We will be continuing our discussion on the
use of the Dirac delta function, and specifically
we will consider the solution of the Schrodinger
equation for the free particle problem. And
we will show that the solution of the Schrodinger
equation contains the uncertainty principle.
However before we do that, let us recapitulate
on what we had done in our previous lecture.
We had said that delta of the Dirac delta
function the integral representation was given
by 1 over 2 pi minus infinity to plus infinity,
and there can be both for plus sign as well
as minus sign i k into x minus x prime into
dk. So, this is known as the integral representation
of the Dirac delta function.
Now, let me introduce a variable p which we
define as p is equal to h cross k, therefore
you will associate this later with momentum,
but let us do this right now just as a mathematical
variable, then from this equation you will
find that dk is equal to dp by h cross, therefore
if I substitute it here I will get delta of
x minus x prime is equal to dk is equal to
dp by d h h cross.
So, this will be 2 pi h cross 2 pi h cross
where h cross is the plank’s constant h
over 2 pi, and this integral we will just
choose the minus sign we could have chosen
the plus sign also, but we will choose the
minus sign here, so this will be e to the
power of minus i p by h cross x minus x prime
into dp. Now, as we know I consider a function
psi of x, so we write down as psi of x is
equal to integral psi of x prime into delta
of x minus x prime dx prime from minus infinity
to plus infinity.
This follows for any well behaved function
psi of x, this follows from the definition
of the Dirac delta function. What I do next
is I substitute for delta of x minus x prime
in this equation.
So, what I will get is what I will get is
psi of x is equal to 2 pi h cross 1 integral
is over x prime, and 1 integral is over d,
so psi of x prime e to the power of minus
i p by h cross x minus x prime let us say.
So this is what I have done if you recollect
that we had these two equations, we have these
two equations, so delta of x minus x prime
was equal to this, and psi of x was equal
to this, so we have substituted for delta
of x minus x prime from the top equation in
the bottom equation.
Then what we do is the limits there are two
integrals minus both limits are from minus
infinity to plus infinity, as we had done
earlier we write 2 pi h cross we split it
into two parts and we define a function a
of p, we collect the x prime part outside,
and write 1 over under root of 2 pi h cross
integral psi of x prime e to the power of
i.
P x prime by h prime dx prime, this limit
is from minus infinity to plus infinity, and
since this is a definite integral, I can remove
the prime from here, then I substitute for
a of p in this equation, then I will get psi
of x psi of x is equal to 1 over under root
of 2 pi h cross, a let me let me take a plus
sign here, so there is a minus sign here,
and a of p e to the power of i p x by h cross
dp minus infinity to plus infinity.
We have two parseval's theorem and in which
these we can similarly prove here, that the
integral minus infinity to plus infinity mod
psi x square dx is equal if this is normalized
then this equation this a p is also normalized,
a of p whole square dp, and as we will show
later we will interpret psi of x mod square
dx as the probability of finding the particle
between x and x plus dx.
We will also interpret that a of p square
dp will be the probability of finding the
x component of the momentum between p and
p plus dp, and we will show from the Fourier
transform pair that if the particle is localized
within a distance of the order of delta x,
then its momentum spread delta p will be such
that delta p delta x is of the order of h
cross. So, this contains this is contained
in the solution of the Schrodinger equation.
So therefore, we will consider an arbitrary
function psi of x, its Fourier transform in
the momentum space, we will represent by this
equation and as I mentioned earlier I can
remove the because this is a definite integral,
so I can remove the prime from here and then
the inverse Fourier transform is given by
this. Now, let me write down in the very second
lecture we had written down the Schrodinger
equation.
The Schrodinger equation the one dimensional
Schrodinger equation for a particle moving
in a field given by V of x is given by i h
cross delta psi by delta t is equal to H psi,
H is the Hamiltonian, and in this case so
this is equal to h psi where h is the Hamiltonian
which is p square by 2 m plus V and we had
in one dimensional case p will be replaced
by minus i h cross delta by delta x.
So, we had shown that this was equal to therefore
minus h cross square by 2 m delta 2 by delta
x square plus V of x. For a free particle
for a free particle the potential energy is
0, V of x is equal to 0 here, so H is given
by just this quantity, therefore the one dimensional
Schrodinger equation for a free particle
so i h cross delta psi by delta t is equal
to minus h cross square by 2 m delta 2 psi
by delta x square. This equation is known
as the 1 dimensional time dependent Schrodinger
equation for a free particle. for a free particle.
We will solve this equation we will solve
this equation we will obtain a rigorously
correct solution of this equation, this is
a partial differential equation so we will
try to solve this equation with the help of
the method of separation of variables.
So, we will assume that psi of x comma t is
a function of x and t, is a function of x
and a function of time, so I substitute it
here, so I will get i h cross, now when I
substitute it here I will get dT by dt, and
psi can be taken because this is partial differentiation
with respect to t I, therefore I will get
psi of x dT by dt, this is equal to minus
h cross square by 2 m this becomes T of t
d 2 psi by dx square.
The variables have still not separated because
the left hand side consists of a function
of x and time, the right hand side also consists
of a function of x and time, the trick is
we divide the entire equation by capital psi,
that is we divide this equation by psi times
T of t, if we do that, i h cross 1 over T
of t dT over dt becomes equal to minus h cross
square by 2 m 1 over psi d 2 psi by dx square.
You must realize that in this equation we
have partial differentials because capital
psi is a function of x and time.
Whereas in this equation I am considering
the differential of a time dependent function,
so the partial differential is represented
by a total differential, now this on the left
hand side is a function of time, and right
hand side is a function of x, so function
of time cannot be equal to a function of x,
unless both of them are equal to a constant
e. So, we say that the method of separation
of variables has worked, and so therefore
we can set the left hand side equal to a constant
and a right hand side also equal to constant.
This e on the right hand side is a number
and I write this e, I can write this as just
p square by 2 m, where p is now a number,
and I will tell you why I have written that,
so I solve each part.
So, let me first solve the time dependent
part time dependent part is i h cross, 1 over
T of t dT over dt is equal to p square by
2 m, so if you integrate this out so you get
log of T log of t is equal to some constant
plus into minus i by h cross p square by 2
m into t.
So, T of t is constant, times e to the power
of minus i by h cross p square by 2 m into
t. This quantity is also sometimes written
as e, so therefore I can write this down as
minus i e t by h cross. Where e here and p
here is a number, m is of course the mass
of the particle. Let me solve the other side
of the equation other part of the equation
namely minus h cross square by 2 m, the x
part 1 over psi d 2 psi by dx square is equal
to p square by 2 m.
So, 2 m 2 m cancels out, so I can cancel this
and this, I take h Cross Square on this side,
so I get d 2 psi by dx square plus p square
by h cross square psi is equal to 0. And the
solution of this equation is trivial is very
simple psi of x is some constant, times e
to the power of I by p i by h cross times
p psi i p by sorry x.
So, the complete solution the complete, so
we have found out that the that we wrote first
that the we assume psi of x comma t, as equal
to psi of x into T of t, and then we said
that we found that T of t was equal to constant,
e to the power of minus I, p square by 2 m
t by h cross i e t by h cross, and psi of
x was equal to e to the power of i by h cross
p x. So, the total solution will be some constant
times e to the power of i by h cross p x minus
e t that is p square by 2 m into t.
Now, what is p; p is a number and p can take
any value from plus infinity to minus infinity,
minus infinity to plus infinity.
So, therefore the most general solution of
the one dimensional Schrodinger equation is
psi x comma t is a superposition of this solution,
so I write this as 1 over i use this as a
constant 1 over 2 pi h cross integral minus
infinity to plus infinity a of p a to the
power of i by h cross times p x minus p square
by 2 m t dp.
So, what I have done is here p goes from minus
infinity to plus infinity, so the most general
solution will be a superposition of all the
solutions, and that is what is known as a
wave packet, so therefore the most general
solution of the one dimensional Schrodinger
equation i h cross delta psi by delta t is
equal to minus h cross square by 2 m delta
2 psi by delta x square, the most general
solution.
Most general solution of the above equation
is given by psi of x comma t, is equal to
1 over under root of 2 pi h cross integral
minus infinity to plus infinity a of p e to
the power of I by h cross p x minus p square
by 2 m into t into dp. Now, I still do not
know what is a of p, so that is determined
by the initial form of the wave function.
So, let at time t equal to 0 psi of x comma
t equal to 0 at t equal to 0, let it be psi
of x, you know if I put c psi t equal to 0,
then it becomes 1 over under root of 2 pi
h cross integral minus infinity to plus infinity
a of p e to the power of 1 by I by h cross
p x dp. this is just a Fourier transform.
So, therefore a of p as we have been discussing
in the earlier lecture this would be inverse
Fourier transform of this function, so this
will be 1 over 2 pi h cross, psi of x, if
this is a plus sign here, and this will be
a minus sign here, dx. So, the recipe is the
following I want to know the initial, I already
know the initial wave function, psi of x,
if I know the initial wave function psi of
x, then from the knowledge of psi of x i can
calculate a of p, once I know a of p, i will
substitute in this equation, and carry out
the integration to find out what is psi of
x comma t.
So, that is the complete solution of the problem,
if psi of x i will interpret, mod psi x square,
as the probability of finding the particle,
between x and x plus dx, and then we will
interpret mod of a p square dp as the probability
of finding the particles momentum between
p and p plus tp.
So, if the function is localized within a
distance of the order of delta x, then its
momentum will be localized with a distance
of the order if h cross pi sigma. Now, let
me give you an example.
Let us calculate let us assume that psi of
x with wave function at t ninety equal to
0, is given by under root of pi sigma 0 square
raise to the power of 1 by 4, so that is square
root of square root. So, let me delete this.
So, I am sorry so this will be 1 by pi sigma
0 square raise to the power of 1 by 4, and
then it is given by e to the power of minus
x square by 2 sigma 0 square, and e to the
power of let us suppose that the particle
is at time t equal to 0, is described by this
wave function. I can tell you that if you
integrate this square of this minus infinity
to plus infinity mod psi x whole square dx
then this will be this is one fourth power.
So, this will be one over under root of pi
sigma 0 square mod psi square will be minus
infinity to plus infinity, e to the power
of minus x square by sigma naught square dx,
because mod of this quantity is unity. And
if you integrate this you get e to the power
of minus alpha x square alpha is 1 over sigma
naught square, so this will be this integral
will be under root of pi divided by alpha
alpha is so much so sigma naught square.
So, this factor will cancel out with this
factor, so this quantity is one. And we say
that this function is normalized Eigen function.
Now, we calculate with this function the corresponding
a of p, so a of p is equal to 1 over under
root of 2 pi h cross, minus infinity to plus
infinity, psi of x, e to the power of minus
i by h cross p x dx. And then what I do is
I substitute for psi of x which is which is
given which is by this expression.
So, I just substituted it there, so I will
obtain I will obtain 1 over under root of
2 pi h cross 1 over pi by sigma 0 square raise
to the power of 1 by 4, and then you will
have psi of x is e to the power of x square
by 2 sigma 0 square, into e to the power of
I by h cross, the p will come from here, p
minus p naught into x dx, the limits are again
from minus infinity to plus infinity.
So, here alpha is equal to 1 over 2 sigma
square, and beta as you can see is equal to
beta x, that will be minus i by h cross p
minus p naught, so if you substitute this
you will get you will get these 2 factors
will come in, e to the power under root of
pi 2 sigma square and this is beta sorry,
so exponential multiplied by exponential beta
square, so therefore this will be beta square
will be minus p minus p naught whole square
by h cross square by 4 into 2 0, so 2 into
sigma 0 square, because alpha is 1 over 2
sigma 0 square.
So, if you work it out and then use then you
will get the following, so you will have the
following relation that a of, i will leave
aside the sum constant, so let us suppose
this constant, this will come out to be e
to the power of minus p naught whole square
sigma naught square by 2 h cross square, and
if you plot a of p mod square, so it will
be Gaussian.
It will be Gaussian with shift with its peak
at p naught, this is p naught and the delta
p will be of the order of h cross by sigma.
So, this is known as the mod a p square is
known as the momentum distribution function,
so this will be mod c square, into e to the
power of minus p minus p naught whole square
sigma naught square by h cross square.
So, if the particle is initially located in
a distance which is of the order of sigma,
then its momentum spread is located within
a distance of the order of h cross by sigma,
so that delta x, delta p is of the order of
h cross. So, the uncertainty principle is
contained in the solution of Schrodinger equation.
So, let me illustrate this with the I will
just go through what I have tried to say.
So, first of all I have the I had mentioned
the integral representation of the delta function,
so first I replaced k by the variable p, so
p i define as h cross k h cross is equal to
h by 2 pi, so dk is equal to dp by h cross.
So, I have here plus minus i p by h cross,
x minus x prime, dp
So, this is the this is again the integral
representation of the delta function, and
because of the delta function f of x is equal
to so much. So, because f of x is equal to
integral delta of x minus x prime f of x prime
so I substitute it here and I will get this
as f of x.
So, thus if thus psi of x is equal to so much,
so you will have if a of p, i define as the
Fourier transform of psi of x, and when I
quietly remove the prime from here, then psi
of x is equal to a of p into the power of
i p x by h cross into dp.
So, psi of x and a f p form a what is known
as the Fourier transform pair, and if this
function is located within a distance of delta
x, and if this function in the p space is
located within a momentum spread of delta
p, then delta x, delta p is of the order of
h cross.
So, if I have a localized wave packet psi
of x which is of the order of delta x,
Then its corresponding momentum spread will
be of the order of h cross by delta x. So,
that the uncertainty principle is contained
in the solution of the wave wave equation.
As I we had discussed last time 2, 3 times
back the 1 dimensional time dependent Schrodinger
equation, so we had i h cross by delta psi
by delta t minus h Cross Square by 2 mu delta
2 by delta x square plus V of x this quantity
is known as the Hamiltonian times psi of x
t, so as Feynman has written where did we
get that equation from nowhere it is not possible
to derive it from anything you know.
It came out of the mind of Schrodinger actually
we did give a heuristic derivation of the
Schrodinger equation, and but that was that
did not that lacked rigor, so we say that
the Schrodinger equation came from the mind
of Schrodinger.
So, this is my Schrodinger equation and for
a free particle V of x is 0, so my Hamiltonian
is p square by 2 m, which is equal to minus
h cross square by 2 m delta 2 by delta x square.
So, this equation is the 1 dimensional time
dependent Schrodinger equation for a free
particle, once again this equation is the
one dimensional time dependent Schrodinger
equation for a free particle and you would
like to solve this equation.
We would like to have this solution the most
general solution of this equation for a free
particle. So, as I had mentioned earlier we
solve this equation by using the method of
separation of variables, if I use the method
and substitute this solution in this equation
then I will get and divide by psi of x t mod
psi of x T capital psi of x t, then I will
get i h cross by T by t of d dT by dt minus
i h cross h cross by 2 m psi of x d.
So, we say that the method of separation of
variables has worked because of the because
left hand side you have the function of time,
and on the right hand side you have a function
of x, both cannot be equal to each other unless
each one of them is equal to a constant, this
is the method of separation of variables,
so we write this constant as e or equal to
p square by 2 m.
So, these 2 are numbers therefore we have
as we have discussed earlier we said this
equal to a constant which I write as p square
by 2 m, here m is the mass of the particle,
so the solution of the space dependent part
if you write this then 2 m, 2 m cancels out
this becomes plus p square by h cross square,
so this space dependent part is psi of x is
equal to so much.
And the time dependent part is e to the power
of minus I, e, t, by h cross, but e is equal
to p square by 2 m, so this is a solution
of the 1 dimensional time dependent Schrodinger
equation for a free particle, what is the
value of p, p take can take any value from
minus infinity to plus infinity.
So, therefore the most general solution of
this equation is an integral like this, so
if I look at the back previous slide this
equation I still do not what is the value
of p; p can take any value from minus infinity
to plus infinity, and therefore since this
is a linear differential equation so the most
general solution of this equation is this,
so this is the most general solution and it
is said that it is it describes what is known
as a 1 dimensional wave packet.
So, this we had discussed earlier if a of
p is given by this, then Fourier’s transform
is given by this, so localized wave packet
its Fourier transform is h cross by there.
So, we have the wave function psi of x i write
it as a of p, e to the power of I by h cross
p f x, in the inverse Fourier transform is
given by this, and because of parseval’s
theorem as I said this integral if this is
1 then this integral is equal to 1.
We will discuss this little later may be in
a turn of two that we will interpret this
is the max burn’s interpretation for the
probability density function, and he describes
that this is the probability of finding the
particle between x and x plus dx, and this
will represent the probability of finding
the momentum between p and p plus dp.
Then we consider a simple Gaussian wave packet,
so at t equal to 0 psi of x is given by this,
this is normalized this factor is such that
mod psi x square dx is 1, if you write down
the corresponding Fourier transform then you
will find that it is given by this, so psi
of x is normalized then a of p is also normalized,
in the sense mod a p square dp is equal.
So, let me write this down so we have actually
we have the numbers now that a of p is equal
to sigma 0 square by pi h cross square, raise
to the power of 1 by 4 e to the power of minus
p minus p 0 whole square sigma square sigma
naught square by h cross square so mod a p
square mod a p square let me leave a little
space this will be sigma 0 square by pi h
cross square under root because 1 by 4 time
plus 1 by and this will become e to the power
of minus p minus p naught whole square sigma
naught square by h cross square.
And if I integrate this from minus infinity
to plus infinity and if I integrate this from
minus infinity to plus infinity, I leave this
is an exercise using the same formula this
will come out to be equal to 1, so a particle
is represented by a Gaussian wave packet,
once I have a of p then I can substitute you
remember that the most general solution of
the Schrodinger equation was psi of x t is
equal to 1 over under root of 2 by h cross
a of p, e to the power of I by h cross p x
minus p square by 2 m into t.
Now, minus infinity to plus infinity so you
see if I know a of p, i can substitute it
here, carry out the integration and I will
get an analytical expression for psi of x
t, so I solved the problem that by knowing
psi of x comma 0, at time t equal to 0, I
found out what I a of p, once I find out what
is a of p, i can substitute it here and carry
out the integration to calculate psi of x
comma t.
So, this is so if the particle is localized
within a distance of sigma naught, then this
momentum spread is of the order of h cross
of sigma naught, so my uncertainty principle
is contained in the solutions of this equations.
So, if I substitute the expression then I
will find that the evolution of the wave function
is something like this, now let me show you
a software to conclude this lecture by showing
you a software we have developed.
And we assume a certain width and we consider
the time evolution of the wave function, so
this is the real part of the wave function
let us do it in 150 steps, so that you can
see this slowly so this is the particle is
described by this kind of let me do even more
slowly, so the particle this is the real part
of the wave function, and this is how it evolves
with time please see this.
So, as the if I know the initial wave function
then the particle is somewhere localized here,
I do not know exactly I can only tell you
a probability distribution and in fact as
this Gaussian wave packet evolves, it spreads
it spreads and this is the uncertainty in
the localization of the particle, so we described
the we described the electron or the proton
or the neutron by a localized function.
The corresponding probability distribution
at t equal to 0 is somewhere here, the corresponding
momentum probability distribution is also
a Gaussian, so you have how will it evolve
with time if a particle is localized within
a distance this, at time t is equal to 0,
as the time progresses as the time progresses
the wave function evolves with time like this.
And at all times you will have the probability
distribution describing the motion of the
particle and the momentum distribution function
satisfying the uncertainty relation, so this
is how the wave packet follows with time.
So, I thought I will show this to you that
and the software that I have just now shown
we will be discussing other examples has been
developed by my colleagues Prem Bhavnani,
Dr. Ravi Varshney and Dr. Vipul Rastogi.
And Dr. Vipul Rastogi is right now in IIT
rookie, and it is in the software is a companion
to a book which entitled basic quantum mechanics
which I had referred to earlier, and we will
show you other examples from this book at
later cources.
So, let me summarize what I have done, first
of all we wrote down that the Schrodinger
equation cannot be derived, and that is given
by i h cross delta psi by delta t is equal
to h psi, now for a free particle.
For a free particle one dimensional free particle
this is equal to minus h cross square by 2
m, delta 2 psi by delta x square. So, what
is the most general solution of this equation
the most general solution the most general
solution of this equation is psi of x comma
t, is equal to I write down a factor 2 pi
h cross for the sake of convenience, and I
integrate the functions like this e to the
power of I by h cross p x minus p square by
2 m t dp.
So, this is the most general solution, now
the only unknown part what my objective is
to determine what is psi of x comma t, the
only unknown part is a of p, so this is from
minus infinity to plus infinity, so what I
say is this you tell me the form of wave function
at t equal to 0, so let us suppose psi of
x comma x at time t equal to 0, I know that.
Let us suppose I know that so that is equal
to psi of x, then this is equal to 1 over
2 pi h cross a of p e to the power of i by
h cross, p x into dp. So, this is now a Fourier
transform relation this limit is from minus
infinity to plus infinity, so then a of p
is given by 1 over under root of 2 pi h cross,
this we have derived and we have psi of x
e to the power of minus i by h cross p x dx,
minus infinity to plus infinity.
So, you tell me the wave function at time
t equal to 0, then using that I will find
out a of p, once I have found out a of p i
will substitute it here and calculate psi
of x of t, this is how, so what is psi of
x comma t, it describes the electron, or the
proton, or the neutron it describes how it
evolves with time, and then we showed that
a particle is described by a localized Gaussian
wave packet, and it propagates with a certain
velocity which is known as the group velocity
of the wave packet.
It propagates as such and in the process there
is also a broadening of the wave packet, so
we have given you a complete solution of the
Schrodinger equation for the free particle.
What we will do next time is consider the
more solutions of the Schrodinger equation
and proceed from there. Thank you.
