>> [Slide 1] Welcome back to Unit 2 of our course.
And this is Lecture 5.
[Slide 2] Now if you will remember,
what we did in Lecture 4 is we took this function N of p
from Lecture 3, you know, which came from this idea
of counting states using the periodic boundary conditions.
So that's this N of p, and what we did was combined it
with the energy momentum relation
to get a density of states.
And what we want to do in this lecture is kind of do the same,
but now we want the number of modes,
which is a slightly different quantity.
And if we remember, the way it's defined is this number
of modes is related to the ballistics conductance,
which depends on density
of states times the average velocity.
And this average velocity involves a numerical factor
that usually depends on this number of dimensions
that we're involved in.
So this in 1-D that average, just have this one here,
2 over pi, and this 1/2.
So this number of modes then is given by this --
this expression with three different constants,
depending on 1-D, 2-D, or 3-D.
Now we already have an expression
for the density of states.
So in principle, all we need to do is multiply it
by the expression for velocity, and we'd be done.
And we would have an expression for M of E.
But instead I'm going to do this in a slightly different way
because it also introduces a very useful relation,
that I want to introduce,
and that relation is a very general relation,
independent of what energy momentum relation we have.
You could have any relation here.
It doesn't matter.
Density of states times velocity times momentum is equal
to N. N is that quantity.
As you know, these functions, they are functions of momentum,
or you could view them as functions of energy
if there's a given energy momentum relation.
So density of states time velocity times momentum is equal
to this N times the number of dimensions.
So let me show you where this comes from,
because this is a very useful relation independent
of the specific energy momentum relation.
And we will use this then to transfer to obtain an expression
for this number of modes.
[Slide 3] Now what we do is we start from this rule we discussed
in Lecture 3 using the periodic boundary conditions,
and remember this doesn't involve any energy
momentum relation.
It just is based on this idea
of discretizing momentum using the requirement that the number
of wavelengths fit in -- fits into your box.
Now we take that quantity and take its derivative, so dN/dp.
So when I take the derivative, what do we get?
Well, these are all constants.
They just stay whatever they were, 1-D, 2-D,
and 3-D, makes no difference.
But because this is p over h to the power of d,
the h to the power d stays as is,
but the p to the power d gives you d times p
to the power d minus 1.
Let's just -- this rule of differentiation,
p to the power d we're differentiating with respect
to p. Now the next point to note is dN/dp you could write
as dN/dE times dE/dp.
And we could multiply both sides with p. Now let's look
at what we have on the left hand side.
We have p. We have dN/dE, which is density of states.
And then we have dE/dp, which is the velocity.
So on the left hand side we have density of states, velocity,
and p. What do we have on the right?
Well, we have the number of dimensions d,
and then if you notice, whatever we have here once we multiply
by p, this p to the power d minus 1 times p,
becomes p to the power d,
and so we basically get back what we started from,
namely the N, and so it's just d times n.
So this is the basic relation that I was talking about.
It's independent of energy momentum relation.
We have made no use of the specific relation at all.
We have said dE/dp velocity,
but we didn't assume any particular form for it.
So now let's move on to what you want to do,
namely use this relation.
[Slide 4] Use this relation to obtain an expression
for the number of modes.
Now the expression that we had before is what I have
stated here.
Modes is proportionate to density
of states times velocity.
Now instead of Dv, I can write Nd divided p.
So I could do the following.
I have this h over 2L right where it is,
and then Dv becomes N times d over p. Next, what I'll do is
for n, I'll use this original expression from here.
So you see we had this p over h to the power d.
And here I have an h over p. So when I multiply it, it becomes p
over h to the power d minus 1.
The 2Ls right there, we have p over h
to the power of d minus 1.
And then I have these constants in here, the 1-D,
2-D, and 3-D constants.
How do I get that?
Well, I took the 1 from here, multiplied it by 2L
because N has this 2L in it.
So 1 times 2L gives me 2L.
Next, I have this 2 over pi and pi WL,
so when I multiply, I get 2WL.
Well, why do I have 4 here?
Well, that's because it's also multiplied
by the number of dimensions here.
So here, this is two dimensions.
So it's 4WL.
And similarly, 1/2 times that is 2 pi over 3AL.
But then it has to be multiplied by the number of dimension.
And so you get 2pi AL.
So finally, you see you have this expression for M.
And you can divide out the 2L all the way
to get the expression for M.
And you'll notice how similar this expression is compared
to the expression for N,
and I'll have more to say about this.
N has this p over h to the power d. M is p over h
to the power d minus 1.
Here, you have 2L.
Here, you have 2W.
So it is as if with modes, one dimension is just 1 here.
I mean in one dimension, d is 1.
So this is 1.
So basically number of modes is 1.
But two dimensions, it's kind
of like the one dimensional result from here.
Instead of 2L, it's 2W.
And for three dimensions, a cross section is,
of course, two dimensional.
And so you kind of have the two dimensional result from here,
pi A. So just notice the similarity.
Whatever is in 1-D here kind of appears in 2-D.
Whatever was 2-D here appears in 3-D, and so on.
So we'll talk more about that.
[Slide 5] So what we have done then is we took this expression
from Lecture 3.
Then this general identity that we talked about
and took this expression for modes from Unit 1 and used
that identity to get an expression for modes
that looks a lot like the expression for N.
And so far we have just done some algebra.
But this is kind of important
because it gives you a totally different physical
interpretation of what M represents.
And for this purpose, what I am going
to do is take this expression and just rewrite it
in a slightly different way.
If d equals 1, then as I said, that amounts to 1.
If d equals 2, then this is like p over h times 2W,
which you could write as W divided by h over 2p.
And similarly, the last one, you could write as A divided by h
over 2p squared, and then a constant in front.
Now what do we gain by doing that?
Well, remember, h over p is what we call the de
Broglie wavelength.
So what this quantity tells you is how many 1/2 de Broglie
wavelengths fit into the width of the conductor.
That is, if you have a two dimensional conductor, length L,
width W, then the number
of modes is basically how many 1/2 wavelengths fit
into that width.
Now if you had a three dimensional conductor,
so then the cross-sectional area is A, then this tells you
like how many 1/2 wavelengths fits in this way
and how many 1/2 wavelengths fits in that way.
And it's a product of the two.
That's what it is.
Area divided by 1/2 wavelength squared.
So this is this idea.
Those of you who are familiar
with electromagnetic wave guides will also recognize
that that's kind of what determines the number
of transverse molds in an electromagnetic wave guide.
I mean that's where the word modes comes from.
But the point I wanted to make is this very important
distinction, this -- of these two ways of looking
at what number of modes means.
The one view is the view we had from Unit 1.
That is, it's like density of states times velocity.
Density of states tells you how many states you have.
Velocity tells you how fast they move, and so the product kind
of determines how much current you can get.
So that's why modes depends on density
of states times velocity.
But a totally different way of thinking of modes is you look
at the cross-section, see how many 1/2 wavelengths fits
into it.
And this is the view that most people are --
who have some familiarity with this would have heard of.
This one is less widely appreciated.
And what I was trying to show is if you make use
of that identity, then you can transform this into this,
which represents two totally different physical views
of what number of modes means.
And incidentally, I should point out that just we're --
the way I've taken M and written it as number of 1/2 wavelengths
that fits into the cross-section,
you could similarly take this function, N, and write it
in a form that looks like how many 1/2 wavelengths fits
into the volume.
That is, if you have a one dimensional solid,
it's like the length divided by 1/2 wavelengths.
Otherwise, it's the area divided by 1/2 wavelength square
or the volume divided by 1/2 wavelength cubed.
So N kind of tells you how many 1/2 wavelengths fit
into the entire volume of the solid,
whereas M tells you how many 1/2 wavelengths fit
into the cross-sectional area of the solid.
[Slide 6] Now, one important consequence
of this is this ballistic resistance
that we talked about in Unit 1.
That is, I had mentioned at that time that if we just look
at this expression, it will tell you that the number
of modes is proportional to the width
or the cross-sectional area.
What that would mean is you would expect
that the ballistic conductance would change linearly with width
or cross-sectional area.
And this is observed.
This was observed back around 1970 or so.
And sometimes people call it the Sharvin resistance.
But the very important development
that came some twenty years later was that in devices
with small cross-section where the number
of modes was relatively small, not thousands but more like tens
and twenties, that's
where people saw this very nice quantization
of the ballistic conductance.
And rather than go linearly, it actually went in steps.
Now how do you understand this?
Well, the idea is that actually M is not quite that quantity,
but rather the integer part of that.
In other words, if this happens to be 2.5,
then it's actually really 2 because the number
of modes is given by, again, how many states are available
in that cross-section, which can only be 2 or 3 or 4.
It cannot quite be 2.5.
So the correct expression is it's the integer part of this.
Normally, you don't worry about it because if it's a big number,
like 10,030, it really doesn't matter if it's 10,031.
And in any case, of course,
if these steps are very close together, you see,
in other words, you always have to average over energy
because one point I mentioned before, although I --
we are not writing it explicitly,
is that whatever we get as a function of energy has
to be averaged with this dF/dE factor.
That is averaged over an energy of kT.
And that tends to, I guess, smooth things out when you're
at higher temperatures.
But at low temperatures especially with conductors
with small cross-section,
this is now a very well-established phenomenon.
It was first seen in semiconductors,
like, I believe it was gallium arsenide.
Since then, people have seen it in all kinds of materials,
including even a small hydrogen molecule.
And this idea that while it is quantized,
you can understand by this view
that M is actually the integer value,
integer part of that quantity.
[Slide 7] So that kind of completes what we were trying to do here,
namely introduce this energy momentum relation,
then introduce the rule for counting states,
and use that to get density of states and density
of -- number of modes.
Next in the next two lectures, what we want
to do is talk a little bit about the electron density and try
to relate our conductivity to the electron density.
That's what we'll be doing next.
Thank you.