固体物理第9章

a
2 a
2nd BZ
2nd BZ
In the reduced zone scheme, one can find different energies at same value of wavevector. Each different energy characterizes a different band. Two wavefunctions at same k but different energies will be independent of each other: the wavefunctions will up of bemade different combination of the plane waves exp[i(k G) r ].
Periodic zone scheme
We can repeat a given Brillouin zone periodically through all wavevector space.
We can translate not only a band from other zones into the first zone, but also a band in first zone into every other zones in the periodic zone scheme.
Electron orbits, hole orbits , and open orbits
k G k
Here k+G is understood to refer to the same energy band as k.
Extended zone scheme

Extended zone scheme

Reducheme
The free electron Fermi surface at 2nd and 3rd zones.
The near free electron Fermi surface at 2nd and 3rd zones.
Holelike Electronlike
Harrison construction of free electron Fermi surface 1) Plot the reciprocal lattice 2) Plot the free-electron Fermi spheres around each lattice point with the radius appropriate to the electron concentration. Any point in k space that lies within at least one sphere corresponds to an occupied state in 1st zone. Points within at least two spheres correspond to occupied states in 2nd zone, and similarly for points in three spheres in 3rd zone, and so on.
2a 2d 2a 2b 3b 2c 3b 3a
2b 2c
3a
2d
1st zone
2nd zone
3rd zone
The free electron Fermi surface in the extended zone scheme.
The free electron Fermi surface in the reduced zone scheme.
The electrical properties of the metal are determined by the shape of the Fermi surface, because the current is due to charges in the occupancy of states near the Fermi surface.
1st Brillouin zone 2nd Brillouin zone 3rd Brillouin zone 4th Brillouin zone
The free electron Fermi gas in a square lattice Mapping of the 1st, 2nd and 3rd Brillouin zone in the reduced zone scheme.
Alkali metals have weak interactions between the conduction electrons and the lattice. Because the alkali metals have only one valence electron per atom, the first Brillouin zone boundaries are distant from the approximately spherical Fermi surface that filled one-half of the volume of the zone. The divalent metals, e.g. Be and Mg, also have weak lattice interactions and nearly spherical Fermi surface. But because they have two valence electrons each, the Fermi surface encloses twice the volume in k space as for the alkalis, i.e. the volume enclosed by the Fermi surface is exactly equal to that of the a zone. But the Fermi surface is nearly spherical which extends out of the first zone and into the second zone.
Reduced zone scheme
It is always possible to select the wavevector index k of any Bloch function to lie within the first Brillouin zone. Prove: suppose k (r ) is a Bloch function, and k k 'G . ik ' r ik r iG r k ' (r ) e uk ' (r ) e e u k ' (r )
Few people would define a metal as ‘a solid with a Fermi surface.’ This may nevertheless be the most meaning definition of a metal one can give today; it represents a profound advance in the understanding of why metals behaves as they do.
Then
ik r k ' ( r ) u ' k ( r )e k ( r )
We can always mapping the band in the reduced zone scheme.

C’
C
C”
A
2 a
A’ 0

a
k
1st BZ
1st zone
2nd zone
3rd zone
The parts of the free electron Fermi surface in the third zone appear disconnected as drawn in the reduced zone scheme. The free electron Fermi surface in the third zone as drawn in the periodic zone scheme.
The Fermi surface is the surface of constant energy F in k space. The Fermi surface separates the unfilled orbitals from the filled orbitals at absolute zero T = 0K.
Constant energy surface in the Brillouin zone of a simple square lattice with nearly free electron approximation. Constant energy surface is discontinuous at Brillouin zone boundary.
Chapter 9
Key points:
Fermi surfaces and metals
• reduced zone scheme and periodic zone scheme
• construction of Fermi surface • electron orbits, hole orbits, and open orbits • cohesive energy of metals
0
k
Construction of Fermi surface
The equation of the zone boundaries is 2kG + G2 = 0 and is satisfied if k terminates on the plane normal to G at the midpoint of G.