Solid State Theory Solution Sheet 3

Solid State Theory
Solution Sheet 3
FS 2013
Prof. M. Sigrist
Exercise 3.1
Lifting the degeneracy of atomic states
The continuous group O(3) has irreducible representations in each odd dimension 2l + 1.
The spherical harmonics Y lm (θ, φ) (m = −l, . . . , l) form a basis for the (2l+1)-dimensional
representations. Within the group O(3), each element which describes a rotation by the
same angle belongs to the same conjugacy class. Therefore, we can focus on the rotations
around the z-axis. A rotation around the z-axis transforms the spherical harmonics
according to
Y lm (θ, φ) ↦→ Y lm (θ, φ + α) = e imα Y lm (θ, φ).
The matrix for such a rotation is diagonal in the basis of the spherical harmonics:
The trace is given by
χ l (α) =
D (l) (α) = diag(e −ilα , . . . , e ilα ). (1)
l∑
m=−l
e imα = 1 +
l∑
e imα +
m=1
l∑
m=1
e −imα
= 1 + eiα − e iα(l+1)
+ e−iα − e −iα(l+1)
= . . .
1 − e iα 1 − e −iα
=
cos(αl) − cos [α(l + 1)]
1 − cos(α)
= sin [ α(l + 1 2 )]
sin( α 2 ) (2)
Now we assume that the atom belongs to a crystal with a crystal field of cubic symmetry.
Due to the reduced symmetry, the representations will in general split into irreducible
representations of O h .
p-orbitals, l=1 From Eq. (2) we obtain for l = 1 the following group character
O h E C 3 (8) C 2 4(3) C 2 (6) C 4 (6)
χ p 3 0 −1 −1 1
There is no need to worry about the inversion, because the parity of the spherical
harmonics is known to be (−1) l . Using the orthogonality relation we find
n Γ1 = 1 (3 − 3 − 6 + 6) = 0,
24
n Γ2 = 1 (3 − 3 + 6 − 6) = 0,
24
n Γ12 = 1 (6 − 6) = 0,
24
n Γ15 = 1 (9 + 3 + 6 + 6) = 1,
24
n Γ25 = 1 (3 − 3 + 6 − 6) = 0.
24
Because the parity for l = 1 is −1 it follows that D 1 ↦→ Γ − 15.
1

d-orbitals, l=2 In the same way we find for l = 2
and
O h E C 3 (8) C 2 4(3) C 2 (6) C 4 (6)
χ d 5 −1 1 1 −1
n Γ1 = 1 (5 − 8 + 3 + 6 − 6) = 0,
24
n Γ2 = 1 (5 − 8 + 3 − 6 + 6) = 0,
24
n Γ12 = 1 (10 + 8 + 6) = 1,
24
n Γ15 = 1 (15 − 3 − 6 − 6) = 0,
24
n Γ25 = 1 (15 − 3 + 6 + 6) = 1.
24
Since the parity is now positive it follows that D 2 ↦→ Γ + 12 ⊕ Γ + 25.
f-orbitals, l=3
This yields
O h E C 3 (8) C 2 4(3) C 2 (6) C 4 (6)
χ f 7 1 −1 −1 −1
n Γ1 = 1 (7 + 8 − 3 − 6 − 6) = 0,
24
n Γ2 = 1 (7 + 8 − 3 + 6 + 6) = 1,
24
n Γ12 = 1 (14 − 8 − 6) = 0,
24
n Γ15 = 1 (21 + 3 + 6 − 6) = 1,
24
n Γ25 = 1 (21 + 3 − 6 + 6) = 1.
24
Because of the negative parity we obtain D 3 ↦→ Γ − 2 ⊕ Γ − 15 ⊕ Γ − 25.
Eigenfunctions of the d-orbitals We know that D 2 ↦→ Γ + 12 ⊕ Γ + 25, where Γ + 12 and Γ + 25
are two- and three-dimensional representations, respectively. It is also known that
the spherical harmonics for l = 2 form a basis for the harmonic, homogeneous
polynomials of order 2. Such a polynomial can be written as
P = c 1 xz + c 2 xy + c 3 yz + c 4 (x 2 − z 2 ) + c 5 (y 2 − z 2 ), (3)
where it is implicitly assumed that x 2 + y 2 + z 2 = 1. Now let us consider a rotation
around the z axis by π/2. Thus, z ↦→ z, x ↦→ y and y ↦→ −x. This yields
xz ↦→ yz, xy ↦→ −xy, yz ↦→ −xz,
x 2 ↦→ y 2 , y 2 ↦→ x 2 , z 2 ↦→ z 2 .
2

We see that the triple {xz, xy, yz} and the pair {x 2 − z 2 , y 2 − z 2 } do not mix under
this transformation.
Therefore, {xz, xy, yz} = { cos φ sin θ cos θ, cos φ sin φ sin 2 θ, sin φ sin θ cos θ } is a basis
for Γ + 25 and {x 2 − z 2 , y 2 − z 2 } = {cos φ 2 sin θ 2 − cos θ 2 , sin φ 2 sin θ 2 − cos θ 2 } is a
basis for Γ + 12. Often, the two-dimensional subspace of Γ + 12 is called the e g subspace
and the three dimensional subspace of Γ + 25 is called the t 2g subspace.
Exercise 3.2
Two-orbital tight-binding model in 2d
a) The Bloch-waves constitute a basis of the (quasi-2-dimensional) Hilbert space of
the system. Since the Wannier functions are the “Fourier-transforms” of the Blochwaves,
they, too, span the whole Hilbert space. Thus, we can write
H =
∑
〈w α (r − r j )|H|w α ′(r − r j ′)〉c † αj c α ′ j
, (4)
′
α, α ′ , j, j ′
which can be split in two terms as
H = ∑ α
H α + ∑ α≠α ′ H α,α ′. (5)
Considering only intra-band hopping terms and restricting the sum to nearest neighbour
terms, we obtain
H α = ∑ j
ε α c † αj c αj + (tx αc † α(j+ˆx) c αj + ty αc † α(j+ŷ) c αj
+ h.c.) (6)
with
ε α =〈w α (r)|H|w α (r)〉, (7)
t x α =〈w α (r − aˆx)|H|w α (r)〉, (8)
t y α =〈w α (r − aŷ)|H|w α (r)〉. (9)
Considering the overlap elements t x/y
α
for both bands, we recognize that
t x p x
= t y p y
, (10)
t x p y
= t y p x
(11)
due to the symmetry properties of the lattice system and the atomic orbitals.
b) If we approximate the Wannier functions (which are orthogonal to each other) by
atomic (hydrogen) states (which are not orthogonal to each other), we choose the
orientation of the orbitals such that
{
positive, x > 0,
sign(w px (r)) =
(12)
negative, x < 0,
{
positive, y > 0,
sign(w py (r)) =
(13)
negative, y < 0.
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=〈w py (r − aˆx)|ε py + ∑ j≠0
=〈w py (r − aŷ)|ε py + ∑ j≠0
Using H j=0 w α (r) = (H kin + V (r))w α (r) = ε α w α (r), we find for the matrix elements
t x p x
=〈w px (r − aˆx)|ε px + ∑ j≠0
V (r − r j )|w px (r)〉 (14)
t y p x
=〈w px (r − aŷ)|ε px + ∑ j≠0
V (r − r j )|w px (r)〉 (15)
t x p y
V (r − r j )|w py (r)〉 (16)
t y p y
V (r − r j )|w py (r)〉. (17)
The main contribution to this matrix element comes from the region between the two
lattice sites where the two orbitals have opposite sign. As ε α < 0 and V (r) < 0, we
obtain that t x p x
= t y p y
> 0. For t y p x
, the orbitals have the same sign and t y p x
= t x p y
< 0.
Performing the Fourier transformation
c αj = √ 1 ∑
e −ik·r j
c αk (18)
N
of the annihilation operators in the Hamiltonian, we obtain
k
H α = ∑ k
ε α,k c † αk c αk
(19)
with
ε px,k = ε + 2t 1 cos(k x a) − 2t 2 cos(k y a) (20)
ε py,k = ε − 2t 2 cos(k x a) + 2t 1 cos(k y a) (21)
where ε = ε px = ε py , t 1 = t x p x
, and t 2 = −t y p x
> 0. The band structure and the Fermi
surface (ɛ F = ɛ for half-filling) are visualized in Fig. 1.
c) We now also take into account the coupling between the different bands. The
nearest neighbour hopping matrix elements between different bands vanishes due to
the symmetry of the orbitals. Therefore, we have to consider next-nearest neighbour
hopping between different bands. In the same way as before, we obtain
H α,α ′ = ∑ j
t + αα
c † ′ α(j+ˆx+ŷ) c α ′ j + t− αα
c † ′ α(j+ˆx−ŷ) c α ′ j
+ h.c. (22)
with
t ± αα ′ =〈w α (r − a(ˆx ± ŷ)|H|w α ′(r)〉 (23)
Due to symmetry properties and the analogue consideration as above, we obtain
t + p xp y
= t + p yp x
= −t − p xp y
= −t − p yp x
≡ t 3 > 0. Performing a Fourier transform of the
Hamiltonian, we obtain
H α,α ′ = ∑ k
−4t 3 sin(k x a) sin(k y a)c † αk c α ′ k . (24)
4

1.0
0.5
0.0
0.5
1.0
1.0 0.5 0.0 0.5 1.0
1.0
0.5
0.0
0.5
1.0
1.0 0.5 0.0 0.5 1.0
Figure 1: The band structure in the absence of interband coupling is visualized by a 3d plot
and contour plot of the two bands. Also shown is the Fermi surface (t 1 = 0.4, t 2 = 0.05).
Defining g k = −4t 3 sin(k x a) sin(k y a), the complete Hamiltonian can be written as
H = ∑ k
( )
c † T ( ) ( )
p xk εpxk g k c pxk
c † p yk
g k ε pyk
c pyk
(25)
such that to diagonalize the Hamiltonian, we have to find the Eigenvalues E ± k
the matrix above determined by the equation
of
The calculation is straightforward and we obtain
(ε pxk − E ± k )(ε p yk − E ± k ) − g2 k = 0. (26)
E ± k = ε + (t 1 − t 2 )(cos(k x a) + cos(k y a)) (27)
√
± (t 1 + t 2 ) 2 (cos(k x a) − cos(k y a)) 2 + 16t 2 3 sin 2 (k x a) sin 2 (k y a).
The resulting band structure and the Fermi surface (ɛ F
plotted in Fig. 2.
= ɛ for half-filling) are
5

1.0
0.5
0.0
0.5
1.0
1.0 0.5 0.0 0.5 1.0
1.0
0.5
0.0
0.5
1.0
1.0 0.5 0.0 0.5 1.0
Figure 2: The band structure in the presence of interband coupling is visualized by a 3d
plot and contour plot of the two bands. Also shown is the Fermi surface (t 1 = 0.4, t 2 =
0.05, t 3 = 0.1).
6