We continue our discussion of the free particle
problem. And then after that, we will discuss
an eigenvalue problem in quantum mechanics
and that is the particle in a one dimensional
box.
We will first discuss the interference experiment.
In our last lecture, we considered the single
slit diffraction experiment.
Now in that experiment, what we did was, that
we assumed that there was a slit of width
b and so the electron was described by a wave
function, which was constant here in this
region, this is my y axis and 0 everywhere
else. So, therefore, we assumed psi b y as
equal to 1 over root b for mod y less than
b by 2 and 0 everywhere else.
If I assume such a wave function, then the
corresponding fourier transform that is a
of p y will be equal to 1 over under root
of 2 pi h cross from minus infinity to plus
infinity psi b y e to the power of i p y y
by h cross d y and this we had integrated;
and we had shown that, this was equal to some
constant times sin beta by beta, where beta
was equal to pi b sin theta by lambda.
And so a p y square d p, a p y mod square
d p y, this is the probability for the electron
or the proton to have the y component of the
momentum between p y and p y plus d p y, that
was would be proportional to sin square beta
by beta square. And this as we all know is
the single slit diffraction pattern. So, therefore,
by confining the electron or the proton to
pass through a slit of width b, this slit
itself imparts a momentum in the y direction;
and the probability distribution of that momentum
is given by this expression.
So, therefore, where will an individual electron
or proton land up on the screen I do not know,
but I can tell I can predict a probability
distribution. And therefore, if we were to
conduct this experiment with a very large
number of electrons or protons or neutrons,
we will roughly know the diffraction pattern.
And from this equation it is obvious that,
smaller the width greater will be the momentum
that is in the y direction; and greater will
be the diffraction, so this is my uncertainty
principle that, delta y is of the order of
b.
So, therefore, you have we have the uncertainty
principle that, delta p y y delta y is of
the order of say h cross or h it does not
matter, it is an order of magnitude relation.
So, delta y is of the order of b. So, delta
p y is of the order of h by b, now before
it entered the slit the electron was coming
in this direction, so the y component of the
momentum was 0. And so, therefore, this slit
imparts a momentum in the y direction, since
the average momentum in the y direction was
0, so p y is of the order of delta p y; so,
this is of the order of h by b from the uncertainty
relation.
So, what is p y as I mentioned that, if this
is the particles momentum and if this is the
angle of diffraction, p y will be equal to
p sin theta. So, this will be p sin theta
will be of the order of h by b. So, sin theta
will be of the order of h by p into b h by
p is the de broglie wavelength, so lambda
by b.
So, smaller the value of b greater will be
diffraction and so, therefore, that is consistent,
smaller the value of b greater will be the
momentum imparted in the y direction so, all
our results are consistent with the uncertainty
relation; smaller the value of delta y greater
will be the value of delta p y and greater
will be the diffraction. And once again the
on the screen you will observe this intensity
distribution something like sin square beta
by beta square, you have a probability distribution,
where it will land up I cannot say I can only
predict a probability distribution. So, physics
has ceased to be deterministic you can only
predict the probability of the arrival of
the electron or the proton in a certain region
of space.
So, let me consider another simple problem
related problem that instead of a single slit
I have two slits. So, let us suppose each
slit is of width b and then this is d by 2
and this is minus d by 2 and the width is
width is b. So, I am sorry. So, the distance
between the slit is d let me let me do this
again let me do this again I have two slits,
one here and one here, the distance between
the two slits is d, so this distance is minus
d by this is d by 2, this is d by 2 and this
is also d by 2.
So, this is my y axis and if I assume the
origin to be here, so for the first slit the
integral will be from minus d by 2 minus b
by 2 to minus d by 2 plus b by 2; and for
this slit, it will be plus d by 2 minus b
by 2 and then d by 2 plus b by 2; and we have
this integral e to the power of i by h cross
p y y and e to the power of i by h cross p
y y multiplied by d y multiplied by d y.
If I carry out this integration and do a little
bit of simplification then you will obtain
sin beta by beta multiplied by cos gamma,
where this is the square of this sin square
beta by beta square is the diffraction pattern
and cos square gamma it is a very simple integration
just very similar to what we had done for
this single slit diffraction pattern.
So, this term is proportional to is this describe
this single slit diffraction pattern and this
is the two point interference pattern. So,
the probability distribution on this will
be something like this, this is the two slit
interference pattern that you will observe,
where will be electron land up you cannot
predict you can predict a probability distribution.
And so, therefore, if the experiment was carried
out with a large number of electrons you will
observe the two slit interference pattern.
So, let me go back to the slides.
So, I first consider the single slit diffraction
pattern and psi of y was non 0 from y equal
to minus b by 2 to plus b by 2, where its
value is 1 over root b. So, I integrate this
and which I did that in detail.
And I obtain sin beta by beta, where beta
is equal to p y b by 2 h cross, h cross is
equal to h by pi, so you will obtain pi b
sin theta by lambda. So, this results therefore,
the probability of the momentum lying between
p y and p y plus d p y will be equal to a
p y square d p y, which is equal to sin square
beta by beta square.
So, this is the diffraction of the electron
and this is the probability distribution.
And this is the angular broadening of a neutron.
So, as I mentioned that, if whether you have
any neutron or proton or electrons, it will
exhibit the same diffraction pattern. And
each particle has a very well as I had mentioned
in the very first lecture has a very well
defined charge, very well defined magnetic
moment, very well defined mass.
So, on the back of your mind you think that,
it is a particle, but then each it it it it
demonstrates the give rise to the diffraction
pattern and I will just now I mention to the
interference pattern. So, therefore, this
is the diffraction pattern angular broadening
of a neutron beam by small slits, that smaller
the width of the slit greater will be the
diffraction pattern and that is consistent
with the uncertainty principle.
So then, we consider the double slit diffraction
pattern. So, you have each slit is of width
b and the distance between the center of the
slit here and the center of the slit here
is d. So, I assume psi b of y is equal to
1 over 2 b under root, which is equal to from
in this region and in this region. So then,
the corresponding momentum distribution function
will be the fourier transform of this. So,
1 over root 2 b I take outside, so this region
this integration is from in this region minus
d by 2 minus b by 2 to minus d by 2 to plus
d by 2 and then from here to here. It is a
very straightforward integration.
If you carry that out you will get and and
square that the probability of the y component
in the momentum lying between p y and p y
plus d p y is the product of the single slit
diffraction pattern and 2 point interference
pattern, where beta is equal to pi b sin theta
by lambda and this is the interference pattern
produced by 2 point sources separated by a
distance d.
So, therefore, the electron or the proton
or the neutron is described by a wave function
which is here and as well as here, it passes
through both the slits simultaneously does
it go through hole number 1 or hole number
2 it goes through both the holes and it interferes
with itself. So, it is described by a wave
function which is present here and here and
it produces an interference pattern a probability
distribution, which looks like this. And it
is this probability distribution that the
quantum that quantum mechanics predicts.
So, here is one of the very beautiful experiment
that I had discussed some time back that,
if you do the double slit interference pattern
in a experiment with 10 electron then, where
will the individual electron will land up?
You cannot predict, you can only predict a
probability distribution.
And, so they will have appear at random on
the screen. The last photograph corresponds
to 70000 electrons and then, slowly the interference
pattern the fringes appear. So, each as you
can see from each of the figure, the photograph
electron is detected only as a total electron
never half of an electron, so you detect either
1 full electron or nothing. So, each spot
here corresponds to the detection of a single
electron which has a certain amount of charge
certain amount of mass and so on. Where will
it land up is given by a probability distribution
and when they when you have a large number
of electrons, then slowly the interference
pattern.
So, in the classical interference experiment,
the intensity will become fainter and fainter,
but the interference pattern will appear in
the form of dark and bright fringes, but the
grain the graininess of the arrival of the
electron is demonstrated in this slide and
which is a very important aspect of quantum
mechanics, that you always detect the whole
electron or the whole proton or the whole
neutron never a fraction of that; but, there
is a probability where will it land up one
cannot predict, but one can predict accurately
the probability distribution. And therefore,
the electron or the proton or the neutron
is described.
So, what is an electron, what is a proton
I do not know, but I can only say that, it
is described by the wave function psi, so
that mod psi square d tau represents the probability
of the finding in the volume element d tau.
And what is psi? psi is the solution of the
schrodinger equation. So, the what is psi,
psi is the solution of the schrodinger equation
H psi. So, that completes the free particle
problem.
Now, in the free particle problem we had found
that, the energy takes the energy of the particle
takes a continuum of values from 0 to infinity
the x component of the momentum takes a continuum
of values from minus infinity to plus infinity;
and e is equal to p square by 2 m. So, p takes
all possible values from minus infinity to
plus infinity and the energy of the particle
can take any value between 0 and infinity.
Now, what I would like to do next is, consider
all the schrodinger equation when the particle
is confined in a box in a one dimensional
box; and we say that, it is in an infinitely
deep potential well.
So, I have an infinitely deep potential well.
So, which I described this that as if the
particle is inside in the very deep potential
well and let let me choose this as the point
origin x is equal to 0 and this as x is equal
to l and the particle is localized somewhere
here and between x is equal to 0 and x is
equal to l, the particle is free.
So that, V of x is equal to 0 for x less than
0 less than l, x lying between; and at the
boundary, there is an infinite jump in the
potential. And so, therefore, the electron
or the proton cannot go outside the well.
So, therefore, the wave function for all points
greater than l and less than 0 should be 0.
So, my boundary conditions are that the wave
function should be 0 at x is equal to 0 must
be 0, at x is equal to l must be 0, so these
are known as the boundary conditions of the
problem.
So, the particle I consider the particle in
a one dimensional box of length l. So, once
again I start with the time dependant one
dimensional schrodinger equation, i h cross
delta psi by delta t is equal to h psi where
h is equal to. So, let me write it again.
So, i h cross delta psi by delta t is equal
to H psi, where H is equal to. So, let me
write it again.
So, i h cross delta psi bt delta t is equal
to H psi of x t and H is equal to p square
by 2 m and that is equal to h cross square
by 2 m delta 2 psi by delta x square plus
the potential energy is assumed to be time
independent V of x psi. So, I want to solve
this equation I solved I have just completed
the solution of the equation, when V of x
is 0 everywhere that is known as the free
particle solution and I obtained the most
general solution describing a wave packet.
Now, whenever the potential energy function
is time independent I can write the solution
as psi of x comma t as equal to psi of x T
of t just as we did in the free particle case.
So, this term will become i h cross psi of
x into d T of d t, because this function depends,
this is known as the method of separation
of variables. And this term becomes psi is
psi of x T of t, so T of t I take outside
T of t and becomes minus h cross square by
2 m d 2 psi by d x square plus V of x into
psi of x.
Next the variables have still not separated
out, because the left hand side contains x
and time and the hand side time also contains
time and x. What I do is, I divide by psi
times T, so I get i h cross 1 over T of t
d T by d t is equal to 1 over psi minus h
cross square by 2 m d 2 psi by d x square
plus V of x psi of x is equal to. Now, the
variables have separated out, the left hand
side is a function of time only. The right
hand side is a function of space only.
So, therefore, both sides must be equal to
a constant, it cannot be a function of x,
it cannot be a function of time, it has to
be a constant because, the left hand side
is a function of time, the right hand side
is a function of x. Hence, both sides must
be equal to a constant. We say that, the method
of separation of variables has worked and
so, therefore, each side is set equal to a
constant. Now, the integration is trivial.
So, let me write down the the time dependant
part. So, I get i h cross T of t d T by d
t is equal to e, so this gives me that 1 over
T d T by d t is equal to e to the power of
sorry sorry i am sorry minus i by h cross
minus i h by h cross that is it. Therefore,
i E by h cross, where E is a constant. So,
therefore, the integral of this equation is
T of t, this is log of t, so if I integrate
this you get log of t is equal to minus i
E t by h cross plus a constant. So, T of t
is equal to constant times e to the power
of minus i E t by h cross. So, this is the
solution of the time independent part time
dependant part.
And the solution of the time space dependant
part will be this is very simple you can write
down if I take psi on this side, so you get
minus h cross square by 2 m d 2 psi by d x
square plus V of x psi of x is equal to E
psi of x. And therefore, if I take if I put
a plus sign here and a minus sign here and
minus sign here and bring this to this side,
so you get d 2 psi by d x square plus 2 m
by h cross square E minus V of x psi of x
is equal to 0, this is known as the time independent
schrodinger equation.
The time dependant part is so much. So, therefore,
the total solution, so we must solve this
equation for a given form of V of x we must
solve this equation and for the next 2 3 lectures
we will assume different forms of V of x and
we will solve this time independent schrodinger
equation and the time dependant part will
be given by this.
So, that the total solution the total solution
psi of x comma t will be equal to psi of x
e to the power of minus i E t by h cross.
First, we will find out psi of x and also
try to find out what are the allowed values
of V, allowed values of E and then, try to
obtain the general solution the most general
solution of the schrodinger equation.
So, till now our analysis is valid that is
we, what is psi of x? psi of x is satisfies
this equation, d 2 psi by d x square plus
2 let me replace m by mu, because for reasons
that will become clear in a moment. So, the
mass of a particle is represented by mu 2
mu h cross square E minus V of x psi of x
is equal to 0, this is a general equation.
The only assumption is that, the potential
energy function is independent of time.
So, let me solve the first example actually
this is the second example. The first example
was the free particle problem the first example
that we had solved was V of x equal to 0 everywhere,
this was the free particle problem that we
had solved in great detail. The second problem
that we are going to solve is sorry sorry,
so the second problem is that we will solve
is the particle in a one dimensional box.
So, we have we consider a box like this, which
is 0 here and l here and so the wave function
psi the wave function has to vanish here,
because the particle cannot escape. So, psi
at x is equal to 0 is equal to 0 and psi at
x is equal to l is also 0.
So, let me try to solve this equation between
0 less than x less than l, V of x is 0, so
the schrodinger equation becomes d 2 psi by
d x square plus k square psi is equal to 0,
where k square is defined to be equal to 2
mu E, mu is the mass of the particle, E is
the energy of the particle and h cross is
of course, the planck's constant divided
by 2 pi.
So, I write down this equation that in the
in the region I have this particle in a one
dimensional box particle in a one dimensional
box between 0 and l, the the schrodinger equation
becomes d 2 psi by d x square plus k square
psi of x is equal to 0, where k square is
equal to defined to be equal to when I write
3 signs here; that means, defined to be equal
to 2 mu by h cross square.
The solution of this equation is very simple
as you all know that, psi of x is equal to
A sin k x plus B cos k x. Now, the wave function
has to vanish at x is equal to 0, because
the particle cannot tunnel can cannot go outside
the well. So, psi of 0 at x is equal to 0,
sin of 0 is 0, cos of 0 is 1, so psi of 0
is 0 and that must be equal to B, so B is
0. So, the the quantity B is 0, so this solution
of my schrodinger equation is equal to A sin
k x.
Then I apply the second boundary condition,
that psi at x is equal to L is equal to 0.
So, B is already 0. So, this will be A sin
k L. Now it has two possible solution, either
A is 0 if A is 0 then, psi of x is 0 everywhere
that is known as the trivial solution, it
does not correspond to any physical situation
the wave function, if the wave function psi
of x is 0 everywhere any value of E is possible.
So, there is no particle anywhere. So, that
solution psi of x equal to 0 everywhere, this
is known as the trivial solution and which
we will neglect.
The other thing that can happen is sin k L
is equal to 0 that is k L is equal to n pi,
where n is equal to 1 2 3 4 I have excluded
the value n equal to 0, because if k is 0
then again then then if k is 0 then psi of
x is 0. So, that is again the trivial solution.
So, n equal to 0 corresponds to again to the
trivial solution. So, we obtain that, the
if the wave function has to vanish at x is
equal to 0 and at x is equal to L, then k
L must be equal to n pi.
So, we have found that for the wave function
to vanish at x is equal to 0 and at x is equal
to L, k L must be a multiple of pi, where
n is 1 2 3 and since E is equal to E is equal
to I am sorry since k square is equal to 2
mu E by h cross square. So, therefore, E is
equal to h cross square k square by 2 mu,
k is equal to n pi by L. So, this is equal
to h cross square n square pi square by 2
mu L square.
So, this becomes where n can take only discrete
values n is equal to 1 2 3. So, therefore,
the energy E of the particle can take only
discrete values and that is E can take only
a set of values E n, which is equal to n square
times this constant, pi square h cross square
by 2 mu L square. When n is 1, so this becomes
E 1 is equal to pi square h cross square by
2 mu L square. So, you will have E 2 is equal
to 4 times E 1, E 3 is 9 times E 1, E 4 is
equal to 16 times E 1 and so on. So, the particle
can take discrete energy levels, this is a
quite a difference from classical mechanics,
where a particle can take a continuum of energy
values. So, you have a particle in a box can
take only a set of discrete energy levels.
And we had shown that, the k L must be equal
to n pi, where n is 1 2 3 therefore, this
tells us that k also can take discrete it
has to take discrete values is equal to n
pi by L; so, these are the discrete values
the values that k can take. The wave function
psi of x we had shown that this was equal
to A sine k x.
So, this will be equal to A sine of n pi by
L into x. So, you will have you will have
let us suppose this is my box in which the
particle is confined and you have the ground
state, this is ground state which has the
value of E 1 and it has a wave function n
is 1, so this will be sine of pi by L into
x. So, at x is equal to 0, it becomes like
this and x become like this and beyond this
it is 0 and beyond this it is 0, this is known
as the ground state wave function.
The second state is 4 times E 1, so this will
be sine of 2 pi L by 2 pi by L x. So, at x
is equal to L by 2, it becomes 0, so it will
be something like this. And the third wave
function will be will be something. So, it
is the first wave function is symmetric about
the point x is equal to L by 2, the second
wave function is anti symmetric with respect
to at x is equal to L by 2, the third function
is symmetric; they will be since the potential
as I will tell you later that, since the potential
energy function is symmetric about the point
x is equal to L by 2 x is equal to L by 2.
So, therefore, therefore the wave functions
are either symmetric or anti symmetric with
respect to the point x is equal to L pi by
2.
So, my wave function is something like this.
So, I will have I will have the wave function,
which is psi of x I write this as, psi n of
x which is equal to A sin of n of n pi by
L into x between 0 less than equal to x less
than equal to L. Let us suppose, I normalize
this wave and 0 everywhere else, 0 for x less
than 0 and 0 for x greater than L.
So, you will have for example, for the second
state it will start from 0, this is the x
equal to 0 part and then you will have like
this, and then it will go like this, the first
wave function will be something like this.
So, this 0 and this is the point x is equal
to L, this is the point x is equal to 0 and
so on for the higher order wave functions
I will so, therefore, we have solved the problem
completely.
And, if I now normalize this wave function
that is minus infinity to plus infinity mod
psi square d x, if I put equal to 1 then,
the wave function is 0 for x greater than
L and less than 0. So, therefore, this will
be equal to A square from 0 to L sin square
n pi by L into x d x, you can easily integrate
this equation. And if you work this out I
leave this as a very small exercise, it come
out to be 2 by L. So, this is known as the
normalization constant.
And so, therefore, for the particle in a box
problem, psi of x is equal to under root of
2 by L sin of n pi x by L, actually you should
write this as you should write this as psi
of x is equal to psi n of x, these are the
discrete wave functions these are the discrete
wave functions corresponding to the particle
in a box problem for 0 less than x less than
L; and the corresponding eigen values are
E is equal to E n is equal to n square pi
square h cross square by 2 mu L square and
in both cases n takes the value 1 2 3 etcetera.
You must remember that, the wave function
was psi n of x the the total wave function
was equal to psi n of x e to the power of
minus i E n t by h cross. Now, the most general
solution of the time dependent schrodinger
equation will be a sum of this, will be the
linear combination C n, psi n of x e to the
power of, where E n is given by this, and
psi n is given by this.
This superposition of different states is
a characteristic of quantum mechanics particle
therefore, let us suppose you superpose only
two states. So, at t equal to 0 let us suppose
at t equal to 0 you can have 1 one over root
2 psi 1 x plus 1 over root 2 psi 2 of x that
means, there is a half probability of finding
it in the first state and the half probability
of finding in the second state you can also
have something like this 1 over 1 over root
3 psi 1 of x plus 1 over root 3 psi 4 of x
plus 1 over I have taken they need not be
psi 5 of x, they need not be all equal, mod
C 1 square as I will explain as I will try
to show tell you we will represent the probability
of finding the particle in the n-th state.
So, therefore, if you make a measurement will
you obtain E 1 or E 2, if this if the particle
is in this state, there is a half probability
of obtaining E 1 and half probability of obtaining
E 2. So, the energy is not known exactly,
it is in a superposed state. So, let me show
you a slide.
And so we consider actually this is example
number 2, particle in a one dimensional box
here, unfortunately instead of capital L I
have done it as small a just with me. So,
the potential energy function is 0 between
x equal to 0 and x is equal to a and it has
an infinite jump at x is equal to 0 and at
x is equal to a, so that the probability of
finding the particle outside the well is 0.
And so, therefore, the wave function must
vanish at x is equal to 0 and at x is equal
to a.
So, in between this region the solution the
schrodinger equation is given by d 2 psi by
d x square plus k square psi is equal to 0,
because V of x is 0, so k square is equal
to 2 mu E by h cross square, where mu is the
mass of the particle. So, the solution of
this equation is psi of x is equal to A sin
k x plus B cos k x. Now, I apply the condition
that psi of 0 is equal to 0.
So, if I if I write x is equal to 0 then,
that is 0, this quantity is 0, this is 1,
so 0 is equal to B. So, therefore, the this
this term goes out, because B is 0, so psi
of x is equal to A sin k x. For the wave function
to vanish at x is equal to a we will have
A sin k a equal to 0 and therefore, k a must
be equal to n pi, where n will take only the
values 1 2 3 4 and 5 and 6 and so on, but
not 0, because 0 will correspond to a trivial
solution.
So, if I square this you get k square a square
is equal to n square pi square. So, therefore,
E is equal to h cross square k square by 2
mu, so the discrete energy levels of the problem
will be equal to n square pi square h cross
square by 2 mu a square; and the corresponding
normalized wave functions are 2 by a sin n
pi sorry this should be x, n pi x by a, there
has to be an x here, sin n pi x by a. These
wave functions are orthogonal to each other
I am sorry there should be also a d x here.
So, the so they are normalized and they are
orthogonal to the any other function, where
delta mn is the kronecker delta function,
where delta mn is 0, when m is not equal to
n and is 1 when m is equal to n.
So, these are the wave functions, this is
the ground state energy for which you have
the wave function psi 1 of x, this is the
first excited state energy, this is the second
excited state and this is the third excited
state.
So, we had started off by solving this equation
this time dependant schrodinger equation and
for the particle in a box problem we wrote
down psi of x t is equal to psi of x e to
the power of minus i E t by h cross. We obtain
these are the eigen functions normalized eigen
functions, and these are the energy eigen
values. So, therefore, the most general solution
is is superposition of this and mod C n square
will represent the probability of finding
the particle in the n-th state.
So, let me let me show you the corresponding.
So, let me consider the infinitely deep potential
well problem and these are the eigen function.
So, this is the x is equal to 0 and this is
x is equal to L, in this particular program
I have taken m is equal to 1 8 3 6 times the
mass of the electron that means it is a proton
and A is 2 angstroms. So, this is the first
eigen value, so much e V. The second eigen
function I put here 2 it will have 1 0. If
I put 3 then, it will have 2 0, this is symmetric
about this about the point x is equal to A
by 2 and then you take 4 in each wave function
and this is anti symmetric.
So, about this, this is the mirror image,
you may see that all the wave functions go
to 0 at the boundaries, x is equal to 0 and
at x is equal to L and then similarly, for
for x is for 5 for the 5th it is they are
alternately symmetric and anti symmetric.
Now, I superpose time wave, so let us suppose
that the first let me let me take a simpler
thing. So, let me take C 1 is equal to 1 and
this I put 0 and C 2 is equal to 1 let me
ok that, the coefficients are not normalized
do you want the program to normalize them,
the wave function has to remain normalized,
so let say yes. So, I am superposing two wave
functions and then, how will it evolve with
time, this is the probability distribution
function as it evolves with time.
Let me again repeat this let me write down
C 1 is equal to 1 and C 2 is equal to 0. So,
that only the first state is getting excited.
How will it evolve with time? It does not
change with time and therefore, each eigen
state is known as a stationary state. If the
particle is in a particular eigen state then,
it will remain in that state for all times
to come.
Let me superpose first 4 states, so it says
C 1 is equal to 1, C 2 let me put C 1 is equal
to 0 and then C 3 equal to 1. So, I have excited
I am I am considering only the third state,
the particle is in only in the third state,
so ok it and then evolve. So, this is also
a stationary state, it does not change with
time. Finally, let me put just superpose the
third and the fourth state. So, let me make
this as 0.707 that is 1 over square root of
2 and this also as 1 over square root of 2
ok this and then evolve and this is the, this
is how the probability distribution is changing
with time.
So, initially it was like this and then it
evolves like this from certain time to certain
time. So, with today I have solved two problems
two very important problems in quantum mechanics;
the first problem is the free particle problem
in which V of x is 0 everywhere, the energy
forms a continuous values and so, therefore,
the general solution was an integral and we
studied the evolution of a wave packet.
The second problem that we studied just now
was the particle in a box problem, in which
the particle is confined within a box and
we obtain discrete energy states and then,
we studied the evolution of the wave function
with time.
