V - Dipartimento di Matematica e Fisica

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Bloch Electrons and Bloch waves
Bloch theorem, its consequences,
density of states, and van Hove
singularities
Electron in a Periodic Potential: Band Theory
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The ions in a perfect crystal are arranged in a regular array, then we are led to
consider the problem of an electron in a periodic potential U(r):
U(r)= U(r+R)
R are the Bravais lattice vectors
Origin of U(r): U(r)= Vion+Velmean
⇒indipendent electron approximation (electron-electron interaction is not included)
mean field
e+
+
+
+
To calculate Velmean is complicated, we just assume that U(r) has the same periodicy of
Vion .
Schrödinger Equation
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The Schrödinger equation for a single electron is:
− h2 2
r  r
r
U
(
r
)
ψ
(
r
)
=
E
ψ
(
r
)
∇
+


2
m


(free electron is a special case with U(r)=0)
Indipendent electrons, each of which obeys a one-electron Schrödinger equation
with a periodic potential, are known as Bloch electrons.
The stationary states of Bloch electrons have the following very important property
as a general consequence of periodicity of the potential U:
the eigenstates of one-electron Hamiltonian with a periodic potential can be chosen
rr
r
r
ψ n kr (r ) = exp(ik r )u nkr (r )
where
r
r
r
u nkr (r ) = u nkr (r + R)
Bloch’s Theorem
for all R in the Bravais lattice
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Bloch’s Theorem
Note the implications:
Bloch’s Theorem
ψ
r
nk
rr
r
r
(r ) = exp(ik r )u nkr (r )
⇒ψ
r
nk
r r r
r
r r
r
(r + R) = exp(ik (r + R)u nkr (r + R)
rr
rr
r
r
= exp(ik R) ⋅ exp(ik r )u nk (r )
rr
r
r
= exp(ik R) ⋅ψ n k (r )
The points r and r+R thus have the same physical properties, the functions differing only
by a phase factor indipendent of r.
rr
r r
r
Consequently
r
r
ψ (r + R) = exp(ik R) ⋅ψ (r )
nk
nk
r r r
rr
rr
r
r
r
r
r
exp(ik (r + R))u nk (r + R) = exp(ik R) exp(ik r )u nk (r )
r r
r
r
then it results that u nk (r + R) = u nkr (r )
The Bloch Theorem is obviously true for an “empty lattice”
rr
r
r
r
U (r ) = 0 ⇒ ψ k (r ) = exp(ik r ) ⋅ 1
Bloch’s Theorem
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•Proof of Bloch’s theorem
For each Bravais vector we define a translation operator
r
r
r
TRv f (r ) = f (r + R)
1.
Since the H is periodic: TR commutes with H
r
r
r
r
r
r
r
r
r v r
v
TR HΨ (r ) = H (r + R)Ψ (r + R) = H (r )Ψ (r + R) = H (r )TR Ψ (r )
⇒ TRv H − HTRv = 0
2.
TR and TR’ commute (the applications of two traslations does not depend from the
order)
r r
r
r
r
TRv TRv ' Ψ (r ) = TRv 'TRv Ψ (r ) = Ψ (r + R + R' )
3.
TRv TRv ' = TRv 'TRv = TRv + Rv '
TR for all Bravais lattice vectors R and H form a set of commuting operators.
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Bloch’s Theorem
Therefore from a fundamental theorem of quantum mechanics we know that the eigenstates of H can be chosen to be simultaneous eigenstates of TR. Then Ψ(r) can be
chosen to satisfy simultaneous eigenstates of all the TR
HΨ = EΨ
r
r
TRr Ψ = c( R)Ψ for each R
Properties of the eigenvalues c(R) of the traslation operators:
r
r r
r
T T Ψ = c( R)TR ' Ψ = c( R)c( R' )Ψ
r r
TRr TRr ' Ψ = TR + R ' Ψ = c( R + R' )Ψ
r r
r r
⇒ c ( R + R ' ) = c ( R )c ( R ' )
r r
R' R
Let us write R with ai, three primitive vectors for the Bravais lattice. We can always write
the c(ai) in the form
r
c(ai ) = exp(2πixi )
for a suitable choice of the xi
r
r
r
r
r
r
r
⇒ c( R) = c(n1 a1 + n2 a 2 + n3 a3 ) = c(n1 a1 )c(n2 a 2 )c(n3 a3 )
r
r
r
= c(a1 ) n1 c(a 2 ) n2 c(a3 ) n3
This is equivalent to:
Bloch’s Theorem
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rr
r
c( R ) = exp(ik R )
provided that
r
r
r
r
k = x1b1 + x 2 b2 + x3 b3
rr
bi a j = 2πδ i , j
where, with the last conditions, the bj are the reciprocal lattice vectors.
Summarizing, we have shown that we can choose the eigenstates Ψ of H so that for
every Bravais lattice vector R,
rr
r
r
r
r
r
r
T R Ψ ( r ) = Ψ ( r + R ) = c ( R ) Ψ ( r ) = exp( i k R ) Ψ ( r )
This is the Bloch’s Theorem
Born-von Karman Boundary Condition
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Which are the possible values of k?
By imposing an appropriate boundary conditions on the wave functions we can
demostrate the k must be real and arrive at a condition restricting the allowed values of k.
Born-von Karman boundary conditions simulate an infinite solid with a finite sample
containing N=N1N2N3 unit cells.
r
r
Ψ(r + Niai ) = Ψ(r)
Using the Bloch’s Theorem:
i =1, 2,3
B.T.
BvK
r
r
r
r
r
r
Ψ (r + N i ai ) = exp(ik N i ai )Ψ (r ) = Ψ (r )
r r
⇒ exp(iN i k ⋅ ai ) = 1 i = 1,2,3
r
If, again, k = ∑ xi bi the above eq. becomes
i
exp(2πiN i xi ) = 1 or
N i xi = mi with mi integral
r
mi r
Therefore k = ∑ bi
i Ni
Born-von Karman Boundary Condition
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The volume ∆k of k-space per allowed value of k
r
r
r
r b1  b2 b3  1 r r r
 = b1 ⋅ (b2 × b3 )
∆k =
⋅
×

N 1  N 2 N 3  N
r r r
r
volume of primitive cell in reciprocal lattice
b1 ⋅ (b2 × b3 ) = N∆k
The above equation asserts that the number of allowed wave vectors in a primitive
cell of the reciprocal lattice is equal to the number of sites in the crystal.
Volume of primitive cell in the reciprocal lattice is (2π)3/ vcell
where vcell=Vcrystal/N
Therefore
r 1 (2π )3 (2π )3
∆k =
=
N v cell
Vcrystal
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Bloch wave’s properties
Bloch’s waves are labelled by quantum numbers:
k (wave vector) ⇒ reflects the traslational invariance of the Hamiltonian
n (band-index) ⇒ energy levels at given k are discrete
i)
ii)
ħk generalizes p (characteristic of the traslation symmetry in the case of a
periodic potential )
k can always be confined in the 1th B.Z. (this because any k’ not in the 1th B. Z.
can be written as k’=k+G if G belongs to the reciprocal lattice ).
r r
r r r
r r
r
r
r
r r
Ψ (r + R) = exp(ik '⋅R )Ψ (r ) = exp(i (k + G ) ⋅ R )Ψ (r ) = exp(ik ⋅ R)Ψ (r )
iii)
the index n appears in the B. T. because for for given k there are many solutions
to the Schrödinger Eq.
r
r
Ψ
Ψ
=
H
(
r
)
E
(
r
)
Hermitian eigenvalue problem
r r r
r
within a fixed volume -> discrete
Ψ (r ) = exp(ik ⋅ r )u (r )
set of solutions (like particle-in-a-box
r
r  h2
r  r
r
problem)
− i∇ + k + U ( r )  u k ( r ) = E k u k ( r )
H k u k (r ) 

(
2m
)

r r
r
with the boundary condition u k (r + R) = u k (r )
Bloch wave’s properties
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iv) k is a continuous parameter in the Schrödinger equation (but allowed values are
given by Born v Karman boundary condition).
The functions En(k) are called “bands”. They are periodic in the reciprocal space:
r r
r
E n (k + G ) = E n (k )
=> they have mimima and maxima
Bloch wave’s properties
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v) The 1th B. Z. is a 3D space. Bands are often plot along lines of particular symmetry,
from k=0 to specific boundaries or along boundaries. Specific directions and points take
specific names taken from group theory.
vi) All functions with the periodicity of the lattice (including uk(r), U(r)) have a Fourier
trasform which contains only the G’s:
r r
r
f (r ) = ∑
c kr exp(ik ⋅ r )
r
k
r r r
r r
f (r + R) = ∑
c kr exp(ik ⋅ (r + R))
r
k
r r
r
r r
r
r
if f (r ) = f (r + R) ⇒ f (r ) = ∑
cG exp(iG ⋅ r )
v
G
vii) Bloch’s waves are stationary states (time-indipendent) of the Schrödinger equation
with a periodic potential. Therefore, the current is not degraded by the fixed lattice ions
(fixed ions are not source of scattering), in contrast to earlier theories of metals (Drude).
Those theories, however, work well provided scattering is interpreted as due not to ions,
but to deviation from perfect periodicity (=> mean free path >>a).
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Density of States
One must often calculate quantities which are an average over the allowed states of the
crystal
r
Q = 2∑r Qn k
if Qn does not depend on spin
()
n ,k
In the limit of a large crystal the allowed values of k get very close together, and the sum
over k may be replaced with an integral (∆K=(2π)3/V):
r
r
Q
dk
q = lim v − >∞ = 2∑ ∫ Qn k
V
(2π )3
n
Qn(k) depends on n and K only through the energy En(k) then in analogy to the free
electron case one can define a density of levels per unit volume g(E) so that q:
()
q = ∫ dEg ( E )Q ( E )
Comparing the two espressions we find that
r
r
dk
g ( E ) = ∑ g n ( E ) with g n ( E ) = ∫ 3 δ ( E − E n (k ))
4π
n
Density of states
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Density of States
the number of allowed wave − vectors

g n (E )dE = (2 / V )× in the n th band , in the energy range
 from E to E + dE

The number of wave-vectors is just the volume of a k-space region with
E<En(k)<E+dE, divided by the volume per allowed wave-vector, ∆k=(2π)3/V
()
r
r
dk 1, if E ≤ En k ≤ E + dE
g n (E )dE = ∫ 3 × 
4π 0, otherwise
This defines two surfaces
Sn(E+dE) and Sn(E) in the k space
and therefore:
Sn(E+dE)
()
r
dS
g n (E )dE = ∫
δk k
3
4π
Sn (E )
Sn(E)
But:
δk(k)
() ()
r
r
E + dE = E + ∇En k δk k
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Density of States
Therefore, by considering the surface integral, we have:
g(E) = ∑
n
dS
1
r = ∑ g n (E )
3
∫
n
∇E n k
S n ( E ) 4π
()
Sn(E) is the surface of constant energy E.
⇒Hence g(E) can be completely determined by the knowledge of En(k).
Since En(k) is periodic there must be at least a maximum and a minimum
⇒∇k En(k)=0 in a finite number di k-points. These points lead to divergencies of the
integrand. In 3D such divergencies are integrable but lead to discontinuities in the slope
of g(E).
The discontinuities are defined van Hove singularities, and can be classified according
to the analytical expression for E(k) in the 3D reciprocal space.
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Singularities
()
( )
r
r
E n k = E n k 0 + ∑ α iα j (k i − k i 0 )(k j − k j 0 )
i, j
Depending on the signs of the α and k0 can be a maximum, a minimum or a saddle point:
M0
 A E < E 0
g n (E) = 
M
 A + B(E − E 0 )1 / 2 E > E 0 1
 A − B(E 0 − E )1 / 2
g n (E) = 
 A
E > E0
E < E0
M2
 A E < E 0
g n (E) = 
M3
1/ 2
 A − B(E − E 0 ) E > E 0
 A + B(E 0 − E )1 / 2
g n (E) = 
 A
E > E0
E < E0
max
E0
saddle point
E0
saddle point
E0
min
E0