We will continue our discussion on the solution
of the one dimensional Schrödinger equation
for a free particle as we had discussed in
our last lecture the electron or the proton
or the neutron is described by a function
psi such that mod psi square d x represents
is proportional to the probability of finding
the electron between x and x plus d x.
So this is the mod psi square is the probability
distribution function and we can always normalize
this wave function such that the integral
from minus infirmity to plus infinity is equal
to one as we had discussed earlier this is
known as the normalization condition and when
this condition is satisfied this quantity
mod psi square d x represents the probability
of finding the particle between x and x plus
d x this is the max born interpretation interpretation
of the wave function. Since this is the probability
distribution function p of x d therefore,
if the electron or the proton is described
by the wave function psi then the expectation
value of x and the average value of x will
be equal to integral from minus infinity to
all limits always all the integral are from
minus infinity to plus infinity x mod psi
square d x. So this is the expectation value
of x and which is usually written like this
as we had mention earlier minus infinity to
plus infinity psi star x multiplies by x and
psi of x d x similarly, the expectation value
this is the expectation value of x similarly,
one can write down the expectation value of
x square. So x will be replace by x square
here so this will be psi star x x square d
x so if I know the wave function psi of x
then in on principle in principle I can evaluate
this integral and obtained expectation value
of x square and the expectation value of x
if we determine this then the uncertainty
in x be defined this is equal to under root
of the expectation value of x square minus
expectation value of x whole square so this
is the uncertainty in the in the.
In the measurement of the quantity x now let
us suppose that the initial wave function
at time t equal to zero. The wave function
describing the electron or the proton at time
t equal to zero is a Gaussian wave function
I have a normalized Gaussian wave function
like pi sigma not square rest to the power
of one by four e to the power of minus x square
by two sigma not square e to the power of
I by h cross p naught x. So the probability
distribution at t equal to zero is a Gaussian
function so p of x d x this is equal to mod
psi x square d x and this will be the square
of the quantity pi sigma not square under
the root because the square of this is under
the root and then the modulus square of this
is one, so this is e to the power of minus
x square by sigma not square. So in order
to find the expectation value of x or x square
I just have to substitute this quantity in
this here and carry out the integration to
evaluate x expectation value of x and expectation
value of x square.
So therefore, we can evaluate if we do that
then we obtain that the expectation value
of x is equal to from integral minus infinity
to plus infinity x times p of x d x so this
will be one over under root of pi sigma not
square x into e to the power of minus x square
by sigma naught square d x and as you can
see this is an odd integrant therefore, this
implies that x is zero the expectation value
of x is zero, so that if you make a large
number of observation of the position x coordinate
of the position of the particle then and you
take the average of these over a large number
of measurements you will get the value zero
sometimes you will get plus something sometimes
you will get minus something but, the average
of that will be zero. Now the expectation
value of x square will be just if you x square
then this integral can be easily worked out
in terms of gamma function because this will
be equal to two times under root of pi sigma
not square integral zero to infinity x square
e to the power of minus x square by sigma
naught square d x you can now substitute x
square by sigma naught square equal to y and
the carry out the integration it is a very
straight forward integration and the final
result will come out to be sigma not square
by two. So therefore the delta x this spread
in the value of x is equal to x square minus
x average whole square under the root this
quantity we have shown this to be zero this
quantity sigma not square by two so we obtained
that for the Gaussian wave packet this becomes
zero by root two this is the this is the uncertainty
in the measurement of x this is the uncertainty
if the the the electron is described a wave
packet. So at t equal to zero. It is somewhere
located in this region therefore, the average
value so this is the origin the average value
is zero but, that is a spread in the wave
function as I will as I had shown you earlier
and this spread in the wave function is described
by the uncertainty delta x and that is equal
to sigma not by root two. So this result we
have obtained delta x is equal to sigma not
by root two so this is an important result
for a Gaussian wave function we next calculate
the expectation value of the momentum operator
and the expectation value and the uncertainty
in the momentum.
Now as we had discussed earlier that the this
general solution of the Schrodinger equation
is given by one dimensional Schrodinger equation
is actually a wave packet one over under root
of two pi h cross minus infinity to plus infinity
a of p e to the power of I by h cross p x
minus p square by two m t into d p. So at
t is equal to zero psi of x comma zero this
will be I substitute t equal to zero so I
obtain under root of two pi h cross minus
infinity to plus infinity a of p e to the
power of I by h cross d x d p thus this psi
of x not is Fourier transform of a of p and
conversely the Fourier transform of a f p
inverse Fourier transform will be given by
as we had discussed earlier this is the inverse
Fourier transform one over two pi h cross
psi of x not x comma zero and now it will
have a minus sign minus I by h cross p x d
x. So if I know psi of x comma zero the recipe
is this as we had discussed earlier the recipe
is this if I know psi of x comma zero, I can
calculate a of p once I calculate a of p I
substitute it this equation carryout this
integration and obtain psi of x of t that
will describe the evolution of the wave packet
with time. So we have we have assumed that
that psi of x comma zero we had psid that
this quantity we assume to be this equal to
1 over under root pi sigma zero square raise
to the power of one by four e to the power
of minus x square by two sigma not square
e to the power I by h cross p naught x.
So if I substitute this expression for psi
of x comma zero I can carry out the integration
using that integral e to the power minus alpha
x square plus beta x is very straight forward
integration and if you do that we will obtain
the following expression for a f p and you
will obtain a of p is equal to sigma 0 square
plus pi please work this out pi h cross square
raise to the power of one by four and the
Fourier transform of Gaussian is Gaussian
minus p minus p naught whole square sigma
naught square divided by 2 h cross square.
So this function the wave function is is is
Gaussian its Fourier transform is also Gaussian
and if you plot this you will find that the
Fourier transform that the a of p function
a of p function is peaked around p equal to
around p naught and with the spread which
is of the order of h cross by sigma naught
then so this is the Fourier transform function
so then if I want to measure the x component
of the momentum or just we say that the p
and the average value of momentum we had discussed
this.
So the average value of momentum will be equal
to minus infinity to plus infinity psi star
p psi d x so p psi will be equal to we have
to replace this by its operator we had discussed
this in detail so minus infinity to plus infinity
psi star minus I h cross I can put a bracket
here delta psi by delta x d and if you carry
out if you substitute for psi this particular
expression this particular expression and
you carry out this integration then you will
find that this will come out to be minus infinity
to plus infinity p of mod a p square d p.
Therefore we can interpret that a of p mod
square d p is the probability of finding the
momentum between p and p plus d p and for
from parcel able theorem if psi square is
normalized then this is also normalized that
is minus infinity to plus infinity a of p
square one p square d p is equal to one if
the wave function is normalized then its Fourier
transform is also normalized. So this is the
expectation value of p similarly, you can
calculate the expectation value of p square
for p square this will be a p square here
and so this will be minus h cross square delta
two by delta x square and if you did the algebra
once again as we had done earlier so this
will come to the p square a of p one square
d p.
We do have the expression for a of p so a
of p as we had derived it a few minutes back
sigma not square by pi h cross square raise
to the power of one by four e to the power
of minus p minus p naught whole square sigma
naught square by two h cross square. So all
that I have to do is therefore, I have the
expression for a of p and analytical expression
I substatute a of p square will be square
of that I substitute it here multiply by p
square and carry out the integration that
will give me p square and for p I just have
to multiply by p and carry out the integration
the the algebra is slightly involve but, it
involve simple evaluation of the integral.
So one fine with a little bit of algebra I
would request all of you to carry out with
algebra so after a bit of algebra which is
just evaluation of the integrals we will obtain
that the expectation value of p that the average
value the x component of the momentum is p
not and the expectation value of p square
will come out to be p not square plus h cross
square by two sigma not square I would request
all of you to work this out it is a very straight
forward algebra but, 1 must do that. So therefore,
delta p is p square average minus p average
square so p average square will be p naught
square if you subtract this from this the
p naught square p naught square cancel out
so delta t will come out to be if you find
out delta p that will be equal to delta p
will be equal to under root of p square minus
p average square so this will be h cross square
by two sigma naught square so.
So we obtained finally, the following expressions
for delta p and delta x delta p we have just
found out that this will be equal to h cross
by root two sigma naught delta x we had found
out to be equal to sigma naught by root two.
So therefore, the uncertainty product this
is known as the uncertainty product 
delta x delta p the sigma not sigma not cancel
out this will become out to be and this is
typical of a Gaussian wave packet the Gaussian
wave packet corresponds to the minimum uncertainty
product so we obtained this.
Now this is actually the uncertainty product
is minimum and 1 can show this analytically
because you may recall that when we derive
when we had derive the uncertainty principle
we had first proved sources in equality and
the inequality was mod f square d tau times
mod g square d tau this was greater than nor
equal to quarter of integral f star g plus
f g star d tau and whole square. Now I can
easily show that if g is a multiple of f that
is let us suppose g g can be complex but,
let us suppose it is a multiple of f therefore,
the time being we can assume that alpha is
a real number let us suppose so this becomes
f square and this g square so g g star so
mod g square becomes alpha square mod f square
where alpha is a constant and I am assuming
this to be a real constant so alpha square
so the left hand side left hand side becomes
alpha square and this also becomes f square
d tau this also becomes f square d tau. So
you will have integral mod f square d tau
whole square 
on the other hand this quantity f star g will
be equal to this will be equal to alpha mod
f square this will be also alpha mod f square
so the sum of the two will be two alpha mod
f square and if I square this it will be 4
alpha square mod f square d tau whole square
the four will cancel out with four. So the
right hand side will also be equal to alpha
square mod f square d tau whole square what
I have done is the following. I had proved
earlier that this inequality is always valid
I had just now shown that if g is a multiple
of f and let this multiplicative constant
be real then the left hand side and the right
hand side become equal so that this inequality
becomes an equality. So therefore, using this
inequality we had proved that delta x delta
p x is greater than equal to half h cross
and we had assumed that we had assumed that
f was equal to f was equal to p psi which
is equal to minus I h cross say d psi by d
x and we assume that g was equal to I x psi.
So if if g is a multiple of f then you have
this if I assume that g is a multiple of f
or f is a multiple of g does not matter either
way so then this tells us that I x psi will
be equal to minus I h cross alpha d psi by
d x 
so the I on both sides cancel out and if I
take psi in the denominator then I will obtain
one over psi d psi by d x is equal to minus
minus h cross alpha and let me put it as x
by sigma naught square just the sigma naught
square is just h cross alpha. So if I integrate
this out so this becomes log psi so psi becomes
equal to e to the power of minus x square
by two sigma not square so the when the wave
function is Gaussian g is g is a multiple
of f and therefore, the inequality becomes
an equality so that for a Gaussian and wave
function delta x delta p becomes equal to
half h cross finally, now we had shown so
this is the uncertainty product so finally,
we.
We have that the wave function I want to study
the time evolution that how I had shown you
a diagram but, maybe we will show this again
today time evolution of the wave function
so that is given by psi of x t is equal to
under root of two pi h cross integral the
limits are from minus infinity to plus infinity
e to the power of I by h cross p x minus p
square by two m t d p minus infinity. So if
I know psi of x comma zero I know a of p and
in fact for the Gaussian wave packet we have
shown that a of p is equal to sigma not square
by pi h cross square raise to the power of
one by four it was a Gaussian momentum distribution
minus p minus p naught whole square sigma
naught square by two h cross square so I know
the function a of p I substitute it here and
carry out the integration of p over p once
again it is a slightly cumbersome integral
but, you just have to repeatedly use you just
have to again use the integral that we had
first written on the first day alpha x square
plus bets x d x from minus infinity to plus
infinity this is equal to square root of pi
by alpha e to the power of beta square by
four alpha. If you do that you obtain an analytical
expression for psi of x comma t and if you
take the modulus of that then you will get
psi of x comma t mod square this is equal
to under root of pi sigma square as the function
of time e to the power of minus x minus p
naught over m into t whole square divided
by sigma square of t where sigma of t after
carrying out the integration you can show
comes out to be sigma naught one plus h cross
square by m square sigma not four sigma not
four into t square raise to the power of half.
So there are two very important points in
this formula the first point is that the centre
of the wave packet is moving with the velocity
p not by m this is the group velocity of the
wave packet the whole wave packet moves with
a certain velocity the average velocity the
group velocity and that group velocity of
the wave packet is given by p not by m and
the second important point is that the width
of the packet is increasing with time so that
initially the packet was the real part of
the wave function was something like this.
At a later time it would have expanded and
it as I will show you this whole thing expands
the delta x increases in fact if you use this
wave function and evaluate x which is equal
to is equal to mod psi x comma t whole square
into x d x from minus infinity to plus infinity.
If you first evaluate mod x expectation value
of x and when you evaluate mod x’s square
then you will find that that the uncertainty
in x will come out to be equal to sigma naught
by under root of two into under root of one
plus h cross square by m square sigma naught
four t square that is the packet expands and
the packet expands because of the dispersion
relation that omega that you see you have
here e is equal to h cross omega and you have
also e is equal to p square by two m 
so h cross omega is equal to p square by two
that is h cross square k square by two m so
your omega is equal to h cross by two m times
k square. So you have a non-linear relationship
if omega is proportional to k as it is indeed
true for electromagnetic waves in free space
there is no broadening of the pulse but, if
there is dispersion what is dispersion when
omega and k has a non-linear relationship
then there is a broadening of the pulse so
the broadening of the pulse is because is
a manifestation of the fact that we have a
dispersive media in which omega k relationship
is not linear. So delta x is so much the delta
p remains the same the delta p is equal to
h cross by under root of 2 that does not change
with time h cross by under root 2 what was
the expression for h cross by under root 2
into sigma naught divided by sigma naught
this does not change with time so that the
uncertainty product the uncertainty product
will change with time and you will have delta
x delta p will be equal to h cross by 2 h
cross by 2 1 plus h cross square by m square
sigma naught 4 into t square. So this is the
uncertainty product for a propagating Gaussian
beam the beam expands along the x direction
along the direction of propagation of the
wave so we have done all the algebra let me
just quickly go through the the my slides
which will tell you the same thing so we continue
with our discussion with the free particle
this is the as you may be called this is the
1 dimensional Schrodinger equation for a free
particle we had discussed the general solution
of that.
So the general solution of this equation is
given by this equation this equation describes
the propagation of a wave packet so how do
we study the evolution of the wave packet
so I put t equal to zero, so in this term
becomes zero so this becomes a p e to the
power of I pi h cross p x.
So the recipe is that I find out a of p by
taking the inverse Fourier transform and if
I take the f p by taking the inverse Fourier
transform and I substitute it in this equation
and I will get psi of x comma t so in a in
a Gaussian wave packet this is the form of
the wave function that is given to us so my
problem is that if I know the wave function
at t equal to zero what is the wave function
at at a later time t in order to do that first
we first showed that the expectation value
of x is zero then we showed that the expectation
value of x’s square is sigma zero square
by two therefore, this is zero. So the expectation
value of x at time t equal to zero at time
t equal to zero is sigma zero by root two.
Then in order to study the evolution of the
wave packet we calculated the a of p the momentum
distribution function so that is the inverse
Fourier transform of psi of x comma zero.
If I substitute this here and carry out the
straight forward integration I will obtain
this equation which is also a Gaussian but,
peaked around p equal to p zero.
And therefore, we can interpret mod psi of
x d square d x as the probability of finding
the particle between x and x plus d x and
a of p mod square d p is the probability of
finding the x component of the momentum of
course, we are considering the one dimensional
case the x component of the momentum between
p and p plus d p
So since a of p is given by this I can then
calculate the expectation value of p if I
if I i would request all of you to carry out
this integration and if I multiply the square
of this function with p and carry out this
straight forward integration this is the average
momentum this is the average momentum with
which the wave packet moves and similarly,
you can calculate p square so you do this
and you obtain this.
So delta p which does not change with time
which does not change with time becomes h
cross by root two sigma naught a so at time
t equal to zero delta p is so much delta x
is so much so delta x delta p is the minimum
uncertainty product and that we had discussed
earlier.
So how do we find the evolution of the wave
packet as I said earlier that the psi of x
t is given by this and if psi of x comma zero
is known.
Then I can find out a of p by taking the inverse
Fourier transform so I will able to find the
a of p I substitute it here and carry out
the integration and once again the integration
is not very straight forward is is little
cumbersome.
But, it is very straight forward if you do
that you will find that mod psi of x t square
is equal to one over root two pi mod root
pi sigma of t and the wave packet itself moves
with the center of the wave packet moves with
the velocity p zero by n. This is the group
velocity of the wave packet and the width
of the Gaussian pulse increases with time
increases with time.
So you will have this wave functions so x
is so much so delta x actually increases with
time because the sigma of t increases with
time.
Therefore, the uncertainty product increases
with time at t equal to 0 it is the minimum
so this is my localized wave packet the electron
is described by this wave packet.
This localized somewhere here somewhere between
the here the localization is within the distance
of the order of sigma naught and the corresponding
momentum distribution is of the order of h
cross by delta x so that the 1 certainty product
is satisfied.
Now this is what I have showed earlier also,
this is the evolution of the wave packet and
let me do it very carefully so the wave packet.
The electron is let me do it more slowly so
let us suppose we put it five hundred here
and so the electron is described by this wave
function so if you ask me the question is
the electron a particle or a wave the answer
is it is neither it is describe by a wave
function so it is somewhere localized where
the wave function is finite you can tell you
can predict a probability distribution of
finding the electron as it evolves with time
and therefore, there is an uncertainty in
the measurement of x there is an uncertainly
in the measurement of p and that is contained
in my uncertainty in in in the solution of
the Schrodinger equation. So therefore, Schrodinger
equation itself contains uncertainty principle
so you see I have here the electron is described
by this wave function actually it is the real
part of the wave function so it is somewhere
localized here and the packet itself broadens
with time and that is because of dispersion.
So therefore, to summarize an electron or
a proton or a neutron is neither a particle
nor a wave it is describe by the wave function
a free particle let us suppose the propagation
is only in the x direction and I neglect the
y and z variation this described by a wave
packet so when I when the electron leaves
the filament it is described by the localized
wave packet which moves with a velocity equal
to p zero by m. So the packet itself moves
forward it is localized somewhere there and
the localization is consistent with the uncertainty
principle so that is what my quantum mechanics
tells me that the electron or the proton or
the neutron is neither a particle nor a wave
it is described by the wave function by the
wave function by a wave function psi which
is a solution of the Schrodinger equation.
Now let me consider a slightly more difficult
problem not difficult little more involved
problem but, I have a Gaussian wave packet
which is propagating in the x direction this
is my x direction and then it has a slit of
width d in the y direction so I am just I
am neglecting z, so the packet is moving in
the x direction. But constructed in the y
direction. So so the wave packet here is a
Gaussian wave packet so I have here therefore,
a wave function psi of x comma y comma zero
at time t equal to zero first the x part at
time t equal to zero is a pi sigma zero square
raise to the power of one by four e to the
power of minus x’s square by two sigma naught
square e to the power of I by h cross p naught
x so this the way Gaussian wave packet. Just
similar to one that we had considered localized
within a distance of sigma naught and propagating
in the x direction but, I now multiply by
a function y function of y which is such that
psi b of y is equal to one over under root
of y sorry one over under root of b I am sorry
for y less than b by two and minus b by two
and zero everywhere else. So I allow this
the wave function to be finite only in this
region and 0zeroeverywhere else this is my
y axis so it is the a Gaussian wave packet
encounters a slit of width b now this wave
function is normalized that is minus infinity
to plus infinity mod psi b of y mod square
d y this is equal to one over b from minus
b by two to plus b by two this is psi b y
square is one over root b so that the square
of that is one over b multiplied by d y so
this is b by two minus minus b by two that
is b by b, so this is one so it is normalized
in the x direction and also in the y direction
so I want to study the evolution of this wave
packet so then what I do is instead of taking
a one dimensional Fourier transform I take
a two dimensional Fourier transform.
So we will have we write this down as that
psi of x comma y comma zero is equal to one
over root two pi h cross whole square then
integral integral a of p x comma p y e to
the power of I by h cross p x x plus p y y
d p x d p y. So therefore, in order to determine
a p x y I take the two dimensional Fourier
transform of psi of x y so I get a p x comma
p y is equal to one over under root of two
pi h cross whole square and then the integral
first integral over x multiplied by e to the
power of minus I by h cross p x x and then
there is a second integral that is psi b of
y e to the power of minus I by h cross p y
y d y this is the this is the Gaussian wave
packet that we had written down that this
this quantity we will substitute here and
this quantity is one over root two b so this
is also from minus infinity to plus infinity
and this is also minus infinity to plus infinity
the x part integral will come out to be the
same thing so, me factors into e to the power
of p minus p naught whole square sigma naught
square by two h cross square so that is the
integral that just similar to one that we
had found out earlier. So actually I can write
down a p of x comma p y so this is a product
to product function a p x and a p y actually
this should be a p x and a p y will be the
integral of that.
So let me evaluate this so a p y is equal
to 1 factor I will take here two pi h cross
please see this this is very it will give
me a very important result psi b of y 
p y y d y but, this is 0 for y greater than
mod b by two. So therefore, this comes out
to be 1 over under root of two pi h cross
one over under root of b and then from minus
b by two to plus b by two let me put this
as e to the power of minus alpha y where alpha
is equal to p y by h cross the integration
of this is very straight forward that is the
integration will be you just have to integrate
this and you will find e to the power of minus
I alpha y divided by minus I alpha from minus
b by two to plus b by two. So this will become
if I do it like this so this will become e
to the power of minus I alpha b by two minus
e to the power of plus I alpha b by two divided
by minus I alpha so there is a minus sign
sitting in the denominator I put it plus sign
here plus sign here and minus sign here multiplied
by two here and multiplied by two here so
this becomes two by two alpha two by alpha
this becomes two by alpha and this will be
e to the power of I alpha b by two minus e
to the power of minus i alpha b by two divided
by alpha divided by alpha divided by two I
so that will be sin of alpha b by two. So
this is equal to this this is equal to this
so I can write this down as I can put a b
factor here sin beta by beta where beta is
equal to alpha b by two where beta is equal
to alpha b by two.
And let me write down what is so we obtain
we obtain a of p y a of p y as one over under
root of two pi h cross one over under root
of b times b sin beta by beta. So this becomes
equal to some constant in front b by two pi
h cross sin beta by beta and what is beta
beta is equal to alpha b by two if you may
recall that alpha was equal to p y by h cross
so this becomes alpha p y b by two h cross
h cross is h over two pi so, that is alpha
sorry alpha is equal to p y by h cross so
alpha is equal to p y by h cross is a factor
two here so the two two cancel out and what
is p y. If you see this if the electron gets
moving in this direction and if this angle
which is the angle of diffraction is theta
and this is the y direction then p y is equal
to p sin theta, so you will have p y is equal
to p sin theta pi will come above and then
h multiplied by b h by p is lambda so you
obtain pi b sin theta by lambda pi b sin theta
by lambda.
So the probability distribution function will
be proportional to that is probability of
the momentum lying between p y and d p y will
be equal to mod a p y square d p y and that
will be proportional to sin square some constant
sin square beta by beta square where beta
is equal to pi b sin theta by lambda.
And this is my single slit diffraction pattern
this is the single slit diffraction pattern
so how do I explain the diffraction phenomenon
as soon as I constrict the electron so this
slit itself imparts a momentum in the y direction
and the probability of the momentum being
lying between p y and p y plus d p y will
be proportional to sin square beta by beta
square and that is the single slit diffraction
experiment. So if I show you to conclude this
lecture so I have the I have the single slit
diffraction experiment in which you have a
Gaussian wave packet in the x direction and
in the y direction you have a single slit
therefore, the probability of the momentum
x component of the momentum lying between
p x and p x plus d p x is given by a Gaussian
function.
On the other hand if I want to find out the
probability distribution function for p y
then I integrate this.
And I obtain sin beta by beta so that the
probability of the y component of the momentum
lying between p y and p y plus d p y will
be proportional to the sin square beta by
beta square into d p y.
So that as soon as you constrict the particle
to pass through the slit the slit itself imparts
a momentum and you obtain the single slit
diffraction experiment.
Where an individual electron will lie I cannot
I can only predict a probability distribution
function so that if the experiment was carried
out with a large number of electrons the single
slit diffraction pattern will be observed
therefore, I have a source of electrons protons
neutrons alpha particles or anything that
you can think of it is the I describe this
as a Gaussian wave packet propagating in the
x direction and then I constrict it within
a distance of the order of b to obtain the
single slit diffraction pattern.
Where an individual electron or the proton
will land up no 1 can predict 1 can only predict
a probability distribution and therefore,
and that probability distribution is the same
as the as the single slit diffraction pattern
as you must have read in your under graduate
optics course.We will continue from this point
onwards in my next lecture thank you.
