In this video, I'm going to complete the
most technically demanding part of the
derivation of a path integral of our
presentation for the propagator that was
introduced in the previous segment.
And here again I show the expression for
this propagator that we derived.
So in, in the following I'm going to be
doing my calculations mostly in one
dimension for the sake of simplicity.
There is no real reason to go to one
dimension so the calculation sort of goes
the same way in two dimensions or three
dimensions or 100 dimensions, it's just
going to one dimensions is just simplifies
notation so it'll be, we don't have to
deal with vector indices.
And also it will allow me to illustrate
certain steps of our derivation in a sort
of an intuitive vectorial way.
So, the second comment I would like to
make is that what we call the propagator
also appears in the literature under the
name of transition amplitude, well, from
an initial point xi to final point xf, in
this case, and a very closely related
object is also called the Green function.
So if you see these expressions in the
literature, they may be actually very much
related to one another, and all refer to
this object [unknown].
Now, the main difficulty in calculating
this object once again, is the presence of
this evolution operator, which is an
exponential of [unknown].
And to calculate this exponential of an
operator is a very tricky business.
So basically the goal of the remaining
calculation, as we discussed, is going to
be to simplify this expression so that it
doesn't really contain any operators.
And we deal with sort of regular
mathematical quantities, not operators.
And in doing this simplification we're
going to rely on two very useful
mathematical, or physical actually, in
this case, two formulas.
So, one of them, the first one listed
here, we already have actually discussed
in the second lecture last week.
So, this is the so called resolution of
the identity operator.
And the fact that it is the identity
operator means that so let's see if we
have an arbitrary wave-function of psi,
and we have to by this operator on this
wave-function of psi.
It is psi, so what we're going to get, is
the full length, x of psi, this matrix
elements, times the [unknown] x, dx.
And this is identically equal to the psi
itself.
So this is the formula number 1.
Another equation which is going to be very
useful, actually, it's a property of the
evolution operator.
It actually relies on the very simple
observation.
So, the evolution operator itself
translates the wave-function from the
initial moment of time.
So, this is time and this is, let's say,
time equals 0.
So, to the wave-function psi of t, at
time, at the final moment of time t equals
t.
So but instead of going directly from 0 to
t, we can do that in sort of two steps.
We first can go to, let's say, certain t1,
which is between 0 and t.
And then go from t1 to t.
So and you know, going from t1 to t at the
response here in the language of this
resolution operator is applying this
operator with a h, which depends on t
minus t1 to the wave-function that we have
here in the middle.
So, on the other hand, the wave-function
in the middle on the pdi of t1 itself can
be represented as the action of the
resolution operator U of t1 on the
original wave-function of psi of 0.
So, therefore what we derive here is that
the evolution operator U of t can be
written as the product of two evolution
operators U of t minus t1 times U of t1.
Now so the particularly simple way of
splitting the time interval into two,
let's, in this case, is just dividing it
in half.
So, let's say again, so we have this time
axis of 0 t, and we just divide it into
two pieces, so from 0 to t2 and t2 to t.
And so therefore, we have the expression
for the evolution operator U of t, is
equal to U of t over 2 times U of t over
2.
So now if we're going to use the well, if
we're going to real, if we're going to
focus on the actual propagator, which is
the matrix element of this evolution
operator between the initial point xi and
the final point xf, what we can do, we
can, well we can write it as xf U of t
over 2, U of t over 2, xi and insert here
the resolution of the identity that we
discussed on the previous slide.
So, if we do so, we get to this
expression, which essentially splits the
original propagator into two propagators.
One goes from xi to x, and the other goes
from point x to the final point x sub f.
This splitting of the regional propagator
into two allows a very useful and
intuitive, illustration that I'm going to
present now which will actually bring us
closer to the notion of a path integral.
So, let's say this is my coordinate x and
this is time t.
So and this is my 0 and final time t.
This is going to be my xi and xf.
And so, what I'm interested in is
calculating the propagator of going from
the initial moment of time, xi, to this xf
at t equals t.
So, this is essentially what I'm
calculating, and this sort of solid line
connecting the two points sort of
corresponds to my G of xi, xf, t.
Now, this equation tells me that I can
take, well, the point in the middle.
So, this is going to be my t over 2 and
what I have to do in order to calculate
this propagator, I have to consider all
possible coordinates in between.
So, let's say, I will go from xi to x and
from x to xf.
This is going to be one path.
Another path is going to be to go to a
different x, like as so, etc., etc.
So, this solid lines are going to be sort
of representing you know, the first and
the second propagator in this interval.
And well, obviously this procedure of
splitting the original time, time interval
and position interval into two can be
continued.
So, what I can do, I can keep on
truncating my, my time interval into
smaller and smaller pieces, and what I'm,
what I will continue doing so I will have
more and more intermediate coordinates
that are going to appear, and therefore, I
will generate all possible trajectories
that are going to cover the entire space.
And this is how path integral or integral
over trajectories really appears, but at
this, at the level of this illustration,
this path integral, this picture is not
very useful for sort of practical
purposes.
To make some progress here, so let me sort
of formalize what I just said and what I
said is simply the fact that, let me just
split, instead of splitting my time
interval into 2 pieces, let it, let's, let
me split it into n pieces and so an n here
is going to be very, very large.
So, that t over n, the each individual at
the time interval is extremely small.
And so, each of this small time interval
was the evolution operator on each of this
time intervals will response to this
exponential, either the bar i or H bar H t
over N.
So this t, t over n, I can make as small
as I want.
So, and if I make it really, really small,
so, instead of writing the full Taylor
series for this exponential, I can write
it as simply two terms, one minus the
first linear term.
And just, I try and take my [unknown]
right here and so basically, approximate
my evolution operator in each of this
dimension intervals as, as so.
So, as a matter of fact, you can say, that
you know, I shouldn't have done this
deviation.
So, because at the end of the day, what I
have actually derived is that either the
power minus i over h bar, h Hamiltonian
times t is equal to 1 minus I over H bar H
t over N to the power N, where N goes to
infinity.
And this is actually, and I'm tryin' to
give definition of exponential.
And this is absolutely correct, this is
all there is, so we're basically just
writing instead of writing the complicated
nonlinear function exponential of an
operator that's a complicated function,
we're just going to write it as a product
of function which are linear in our
[inaudible] doing it.
And as we, as we do so in between each of
these terms in this product, I'm going to
insert my favorite resolution of the
identity.
So, this dx, x, x.
And well, since I have many x's now, so
I'll have to label them differently, so
there's going to be x1, x2, x3, which will
sort of correspond to this intermediate
coordinates that I have in, in my
projections.
Now, if we put everything together, we get
this complicated-looking expression for
the propagator, which is obtained by,
well, sort of sandwich, sandwiching this
product of evolution operators between xi
and xf.
And here, I'm also going to be using the
notation x0 is the initial one and x sub N
is going to be the final one.
So, this will allow me to write the, this
propagator in a more compact way.
And also as I mentioned so in between of
each of these sort of small evolution
operators for a small time, t over N or
delta t.
So, I insert this resolution identity but
I have to sep distinguish different
instances where I have to do so and
therefore, I have so many different
integrals.
Now the reason again on doing this is
because instead of dealing with this
complicated exponential, I now can deal
with this linear matrix elements of, of
the Hamiltonian, which can be calculated
in a rather straightforward way.
So just to summarize what, what remains to
be calculated if you just look at this
expression is, well simply the matrix
element between x of k plus 1 and x of k
and the matrix element of the Hamiltonian.
So, all these guys are, you know, i, H
bar, m and tr are numbers, so they don't
really act on this on this function.
And the Hamiltonian, itself, of course is,
is just basically the operator [unknown]
wanting to the energy [unknown] because
the kinetic energy part and the potential
energy part.
And so, apart from this matrix element, I
have to calculate two other matrix
elements, which essentially are the last
pieces of the puzzle that requires to be
completed in order to get this [unknown].
And so, and the remaining goal of this
video, I'm going to be calculating
basically these guys.