Download Electrons and Bandstructure

Electrons and Bandstructure
• Velocity of Electrons
– group velocity vg=dω/dk=1/h dE/dk
• Equation of Motion
– same as for free electrons: h
v=
1
!E
h
dk
=F
dt
– in an electric field E and a magnetic field B
h
dk
= !e( E + v " B)
dt
1
$
= !e % E + #E " B&'
h
1
Creation of Holes
Vacant orbitals in otherwise fully occupied bands are commonly treated as
holes. A hole acts in applied electric and magnetic fields as if it has a positive
charge of +e. The reason why is given in the next few slides.
Point 1: The wave-vector kh of the hole is -ke.
The total wavevector of the electrons in a filled band is zero: Σk=0. This result
follows from the geometrical symmetry of the Brillouin: every fundamental
lattice type has symmetry under the inversion operation r->-r about any lattice
point; it follows that the Brillouin zone of the lattice also has the inversion
symmetry. If the band is filled all pairs of orbitals k and -k are filled, and the
total wavevector is zero.
If an electron is missing from an orbital of wavevector ke, the total wavevector
of the system is -ke (due to the uncompensated orbital at G), and this
wavevector is attribution to (the creation of) a hole. This result is surprising:
the electron is missing from ke and the position of the hole is usually indicated
graphically as situated at ke, as in the slide. But the true wavevector kh of the
hole is -ke, which is wavevector of the point G if the missing electron was
taken from point E.
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Energy Bands for Holes
Point 2:
Eh (kh ) = ! Ee (k e )
Let the zero of energy of the valence band be at the top of the band. The lower
in the band the missing electron lies, the higher the energy of the system. The
energy of the hole is opposite in sign to the energy of the missing electron,
because it takes more work to remove an electron from a low orbital than from
a high orbital. Thus if the band is symmetric [as it invariably is], then
Ee (ke ) = Ee (! ke ) = !Eh (kh ) = !Eh (!k h )
The graph on the slide shows a band scheme to represent the properties of a
hole. This hole band is a helpful representation because it appears right side
up.
Point 3: The velocity of the hole vh is equal to the velocity of the missing
electron ve. We see this from the hole band picture where
!Eh (kh ) = !Ee (ke ) and so v h (k h ) = v e (ke )
3
Holes as Particles with Charge +e
• Since kh=-ke and vh=ve, we have
dke
= !e(E + v e " B)
dt
dk
h h = +e (E + vh " B)
dt
h
This charge of +e is very important, and explains the positive coefficient in
experiments which measure the Hall effect. In a Hall effect experiment where
current is flowing along a bar in the x-direction, a magnetic field in the ydirection generates a voltage in the z-direction [it is just about qv×B. For most
materials the Hall coefficient is negative (indicating the current is carried by
negative charges: ie electrons) but in some materials (eg. ) holes carry the
current. This is one of the major success of the band theory to be able to
explain this result.
With the application of the electric field E
in the positive x-direction, the electrons
increase their wavevector in the negative
x-direction (since F=-eE). This means the
vacant orbital also moves to the left, and
since kh=-kh this is equivalent to the hole
moving to the right.
4
Effective Mass
• The time-derivative of the group velocity leads to the
expression
'1
! 1 d 2 E $ dv
1
1 d2 E
F=# 2 2 &
and so
= 2
m * h dk 2
" h dk % dt
– Effective masses can be very large (tight-binding), very
small (near band edge), and even negative.
1 dE
vg =
h dk
dv g 1 d 2 E dk
=
dt
h dk 2 dt
1 d 2E F
dk
=
(since
h
= F)
h dk 2 h
dt
1 d 2E
= 2
2 F
h dk
tight-binding: low curvature=large effective mass (seen in the low probability
of electrons hopping between atoms)
Explain negative effective mass: consider (a beam of) incident electrons with
wavevector slightly less than the Bragg condition. The beam will propagate
through the crystal as in the left-hand figure. Now consider the application of a
voltage V which slightly increases the energy (and wavevector) of the incident
electrons. With the Bragg condition satisfied the electron has its linear
momentum reversed, and thus in the framework of -eE=F=ma, the electron has
behaving as though its mass was negative. Can’t just consider the external
force and the electron, must consider the role of the lattice as well.
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