1 lecture viii band theory dr hab. ewa popko. 2 band theory the calculation of the allowed electron...
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Lecture VIII
Band theory
dr hab. Ewa Popko
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Band TheoryThe calculation of the allowed electron states in a solid is referred to as band theory or band structure theory.
To obtain the full band structure, we need to solve Schrödinger’s equation for the full lattice potential. This cannot be done exactly and various approximation schemes are used. We will introduce two very different models, the nearly free electron and tight binding models.
We will continue to treat the electrons as independent, i.e. neglect the electron-electron interaction.
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Bound States in atoms
r4
qe = )r(V
o
2
Electrons in isolated atoms occupy discrete allowed energy levels E0, E1, E2 etc. .
The potential energy of an electron a distance r from a positively charge nucleus of charge q is
-8 -6 -4 -2 0 2 4 6 8-5
-4
-3
-2
-1
0
F6 F7 F8 F9
r
V(r)E2
E1
E0
r
0
Increasing Binding Energy
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Bound and “free” states in solids
-8 -6 -4 -2 0 2 4 6 8-5
-4
-3
-2
-1
0
F6 F7 F8 F9
r
-8 -6 -4 -2 0 2 4 6 8-5
-4
-3
-2
-1
0
F6 F7 F8 F9
r
-8 -6 -4 -2 0 2 4 6 8-5
-4
-3
-2
-1
0
F6 F7 F8 F9
r
V(r)E2
E1
E0
The 1D potential energy of an electron due to an array of nuclei of charge q separated by a distance a is
Where n = 0, +/-1, +/-2 etc.
This is shown as the black line in the figure.
n
2
naro4
qe = rV
)(
r
0
0
+ + + + +aNuclear positions
V(r) lower in solid (work function).
Naive picture: lowest binding energy states can become free to move throughout crystal
V(r)Solid
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Energy Levels and Bands Isolated atoms have precise allowed energy levels.
In the presence of the periodic lattice potential bands of allowed states are separated by energy gaps for which there are no allowed energy states.
The allowed states in conductors can be constructed from combinations of free electron states (the nearly free electron model) or from linear combinations of the states of the isolated atoms (the tight binding model).
+E
+ + + +position
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Influence of the lattice periodicityIn the free electron model, the allowed energy states are
where for periodic boundary conditions
nx , ny and ny positive or negative integers.
)(2
2222
zyx kkkm
E
L
nk
L
nk
L
nk z
zy
yx
x
2;
2;
2 E
k0
-5
-4
-3
-2
-1
0
r
E
Periodic potential
Exact form of potential is complicated
Has property V(r+ R) = V(r) where
R = m1a + m2b + m3c
where m1, m2, m3 are integers and a ,b ,c are the primitive lattice vectors.
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Waves in a periodic lattice
Recall X-ray scattering in Solid State:
n= 2asin
Consider a wave, wavelength moving through a 1D lattice of period a.
Strong backscattering for n= 2a
Backscattered waves constructively interfere.
Wave has wavevector k = 2
a
Wave moving to right
Scattered waves moving to left
Scattering potential period a
1D lattice: Bragg condition is k = n/a (n – integer)
3D lattice: Scattering for k to k' occurs if k' = k + G
where G = ha1 + ka2 + la3 h,k,l integer and a1 ,a2 ,a3
are the primitive reciprocal lattice vectors
k
k'
G
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Real and Reciprocal Lattice Spaces
• R for a crystal can be expressed in general as: R=n1a1+n2a2+n3a3 where a1, a2 and a3 are the primitive
lattice vectors and n1,n2 and n3 are integers
• Corresponding to a1, a2 and a3 there are three primitive reciprocal lattice vectors: b1, b2 and b3 defined in terms of a1, a2 and a3 by:
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Bragg scattering & energy gaps1D potential period a. Reciprocal lattice vectors G = 2n /a
A free electron of in a state exp( ix/a), ( rightward moving wave) will be Bragg reflected since k = /a and a left moving wave exp( -ix/a) will also exist.
In the nearly free electron model allowed un-normalised states for k = /a are
ψ(+) = exp(ix/a) + exp( - ix/a) = 2 cos(x/a)
ψ(-) = exp(ix/a) - exp( - ix/a) = 2i sin(x/a)
+E
+ + + +position a
N.B. Have two allowed states for same k which have different energies
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Reciprocal lattice
Use of reciprocal lattice space: Wave vectors k for Bloch waves lie in the reciprocal lattice space.
Translation symmetry=> a Bloch wave can be characterized by two wavevectors (or wavelengths) provided they differ by a reciprocal lattice vector!
Example in 1D:
Suppose k’=k+(2/a) then k(x)=exp(ikx)u(x)
and k’(x)=exp(ik’x)u(x)=exp(ikx)exp(i2x/a)u(x) =exp(ikx)u’(x)
essentially have the same “wavelength”
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Cosine solution lower energy than sine solutionCosine solution ψ(+) has maximum electron probability density at minima in potential.
Sine solution ψ(-) has maximum electron probability density at maxima in potential.
Cos(x/a) Sin(x/a)
Cos2(x/a)
Sin2(x/a)
In a periodic lattice the allowed wavefunctions have the property
where R is any real lattice vector.
22)()( rRr ψψ
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Magnitude of the energy gapLet the lattice potential be approximated by Let the length of the crystal in the x-direction to be L. Note that L/a is the number of unit cells and is therefore an integer. Normalising the wavefunction ψ(+) = Acos(x/a) gives
so
Solving Schrödinger’s equation with
)/2(cos)( 0 axVxV
1)/(cos22
0 dxaxA
L
21
2
LA
)()( ψψ EH
)()()/2cos(2 02
22
EaxVxm
220
22 Vmk
E
)/2(cos)( 0 axVxV
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Gaps at the Brillouin zone boundaries
At points A ψ(+) = 2 cos(x/a) and E=(k)2/2me - V0/2 .
At points B ψ(-) = 2isin(x/a) and E=(k)2/2me + V0/2 .
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Bloch StatesIn a periodic lattice the allowed wavefunctions have the property
where R is any real lattice vector.
Therefore where the function (R) is real, independent of r, and dimensionless.
Now consider ψ(r + R1 + R2). This can be written Or
Therefore
(R1 + R2) = (R1) + (R2)
(R) is linear in R and can be written (R) = kxRx + kyRy + kzRz = k.R. where
kx, ky and kz are the components of some wavevector k so
(Bloch’s Theorem)
22)()( rRr ψψ
)()( . rRr Rk ψψ ie
)()( )(21
21 rR ψψ RRRr ie
)()( )( rRr Rψψ ie
)()()( )()(2
)(21
211 rRrRRr RRR ψψψ iii eee
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(Bloch’s Theorem)
For any k one can write the general form of any wavefunction as
where u(r) has the periodicity ( translational symmetry) of the lattice. This is an alternative statement of Bloch’s theorem.
)2()()( . rr rk ueiψ
)1(ψ(r)eR)ψ(r ik.R
Alternative form of Bloch’s Theorem
Re [ψ(x)]
x
Real part of a Bloch function. ψ ≈ eikx for a large fraction of the crystal volume.
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Bloch Wavefunctions: allowed k-statesψ(r) = exp[ik.r]u(r)
kp
Periodic boundary conditions. For a cube of side L we require
ψ(x + L) = ψ(x) etc.. So
but u(x+L) = u(x) because it has the periodicity of the lattice therefore
Therefore i.e. kx = 2 nx/Lnx integer.
Same allowed k-vectors for Bloch states as free electron states.
Bloch states are not momentum eigenstates i.e.
The allowed states can be labelled by a wavevectors k.
Band structure calculations give E(k) which determines the dynamical behaviour.
L)u(xeL)u(xe xikL)(xik xx
xikL)(xik xx ee
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Nearly Free Electrons
Need to solve the Schrödinger equation. Consider 1D
write the potential as a Fourier sum
where G = 2n/a and n are positive and negative integers. Write a general Bloch function
where g = 2m/a and m are positive and negative integers. Note the periodic function is also written as a Fourier sum
Must restrict g to a small number of values to obtain a solution. For n= + 1 and –1 and m=0 and 1, and k ~ /a
E=(k)2/2me + or - V0/2
(x) E = (x) (x) V + x
2m
-2
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ψψ
G
iGxGeVxV )(
g
igxg
ikxikx eAeruex )()(ψ
Construct Bloch wavefunctions of electrons out of plane wave states.
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Tight Binding Approximation
NFE Model: construct wavefunction as a sum over plane waves.
Tight Binding Model: construct wavefunction as a linear combination of atomic orbitals of the atoms comprising the crystal.
Where (r)is a wavefunction of the isolated atom
rj are the positions of the atom in the crystal.
) ( c = )( j
jj
r-r r ψ
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Molecular orbitals and bondingConsider a electron in the ground, 1s, state of a hydrogen atom
The Hamiltonian is
Solving Schrödinger’s equation :
E = E1s = -13.6eV
o
2
4e = where
RadiusBohr theis a where a 1
= (r) i.e. oo e ar/-3/2 o
r
- 2m
- = H
22
+
E1s
V(r)
(r)
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Hydrogen Molecular Ion
Consider the H2+ molecular ion in which
one electron experiences the potential
of two protons. The Hamiltonian is
We approximate the electron wavefunctions as
and
|R - r|-
r -
2m
- = )rU( +
2m
- = H
2222
] + A[ |)] R - r(| + )r([ A = )r( 21 ψ
] B[ |)]R - r(| )r([ B = )r( 21 ψ
p+ p+
e-
R
r
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Bonding andanti-bonding states Solution:
E = E1s – (R) for
E = E1s + (R) for
(R) - a positive function
Two atoms: original 1s state
leads to two allowed electron
states in molecule.
Find for N atoms in a solid have N allowed energy states
)r(ψ
)r(ψ)r(ψ
-6 -4 -2 0 2 4 6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
r
-6 -4 -2 0 2 4 6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
r
V(r)
2)r(ψ
)r(ψ
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The tight binding approximation for s states
)aos(kc 2- -)e ae a( - - = k)( E xxx k i-k i
+ + + + +aNuclear positions
Solution leads to the E(k) dependence!!
1D:
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E(k) for a 3D lattice Simple cubic: nearest neighbour atoms at
So E(k) = 2(coskxa + coskya + coskza)
Minimum E(k) = 6for kx=ky=kz=0
Maximum E(k) = 6for kx=ky=kz=+/-/2
Bandwidth = Emav- Emin = 12
For k << acos(kxx) ~ 1- (kxx)2/2 etc.
E(k) ~ constant + (ak)2/2
c.f. E = (k)2/me
),,();,,();,,( a000a000a
-4 -2 0 2 4-18
-16
-14
-12
-10
-8
-6
-4
-2
0
F1
k [111] direction
/a/a
E(k)
Behave like free electrons with “effective mass” /a2
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Each atomic orbital leads to a band of allowed states in the solid
Band of allowed states
Band of allowed states
Band of allowed states
Gap: no allowed states
Gap: no allowed states
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Independent Bloch states
Bloch states
Let k = k ́ + G where k is in the first Brillouin zoneand G is a reciprocal lattice vector.
But G.R = 2n, n-integer. Definition of the reciprocal lattice. So
k is exactly equivalent to k.
)()( rRr k.Rψψ ie
)(ee)( Gi rRr .Ri.Rk ψψ
)(e)( and 1e iiG rRr .Rk.R ψψ .Rkk.R ii ee
-4 -2 0 2 4-18
-16
-14
-12
-10
-8
-6
-4
-2
0
F1
k [111] direction
/a/a
E(k)
The only independent values of k are those in the first Brillouin zone.
Solution of the tight binding model is periodic in k. Apparently have an infinite number of k-states for each allowed energy state.
In fact the different k-states all equivalent.
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Reduced Brillouin zone schemeThe only independent values of k are those in the first Brillouin zone.
Results of tight binding calculation
Results of nearly free electron calculation
Discard for |k| > /a
Displace into 1st B. Z.
Reduced Brillouin zone scheme
-2/a
2/a
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Extended, reduced and periodic Brillouin zone schemes
Periodic Zone Reduced Zone Extended Zone
All allowed states correspond to k-vectors in the first Brillouin Zone.
Can draw E(k) in 3 different ways
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The number of states in a bandIndependent k-states in the first Brillouin zone, i.e. kx < /a etc.
Finite crystal: only discrete k-states allowed
Monatomic simple cubic crystal, lattice constant a, and volume V.
One allowed k state per volume (2)3/V in k-space.
Volume of first BZ is (2/a)3
Total number of allowed k-states in a band is therefore
.etc,....2,1,0,2
xx
x nL
nk
N
aV
Va
3
3322
Precisely N allowed k-states i.e. 2N electron states (Pauli) per band
This result is true for any lattice:
each primitive unit cell contributes exactly one k-state to each band.
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Metals and insulatorsIn full band containing 2N electrons all states within the first B. Z. are occupied. The sum of all the k-vectors in the band = 0.
A partially filled band can carry current, a filled band cannot
Insulators have an even integer numberof electrons per primitive unit cell.
With an even number of electrons perunit cell can still have metallic behaviourdue to band overlap.
Overlap in energy need not occurin the same k direction
E
k0 a
EF
Metal due to overlapping bands
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Full Band
Empty Band
Energy Gap
Full Band
Partially Filled Band
Energy GapPart Filled Band
Part Filled Band EF
INSULATOR METAL METAL or SEMICONDUCTOR or SEMI-METAL
E
k0 a
EF
E
k0 a
E
k0 a
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Bands in 3D
In 3D the band structure is much more complicated than in 1D because crystals do not have spherical symmetry.
The form of E(k) is dependent upon the direction as well as the magnitude of k.
Figure removed to reduce file size
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