[Slide 1] Welcome back to the first unit of our course,
and we are now on the fourth lecture.
[Slide 2] You see what we did in the last lecture is we obtained an
expression for the current.
See this current depends on f1 minus f2,
and that is a point I explained physically earlier,
that current flows only because the 2 contacts have 2 different
Fermi functions.
And one wants to fill it up.
The other wants to empty it.
And the current that flows depends on the difference.
And then there is this factor here, this conductance,
which tells you how easily electrons can flow
through the channel.
And that depends on the density of states and the time
that it takes for an electron to get from left to right.
Now what we'll do in this lecture is we'll obtain an
expression for the conductance.
That is this current divided by voltage.
What we'll show is that for low voltages,
you can do a little approximate expansion here
and get this expression for conductance
where this overall conductance is
like this conductance function, but averaged
over energy according to this relation.
See? Notice what enters this relation is the derivative
of the Fermi function.
Now what we had talked about earlier, if you remember,
probably in the second lecture, is this Fermi function.
And Fermi function, as you know, is like
there is this electrochemical potential,
and above the electrochemical potential, it goes to zero.
Below the electrochemical potential, it goes to 1.
Okay? So that's this Fermi function.
Now what about the derivative of this?
That's this df/dE.
Well the derivative looks something like this.
Right around zero, the derivative is the highest that's
where it's changing the most.
That's this peak value.
Okay? And that peak value,
I guess what you have plotted here is 4kT times the
del f del E. So the peak value
of this derivative is actually 1 over 4 k. Okay?
And as you go away from the electrochemical potential,
it gradually goes to zero.
And the range over which it goes to zero is roughly 4kT.
In fact what I'll -- what will show later that the area
under this curve is actually 1.
So roughly speaking, you can think of it as if the peak is
1 over 4kT, the width is 4kT and so the area is about 1.
Okay? Now so how do I get from this current expression
to this conductance expression?
[Slide 3] So what we do is we start from this current expression here,
and we'll have this f1 minus f2.
And f is the Fermi function.
So what's the difference between f1 and f2?
Well, basically f1 is the Fermi function in contact 1,
which depends on mu 1, because the Fermi function,
as you can see, depends on mu.
So we could think of f1 as like the Fermi function with E
and comma mu 1, as if it's the function
of 2 variables, E and mu 1.
And the f2 you could think of as E comma mu 2.
Now -- now what you can do is the Taylor expansion,
that is you could write the difference between these two,
assuming that mu1 and mu2 are not too different.
You can write this difference as the derivative partial of f
with respect to mu, evaluated at the average value.
mu zero is that average of mu1 and mu2,
so at that value whatever it is times mu1 minus mu2.
Now what take a little thinking, and I'll explain
that in a minute, is that partial of f with respect
to mu is actually equal to the partial of f
with respect to energy.
So instead of this, the del f del mu,
you can write minus del f, del E.
So let me explain why.
So the way to think about this is when you think
of this f you could write it in the form 1 divided by 1 plus E
to the power of x, where x is E minus mu over kT.
Now so what is partial of f with respect to mu?
Well, you could use this chain rule.
That's equal to df/dx times partial of x with respect to mu.
Similarly when you look at partial of f with respect
with E, again the chain rule, the first step is df/dx,
but then it's partial of x with respect of E.
So when you compare these two things,
you can see that they both have df/dx in the beginning.
What makes them different is the second term.
So what's this second term?
Well, that's x. So when you look at partial of x with respect
to mu, what you get is minus 1 over kT.
Whereas when you look at partial of x with respect to energy,
what you get is plus 1 over kT.
So basically then these two partial derivatives are just
negative of each other.
And so instead of del f del mu,
you can write minus del f del E. Okay?
So with this result, let's carry on to the next slide.
[Slide 4] We have written f1 minus f2 as the partial of f
with the respect energy times mu1 minus mu2.
And then we note that the difference
between the 2 electrochemical potentials,
that's the applied voltage, qV.
So we put that in.
So we now use this result and go back to our original one here.
So instead of f1 minus f2, we put in this partial times qV.
And the qV is a constant.
It's independent of energy, so it can be pulled out of that.
And so you could write, you see the q, I guess cancels this q,
and so current divided by voltage,
you could write as this quantity.
So that's this conductance formula
that we were talking about.
That current divide by voltage,
so this is the general current formula, and what we did
in getting from here to here is use the Taylor series
expansion idea.
That f1 minus f2 can be written
as del f del E times the applied voltage,
and that applies only
if the applied voltage is small compared to kT.
So using that idea, you get this expression for the conductance.
And what it tells you is that the measured conductance,
this current over voltage, is like an average
of this conductance function in the sense that you multiplied
by del f del E, this quantity which I had shown before,
which has a peak right around E equal to mu, and then dies out.
So when you find the observed conductance,
it is largely determined by the value
of the conductance at E equal to mu.
In fact at zero temperature, it is essentially just the --
this conductance at that energy, whatever that is,
because at zero temperature,
this function is very strongly peaked,
in the since that the peak value is 1 over kT.
Width is kT.
So as T tends to zero, it becomes infinitely tall
and rather thin, like a delta function.
See? In fact what you can show is that the area
under this curve is 1, so that those of you who are familiar
with delta functions will recognize
that this quantity del f del E,
as you let the temperature turn to zero,
it actually becomes a delta function.
You see? So the area under this curve,
the way you can show it's 1 is take del f del E
and integrate over energy.
So this is a derivative you're integrating,
so basically you get f zero, and you put in the two limits,
minus infinity and plus infinity.
And there's a minus sign there,
which is why I have reversed the -- I have put the plus infinity
over there and the minus infinity over here,
and the Fermi function at minus infinity is 1,
at plus infinity is zero and so that's 1.
So the area under that curve is 1.
So the point to note then is that the measured conductance is
like an average of this conductance function.
So when you're thinking about it, we'll often not carry
around this whole entire integral.
We'll just work with the conductance function
at a particular energy because understanding is,
once you understand it at a particular energy,
if it's zero temperature, that's the answer anyway.
What you want is the conductance at the Fermi energy.
On the other end, if you're at nonzero temperatures,
then you have to average over energy.
[Slide 5] Okay? So with that understanding then, we are now ready
to move on to the next step.
That is to let us take this conductance function
and applied it apply it to ballistic conductors.
Thank you.