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Landauer began tackling the problem discussed in section 5.1 as early as 1957. By now his method is considered the most widely used, if not the most powerful. Concerns have been raised as to the application of this method to conduction of electrons suffering inelastic collisions, but we will not deal with such issues since the continuous shrinking of devices has made such events rare. If one could summarize Landauer’s work in one sentence, one would say that if electrons are treated as waves then the conductive properties of a system are essentially its wave transmission properties. We will first tackle a strictly 1-dimensional problem to show this, and then we will generalize to 3-dimensional systems where conduction occurs still along 1 dimension but the device is a 3-dimensional object with 1 or 2 of its dimensions in the nanometric range.

Landauer envisioned a channel connecting two thermodynamic reservoirs which are kept at a constant potential (not potential energy) difference V, see figure 5.6. Then for the Fermi level difference of the reservoirs (which represent the leads to the device) we have eV = AEF. For the leads or electrodes to behave as thermodynamic reservoirs, they must be large enough

Conduction as envisaged by Landauer

FIGURE 5.6 Conduction as envisaged by Landauer: electrons flow along a channel between two thermodynamic reservoirs separated by an energy eV, where V is the applied voltage.

to keep their Fermi levels constant while electrons are ejected from the left reservoir and thrown to the right reservoir and vice versa after being accelerated through the channel.

We first consider a very simplified model in which the channel can be thought of as a wire of negligible radius or thickness. The net current will be the difference of the current from the left reservoir traveling to the right minus the opposite one, see figure 5.6. Every electron is considered as a wave of the form e'kx traveling either from left to right with a positive wavenumber к or from right to left with a negative k. The current carried by one electron can be calculated by multiplying the current density / = nev by an infinitesimal cross section A, where n=the electron density of a single electron is f/AL with/the Fermi- Dirac distribution. Then this product must be summed over all electrons, i.e. over all occupied states. We have

Assuming that all the electrons emerging from the reservoir reach the other end

where v(A) is the velocity of the electrons in the channel and fr are the equilibrium Ferni-Dirac distributions in the left and right reservoir respectively and the factor 2 comes from the spin of the electron. Note that the second term should actually have /с < 0 in the summation, but we have taken care of this by the minus in front of the summation. Turning the sums into integrals (i.e. multiplying by (L/2n) we get

As usual we can assume parabolic bands E(k) so that

Using the above relation and changing the variable of integration from A: to E we get

Hence assuming conduction at the Fermi level and T —» 0 the current is (changing Ti to h)

which becomes

Note the change from Й to h in going from 5.21 to 5.22 to transform to the original result. Equation 5.23 means that the conductance of the wire G defined by I = GV is

Had we not assumed that all electrons emerging from the reservoirs reach the opposite reservoir a factor T(£) equal to the fraction of electrons transmitted through the channel, would have appeared in the equations leading to 5.24 and we would then get

on the assumption that again conduction occurs very near Ep (as we have seen in the previous chapter) and that T(£) can be represented by an average, equal to T(Ep). This assumption will be lifted immediately below and a more rigorous proof will be given. A further assumption is that the transmission coefficient from left to right T~*(E) and the transmission coefficient from right to left £*“(£) are equal. We will not prove this. Note that we did not have to assume anything about the Fermi level in the channel, which in any case is not properly defined. Only the Fermi levels of the reservoirs enter the derivation which are properly defined. A final point, that we wish to make, is that due to the low dimensional character of the system the conductivity is not properly defined and we can only talk about the conductance of the system.

We will now deal with the realistic case, where the conductor - channel is a three dimensional object that has width and depth see figure 5.7a. To simplify matters, we will use Cartesian geometry and coordinates. At thermodynamic equilibrium there is a common Fermi level for both reservoirs and the channel. When a positive voltage difference V is applied between the two leads - reservoirs the Fermi level of the right reservoir is lowered by eV and a net flow of electrons from the left one to the right one is established. Let this voltage difference V produce a potential energy distribution U(x,y,z) which can be separated in parallel (along the channel) and perpendicular components, see figure 5.7b. Note that since we will need both the terms potential

Same as 5.6 but in three dimensions; (a) the reservoirs are the leads and the channel is the device, and (b) the variation of potential energy along the channel and in the reservoirs-leads

FIGURE 5.7 Same as 5.6 but in three dimensions; (a) the reservoirs are the leads and the channel is the device, and (b) the variation of potential energy along the channel and in the reservoirs-leads.

energy and voltage or potential, we assign now the symbol U to the former and V to the latter.

Then the wavefunction can be put into a product form in accordance with the discussion in section 5.2

The wavefunctions appearing in equation 5.27 are, of course, the envelop functions. As we have described in section 5.2b, the eigenvalues of this low-dimensional system can be written as E(kx,l,m), see equation 5.16.

Let fi and /r be the number of electrons of a given spin in the left and right reservoir respectively in a given state E(kx,l,m). Then fi and /r are given by the usual Fermi-Dirac function. We assume that the Fermi level of the left reservoir Ep is unaltered and the Fermi level of the right reservoir is lowered by eV. Furthermore the corresponding densities are 2 fi/Vol and 2/r/Vo/ where Vol is the volume of our channel.

Again as in the simplified 1-dimensional case, the current (not current density) is the difference of currents ejected from the left and right reservoirs or algebraically

Let Ep denote the common unperturbed Fermi level then the Fermi level of the left reservoir is EF and of the right reservoir Ep - eV. We have for 1щ, following the steps of the 1-dimensional case,

and for I pi similarly

In both 5.29 and 5.30 we have included a factor of 2 for spin. As in the elementary case treated above we have divided by L only (instead of Vol) because the cross section A has been absorbed in the current I. Note also that the transmission coefficient depends only on the energy of a particle normal to the barrier which in this case is Ex. Furthermore kx> 0 in 5.29 picks up only the electrons travelling from left to right and kx < 0 in 5.30 picks up only the electrons travelling from right to left.

The total current is according to 5.28

Note that in 5.31 we have reversed the sign of the inequality kx < Oso as to have v(kx) as a common factor. This amounts to having a minus sign in 5.28 with both currents being positive.

We can now change the order of the summations in 5.31 and perform the sum over l, m first, taking v(kx) and T(EX) as common factors since they do not depend on (l,m). Then we will get factors in 5.31 of the form

and of the form

Flence equation 5.31 for the current becomes

As a final step we can turn the summation over kx into an integration. Remembering to

,. ., - 2k divide by — we 8et

To reach equation 5.33 we have changed the integration from kx to Ex- Note that we get the 1-dimensional result if we substitute Ь = h/(27t).

This is the celebrated Landauer’s formula in a more general form. A close inspection of the required steps for its derivation show that it is not necessary that the eigenstates denoted by the indices (l,m) are localized in the 2-dimensions perpendicular to the direction of propagation as we initially assumed. They can also be extended states in they and z directions. Of course the value of Nr and Nl will depend critically on such an issue. In the next part, devices, we will see that the parts of the transistors which correspond to the reservoirs of our abstract so far theory, the source and the drain, always have 1 dimension (the width) large enough to be considered macroscopic.

Taking the limit T —> 0, i.e. working at very low temperatures, we get immediately Ohm’s law. To see this observe that the functions Nr, N/.become step functions for metals with the discontinuity occurring at the corresponding Fermi level of each reservoir, pi = Ep for the left one and p2 = Ep-eV for the right one. Denoting by ANrl the difference appearing in

5.33 and using a first order Taylor expansion for this difference as a function of V we have

The minus in the second step of equation 5.34 arises because the NL and NR are functions of Ex - p. Remembering that the derivative of a step function is a delta function and using

5.34 in 5.33 we get

A more general expression than 5.35 can be obtained if we assume that T(EX) is not independent of (l,m), i.e. of the transverse quantum states from which the electrons originate. Then we should write T;>m(Ex) instead of simply T(EX). Note that the transmission coefficient is still a function of the energy normal to the barrier which is Er Then starting from equation 5.31, instead of 5.33, and using equation 5.34 we get

Hence the conductance - which is defined by I = GV - is

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