Solid State Theory
Solutions Sheet 7.
Exercise 1.
FS 2014
Prof. Manfred Sigrist
Bound and Antibound States in One Dimension
We consider a one-dimensional periodic chain of N atoms with nearest neighbor hopping where one
hopping element is different from the others. The lattice constant is set to unity. As we will see, this
leads to the formation of a bound state and an antibound state in addition to a continuum of states.
This should show that modifications of the hopping elements (as e.g. originating from electron-phononcoupling) lead to a modification of the energy bands of the electrons.
We describe this by the Hamiltonian
H = −t
X
(ˆ
c†j cˆj+1 + cˆ†j+1 cˆj ) − ∆t(ˆ
c†0 cˆ1 + cˆ†1 cˆ0 ),
(1)
j
where t > 0 and ∆t > 0. Show that the spectrum of this Hamiltonian has a bound state below and an
antibound state above the energy band with energies
Eb/ab = ∓2t ∓
∆t2
.
t + ∆t
(2)
Hint. Rewrite the Hamiltonian in k-space and use a general ansatz for a single particle wavefunction in
k-space. Then find the conditions for this state to be an eigenstate.
Solution.
Expressing the real-space operators cˆj through momentum-space operators cˆk via
1 X ikj
e cˆk
cˆj = √
N k
(S.1)
and obtain the Hamiltonian
X
1 XX
0
(−∆t)(eik + e−ik )ˆ
c†k cˆk0 .
H=
(−2t cos k)ˆ
c†k cˆk +
N
0
k
k
(S.2)
k
We make the most general ansatz for a state in momentum-space
X
|Ψi =
Aq cˆ†q |0i
(S.3)
q
This state should satisfy the Schr¨
odinger equation
H|Ψi = E|Ψi.
(S.4)
From this equation, we obtain
X
X
1 XX
(−2t cos k)Ak cˆ†k |0i +
(−∆t)Aq (eiq + e−ik )ˆ
c†k |0i =
EAk cˆ†k |0i.
N
q
k
k
(S.5)
k
In order that |Ψi is an eigenstate, the condition
1 X
(−∆t)Aq (eiq + e−ik ) = (E + 2t cos k)Ak
N q
1
(S.6)
has to be fulfilled for all k, which can be rewritten as
(−∆t)
1 X
(−∆t)e−ik 1 X
Aq eiq +
Aq = Ak .
E + 2t cos k N q
E + 2t cos k N q
(S.7)
We once sum this equation over k and once multiply this equation by eik and obtain the set of
equations
1 X
1 X (−∆t)e−ik 1 X
1 X
1 X (−∆t)
Aq eiq +
Aq =
Ak
N
E + 2t cos k N q
N
E + 2t cos k N q
N
(S.8)
1 X (−∆t)
1 X
1 X ik
1 X (−∆t)eik 1 X
Aq eiq +
Aq =
e Ak .
N
E + 2t cos k N q
N
E + 2t cos k N q
N
(S.9)
k
k
k
k
k
By introducing C0 =
equation
P
k
Ak and C1 =
1 P (−∆t)e−ik
k
N
cos k
P E+2t
(−∆t)
− N1 k E+2t
cos k
P
k
k
Ak eik , we can write this set of equations as a matrix
(−∆t)
k E+2t cos k
P
(−∆t)eik
1
k E+2t cos k
N
− N1
1−
1−
P
! C0
0
=
.
C1
0
(S.10)
As the sum runs in the range [−π, π] only the even part of the integrand survives and we have
a matrix equation of the form
C0
a b
0
(S.11)
=
b a
C1
0
with
1 X (−∆t) cos k
,
N
E + 2t cos k
k
1 X (−∆t)
b=−
.
N
E + 2t cos k
a=1−
(S.12)
(S.13)
k
In order to have a solution, the determinant of the matrix above should vanish, i.e. a2 − b2 = 0,
from which follows that a = ±b. We now go to the continuum limit such that
Z π
dk (−∆t) cos k
a=1−
,
(S.14)
−π 2π E + 2t cos k
Z π
dk (−∆t)
b=−
.
(S.15)
2π
E + 2t cos k
−π
and we obtain the condition
Z
π
1 = (−∆t)
−π
dk ±1 + cos k
2π E + 2t cos k
For |E| > 2t, this integral can be calculated using the integral
( 1
Z π
√
x>1
dq
1
2
= 2 x −1
1
− 2√x2 −1 x < −1
0 2π x + cos q
(S.16)
(S.17)
and we obtain for E > 2t
∆t
(−1 +
1=
2t
2
r
E ∓ 2t
)
E ± 2t
(S.18)
which has a solution only for the lower sign and we obtain
E = 2t +
∆t2
.
t + ∆t
(S.19)
In the same way, we find for E < −2t a solution
E = −2t −
Exercise 2.
∆t2
.
t + ∆t
(S.20)
Peierls’ Instability in One Dimension
We consider a one-dimensional chain with nearest-neighbor hopping where the position of the atoms is
not fixed. The Hamiltonian is thus given by a (renormalized) hopping and an elastic part:
H=
X †
X δu2j
(cj+1,s cj,s + h.c.)(−t + αδuj ) + λ
2
j,s
j
(3)
where δuj = uj+1 − uj and uj is the displacement of the atom at site j from its equilibrium position.
λ > 0 is a measure of the stiffness of the system and α > 0 is the coupling constant. We set the lattice
constant a = 1.
In the following, we consider the half filled case (one electron per site) and make for δuj the ansatz
δuj = u0 cos(qrj ),
(4)
with rj = j denoting the position of the lattice site.
(a) Calculate for q = π the eigenenergies and the eigenstates of the system and the density of states.
Hint. Write the electronic part of the Hamiltonian in the Form
X
H=
c†ks Hk cks ,
(5)
|k|<π/2,s
where c†ks = (c†ks , c†k+πs ) and Hk is a 2 × 2 matrix which can be written in terms of Pauli matrices.
The diagonalization is then just a rotation in the space of these matrices. Note that the sum now
only runs over a reduced Brillouin zone, k ∈ [− π2 , π2 ].
(b) Show that in this one-dimensional system, there is always a finite u0 that minimizes the total energy.
Hint. Show it for large λ and small u0 by using the elliptic integral of the second kind,
Z ϕp
E(ϕ, k) =
1 − k 2 sin2 αdα
(6)
0
and its series expansion
4
π p
1
1
E( , 1 − k 02 ) = 1 + (log 0 − )k 02 + O(k 04 ).
2
2
k
2
(7)
P
(c) Show that the density of electrons per site, ρj = s hc†js cjs i = 1 for all i but the bond density,
P
ρ˜j = s hc†js cj+1s + c†j+1s cjs i oscillates with position i. Discuss also the limits λ → 0 and λ → ∞
for α = t.
3
Solution. For the eigenstates of our system we first only consider the electronic part of the
Hamiltonian,
X †
Hel =
(cj+1s cjs + h.c.)(−t + αδuj ).
(S.21)
j,s
We change to momentum space by introducing the corresponding electronic operators,
1 X −ikrj †
c†js = √
e
cks .
L k
(S.22)
While the first part of the sum obviously yields
X
Hdiag =
(−2t cos k)c†ks cks ,
(S.23)
k,s
more care has to be taken for the calculation of the second term: by using
δuj = u0 cos qj.
we find
Hoffd =
(S.24)
i
αu0 X h i(k−q)
(e
+ e−ik )c†ks ck−qs + (e−ik + ei(k+q) )c†ks ck+qs .
2
(S.25)
k
(a) We now want to consider the half filled case, i.e. n = 1. In this case, the Fermi energy is
exactly F = 0 and we have a nesting vector q = π. This results in a periodicity of twice
the lattice constant, the unit cell is doubled (a → 2a) and thus the first Brillouin zone is
folded back. In particular, notice that the doubling means we now measure rj in units of
2a. We can thus write the Hamiltonian as
X0 †
(S.26)
H=
cks Hk cks
k,s
where
Hk =
−2t cos k −2iαu0 sin k
2iαu0 sin k
2t cos k
(S.27)
P
and the primed sum 0 only runs over the reduced Brillouin zone, k ∈ [− π2 , π2 ].
We can now write the Hk in terms of Pauli spin-matrices,
Hk = σy 2αu0 sin k − σz 2t cos k = d · σ.
(S.28)
The diagonalization of this matrix thus corresponds to a rotation of the vector d =
(0, 2αu0 sin k, −2t cos k) around the x-axis. The angle of rotation is given by
αu
0
ω(k) = − arctan
tan(k) .
(S.29)
t
We thus immediately find the eigenenergies which are just the length of the vector:
q
(S.30)
ξk± = ±2 t2 cos2 k + α2 u20 sin2 k.
With this trick, it is also straight forward to find the transformation matrix: Since a
rotation in the spin space is in general given by
Uω = e−iσ·ω = cos
4
ω
ω
− iˆ
ω · σ sin
2
2
(S.31)
with ω = |ω| and ω
ˆ = ω/ω, we find for our new operators
bks−
−i sin ω2
cks
cos ω2
=
.
−i sin ω2
cos ω2
bks+
ck+πs
Eventually, we can now calculate the density of states using
s
!
ξ 2 − 4t2
k = arcsin
4α2 u20 − 4t2
(S.32)
(S.33)
which follows directly from eq. (S.30). Thus the density of states yields
ξ
1 ∂k 2 ρ(ξ) = 2 · 2 = p
2
2
2
2
2
2π ∂ξ
π
(ξ − 4t )(ξ − 4α u0 ) (S.34)
with square root singularities at the band edges. The results of this part are summarized
in Fig. 1.
ρ(ε)
ε(k)
2
k
−π
-
π
π
2
2
π
ε
-2
Figure 1: Combined plot of the dispersion and the density of states of the one-dimensional
chain with no fixed positions for the atoms. On the left side, the arrows indicate the folding
of the Brillouin zone. On the right side, one sees the square root singularities of the density of
states,typical for one-dimensional systems, occurring at the band edges.
(b) To calculate the total energy we first again consider the electronic part. For T = 0, only
the lower band is filled and thus the energy (per length) yields
Z
2 X0
1 π/2
Eel =
ξk− dk
(S.35)
ξk− =
L
π −π/2
k
Z
q
4 π/2
= −
t2 cos2 k + α2 u20 sin2 kdk
(S.36)
π 0
r
Z
α2 u2
4t π/2
= −
1 − (1 − 2 0 ) sin2 kdk
(S.37)
π 0
t
!
r
4t
π
α2 u2
= − E
, 1− 20
(S.38)
π
2
t
5
where the first factor of two comes from the spins and in the last line we have used the
elliptic function of the second kind as given on the exercise sheet.
For large λ the displacement of an atom√costs a lot of energy and we only expect a very
small u0 so that we can expand in k 0 = 1 − k 2 = αu0 /t. The total energy is then
2
2α
4t
1
λ
4t
log
−
−
u20 .
(S.39)
Etot ≈ − −
π
tπ
αu0 2
2
We see that for small enough u0 the expression in the bracket becomes positive and the
total energy is reduced with respect to the energy without any displacements.
(c) We now want to calculate
P
P
0
(i) ρj = s hc†js cjs i = L2 k,k0 e−i(k−k )rj hc†ks ck0 s i
P
P
0
0
(ii) ρ˜j = s hc†js cj+1s i = L2 k,k0 e−i(k−k )rj eik hc†ks ck0 s i.
where the first one is the average electron density per atom and the second one is a measure
for the electron density on the bond between j and j + 1.
Before continuing we note that the ground state of the system is given in terms of the
new operators bkα and the only expectation values not vanishing are hb†ks− bks− i. Since the
original operators are given as
ω(k)
ω(k)
bks− + i sin
bks+
2
2
ω(k)
ω(k)
= i sin
bks− + cos
bks+ ,
2
2
cks = cos
ck+πs
(S.40)
(S.41)
this means that the only non vanishing matrix elements in terms of the original operators
are
ω(k)
δk,k0
2
ω(k)
ω(k)
cos
δk,k0
hc†ks ck0 +πs i = i sin
2
2
ω(k)
hc†k+πs ck0 +πs i = sin2
δk,k0
2
ω(k)
ω(k)
hc†k+πs ck0 s i = −i sin
cos
δk,k0
2
2
hc†ks ck0 s i = cos2
(S.42)
(S.43)
(S.44)
(S.45)
(S.46)
We can thus write the density per site as
ρj
=
=
=
2 X −i(k−k0 )rj †
e
hcks ck0 s i
L 0
k,k
2 X0
ω(k)
ω(k)
2 ω(k)
2 ω(k)
+ sin
+ i sin
cos
(1 − 1)
cos
L
2
2
2
2
k
2 X0
=1
L
(S.47)
(S.48)
(S.49)
k
For the ‘density per bond’ we can again split the sum into parts with equal momentum
6
and parts with momentum differing by π. For the first ones we find
ρ˜const
=
j
=
=
2 X0 ik †
e hcks cks i + ei(k+π) hc†k+πs ck+πs i
L
k
X
0 ik ω(k)
2
ω(k) e cos2
− sin2
L
2
2
k
2 X0
2 X0 cos k cos ω(k)
cos k cos ω(k) + i sin k cos ω(k) =
L
L
k
(S.50)
(S.51)
(S.52)
k
In the last step we have used that cos ω(k) is an even function in k (cf. eq. (S.29) and thus
sin k cos ω(k) is an odd function,thus vanishing when summed over.
For the latter part we find
2 X0 iπrj ik †
e e hck+πs cks i + e−iπrj ei(k+π) hc†ks ck+πs i
L
k
2 X0 ik
ω(k)
ω(k) −iπrj
=
e cos
sin
e
− eiπrj
L
2
2
k
2 sin(πrj ) X0
i cos k sin ω(k) − sin k sin ω(k)
= −
L
k
2 sin(πrj ) X0
=
sin k sin ω(k)
L
ρ˜osc
=
j
(S.53)
(S.54)
(S.55)
(S.56)
k
Here, we used the fact that ω(k) is odd and hence is sin ω(k). Putting back in the units
(rj → rj /2a) one identifies the oscillating behaviour.
So we indeed see that the bond density is composed of a constant term and an oscillating
term. We can now analyze this density in the two limits:
• λ → ∞: In that case, the displacement of the atoms will go to zero, u0 → 0 and
thus ω → 0, too. Consequenty, the oscillating term vanishes while the constant term
becomes,
2
ρ˜j = .
(S.57)
π
• λ → 0: In this case, displacement of an atom does not cost any energy and the total
displacement of two atoms becomes u0 = min{ αt , 1}. For the case where α ≥ t, this
corresponds to a gap of 4t and the dispersion is completely flat, the electrons are
localized. In that case ω = k and the bond density is given as
1
ρ˜j = ρ˜const
+ ρ˜osc
j
j = (1 + sin πrj )
2
which means that the bond charge is alternatively 0 or 1.
7
(S.58)
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