56 CHAPTER 2.

SECOND QUANTISATION

2.4.2 Answers
1. (a) Making use of the commutation relations for bosons, one finds

a† aa = a(a† a − 1), a† aa† = a† (1 + a† a)

from which the results follow. Using these results, one finds that, providing a|αi =
6 0,

a† a a|αi = a(a† a − 1)|αi = (α − 1)a|αi

a† a a† |αi = a† (1 + a† a)|αi = (1 + α)a† |αi

(b) If |αi is a normalised eigenstate of a† a with eigenvalue α, the norm of state

created by the action of the creation operator is given by
p p √
||a† |αi|| ≡ hα|aa† |αi = hα|a† a + 1|αi = α + 1.

Similarly, the norm of state created by the action of the annihilation operator is
given by
p √
||a|αi|| ≡ hα|a†a|αi = α.

Therefore, if we define |α + 1i and |α − 1i as the normalised eigenstates of the

operator a† a with eigenvalue α + 1 and α − 1 respectively, one finds
√
a|αi = α|α − 1i
†
√
a |αi = α + 1|α + 1i

Defining as the vacuum |Ωi the normalised state that is annihilated

√ by† the operator
n
a, an application of the result above shows the state |ni = (1/ n!)(a ) |Ωi to be a
normalised eigenstate of a† a with eigenvalue n.
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2. The kinetic energy operator is diagonal in the momentum basis. Following the anal-
ysis in
Pthe text, the corresponding second quantised one-body operator is given
R by
p2 † −1/2 L
T̂ = p 2m ap ap . Transforming to the coordinate representation, ap = L 0
dxeipx/~a(x),
one obtains
Z L Z L
1 X p2
T̂ = dy dxeip(x−y)/~a† (y)a(x).
L p 2m 0 0

Expressing factor p2 as a derivative of the exponential factor, and integrating by

parts, one obtains the required result.
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2.4. PROBLEM SET 57

3. A Hamiltonian which is translationally

P invariant is easily diagonalised in the Fourier
† 1 ik·m †
representation. Setting cmσ = N d/2 k e ckσ , the Hamiltonian takes the form

B.Z. X
X
Ĥ (0)
= ǫk c†kσ ckσ ,
k σ

P P
where ǫk = −t i=x,y,z eik·êi = −2t i=x,y,z cos(k · êi ), with the sum running over
neighbouring lattice vectors êi , and the lattice spacing is taken to be unity.
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4. Substituting the definition of the spin raising and lowering operators using the
Holstein-Primakoff transformation, the commutator is obtained as

 + a† a 
1/2 1 z}|{ 1/2  
†
1 + − aa † a† a † a† a
[Ŝ , Ŝ ] = 1 − aa 1− −a 1− a
2S 2S 2S 2S
   
a† a † a† a a† a† aa a† a
= 1− +a a 1− − a† a + =1− .
2S 2S 2S S

With S z = S − a† a, we obtain the required commutation relation [Ŝ + , Ŝ − ] = 2S z .

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5. By symmetry, the maximal exchange energy that can be recovered is obtained when
the spins are maximally anti-aligned, i.e. at 120o to each other. Using the spin

(a) (b)

orientation of a single triangle, the two-dimensional triangular lattice can be tessi-

lated with all spins aligned at 120o to the neighbours. Notice that this configuration
represents just one of an infinite degenerate manifold of ground states obtained by
global rotation of the spins.
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58 CHAPTER 2. SECOND QUANTISATION

6. To confirm the validity of the Bogoluibov transformation let us consider the com-
mutation relations of α:27
   
α, α† = cosh θ a + sinh θ a† , cosh θ a† + sinh θ a
   
= cosh2 θ a, a† + sinh2 θ a† , a = cosh2 θ − sinh2 θ = 1.
as required.
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7. (a) Making use of the equation of motion iṠi = [Ŝi , Ĥ], and the commutation
relation [Siα , Sjβ ] = iδij ǫαβγ Siγ , we obtain
 
iṠi = JiŜi × Ŝi+1 + Ŝi−1

(b) Interpreting the spins as classical vectors, and applying the Taylor expansion
Si+1 = Si + a∂Si + (a2 /2)∂ 2 Si + · · ·, we obtain the classical equation of motion
Ṡ = Ja2 S × ∂ 2 S.
√

Substituting, we find that S = c cos(kx − ωt), c sin(kx − ωt), S 2 − c2 , satisfies
√
the equation of motion with ω = J(ka)2 S 2 − c2 .

Figure 2.14: Spin-wave dispersion.

(c) The corresponding spin wave solution has the precessional form shown in Fig. 2.14.
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27
More formally, one may prove the relation as follows: Suppose we define a two component operator
a = (a, a† ). If a obeys bosonic commutation relations, [aµ , a†ν ] = gµν where the diagonal matrix has
elements g = diag(1, −1). If we define an operator transformation Λ such that αµ = Λµν aν (summation
convention implied), then the condition that the commutation relations are preserved requires
!
gµν = [αµ , α†ν ] = Λµη Λ∗νγ [aη , a†γ ] = Λµη Λ∗νγ gηγ ,
i.e. the admissable transformations fulfil the condition that gµν = Λµη Λ∗νγ gηγ . Such transformations, that
preserve the “metric” g = diag(1, −1), belong to the group of Lorentz transformations. In the context of
bosonic operators, they are termed Bogoluibov transformations. Note that, for fermionic systems, with
c = (c, c† ), [cµ , c†ν ] = δµν . In this case, we require that δµν = Λµη Λ∗νγ δηγ , i.e. Λ belongs the class of
Unitary transformations: Λ† Λ = 1

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2.4. PROBLEM SET 59

8. Defining the total spin on a triad Ĵn = Ŝn−1 + Ŝn + Ŝn+1 , the Hamiltonian can be
recast in the form
N
X
ĤMG = |J| P̂3/2 (n − 1, n, n + 1),
n=1

where P̂3/2 (n − 1, n, n + 1) = (Ĵ2n − 3/4)/3 annihilates any state with total spin
J = 1/2 of the triad. Since in any three sites, two of the spins are in a singlet,
there can be no components of J = 3/2 on any triad. Therefore the dimer states
are eigenstates of zero energy. Now since P̂3/2 is positive definite, these states must
be the ground states.
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9. (a) Since each unit cell is of twice the dimension of the original lattice, we begin by
recasting the Hamiltonian in the sublattice form
N/2
X      Nks α2
Ĥ = −t (1 + α) a†mσ bmσ + h.c. + (1 − α) b†mσ am+1σ + h.c. + .
m=1,σ
2
p P
Switching to the Fourier basis, am = 2/N k e−2ikm ak (similarly bm ), where k
takes N/2 values uniformly on the interval [−π/2, π/2], the Hamiltonian takes the
form
Nks a2 2
Ĥ = α
2   
X † 0 (1 + α) + (1 − α)e−2ik akσ
−t ( akσ b†kσ ) .
(1 + α) + (1 − α)e2ik 0 bkσ
kσ

Diagonalising the 2 × 2 Hamiltonian, we obtain the spectrum

 1/2
ǫ(k) = ±2t cos2 k + α2 sin2 k .

Reassuringly, in the limit α → 0, we obtain the cosine spectrum of the undistorted

problem, while in the limit α → 1, pairs of monomers become decoupled and we
obtain a massively degenerate bonding and antibonding spectrum.
(b) According to the formula given in the text, the total shift in energy is given by

Nks α2
δǫ = −2t(a1 − b1 ln α2 )α2 +
2
Maximising the energy gain with respect to α, one finds that the stable configuration
is found when
 
2 a1 Nks
α = exp −1−
b1 4tb1

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60 CHAPTER 2. SECOND QUANTISATION

(c) If the number of sites is odd, the Peirels distortion is inevitably frustrated — a
configuration that starts ABABAB must finish as BABABA. The result is that the
polymer chain must accommodate a topological excitation. The excitation is said
to be topological because the defect can not be removed by a smooth continuous
deformation — it is like a dislocation line in a crystal. Its effect on the spectrum of
the model is to introduce a state that lies within the band gap of the material.
The consideration of an odd number of sites forces a topological defect into the
system. However, even if the number of sites is even, one can create low energy
topological excitations of the system either by doping (see fig. 2.13c), or by the cre-
ation of excitons, particle-hole excitations of the system. Indeed, such topological
excitations can dominate the transport properties of the system.
As a footnote, one should add that the particular model considered above is some-
what over-simplified. It seems likely that Coulomb interactions play a dominant
role in driving the Peirels instability in Polyacetylene. However, the qualitative
interpretation of the existence of topological excitations is born out by experiment.
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10. (a) Using the commutation relation for bosons, one finds.

[Ŝ + , Ŝ − ] = a† b b† a − b† a a† b = a† abb† − aa† b† b

= a† a(b† b + 1) − (a† a + 1)b† b = a† a − b† b = 2Ŝ z .

(b) Using the identity

1 + −  1 1 † †

Ŝ2 = (Ŝ z )2 + Ŝ Ŝ + Ŝ − Ŝ + = (n̂a − n̂b )2 + a b b a + b† a a† b
2 4 2
1 1
= (n̂a − n̂b )2 + n̂a n̂b + (n̂a + n̂b ) ,
4 2
one finds that
 
Ŝ2 |S, mi = m2 + (S + m)(S − m) + S |S, mi = S(S + 1)|S, mi

as required. Similarly, one finds

1
Ŝ z |S, mi = (na − nb )|na = S + m, nb = S − mi
2
1
= [(S + m) − (S − m)] |S, mi = m|S, mi,
2
showing |S, mi to be an eigenstate of the operator Ŝ z with eigenvalue m.
As with the Holstein-Primakoff representation, the Schwinger boson represents yet
another representation of quantum mechanical spin. Which representation is most
convenient for the analysis of quantum spin models depends sensitively on the nature
of the microscopic Hamiltonian.
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2.4. PROBLEM SET 61

11. (a) Using the fermionic anticommutation relations, one finds

[Ŝ + , Ŝ − ]− = [f † , f ]− = f † f − f f †
= 2f † f − 1 = 2Ŝ z .

(b) Using the fact that the number operators on different sites commute, one finds
P P
+ − † iπ nj −iπ nl † −iπnm †
Ŝm Ŝm+1 = fm e j<m e l<m+1 fm+1 = fm e fm+1 = fm fm+1
† −iπnm †
where here we have made use of the fact that, for fermionic particles fm e ≡ fm .
(c) The fermion representation is simply obtained by substitution.
(d) With Jz = 0, the spin Hamiltonian assumes the form of a non-interacting tight-
binding Hamiltonian
J⊥ X †

Ĥ = − fn fn+1 + h.c. .
2 n

This Hamiltonian, which has been encountered previously, is diagonalised in the

Fourier space after which one obtains the cosine band dispersion.
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Concepts in Theoretical Physics