To calculate the remaining metrics elements, in particular, the metrics
element the first metrics element that we saw in the previous slide.
So let me recall a distinction between the [inaudible] vector of psi used in the
direct notations, and the wave function, psi of x.
As it appears, let's say, in the Schrodinger equation.
So the former really represents an abstract state of the system which is sort
of unrelated to our choice of describing it.
So, it sort of doesn't rely on the particular choice of the coordinates.
Well psi of x is a specific what we call representation, which is much like
coordinates of a vector. So, the relation between the two is the
full length. So it's basically matrix elements of psi
of x is a matrix element between x and psi.
You know, and both are abstract vectors and some Hilbert space.
Now a particular example of this wave function wave for example is this plane
wave which just grabs the particle. Moving with a certain momentum.
I don't want to discuss too much the choice of the coefficient at this stage,
we're going to discuss it later in the course.
In any case, we definitely have seen this function before and we know it describes a
particle with a given momentum. Now the question that you're really asking
now in relation with this matrix element It was a, what is, what is the wave
function of a particle with the given coordinate, let's say x pri.
So, in order to answer this question we need to remind ourselves to introduce for
those of who[UNKNOWN] know the notion of the Dirac delta-function which basically
is exactly this. Matrix elements we're after.
So in order to introduce it, let me recall a very few, very simple facts from the
usual linear algebra. So let's say if we have a finite
dimensional vector space with some cartesian coordinate system and basis
vectors, let's say e1 E2 well in principle it's every 3 or 4 so the Cartesian
Coordinate system implies that we, the product, the dot product of two basis
vectors, e sub i and e sub g is equal to zero if they are different.
Which means they are orthogonal to each other or is equal to 1 if this is the same
vector. Which means that the, normal trajectory is
equal to 1. And, so the, symbol, which sort of
summarizes, this, relation, is the chronicer delta symbol, delta IG, which is
defined as this. Of course if we're going to calculate the
sum, over all let's say i's, of delta IG, we're going to have the sum equal to one,
just because all of them are equal all, all these terms are equal to zero apart
from the one we, where I is equal to G. Now the difference between the sort of
simplistic picture of the usual basis vectors and the usual linear vector space,
and the basis vectors in quantum mechanics in, i-, i-, is that the basis we're
dealing now is sort of continual base. So the basis vectors are really sort of
some [unknown] vectors which are labeled by coordinates, and there are infinitely
many coordinates and there's sort of a continuum space of this coordinates.
But if we sort of use the analogy between, let's say e i, and x, and e g, and x
prime, so this product is much like a [unknown] product of these two vectors.
So using this analogy we can demand, therefore that if x is not equal to x
prime, this measured [unknown] and therefore, this delta function, which is
sort of a continuum version of this delta symbol, are equal to zero.
And if x is equal to x prime and something else and so there's something else that is
determined by a condition which is very similar to this condition but now instead
of summing over a discreet set of vectors by i, we have to integrate over a
continuum set of vectors labeled by x. And so these are all possible coordinates
in one dimensional space going from minus infinity to plus infinity.
And the corresponding the corresponding delta function, the integral of the
corresponding delta function is equal to one.
And so therefore, if we try to visualize this delta function, it's something which
is equal to zero everywhere. So this is our delta of x, and this is x.
So we have something which is equal to zero everywhere apart from just one single
point where this function is so large, it has such a huge high peak that the
integral collected from this single peak gives us a finite result.
And well, this sort of implies that the value of this delta function at x equals
x-prime is, is infinity. And these, conditions are the canonical
conditions that determine the Dirac delta function.
Now, this Dirac delta function has many, properties that can be discussed.
So, I don't have time for that now but let me just mention one important property,
which is essentially the Fourier transform or the delta function.
So we, we, know that any reasonable function and, in this case, actually,
unreasonable function too can be expanded in terms of these sort of, plain wave like
functions. And this expansion is called the Fourier
transform so if you the Fourier transform, the DIrac delta function.
It will look like that so essentially an integral from minus infinity to plus
infinity of dk over 2 pi either for ikx. So these are well, things that we need now
to calculate the remaining matrix elements.
So, one of them in some sense have already calculate.
So let me just put everything together and clean up this space a little bit.
So, one matrix element that we can immediately write down a by analogy with
this relation we just arrived is the matrix element between the neighboring
points x sub k and x sub k plus 1, which can be written just in one to one
correspondence with this result. We just replaced x with xk plus 1 and x
prime with x sub k and we get this delta function of the difference which we'll
also call later on delta x. Now, in a very similar way.
We can now calculate, also, the, matrix element of the, potential, v of x.
So since v has the potential depends on the, on the coordinate.
And this vector, this state vector is the eigenstate of the coordinate.
Means that v of x, when acting on the x sub key just gives us.
The number v of x of k which can be factored out and the remaining matrix
element we just calculate it so the result is simply this.
So a slightly less trivial matrix element appears from the kinetic energy where we
have to calculate the matrix element of the p squared over 2m and So the x of k is
not and eigenfunction of momentum p. So in order to calculate it, do we can use
again the resolution of the identity that we discussed before, but now formulated
for the momentum rather than for coordinates.
So this guy is equal to 1. And we can insert this resolution of the
identity in between this operator p squared over 2m and x sub k, and if we do
so we get this integral. So here are placed the dummy variable of
integration. Instead of writing it as p here, I just
write it as p sub k, and, but it's the same resolution of the identity.
And so here now, so the, momentum operator, this kinetic energy, does act on
its eigenstate. So therefore this becomes simply P
sub-case squared over 2m, which is a number that can be factored out.
And there are many matrix elements this guy and this guy has to be compared, has
to be compared with this expression with the plain wave.
So these are simply plain waves. And so if we put everything together for
this room, we get the following results with the matrix element of the kinetic
energy. So which involves an integral o over a
momentum. In this case p of k is the momentum.
So in order to write all measuring supplements in a similar fashion we can
use this now this relation for the delta function which with the fourier transform
put the delta function and just replace key with a p over e bar.
So if we do so, we get essentially the same integral but now, integrate it over a
momentum and the exponential here will, will i over each bar, p sub p times x.
So but it's essentially the same the same expression for the delta function and we
can use it here to write the matrix element in terms of this in terms of this
integral. So you see that all three metrics elements
that appear in the in our previous calculation can be written in a very
similar fashion. Now, the good news is that we have reached
our final slide and we are now in the position to put everything together and
well, get to the main result of Feynman. We have done in the previous slide, we
have calculated the, the matrix elements that we need to, to describe the
propagator and so these are these matrix elements.
So we write them sort of in the uniform fashion.
So the, there is this one coming from his term and there is a kinetic and potential
energy from the cumulitonian. Now why, let's recall why do we even
bother you know to think about this [unknown] element.
So, where it comes from is really the this sort of picture of the propagators.
So remember we want to find the propagator from an initial point xi or 0 to the final
point xf and a time t. And so what we did, we split the time
interval from 0 to t into n a very, very small time intervals delta t and in doing
so we have represented our initial propagature as a product essentially of
this small propagature. Going let's say from this point to x1,
from x1 to x2 etc., etc. And so basically to calculate this, this
many matrix elements you know to calculate the 0 without the delta t, we needed to
calculate this guy and so we have the answer.
So this is the expression. And now we can just use it.
So notice the good thing about this expression.
Well, it looks a little bit cumbersome, but the good thing about this expression
is that it doesn't have any operators any more.
The operators are gone. There are no hats here.
There are only numbers, the regular numbers we're used to.
So another thing we should, we can notice is that this time interval, delta t, is
very small, because we try and keep our Interval from 0 to 10 to n pieces and n
goes to infinity, so this guy is really, really small.
At this point what we can recall is the[UNKNOWN] Taylor expansion for an
exponential of some small number. So if we have a small number, let's say
epsilon much smaller that 1. So the exponential can be written as 1
plus epsilon, we can neglect the other terms.
So what Feynman does at this, at this stage, he notices that, well this term
here is a small, much like, the epsilon here.
And he uses this, Taylor Expansion sort of backwards.
So instead of writing the exponential as a liner.
Well, 1 plus a linear term. He write this 1 minus a linear term, is an
exponential. So, and, This is the, result that he has.
So the reason he does that is because now the integral that we get is very
convenient. Because this is really the Gaussian
integral that we know how to calculate so you see we have this, we have an integral
over momentum, we have a programmatic term and a linear term and this is something we
have seen before. So if we do the calculation of the
integral so what we'll going to get is this.
So just, you know, the result of exponential m over 2 delta x squared over
delta t squared. But now let's look again at this picture
which shows the trajectory. So you see these are these trajectories
that, various trajectories that go from the initial point to the final point.
And so here, this term sort of corresponds to a single segment, let's say this guy.
Where a delta t is this small dimension. And delta x is really the change in the
coordinate. Well, basically, you may think about the
distance that the particle propagates in this time delta t.
So delta x or delta t simply, can interpret as simply the velocity of the
particle. So therefore, if we write it, so we, we
can see that this matrix element is nothing but either the power I over H, MV
squared over two, so basically the kinetic energy minus the potential energy times
Delta T. So and this is something which has a name
in classical physics, so maybe some of you know it, and this Name is called the
Lagrangian. Let me call it L sub k.
And so in these notations we can write the result for the matrix element of U of
delta T is simply E to the power of I over H bar.
L sub k times delta t. So using this result we can now rewrite
our main object this propagature as follows.
So we're going to have a bunch on integrals over x1, this point, this point,
this point. Uih, and of this exponentials of this
Lagrangians. First point it was at the first segment,
second segment thir segment etcetera. So notice that eseentially these integrals
over X1, X2 etcetera sort of correspond to moving aroudn these poitns tehse
intermediate points between Xi and Xf. The endpoints.
Are completely fixed. But the intermediates intermediate points
are free. And these integrals essentially symbolize
this freedom. So now the last step here in the
continuation is you have to recognize that, so here the product of exponential,
it can be written of the exponential of the sum, i over h bar.
L sub k delta t. And as delta t the interval, time interval
goes to zero, we can simply write it as e to the power i over h bar, an integral
well from 0 to t l d t. And the integral of the Lagrangien over
time is called is known as classical action.
So what we got is essentially the main result that the propogator can be written
as an integral over all possible trajectories, all possible intermediate
coordinates of the weighted with this exponential of the classic election.
So this is the result that we sort of advertised in the very beginning.
And Feynman introduced essentially a symbol, a notation, instead of writing a
bunch of these Intermediate step, so if you can introduce a single symbol which is
this capital curly d of x of t which is called the path integral.
So, it's just a symbol. But it turns out that it's actually a very
convenient symbol. So now our derivation is over so this is
again, this is the main result, so maybe let me just write it.
Sort of underline it again, so this is our main result.
And so, I know it's been a very long derivation, but as we'll see in the next
lecture, so this result is extremely useful in that, it will allow us to get,
well from now on, without extraordinary calculations, extremely profound and
interesting physical facts. So we're going to discuss the
correspondence between classical and quantum mechanics.
We're going to discuss so called quantum localization, we're going to discuss a
very interesting process of quantum dissipation.
So there are a lot of cool things that can be studied now with this object.
But well, it did take us a while to derive it, so hope some of you got the main
points of this derivation.