if I integrate a function like this 1 over
2 pi. Let us suppose from minus g consider
the definite integral minus g to plus g e
to the power of plus i k x or minus i k x
it does not matter multiplied by t k. So I
carry out the integration over k and the function
is e to the power of i k x the integral is
trivial its simple 1 over 2 pi and if I integrate
this with respect to k, so I get e to the
power of i k x divided by i x from minus g
to plus g.
So you will have 1 over 2 pi 1 over 2 pi and
i x will in the denominator, let me put x
outside I here e to the power of i, k becomes
g. So I g x minus e to the power of minus
i g x I multiply by 2 and divide by 2. So
this quantity this quantity e to the power
of I g x minus i g x, this quantity is just
sin g x as you will wish, so this becomes
1 over pi x sin g x. So therefore, sin g x
over pi x this quantity is just this integral
is just this integral and therefore, we have
a very important relation.
We have just now shown we have just now shown
that that sin g x over pi x, let me leave
some space this is equal to 1 over 2 pi 1
over 2 pi minus g to plus g e to the power
of i k x dk. So this I have shown just now
by evaluating the integral on the right hand
side and this follows immediately. But I have
just now shown before that that in the limit
of g tending to infinity this become the Dirac
delta function delta of x. So if I make limit
as g tends to infinity so this becomes 1 over
2 pi minus g to plus g actually you can take
either plus sign or minus sign it does not
really matter plus minus i k x dk. So this
is known as the integral representation of
the Dirac delta function, this is known as
the integral integral representation representation
of the Dirac delta function of the delta function.
So if I shift the origin if I replace x by
x minus x prime, so I get delta of x minus
x prime this is equal to 1 over 2 pi sorry
this because, I have put g equal to infinity
I am sorry this will be minus infinity to
plus infinity, so you will have because if
I make g tend to infinity this becomes minus
infinity to plus infinity. So I should have
included that minus infinity to plus infinity
e to the power of either a plus sign or a
minus sign it does not really matter either
1 e to the power of i k x minus x prime dk.
This equation is an extremely important equation,
it is known as the integral representation
of the Dirac delta function. integral representation
of the Dirac delta function.
So let me let me show this in the form of
slide, that this is for example, the function
sin of g x minus 2 divided by pi of x minus
2 for g is equal to phi5 the function is spread
out. If I increase the value of g it is symmetric
around the point x is equal to 2 it becomes
sharply peaked and the and and the width becomes
extremely small. In the limit of but the area
under the curve is always 1 in the limit of
g t becoming a very large number this therefore,
become this has an infinite value of x is
equal to 2. But the area under the curve remains
unity and therefore, it has all the properties
of the Dirac delta function. So we have so
the area under the curve is equal to 1 for
all values of g and therefore, we have delta
of x minus a is equal to limit as g tends
to infinity of sin g x minus a divided by
pi of x minus a.
So, we just now showed this that the area
under the curve is 1 and it becomes more and
more sharply peaked as g becomes large. So,
now I have just now shown that if I integrate
this then I get the left hand side, if I carry
out the integration on the right hand side
which is very simple then I will get the left
hand side. Therefore, if I make now g tends
to infinity so, if I make g tend to infinity
so, the limits will become from minus infinity
to plus infinity so, I have delta of x minus
a is equal to so much.
So this is known as the integral representation
of the Dirac delta function so let me now
use this.
So I have here e to the power of delta x minus
x prime is equal to 1 over 2 pi 1 over 2 pi
e to the power of plus or minus so, let me
take the plus sign i k x minus x prime dk
and the limits are from minus infinity to
plus infinity. Now I had told you earlier
that I consider a well behaved function f
of x, any arbitrary well behaved function
f of x and if I integrate this over the delta
function that is, if I multiply this f of
x by the delta function and integrate over
x prime, then I will get the value of the
function at f of x. Now for this delta of
x minus x prime i substitute this expression
so, I write this down as please see this 1
over 2 pi 2 integrals 1 integral is over x
prime the other integral is over k so, both
limits are from minus infinity to plus infinity
e to the power of i k x minus x prime f of
x prime dx prime.
Now I separate the ones which contain x prime
so you rem[ember]- so you see this and then
dk also dk from this. So, I separate the ones
from which contain only x prime and I define
a function F of k, I define this function
as 1 over root 2 pi, please see this minus
infinity to plus infinity f of x prime, which
is this e to the power of minus i k x prime
which is this and dx prime.
And then what I do is what I do is that I
then write down that f of x f of x will be
equal to I have taken 1 over root 2 pi here.
So I can write down here 1 over root 2 pi,
then F of k e to the power of i k x dk these
limits are also from minus infinity to plus
infinity. These two equations describe what
is known as the Fourier integral theorem,
the function F of k is the Fourier transform
of f of x and this is the inverse Fourier
transform. So in this equation since x prime
is a definite integral I can remove the prime
on the x.
And therefore, I will obtain I can write down
you see this I define a function F of k, which
is defined to be equal to 1 over root 2 pi
from minus infinity to plus infinity f of
x e to the power of minus i k x dx. Then f
of x is known as the inverse Fourier transform
of F of k so 1 over root 2 pi minus infinity
to plus infinity F of k e to the power of
plus i k x now the integral is over k.
These 2 equations constitute what is known
as the Fourier integral theorem Fourier and
is a very important concept in mathematical
physics. If I have a plus sign here then it
will be a minus sign here, if I have a minus
sign here I will have a plus sign here, both
are extensively used in, so this is one of
the very important relationship that I have
derived. Now let me take let me give you a
simple example so, let me assume that f of
x is something like a into e to the power
of minus x square by 2 sigma square x square
by 2 sigma square. So, it is a Gaussian function
a Gaussian function whose width is about sigma
something like this is the f of x function
this width is of the order of sigma.
Then its Fourier transform will be F of k
will be equal to A over root 2 pi A root 2
pi is should be there. So, then you will have
e to the power of minus x square by 2 sigma
square e to the power of i k x dx all limits
are from minus infinity to plus infinity.
And then we introduce then we have the formula
that we had used that e to the power of, this
is a very important formula that we will be
using quite extensively that, e to the power
of minus alpha x square plus beta x is equal
to square root of pi by alpha into e to the
power of beta square by 4 alpha.
So here you can see that alpha is equal to
1 over 2 sigma square and beta is equal to
i k.
So you will have so I will have alpha as 1
over 2 sigma square and beta is equal to i
k F of k will be equal to as I had written
down a moment back A under root of 2 pi e
to the power of minus x square by 2 sigma
square e to the power of i k x into dx. So,
from minus infinity to plus infinity and therefore,
you will have A by root 2 pi multiplied by
root pi divided by alpha so that root pi times
2 sigma square beta is i k so, square of beta
beta square is minus k square minus k square
and k square.
Then 2 sigma square by 4 so, k square sigma
square by 2. So, this is just a constant but
the main thing that if you plot this function
F of k in the k space then that is also a
Gaussian e to the power of minus k square
by sigma square but the width is now 1 over
sigma. Therefore, if a function if a function
I had just shown that f of x has a width of
the order of sigma then its Fourier transform
has a width of the order of 1 by sigma. So
the if f of x is localized within a distance
which is delta x then, F of k will be localized
within a distance in the k space delta k,
which is 1 over sigma so that delta k delta
x is of the order of 1, this is a very important
relationship and from which we will prove
the derive the uncertainty principle.
So let me go back to my slides. So, this is
my integral representation of the Dirac delta
function. So, any arbitrary well behaved function
if I integrate by after multiplying by delta
of x minus x prime so, I will get f of x.
So, then I substitute for delta of x minus
x prime from this equation and we obtain this
double integral, this integral is known as
the Fourier integral theorem. I can have either
of this sign I can have a plus sign here or
a minus sign here.
So what I have just now shown is that f of
x is equal to 1 over 2 pi e to the power of
plus minus i k x minus x prime f of x prime
dx prime dk. I now define the Fourier transform
of f of x and because of this 2 pi factor
here, I spit it into 2, so that the equations
becomes symmetric 1 over root 2 pi k. So,
f of x e to the power of minus i k x, I have
dropped the prime and then I and then I substitute
it back then f of x is equal to 1 over root
2 pi F of k e to the power of i k x d k. So
this function F of k is known as the Fourier
transform of the function f of x and this
is the inverse transform relation and these
2 together is often called as the Fourier
integral theorem.
So as I mentioned just now that let me take
a Gaussian function f of x is equal to A exponential
minus x square by 2 sigma square. So, I substitute
it here and calculate its Fourier transform
so I substitute it here and here, you again
use the same relation with alpha is equal
to 1 over 2 sigma square and beta is equal
to minus i k so, this square of that is minus
k square sigma square by 2. So we say that
the Fourier transform of a Gaussian function
is a Gaussian function.
And therefore, this is the function f of x
this is the Gaussian function in the x space
and its Fourier transform is also a Gaussian
function. So, what we have derived is that
the Fourier transform of a Gaussian is a Gaussian,
the main thing is that its width something
like full width at half maximum is of the
order of sigma and its half width. If the
localization of the function F of k in the
k space is of the order of 1 over sigma, so
that you have delta x delta k is of the order
of 1 delta x delta k is of the order of 1.
Now let me let me now derive one more relation
that I have this Fourier transform and I will
show that if f of x is normalized then F of
k is also normalized and the proof is very
simple and we will again use the Dirac delta
function.
So I have here let us suppose F of k f of
k is equal to is equal to 1 over under root
of 2 pi, f of x e to the power of minus i
k x dx, all limits are from infinity. Therefore,
your f of x is equal to 1 over under root
of 2 pi, F of k e to the power of i k x dk
this is all limits are from minus infinity
to plus infinity and if I take the complex
conjugate so, f star of x. If the complex
conjugate of f of x is equal to the complex
conjugate of this equation so, 1 over root
2 pi F star k f star k e to the power of minus
i k x d k.
Now what I want to evaluate is integral from
minus infinity to plus infinity mod f square
x mod f square x dx. So, this is equal to
f star times f into dx so, please see this,
this this is slightly cumbersome but very
straightforward. So what I will like to substitute
is f star from here f from here and substitute
it and then carryout the integration over
x. So, now you must be very careful because
this k should not get confused with this k
so I will quietly put a prime here k prime
here k prime here then everything is all right.
So you will have so if I substitute this f
star times this 1 over 2 root 2 pi into 1
over root 2 pi. So, this will become 1 over
2 pi and there are 3 integrals 1 integral
here over x 1 over k prime and 1 over k so,
all limits are from minus infinity to plus
infinity. So you will have first write down
F of k prime f star of k e to the power of
e to the power of i x, then k prime minus
k k prime minus k into dx dk d k prime, dx
comes from here dk comes from here and dk
co[mes]- prime comes from here.
Now let me then collect all those terms which
involve x this does not involve x, this does
not involve x, this involves x and this involves
x. So, I carry out the integration over x
so, what I do is I collect this term, I collect
this term, and I collect this term and carryout
the integration over x so, if I do that if
I put it at the top.
Then what I will find is 1 over 2 pi minus
infinity to plus infinity e to the power of
i x k prime minus k dx, this is equal to I
can write delta k minus k prime or k prime
minus k, it does not really matter k minus
k prime. So I have carried out 1 integration
over x and now what I do is I carry out the
integration over k prime. So, I collect all
the terms which involve k prime which is F
of k prime which is F of k prime dk prime
and delta of k minus k prime. So, I write
this down so integral F of k prime F of k
prime then delta of k minus k prime and dk
prime and of course, this limit is also from
minus infinity to plus infinity. So, this
is just F of k so I substitute it here and
I get the relation that minus infinity to
plus infinity mod f square dx is equal to
minus infinity to plus infinity F star of
k multiplied by F of k multiplied by dk. So
that we get the very important relation that
this is known as parseval's theorem, that
minus infinity to plus infinity the mod square
of the function and the Fourier transform
are equal. So if one of them is normalized
one of them is normalized the other will be
automatically normalized
So we obtain the very important relation that
that mod f square dx is equal to mod of f
square dk. This is known as the parseval's
theorem. So this is what I have tried to show
here any arbitrary well behaved function,
if I square this and integrate over dx in
the square of that is equal to 1. Therefore,
if this function if a is such that if mod
f square dx the curve is 1 then mod f k square
the under that function will also be equal
to 1.
So, if this is what parseval's theorem is,
that if 1 of them is equal to 1, then the
other one is this is also known as the normalization
condition, this is also known as the normalization
condition.
Till now I have done between x and k, there
is no reason why we cannot consider a time
dependent function, and consider its Fourier
transform. So let us consider a time dependent
function f of t.
Let us let us consider a time dependent function
which I write as f of t and consider its Fourier
transform, so the Fourier transform I define
as f of omega is defined to be equal to 1
over root 2 p from minus infinity to plus
infinity e to the power of i omega t dt. This
is known as the frequency spectrum and I will
tell you in a moment why this is known as
a frequency spectrum. Then the Fourier integral
theorem tells us that, if I take the inverse
Fourier transform then f of t will become,
equal to 1 over root 2 pi 1 over root 2 pi
F of omega f of omega e to the power of minus
i omega t d omega minus i omega t d omega.
Now please see this what is an integral, an
integral is nothing but a sum. So, this is
a sum over all the frequencies so, this quantity
F of omega is known as the frequency spectrum
of the time dependent function f of t. For
example, if the time dependent function has
only one1 frequency then the corresponding
F of omega is a Dirac delta function. So if
the frequency spectrum is has only one frequency
let us suppose delta of omega minus omega
0 then, if I substitute that here carryout
the integration then f of t becomes equal
to 1 over under root of 2 pi e to the power
of minus i omega 0 t.
So this is a monochromatic wave this is a
monochromatic wave so, if I substitute this
f of t function in in in in here, then I will
get F of omega f of omega will be 1 over 2
pi 1 over 2 pi e to the power of e to the
power of i of omega minus omega 0 into t into
dt. So, this is the definition of the Dirac
delta function this is one of the definitions
of the Dirac. All limits are from minus infinity
to plus infinity so, if the if a time dependent
signal has only 1 frequency, then the corresponding
frequency spectrum is a Dirac delta function
and the corresponding time dependent function
is a harmonic wave. So, it is if I take the
real path it is Cos omega 0 t from from from
t equal to minus infinity to t equal to plus
infinity.
So you have here let me consider a simple
function let me consider a simple function
e to the power of minus t square by 2 tau
square e to the power of minus i omega 0 t.
So let us suppose that my time dependent function
f of t is something like this, if you take
the real path then it will be something like
a Gaussian time dependent function multiplied
by Cos omega 0 t. So it is a Gaussian pulse
of duration tau of duration tau so, it will
look something like this. something like this
So you will have a Gaussian pulse localized
within a time interval of the order of tau.
So let me take calculate its its its Fourier
transform, the Fourier transform will be F
of omega, F of omega will be equal to 1 over
root 2 pi A I can take outside so, integral
f of t e to the power of I omega t dt. So
I will write it down integral e please do
this carefully minus t square by 2 tau square
and then e to the power of i so, this will
be e to the power of iI omega minus omega
0 multiplied by t times d t.
So here again I again have the relation like
e to the power of minus alpha x square plus
beta x where x is replaced by t so, my alpha
is equal to 1 over 2 tau square and beta is
equal to I times omega minus omega naught.
So the evaluation of this integral these limits
are of course, minus infinity to plus infinity.
So this will be please see this carefully
1 over root 2 pi 1 over root 2 pi under root
of pi by alpha that is under root of pi times
2 tau square multiplied by beta square, that
is exponential beta square beta square by
4 alpha, that is minus omega minus omega 0
whole square by 4 alpha is 2 tau square 1
over 2 tau square. So 2 pi 2 pi cancelled
out.
So I will get the frequency spectrum F of
omega will come out to be A times tau, this
is an unimportant constant factor but once
again the Fourier transform of the Gaussian
function is a Gaussian function minus omega
minus omega 0 whole square by tau square by
2 this is a square here. So this function
is picked around omega is equal to omega 0
and it falls by a factor of 2 when omega the
del the frequency is spread is of the order
of 1 over tau.
So the duration of the pulse is of the order
of tau the frequency width of the pulse is
of the order of 1 over tau. So I get this
uncertainty relation delta omega delta t is
of the order of 1, this is known as the mono
chromaticity of the pulse smaller the delta
variable omega, the greater is the sharpness
of the frequency. So t tau is something like
the, but we did in optics as the coherence
time larger the coherence time smaller is
the spectral width and this we have all studied
in optics and therefore, if you have a very
long coherent pulse then its frequency spread
is extremely small.
So let us suppose that I have the wavelength
is equal to say 6 000 armstrong in the yellow
region 6,000 armstrong. So you will have this
is equal to 6 into minus 10 to the power of
minus 7 meters and omega will be equal to
2 pi times nu, nu is c by lambda. So this
will be 2 pi into 3 into 10 to the power of
8 into 6 into 10 to the power of minus 7,
so this will come out to be about 3 into 10
to the power of 15 hertz 3 into 10 to the
power of 15 hertz.
And let us suppose I have a 1 nanosecond of
duration pulse so, the frequency spread delta
omega is of the order of 1 over tau. So this
is of the order of 10 to the power of 9 hertz,
which is about a millionth of the frequency
so, delta omega by omega is a very small number
so delta omega by omega will be equal to 10
to the power of 9 divided by 3 into 10 to
the power of 15 so, this is about 10 to the
power of minus 6 in fact less than 10 to the
power of minus 6. So it has a very, very small
width an almost monochromatic so, let me do
this again.
So let me have a we consider an optical pulse
with Gaussian envelope. So the duration of
the pulse is of the order of tau and the central
frequency is around omega 0 is at omega 0.
The spectrum the Fourier transform pulse is
given by of omega equal to so much and if
you substitute this here, we have shown that
the frequency spectrum is so much. The spectral
width is of the order of tau.
So, this is something like a let us suppose
we consider a Gaussian pulse which corresponds
to 1 micro meter wavelength and it is a 20
second 20 fem to second pulse. This is actually
the how the electric field will look like.
And its corresponding Fourier transform will
be very sharply peaked the Fourier transform
of a Gaussian is a Gaussian will be extremely
small delta omega will be very small.
And we will have the uncertainty relation
that delta omega delta t is of the order of
1.
And we can consider similarly, another time
dependent function where tau c, where where
we have f of t is 0 for all times which is
greater than plus half t c or minus half t
c. So tau c is the duration of the pulse.
So this is known as the coherence time and
in between it is A into e to the power of
minus i omega 0 t.
If you take the calculate the Fourier transform,
it is a very simple exercise it will be very
sharply peaked around omega equal to omega
0, in this particular case I have assumed
omega 0 tau c is of the order of 100. So longer
the value of tau c greater will be the sharpness
of F of omega and greater will be the mono
chromaticity of the pulse.
So for example, the example that had just
now considered for omega 0 is equal to 3 into
10 to the power of 15 hertz, the corresponding
wavelength is 6000 armstrong. If I consider
a 1 nano second pulse then the corresponding
frequency 6,000 will be 1 over tau c 1 over
tau c is 10 to the power of 9 hertz and so
the spectral purity of the pulse is point
3 into 10 to the power of minus 6. Now so
I summarize this part of the lecture by saying
that if I had any time dependent function
or a space dependent function.
So I have f of x I write its Fourier transform
as F of k and I write this F of k, I define
this as equal to 1 over root 2 pi minus infinity
to plus infinity f of x e to the power of
minus i k x into dx. Then its inverse Fourier
transform will be f of x will be 1 over under
root of 2 pi F of k e to the power of i k
x dk. I can replace x by time and then k with
omega so, that is also possible and we can
time dependent function and write down the
corresponding frequency spectrum of the pulse.
If the function f of x has a localization
of the order of delta x then F of k the localization
is delta k. Then in general delta k delta
x is of the order of 1 similarly, I can define
for instead of x as I have just now mentioned
this is t and then this will be the omega
spectrum and if the time duration of the pulse
is delta t. Then its corresponding frequency
spectrum is delta omega and that is 1. So
this complete so you can carry out this with
other examples, that you take an arbitrary
well behaved function f of x calculate its
Fourier transform show that, if it is localized
within a distance of the order of delta x,
then its Fourier transform will be localized
within a distance of the order of 1 over delta
x thank you.
