Brillouin partial filling

Wilson ((>85) has pointed out that if a Brillouin zone is full, the electrons occupying the states of this zone can make no contribution to the electric current. This fact follows from the definition of the zone as a region enclosing all reduced wave vectors. Imagine all electrons of Figure 6 shifted Akx by an external field. The electrons in states within Akx of the zone boundary are reflected to the opposite zone boundary, so that the zone remains filled and there is no transfer of charge. This observation permits a sharp distinction between metallic conductors, semiconductors, and insulators. Because of the high density of states in a band, a crystal with partially filled bands is a metallic conductor. If all occupied zones (or bands) are filled, the crystal is a semiconductor if Ef kT, is an insulator if kT. 

The d-like band is partially filled up to 5/6. The -ir-like band is occupied up to 1/3. The Peierls instability in the quasi one-dimensional ir-like band causes a superstructure with the period of 3c. This reduces the first Brillouin zone to 1/3... 

Figure 8.01. Occupied states and band structures giving (a) an insulator, (b) a metal or semimetal because of band overlap, and (c) a metal because of partially filled band. In (b) the overlap need not occur along the same directions in the Brillouin zone. If the overlap is small, with relatively few states involved, it is called a semimetal.

This is certainly one of the most important design considerations. Partial filling of the Brillouin zone is necessary for metallic conduction. In fact, even in the early TCNQ research at duPont in the 1960 s, higher conductivities were achieved from complex TCNQ salts than with simple salts [88-90] (Table 3). The last two entries in Table 3 are really semiconductors with an activation energy for conduction, but the room-temperature resistivities are quite low. It is clear, however, that partial CT is essential it is also clear that mixed valence (in the sense of the Robin-Day [91] class IIIB) must also be achieved, i.e. one cannot have discrete valences at discrete sites (as, e.g. in CS2TCNQ3, which is a complex salt, but has low conductivity because it has a ISCA3D lattice). 

The problem, then, is to calculate an accurate average energy for each of the filled bands, the average being over the first Brillouin zone, and then multiply each such average energy by the number of electrons in that band, per unit cell. The sum of these is the Hiickel total energy per unit cell. (This procedure assumes that the bands are not partially filled.)... 

Fig. 17.1S A possible Fermi surface obtained by partially filling the band structure indicated in Fig. 17.13. Because the example is two-dimensional the Fermi surface is here a line. For simplicity the zone edge which runs from the x to the y axis has been shown as a straight line. Really, it should be more faceted, as a cross-section of any of the Brillouin zones shown in Fig. 17.11 demonstrates.

It has been more recently proposed, on the basis of band structure calculations including HOMO bands, that both the LUMO and the HOMO bands of the Ni(dmit)2 acceptor molecules are partially filled and also that the multi-Fermi surface of the TTF[Ni-(dmit)2]2 system exhibits some two-dimensional character at the F point of the Brillouin zone [19]... 

Space) is divided as usual into Brillouin zones (BZ). The 1st BZ extends from q = -7t Iau to +7TIau (Fig. 9.8a). When the amplitude V of the potential is weak and can be treated as a perturbation, the band splits at the boundary of the 1st BZ. The band gap is then 2 V. When the one-dimensional system has two electrons per unit cell, then n = Ijau- Then, kp = 7t jau, the band is thus filled and the onedimensional system is an insulator. When however n < Ifau, e.g. n = Ijau and therefore kp <7t ja, for the example kp = OSnIum, the system becomes metallic. This is the simplest picture of an intrinsic semiconductor or a metal. N < 2/um corresponds to a partial and n=llau full CT degree 8 per unit cell in an organic metal (CT complex, class 3), independently of how many molecules are in the unit cell. If it contains e.g. two acceptors and one donor (as in (DCNQI)jCu ), or two donors and one acceptor (as in (Fa)2 PF ), then for n = l/aM the CT degree of the... 

The bivalent metals possess two valency electrons for each atom, and since these should fill the lowest zone entirely, the conductivity might have been expected to show a very sharp fall. While it is quite true that group 1 contains the best conductors known, the power of other metals to carry current is also quite considerable. What has to be postulated, and what can be supported by semi-quantitative calculations, is that with certain atoms the three-dimensional Brillouin zones for the p electrons partially overlap with those of the s electrons, so that there can be a response to an accelerating field by a passage from the one tsrpe of zone to the other. The... 

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