55
Last updated November 11, 2018 UNIVERSITY COLLEGE LONDON DEPARTMENT OF PHYSICS AND ASTRONOMY 1
Last updated November 11, 2018
UNIVERSITY COLLEGE LONDON
DEPARTMENT OF PHYSICS AND ASTRONOMY
1
PHASM/G048 CONDENSED MATTER THEORYProf. Andrew Green and Dr. Frank Krüger
Please send any comments and corrections to [email protected] [email protected] Please do not distribute without permission.
These notes are intended as an aid to revision and are a supplement to, not a substitutefor your own notes taken in lectures. They are continually being revised, added to and
corrected.
2
Contents
1 From the Harmonic Oscillator to Phonons 51.1 The Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Two Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 The Harmonic Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3.1 Diagonalising by Fourier Transform . . . . . . . . . . . . . . . . . . 91.3.2 Raising and Lowering Operators . . . . . . . . . . . . . . . . . . . . 13
1.4 The Elastic String . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5 Canonical Quantization [Aside] . . . . . . . . . . . . . . . . . . . . . . . . 19
2 Second Quantization 212.1 Identical particles/Many-particle states . . . . . . . . . . . . . . . . . . . . 212.2 Many-particle Basis States . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 Occupation Numbers and Fock Space . . . . . . . . . . . . . . . . . . . . . 232.4 Creation and Annihilation Operators . . . . . . . . . . . . . . . . . . . . . 242.5 Transformation Between Bases . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.5.1 The Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 262.5.2 Transforming (Anti-)Commutation Relations . . . . . . . . . . . . . 272.5.3 Transforming Between x- and p-Bases . . . . . . . . . . . . . . . . . 27
2.6 Single- and Two-Particle Operators . . . . . . . . . . . . . . . . . . . . . . 282.6.1 Single-Particle Operators . . . . . . . . . . . . . . . . . . . . . . . . 282.6.2 Two-Particle Operators . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.7 Diagonalizing Quantum Hamiltonians . . . . . . . . . . . . . . . . . . . . . 312.7.1 Unitary transformations . . . . . . . . . . . . . . . . . . . . . . . . 312.7.2 Bogoliubov Transformation . . . . . . . . . . . . . . . . . . . . . . 32
3 The Weakly-Interacting Bose Gas 353.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 Mean Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.3 Landau’s Critical Superfluid Velocity . . . . . . . . . . . . . . . . . . . . . 373.4 Superfluid Fraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4 Quantum Magnets 394.1 The Heisenberg model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2 Holstein-Primakoff transformation . . . . . . . . . . . . . . . . . . . . . . . 414.3 The Heisenberg ferromagnet . . . . . . . . . . . . . . . . . . . . . . . . . . 42
4.3.1 The Groundstate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3
4.3.2 Excitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.3.3 Thermal Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.4 The Heisenberg Anti-ferromagnet . . . . . . . . . . . . . . . . . . . . . . . 454.4.1 The Néel state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.4.2 Holstein-Primakoff transformation . . . . . . . . . . . . . . . . . . . 454.4.3 Bogoliubov transformation . . . . . . . . . . . . . . . . . . . . . . . 474.4.4 Fluctuations of the Antiferromagnet . . . . . . . . . . . . . . . . . 474.4.5 Haldane’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . 484.4.6 Jordan-Wigner Transformation . . . . . . . . . . . . . . . . . . . . 494.4.7 Integer Spin Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
5 The Renormalization Group 535.1 The 1d Classical Ising Model . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4
Chapter 1
From the Harmonic Oscillator toPhonons
In this chapter, we start with a review of the harmonic oscillator and its description usingraising and lowering operators. We note in particular how all of its properties - expectationsof operators in arbitrary states or even thermal distributions - can be calculated in terms ofthese raising and lowering operators without ever having to write down the wavefunctionexplicitly.
Next, we will proceed to couple two such oscillators and then a whole chain. Thelatter provides a simple model for quantum oscillations of a lattice - known as phonons. Inworking through this problem, we will have introduced the central ideas of quantum fieldtheory. These will be formulated as a book-keeping tool to describe many-particle quantumsystems in chapter 2 and we will use them throughout the course.
1.1 The Harmonic OscillatorConsider a quantum particle whose motion is described by the Hamiltonian
H =p2
2m+
1
2mω2x2. (1.1)
We solve this by introducing raising and lowering operators
H = ~ω√mω
2~
(x− i p
mω
)︸ ︷︷ ︸
a†
√mω
2~
(x+ i
p
mω
)︸ ︷︷ ︸
a︸ ︷︷ ︸complete square
− i~ωmω
2~1
mω[x, p]︸ ︷︷ ︸
allow for x,p not commuting
= ~ω(a†a+ 1/2). (1.2)
The constant ~ω/2 is the zero-point energy. It follows directly from the Heisenberg un-certainty principle, since the classical groundstate is forbidden as it would give perfectknowledge about both the position and momentum of the operator.
5
Commutation Relations:The definitions of a and a† given above (a/a† =
√mω/(2~)
(x± i p
mω
)) imply the commu-
tation relations[x, p] = i~ ⇔ [a, a†] = 1. (1.3)
The Quantum groundstate is given by
a|0〉 = 0.
We could in principle express a as a differential operator (in real space, for example) andsolve for a (real space) wavefunction. The beauty of raising and lowering operators is thatwe can calculate every single property of the state without ever having to obtain an explicitrepresentation in this way.
Excitations are created by the action of the raising operator on the groundstate. The state
|n〉 =(a†)n√n!|0〉
is a state that contains n excitations (the action of the number operator n = a†a on thisstate is n|n〉 = n|n〉). It has energy ~ω(n + 1/2). The action of the raising and loweringoperators on this state are
a|n〉 =√n|n− 1〉,
a†|n〉 =√n+ 1|n+ 1〉.
These follow directly from the commutation relations, Eq.(1.3).
Calculations with Raising and Lowering Operators:We can calculate all of the observable properties of the harmonic oscillator using the aboverelations and without ever having to obtain explicit expressions for the wavefunction. Thefollowing are a few instructive examples:
a) Average in the nth excited state:
〈x2〉 = 〈n|x2|n〉
=~
2mω〈n|(a+ a†)2|n〉
=~
2mω〈n|(a2 + a†a† + aa† + a†a)|n〉
Using 〈n|a2|n〉 ∝ 〈n|n− 2〉 = 0 and 〈n|a†a†|n〉 ∝ 〈n|n+ 2〉 = 0
=~
2mω〈n|(aa† + a†a)|n〉
Using aa† = 1 + a†a
=~
2mω〈n|(2a†a+ 1)|n〉
=~mω
(n+ 1/2)
6
b) The partition function :
Z =∑n
e−βεn〈n|n〉
=∑n
e−β~ω(n+1/2)
=1
1− e−β~ω e−β~ω/2
NB: The zero point term e−β~ω/2 is usually omitted as it does not contribute to observables.
c) Finite temperature average of the number operator:
〈〈n〉〉 =
∑n e−βεn〈n|n|n〉∑
n e−βεn〈n|n〉
=
∑n n e
−β~ωn
Z=− 1
~ω∂βZZ
=1
eβ~ω − 1= nB(~ω)
d) The finite temperature average of arbitrary operator, 〈〈θ〉〉, follows similarly: i. write theoperator in terms of numer operators n using commutation relations; ii. calculate averagesas above (often this amounts to the replacement n → nB although sometimes a bit morecare is required).
Problems:Q1. What is 〈x4〉 in the nth excited state of the harmonic oscillator?Q2. What is 〈〈x4〉〉 for a thermal distribution of harmonic oscillators?
1.2 Two Coupled OscillatorsUp to now this has hopefully been revision. Now lets turn to something new. We willconsider a simple problem of two coupled quantum harmonic oscillators. In doing so, wewill see many of the techniques that underpin quantum field theory.
Consider two coupled particles whose motion is described by the Hamiltonian
H =p2
1
2m+mω2
2x2
1 +p2
2
2m+mω2
2x2
2 +κ
2x1x2, (1.4)
which captures the dynamics of the following configuration:
x1 x2
m m
7
Sum and Difference Coordinates:Our physical intuition about this situation suggests two types of periodic motion or normalmodes:
m m
symmetric normal mode
m m
anti-symmetric normal mode
Accordingly, we make a change of variables to the mean displacement and difference indisplacement;
X =x1 + x2√
2and x =
x1 − x2√2
,
with conjugate momenta
P =p1 + p2√
2and p =
p1 − p2√2
.
In the position representation, p1 = −i~∂x1 and p2 = −i~∂x2 . The expressions for themomenta P and p — in particular the factors of 1/
√2 — follow from these.
Preservation of Commutation Relations:These definitions imply the commutation relations
[x1, p1] = [x2, p2] = i~ ⇔ [X, P ] = [x, p] = i~.
The requirement that the commutation relations take this form — i.e. that their form ispreserved by our change of variables — can be used to fix the expressions (including thefactors of 1/
√2) for P and p.
[X, P ] = [x1 + x2√
2,p1 + p2√
2] = [x1, p1]/2 + [x2, p2]/2 + [x1, p2]/2︸ ︷︷ ︸
=0
+ [x2, p1]/2︸ ︷︷ ︸=0
= i~.
[x, p] = [x1 − x2√
2,p1 − p2√
2] = [x1, p1]/2 + [x2, p2]/2− [x1, p2]/2︸ ︷︷ ︸
=0
− [x2, p1]/2︸ ︷︷ ︸=0
= i~.
Substituting into the Hamiltonian( 1.4), we find
H = =1
4m(P + p)2 +
mω2
4(X + x)2 +
1
4m(P − p)2 +
mω2
4(X − x)2 +
κ
4(X + x)(X − x)
=P 2
2m+
1
2
(mω2 +
κ
2
)X2 +
p2
2m+
1
2
(mω2 − κ
2
)x2
We have reduced the motion to two independent harmonic motions with masses 2m andm/2, and frequencies
ω± = ω
√1± κ
2mω2.
We may introduce creation and annihilation operators to calculate the properties of this2-particle system.
8
This method of separating — or diagonalising — a Hamiltonian into independent har-monic oscillators is more general. As we shall see in the next section, it is related to aFourier transform for a translationally invarient system; for a 2-site chain, the allowedFourier components have wavevector 0 and 2π/2a — we have used a Fourier transform todiagonalize our Hamiltonian.
Problems:For the system of two coupled harmonic oscillators described above:Q1. What is the r.m.s. zero-point fluctuations in the position of the 1st particle?Q2. What is the thermal expectation of x1 − x2 at temperature T?Q3. What is the heat capacity (∂T 〈〈H〉〉) at temperature T?
1.3 The Harmonic ChainNext, we consider a harmonic chain consisting of quantum particles whose Hamiltonian isgiven by the elastic energy due to coupling to nearest neighbours.
m mm m
xn xn+1 xn+2xn1
The Hamiltonian is given by
H =N∑n=1
[p2n
2m+
1
2κ(xn − xn−1)2
](1.5)
and we will use periodic boundary conditions corresponding to taking x0 = xN . Oursolution is going to follow essentially the same steps as we used for two coupled oscillators.
1.3.1 Diagonalising by Fourier Transform
The first step is to reduce the problem to a set of independent harmonic oscillators. Weachieve this by using a discrete Fourier transform. Many problems in physics are transla-tionally invariant (or can be treated as such) and in these cases the Fourier transform takesus much of the way towards diagonalising our system.
The Fourier components of the particle displacements are given by
xk =1√N
∑n
eiknaxn with inverse xn =1√N
∑k
e−iknaxk, (1.6)
for the position operators and
pk =1√N
∑n
e−iknapn with inverse pn =1√N
∑k
eiknapk, (1.7)
for the momentum operators. N is the length of the chain, a is the separation of theparticles, k = 2π
Namk and mk is an integer taking values from −N/2 to N/2. We could
9
have written this in terms of cos and sin rather than taking the real part, but I prefer tokeep things closer to the Fourier transform. Since xn is real (or has only real expectations)xk = x∗−k. This fixes the number of degrees of freedom to be correct — otherwise therewould appear to be twice as many components of xk due to the real and imaginary parts.Note the opposite sign of wavevector k in the transformations of x and p. The reasons forthis will become clear shortly.
The restriction of k to values 2πNamk with mk an integer between −N/2 and N/2 occurs
because a wavevector k′ = k + 2π/a leads to the same displacements of the particles;eik′na = eikna+2πn = eikna. This is known as the first Brillouin zone.
The Hamiltonian expressed in terms of these transformed operators is given by
H =∑k
[pkp−k2m
+ 2κ sin2(ka/2)xkx−k
]. (1.8)
Except for the presence of both pk and p−k (and xk and x−k) this a Hamiltonian for oneindependent oscillator for each value of k. We can make it look exactly like independent har-monic oscillators by separating xk and pk into parts that are symmetric and anti-symmetricin k;
xck =xk + x−k√
2and xsk =
xk − x−ki√
2, and pck =
pk + p−k√2
and psk =pk − p−ki√
2.
As implied by the notation, these are the coefficients of cos(ka) and sin(ka) in a real Fourierexpansion. In terms of these
H =∑
k>0,σ=c,s
[pσk p
σk
2m+ 2κ sin2(ka/2)xσk x
σk
](1.9)
The system has been reduced to a set of independent harmonic oscillators with frequenciesgiven by mω2
k/2 = 2κ sin2(ka/2) or ωk = 2√κ/m sin(ka/2), which in the limit of small
frequencies reduces to ωk ≈ ka√κ/m.
Commutation relations:Just as in the case of the two coupled oscillators, the commutation relations are preservedby our Fourier transformation;
[xn, pm] = i~δn,m ⇔ [xp, pq] = i~δp,q
10
Details of Fourier Transform
A. Forwards and Backwards Transforms
This can be verified using the identity
δn,m =1
N
∑k
eik(n−m)a
which is the discrete equivalent of the integer representation of the deltafunction δ(x − y) =
∫∞−∞ dke
ik(x−y)/(2π). Substituting into the Fouriertransformed operators, we find
xn =1√N
∑k
e−iknaxk
=1
N
∑k
e−ikna(∑
m
eikmaxm
)=
1
N
∑k,m
e−ik(n−m)axm
=1
2
∑m
δn,mxm
= xn
We can also verify the relationship between the forward and backwardtransforms using the wavevector version of the delta function summationformula
δk,q+G =1
N
∑k
e−i(k−q)na,
where G = 2π×integer/a is a reciprocal lattice vector. This extra contri-bution of G effectively allows us to fold back contributions of wavevectorsoutside of the Brillouin zone into the Brillouin zone, since they give thesame displacements. The calculation takes the form
xk =1√N
∑n
eiknaxn
=1
N
∑n
eikna
(∑q
eiqnaxq
)
11
=1
N
∑k,n
e−in(q−k)axq
=1
2
∑k
δq,kxq
= xk
B. Fourier transforming the Hamiltonian
The Fourier transformation of the Hamiltonian is most easily achieved by splittingit into the kinetic and potential energy parts.i. Kinetic Energy
H =∑n
p2n
2m
=1
2m
∑n
(1√N
∑k
eiknapk
)(1√N
∑q
eiqnapq
)︸ ︷︷ ︸
NB: different dummy variables
=1
2m
∑k,q
1
N
∑n
ei(k+q)na
︸ ︷︷ ︸=δk,q
pkpq
=1
2m
∑k
pkp−k
ii. Potential Energy
H =κ
2
∑n
(xn − xn−1)2
=κ
2
∑n
(1√N
∑k
e−ikna(1− eika)xk)(
1√N
∑q
e−iqna(1− eiqa)xq)
︸ ︷︷ ︸NB: different dummy variables
=κ
2
∑k,q
1
N
∑n
e−i(k+q)na
︸ ︷︷ ︸=δk,q
(1− eika)(1− eiqa)xkxq
=κ
2
∑k
(1− eika)(1− e−ika)xkx−k
= 2κ∑k
sin2(ka/2)xkx−k
12
B. Fourier transforming the commutation relations
The Fourier transform of the commution relations proceeds in a very similar mannerto the Fourier transform of the Hamiltonian
[xp, pq] =1
N
[∑n
eipnaxn ,∑m
e−iqmapm
]=
1
N
∑n,m
ei(pn−qm)a [xn, pm]︸ ︷︷ ︸=i~δn,m
= i~1
N
∑n
ei(p−q)na
= i~δp,q
1.3.2 Raising and Lowering Operators
Just as in the case of a single harmonic oscillator, we can now calculate properties of theharmonic chain by introducing ladder operators for each k-mode. These create or destroyquanta of energy in each of the normal modes of oscillation of the harmonic chain. Theresulting Hamiltonian is given by
H =∑k
~ωk(a†kak + 1/2)
ωk = 2√κ/m sin(ka/2) (1.10)
with
ak =
√mωk2~
(xk + i
p−kmωk
), and a†k =
√mωk2~
(x−k − i
pkmωk
). (1.11)
Notice the opposite momentum labels in xk and p−k, and the relative signs of k betweena and a†. This is required to obtain the usual commutation relations for the raising andlowering operators;
[ak, a†q] = δk,q (1.12)
Checking the commutation relations
[ak, a†q] =
m√ωkωq
2~
[(xk + i
p−kmωk
),
(x−q − i
pqmωq
)]
=m√ωkωq
2~
i
mωk[p−k, x−q]︸ ︷︷ ︸
=−i~δk,q
− i
mωq[xk, pq]︸ ︷︷ ︸=i~δk,q
= δk,q
13
Particle Interpretation/Fock Space:Just as for the harmonic oscillator, armed with these tools we can describe the state of theharmonic chain in terms of the number of quanta — or occupation number nk — of eachmode k. A general state may be written
|ψ〉 = |nq1 , nq2 , nq3 ...〉 =∏q
(a†q)nq√nq!|0〉,
where |0〉 is the vacuum state with no quanta in any mode. The collection of all such stateswith arbitrary occupations forms a space known as Foch space.
Observables:One can evaluate the observable properties of the harmonic chain by first translating intoak and a†k. For example, the mean squared displacement of a particle in the chain at finitetemperature is given by
x2 =1
N
∑n
〈〈x2n〉〉
Fourier transform - allows us to use the properties of independent k-modes
=1
N
∑k
〈〈xkx−k〉〉
Express in terms of ladder operators
=1
N
∑k
~2mωk
〈〈(ak + a†−k)(a−k + a†k)〉〉
=1
N
∑k
~2mωk
〈〈(aka−k︸ ︷︷ ︸=0
+ aka†k︸︷︷︸
=a†a+1=nk+1
+ a†−ka−k︸ ︷︷ ︸=n−k
+ a†−ka†k︸ ︷︷ ︸
=0
)〉〉
=1
N
∑k
~2mωk
(〈〈nk〉〉+ 〈〈nk〉〉+ 1)
=1
N
∑k
~mωk
(1
e~ωkβ − 1+ 1/2
).
Zero-point Energy and Normal ordering:Just as for the harmonic oscillator, the Hamiltonian for the harmonic chain has a constantterm ~ωk/2 for each wavevector. This implies a minimum energy proportional to N ,
HZero-point =∑k
~ωk/2.
This energy cannot usually be extracted (except by changing boundary conditions c.f. theCasimir effect) and so is often ignored in field theory. For an infinite system the zero-pointenergy → ∞. In the continuum limit, the situation is even more severe and the energydensity → ∞. This is the first infinity that we have encountered. Quantum field theoryis plagued by them. To deal properly with all of the infinities that occur, one needs therenormalisation group. We will touch upon this towards the end of the first part of thecourse.
14
The infinite zero-point energy can be dealt with rather straightforwardly by normalordering. The infinite zero-point energy arises from the form of the Hamiltonain H =∑
k ~ωk(a†kak+ aka
†k)/2, when we use the commutation relations to reorder the second term
so that it can be written in terms of number operators. Similar infinities can arise frominteraction terms — terms higher than quadratic order — in Hamiltonians. The trick todealing with these is to find a consistent way to ignore them! Basically, we define theoperation of normal ordering to consist of ordering the creation and annihilation operatorsso that the creation operators are on the left and the annihilation operators on the right:
: a†kak + aka†k := 2a†kak (1.13)
Applied to quadratic terms, normal ordering essentially ignores zero-point energy. Appliedto interaction terms, it ignores self-interaction of particles that can lead to other unphysicaldivergences.
Problems:Q1. What is the r.m.s. displacement of a mass in the harmonic chain in the groundstate?Q2. What is the r.m.s. speed of a particle in the harmonic chain in the groundstate?Q3. What are the r.m.s. displacement and speed at temperature T?Q4. What is the heat capacity of the harmonic chain?
1.4 The Elastic StringLet us consider the limit in which the separation of masses in our chain becomes zero whilemaintaining a fixed length of chain L and mass density:
N →∞L = Na fixedm→ 0 such that ρ = m/a fixedκ→∞ such that τ = κa fixedn→ x/a position along string, x ∈ [−L/2, L/2]
xn → φ(x)|x=na displacement fieldpn → aΠφ(x)|x=na momentum field
lima→0
∑n
=1
a
∫dx
lima→0
(xn+1 − xn) = a∂xφ. (1.14)
With these replacements, the Hamiltonian and Commutations Relations become
H =
∫dx
1
2
[1
ρΠ2φ + τ
(∂xφ)2]
[φ(x, t),Πφ(y, t)
]= i~δ(x− y) (1.15)
This field theory is the same as that of the one-dimensional Klein-Gordon model of rela-tivistic quantum particles. We shall see how the relativistic dispersion arises later. Notice
15
that the commutation relations are expressed at equal times, since the field operators —the displacement and momentum fields — are presented in the Heisenberg picture.
Heisenberg, Schrödinger and Dirac/Interaction Pictures:In the Schrödinger picture of quantum mechanics, operators (such as position andmomentum) are fixed and time-dependence is carried by the state or wavefunction.An alternative picture — due to Heisenberg — is to keep the state fixed in timeand allow the operators to vary. The relationship between these two pictures can beunderstood by insisting that physical observables are unaffected by our notation.
In the Schrödinger picture dt|ψ〉 = iH|ψ〉 ⇒ |ψ(t)〉 = ei∫dtH|ψ(0)〉. We can use this
to deduce the time-dependence of the expectation of an operator Θ as follows:
〈Θ〉(t) = 〈ψ(t)|ΘS|ψ(t)〉 = 〈ψ(0)| e−i∫dtHΘSe
i∫dtH︸ ︷︷ ︸
ΘH
|ψ(0)〉
The subscripts S and H on Θ indicate Schrödinger and Heisenberg versions respec-tively. Their relationship is ΘH = e−i
∫dtHΘSe
i∫dtH. The equation of motion for ΘH
can be deduced by taking a time derivative of this equation to be
dtΘH = i[ΘH , H]
The Interaction Picture is another alternative (that we will not use in this course)often used in field theory. This separates the Hamiltonian into bare and interactionparts, H0 and HI , respectively. The time dependence arising from the bare partsis carried by the operators and the additional time dependence arising from theinteractions is carried by the states. This embodies the notion that interactions maycause particles/systems to change their states.
16
Fourier Transform (finite string):As in the case of the chain, the elastic string is diagonalised by a Fourier transform. Ap-plying the rules established above for the continuum limit, these are given by
φ(k, t) =1√L
∫dxeikxφ(x, t) and Πφ(k, t) =
1√L
∫dxe−ikxΠφ(x, t) (1.16)
with k = 2πnk/L and nk an integer in the range [−∞,∞]. The inverse transforms are
φ(x, t) =1√L
∑k
e−ikxφ(k, t) and Πφ(x, t) =1√L
∑k
eikxΠφ(k, t). (1.17)
The corresponding Fourier transformed commutation relations are given by[φ(p, t), Πφ(q, t)
]= i~δp,q (1.18)
The relationship between the forward and backwards transformations and the transformedcommutation relations can be verified by using the summation/integral representation ofthe delta function
1
L
∞∑k=−∞
eik(x−y) = δ(x− y)
∫ L/2
−L/2dxe−ix(q−p) = 2πδ(p− q) = Lδp,q. (1.19)
As before, I use the slight abuse of notation δp,q to indicate δnp,nq , where p = 2πnp/L.
Using these rules for the Fourier transformation of the Hamiltoiain can be written
H =1√L
∑k
[1
2ρΠφ(k)Πφ(−k) +
κ
2k2φ(k)φ(−k)
].
Notice that the mode frequencies become ω = k√κ/ρ — i.e. a linear, relativistic disper-
sion with effective speed of light√κ/ρ.
Fourier Transform (infinite string):In the case of the infinite chain, these transformations become
φ(k, t) =
∫ ∞−∞
dxeikxφ(x, t) and Πφ(k, t) =
∫ ∞−∞
dxe−ikxΠφ(x, t) (1.20)
The inverse transforms are
φ(x, t) =
∫ ∞−∞
dk
2πe−ikxφ(k, t) and Πφ(x, t) =
∫ ∞−∞
dk
2πeikxΠφ(k, t). (1.21)
The corresponding Fourier transformed commutation relations are given by[φ(p, t), Πφ(q, t)
]= i~δ(p, q) (1.22)
17
The relationship between the forward and backwards transformations and the transformedcommutation relations can be verified by using the summation/integral representation ofthe delta function ∫ ∞
−∞
dk
2πeik(x−y) = δ(x− y)∫ ∞
−∞dxe−ix(q−p) = 2πδ(p− q). (1.23)
Notice that I have made a choice here about where I put the factors of 2π from the integralrepresentation of the δ-function. The conventional physics choice is as above, so thatwavevector integrals always come with a factor of 1/2π.
The Hamiltonian takes the same relativistic form as the finite string, with the discretesummation over wavevectors becoming an integral;
H =
∫dk
2π
[1
2ρΠφ(k)Πφ(−k) +
κ
2k2φ(k)φ(−k)
].
Problems:Q1. Verify the k-space commutation relations of the finite-length elastic string.Q2. Verify the k-space Hamiltonian of the finite-length elastic string.Q3. Verify the k-space commutation relations of the infinite elastic string.Q4. Verify the k-space Hamiltonian of the infinite elastic string.
18
1.5 Canonical Quantization [Aside]Some of you may be doing a quantum field theory course, where the idea of field operatorsis introduced by canonical quantization. This aside is given to show the connection betweenwhat we are doing here and that approach. It isn’t an examinable part of the course, buthopefully will serve to give you greater insight about the connections between the thingsthat you are learning.
Single harmonic OscillatorThe classical Lagrangian for the harmonic oscillator is given by
L =mx2
2− 1
2mω2x2/2
and from it we can obtain the both the momentum conjugate to x — px = ∂L/∂x and theHamiltonian
H = xpx − L =p2x
2m+
1
2mω2x2.
Canonical quantization imposes commutation relations between conjugate fields — in thiscase the position and momentum;
[x, p] = i~.
Quantum Elastic RopeFor the quantum elastic rope, the Lagrangian is given by
L =
∫dx
(1
2ρφ(x)2 − 1
2κ(∂xφ(x))2
)from which w obtain the conjugate momentum Πφ(x, t) = δL/δφ(x, t) = ρφ(x, t) andHamiltonian
H =
∫dxφΠφ − L =
∫dx
(1
2ρΠ2φ −
1
2κ(∂xφ(x))2
)Quantization involves imposing canonical commutation relations[
φ(x, t), Πφ(y, t)]
= i~δ(x− y)
.
Complex Quantum Elastic RopeThe displacement field in the cases above was chosen to be along the chain/rope. Infact, the model of transverse fluctuations takes precisely the same form. If we allow fortransverse fluctuations in both transverse directions of the rope, we can encode this byallowing the real-space displacement field φ(x, t) to be complex — the real and imaginaryparts describing the different components of the transverse displacement. The Langrangianin this case is precisely that obtained for the Klein-Gordon field;
L =
∫dx
(1
2ρ ˙φ(x)φ(x)− 1
2κ(∂xφ(x))(∂xφ(x))
)(1.24)
19
from which w obtain the conjugate momentum Πφ(x, t) = δL/δφ(x, t) = ρ ˙φ(x, t) andΠφ = δL/δφ(x, t) = ρφ(x, t) and Hamiltonian
H =
∫dx(φΠφ +˙φΠφ
)− L =
∫dx
[1
2ρΠφΠφ −
1
2κ(∂xφ(x))(∂xφ(x))
]Quantization involves imposing canonical commutation relations[
φ(x, t), Πφ(y, t)]
= i~δ(x− y).
Of course, in the case of the elastic rope, the fields that we are quantizing correspond todisplacements and momenta and we are simply imposing the commutation relations [x, p] =i~. In the case of the Klein-Gordon model, the situation is somewhat different. The Euler-Lagrange equations arising from Eq.(1.24) are the Klein-Gordon equation, φ−d2
xφ = 0. Thisis a relativistic analogue of the Schrödinger equation obtained from energy conservation,where first quantization (changing x and p to operators and imposing [x, p] = i~ ) hasbeen used already identifying the energy with a derivative with respect to time and themomentum with a derivative with respect to position. Changing the Klein-Gordon fieldinto a field operator is therefore a second quantization step. The insight is that one ofthe quanta of this field operator will have a wavefunction that obeys the Klein-Gordonequation. This is at first a difficult concept to appreciate. In the next section we shall seehow the idea of second quantization can be alternatively introduced as a book keeping toolto keep track of many-body wavefunctions. This effectively leads to the same constructionas obtained by this second quantization trick, and essentially provides a justification for it.
20
Chapter 2
Second Quantization
In the first chapter, we saw how creation and annihilation operators are a very convenientway of ealing with systems of many harmonic oscillators — armed with their commutationrelations, we can calculate observable properties without ever having to write down awavefunction. The idea of second quantization — and field theory in general — is to usethis machinary to study many-body quantum systems1. The essence is to interpret thecreation of a quantum of energy by the operation of a† as the creation of a particle of aquantum matter field;
a†k − Creates a particle with momentum k i.e. in a plane wave statea†x − Creates a particle at the point x i.e. in a delta-function wavefunction state at x.
2.1 Identical particles/Many-particle statesMany-body quantum mechanics in general — and theoretical condensed matter in par-ticular — is concerned with the collective quantum behaviour of many identical quantumparticles (for example electrons or phonons in a solid). In this chapter, we will review theformalism required to describe such systems. Consider a classical system of N identicalparticles. This might have a Hamiltonian
H =N∑i=1
p2i
2mi
+∑i,j>i
V (xi − xj) +N∑i=1
U(xi).
This is (first) quantized by taking the position and momentum variables to be operatorssatisfying commutation relations
[xi, pj] = i~δi,j,
where i and j are particle labels.
Solution for Non-interacting, Distinguishable ParticlesWhen V (xi − xj) ≡ 0 and the particles are distinguishable, the Schrödinger equation is
1For relativistic systems — such as particles described by the Klein-Gordon equation — it is essential,since the energy density and hence particle number depend upon the inertial frame of reference.
21
separable. The solution is a product of single-particle solutions;
ψ(x1, x2, ...xN) = ψα1(x1)ψα2(x2) ... ψαN (xN)e−iEt/~
ψα(x) single particle state with energy EαE = Eα1 + Eα2 ... EαN total energy (2.1)
Identical Particles and ExchangeInvariance of observable properties under the exchange of indistinguishable particles placesimportant constraints upon the wavefunction. Let us define the permutation operator asfollows: Pij permutes particles i and j so that
Pijψ(x1, ... xi ... xj ... xn) = ψ(x1, ... xj ... xi ... xn).
For indistinguishable particles, Pij should not change the results of measurements, i.e.expectations of operators are the same. Let us consider explicitly the expectation of anoperator Θ and the action of Pij upon it.
〈ψ|Θ|ψ〉 invariant under |ψ〉 → eiΦ|ψ〉PijPji = 1
⇒ Pijψ(x1, ... xi ... xj ... xn) = ±ψ(x1, ... xj ... xi ... xn)
+ Bosons− Fermions
We will see presently the effects that this has upon our construction of many-particlewavefunctions.
2.2 Many-particle Basis StatesIn order to describe our many body system, we require a suitable set of basis states thatsatisfy the symmetry or anti-symmetry identified above. Let us start with a complete,orthonormal set of single-particle basis states, ψα(x). From these, we can construct a manyparticle state as follows:
ψ(x1, , xN) = N∑P
(±)Pψα1(x1) ... ψαN (xN) (2.2)
(±) for bosons/fermionsP = order of permutationα1, α2, ... αN some permutation of state labelsN Normalization (2.3)
Note that here we sum over ALL permutations. Some other texts (such as that of ProfChalker) sum over distinct permutations — i.e. if some of the labels are the same thenswapping them isn’t counted. Permutations are divided into even and odd depending uponthe number of pairwise swaps that are required to perform them. An even number givesan even permutation and an odd number gives an odd permutation.
22
Normalization
1 =
∫dxi ...dxN |ψ(x1, ... xN)|2 (2.4)
= |N |2∑Pα
∑Pβ
(±)Pα+Pβ
∫dx1ψ
∗α1
(x1)ψβ1(x1)︸ ︷︷ ︸=δα1,β1
...
∫dxNψ
∗αN
(xN)ψβN (xN)︸ ︷︷ ︸=δαN ,βN
(2.5)
= |N |2∑Pα
(±)2Pα = |N |2N ! (2.6)
From which we deduce that N = 1√N !.
NB in the case where permutations are restricted to distinct permutations, the sumover permutations is different and the the result is 1 = |N |2N !/(n1!n2!...), where niis the number of times that a distinct single particle state ψαi appears in the state.ni! gives the permutations of these.
Slater Determinants and PermantentsThe wavefunction for fermions may be written in the following form:
ψ(x1 xN) =1√N !
∣∣∣∣∣∣∣∣ψα1(x1) ... ... ψα1(xN)ψα2(x1) ... ... ...... ... ... ...
ψαN (x1) ... ... ψαN (xN)
∣∣∣∣∣∣∣∣ . (2.7)
The minus signs (−)P required for fermions are taken care of by the minus signs comingfrom the determinant. Note that the determinant is zero if any two or more of the single-particle states are the same. The determinant — known as a Slater determinant— thereforeproperly encodes the Pauli principle. An ordering convention is needed to fix the overallsign of the wavefunction. For bosons, we may construct a similar representation in terms ofa permanent. This is essentially the same object as a determinant but without the minussigns.
2.3 Occupation Numbers and Fock SpaceThe states that we identified as many-particle basis states can be specified completely bythe number of particles in each single-particle basis state. These are usually denoted
|n1, n2, ...〉,
where ni is the number of quanta/particles in the state i. ni takes values 0, 1 for fermionsand values 0, 1, 2, ...,∞ for bosons. This notation assumes: i. a particular identification ofsingle-particle states. In condensed matter, we typically use either momentum or plane wavestates, or an orthonormal set of states constructed from orbitals on particular atoms, i.eposition states (known as Wanier states); ii. that the states are appropriately symmetrizedor anti-symmetrized for bosons or fermions, respectively.
23
Fock Space: is the set of states with all possible combinations of the occupation numbers.
The Vacuum State: is the state in which none of the particle states are occupied. It iswritten as |0〉 and normalized 〈0|0〉 = 1.
Comparison with the Harmonic Oscillator:This structure is very similar to that of the eigenstates of the harmonic oscillator andharmonic chain considered previously. In the case of the harmonic chain, for example, ageneric eigenstate could be written |nk1 , nk2 , ...〉 describing a state with nki quanta in thekthi wavevector mode. These modes are bosons and are given the name phonons whenapplied to vibrational modes of a crystal lattice. We have been using Fock space all alongto describe the harmonic oscillator and its derivatives.
2.4 Creation and Annihilation OperatorsSince the structure of Fock space discussed above is so reminiscent of of the state spaceof the harmonic oscillator, it is natural to expect that we can also find operators that arethe analogue of the ladder/creation and annihilation operators. These operators will allowus to navigate Fock space and, moreover, to calculate properties of our many-body systemwithout ever having to write down its wavefunction.
The basic notion is to identify creation operators c†a that create a particle in the single-particle state |ψa〉. We will then manipulate the resulting states to determine the propertiesof the operators c†a. A many-particle state can be constructed from the action of many suchoperators,
c†a1 ... c†aN|0〉 = (n1!n2! ...)1/2|n1, n2, ...〉. (2.8)
This corresponds to a state described by the real-space, many-body wavefunction
ψ(x1, ...xN) = 〈x1, ...|(c†1)n1 ... (c†N)nN |0〉 =1√N !
∑P
(±1)Pψa1(x1) ... ψaN (xN) (2.9)
Hermitian Conjugation and Annihilation of the VacuumBy definition, annihilation operators are given by the Hermitian conjugate of the creationoperators. Their action on the vacuum is as follows:
c†q|0〉 = 0 and 〈0|c†a = 0.
24
Proof As for the harmonic oscillator
c†a|0〉 state with one particle in|ψa〉Normalization ⇒ 〈0|cac†a|0〉 = 1
⇒ 〈0| = 〈0|cac†a and |0〉 = cac†a|0〉
General state |φ〉 other than the vacuum⇒ c†a|φ〉 has more than one particle⇒ 〈0|cac†a|φ〉 = 0
⇒ |0〉 = ca(c†a|0〉) = ca|na = 1〉
Also 〈0|c†a|φ〉 = 0 = 〈φ|ca|0〉⇒ 〈0|c†a = ca|0〉 = 0
Commutation and Anti-commutationConsidering the action of the permutation operator on a two-particle state, we have
c†ac†b|0〉 = ±c†bc†a|0〉
for bosons and fermions, respectively. This implies that[c†a, c
†b
]= 0 = c†ac
†b − c†bc†a Bosons (2.10)
c†a, c†b
= 0 = c†ac
†b + c†bc
†a Fermions (2.11)
and (after Hermitian conjugation)
[ca, cb] = 0 Bosons (2.12)ca, cb = 0 Fermions (2.13)
In order to reproduce the values of inner products of our general multi-particle state,Eq.(2.9), we must take2 [
ca, c†b
]= δab Bosons (2.14)
ca, c†b
= δab Fermions (2.15)
Number Operators The action of the creation operators on Fock space is determined by thenormalization of Eq.(2.8), which implies
c†l |n1, n2, ... nl, ...〉 = (±)n1+ ...+nl−1√nl + 1|n1, n2, ... nl + 1, ...〉
The negative signs in the fermionic case come from commuting c†l through all of the creationoperators to the left. Note also that for fermions (c†l )
2 = 0 which implies that the right
2For example, consider the norm of the state |n1〉 = c†1|0〉. This is given by 1 = 〈n1|n2〉 = 〈0|c1c†1|0〉 =±〈0|c†1c1|0〉+ 〈0|0〉.
25
hand side of the above expression is zero when nl = 1. The action of the annihilationoperators can be considered similarly and we find
cl|n1, n2, ... nl, ...〉 = (±)n1+ ...+nl−1√nl|n1, n2, ... nl − 1, ...〉
This is zero for both bosons and fermions if nl = 0. That is, cl annihilates a state propor-tional to the vacuum of the orbital l.
From these results, we deduce that
c†l cl|n1, n2, ... nl, ...〉 = nl|n1, n2, ... nl, ...〉.
2.5 Transformation Between BasesIt is often useful to address different physical questions with reference to different singleparticle bases (such as the position and momentum bases). We need then to understandhow to transform our creation and annihilation operators between such bases. We havealready seen an example of this for the harmonic chain. Let us think about this moregenerally for a moment.
2.5.1 The Transformation
Consider two orthonormal sets of basis functions |χa〉 and |ψa〉. The transformationbetween them is acheived as follows3:
|χa〉 =∑b
Uab|ψb〉 (2.16)
Uab = 〈ψb|χa〉 (2.17)
The transformation is unitary (UU † = 1), as can be seen from the following simple calcu-lation:
UabU†bc =
∑b
〈ψb|χa〉(〈ψb|χc〉)† =∑b
〈χc |ψb〉〈ψb|︸ ︷︷ ︸1
χa〉 = 〈χc|χa〉 = δac.
Next, let us determine the corresponding transformation rule for the creation and annihi-lation operators.
c†a creates a particle in |ψa〉,d†a creates a particle in |χa〉.
From the relationship between the orbitals, (2.17), we deduce that
d†a =∑b
Uabc†b and by Hermitian conjugation da =
∑b
cb(U†)ba (2.18)
3NB. the transformation matrix used here is the transpose of that used by Prof Chalker in his notes
26
2.5.2 Transforming (Anti-)Commutation Relations
Both commutation and anticommutation relations are preserved by this transformation4
Bosonic Commutation Relations:[da, d
†b
]=∑cd
[cc(U
†)ca, Ubdd†d
]=∑cd
Ubd(U†)ca
[cc, c
†d
]︸ ︷︷ ︸
=δcd
=∑c
Ubc(U†)ca = δab
Fermionic Anti-Commutation Relations:da, d
†b
=∑cd
cc(U
†)ca, Ubdd†d
=∑cd
Ubd(U†)ca
cc, c
†d
︸ ︷︷ ︸
=δcd
=∑c
Ubc(U†)ca = δab
The invariance of the commutations relations between position and wavevector basis thatwe noted in the case of the harmonic chain is just a special case of the invariance undergeneral unitary transformations noted here.
2.5.3 Transforming Between x- and p-Bases
The position and momentum basis are often the most convenient for the calculation ofphysical observables in translationally invariant systems. It turns out that the transfor-mation of states between these bases is simply a Fourier transform. The commutationrelations [x, p] = i~ imply that the momentum operator is given by i~∂x in real space 5 andthe following overlap between position and momentum eigenstates:
〈x|k〉 =eix.k√2π~
= Uk,x.
Using (2.18) the transformation from a position space creation operator to a momentumspace one is given by
d†k =
∫dke−ix.k√
2π~c†x.
The condition that the transformation be unitary simply reduces to the integral represen-tation of the δ-function;∫
dxUk,xU†x,q =
∫dxei(k−q).x
2π~= δ(k− q)/~.
4An alternative notation uses [A,B]− = [A,B] and [A,B]+ = A,B. Using this the equations lookidentical with the usual ± indicating where bosonic and fermionic versions differ
5and that the position operator is given by −i~∂k in momentum space
27
Normalization of Momentum States and Fourier Transform ConventionsFourier transformations and transformations between position and momentum bases havevarious factors of 2π (and ~) flying around. The easiest way to remember these is to keepa factor of 1/(2π~) for each momentum in the measure.
Discrete:
c†k =1√N
∑x
e−ik.x c†x c†x =1√N
∑k
eik.x c†k x = a(nx, ny, nz) k = 2π(l,m, n)/a
Continuous, Finite Length:
c†k =1√V
∫dxe−ik.x c†x c†x =
1√V
∑k
eik.x c†k k = 2π(l,m, n)/a
Continuous, Infinite Length:
c†k =
∫dxe−ik.x c†x c†x =
∫dk
2π~eik.x c†k
The astute will notice a slight difference compared with the previous page. This is becausemomentum states can be defined with 2 different normalizations. On the previous page
〈x|x〉 = V and 〈k|k〉 =V
2π~
whereas on this page we have used
〈x|x〉 = V and 〈k|k〉 = V.
The point is that one cannot really talk about a plane wave state for a single particle asthe prefactor would go to zero (the probability density would be spread out uniformly overthe whole of space). Instead, one defines the state to have a certain density of particles perunit volume and one must decide what that density is.
2.6 Single- and Two-Particle Operators
2.6.1 Single-Particle Operators
We are going to build up gradually to a way of writing general single particle operators ina second quantised form - i.e. in terms of creation and annihilation operators.
Kinetic EnergyThe classical kinetic energy for a collection of N particles takes the form
HKE =N∑i=1
p2i
2mi
28
i.e. the sum of the kinetic energy of each particle. The way in which we write this insecond quantized form is to use the number operator in the momentum basis to count thenumber of particles with each momentum;
HKE =∑k
(~k)2
2mnk.
The expectation of this in a momentum Fock state |nk1 , nk1 , ...〉 reduces to
〈HKE〉 =∑k
〈nk1 , nk1 , ...|(~k)2
2mnk|nk1 , nk1 , ...〉︸ ︷︷ ︸
when k=ki⇒nk|nk〉=nk|nk〉
=N∑i,1
(~ki)2
2mnki〈nk1 , nk1 , ...|nk1 , nk1 , ...〉
=N∑i=1
(~ki)2
2mnki
General Case:In general, an operator Θ can be expressed in second quantized form by using the numberoperator in the eigenbasis of Θ.
Assume Θ has eigenstates |ψα〉 and eigenvalues Θα such that
Θ|ψα〉 = Θα|ψα〉, or Θα = 〈ψα|Θ|ψα〉.
The first quantized expectation for an N particle system is given by
〈Θ〉 =N∑i=1
Θαi .
Following the same procedure as for the kinetic energy, this can be written
Θ =∑α
Θαnα =∑α
Θαc†αcα.
General Case in Arbitrary Basis:Of course, we may wish to write our operator Θ in a basis other than its eigenbasis. Usingcreation and annihilation operators in an arbitrary basis, it can be written as
Θ =∑ab
〈ψa|Θ|ψb〉c†acb. (2.19)
29
This can be shown as follows:
• Relate bases by Uαa = 〈ψa|ψα〉, which implies c†α = Uαbc†b and cα = U †bαcb from
our definition in Eq.(2.18) or alternatively c†a = U †aαc†α and ca = cαUαa.
• Substitute in general expression
Θ =∑abαβ
〈ψa|Θ|ψb〉U †aαUβbc†αcβ
=∑abαβ
U †aα〈ψa|Θ|ψb〉Uβbc†αcβ
=∑abαβ
〈ψα|ψa〉〈ψa|︸ ︷︷ ︸∑a |ψa〉〈ψa|=1
Θ |ψb〉〈ψb|ψβ〉︸ ︷︷ ︸∑b |ψb〉〈ψb|=1
c†αcβ
=∑αβ
〈ψα|Θ|ψβ〉︸ ︷︷ ︸Θ|ψβ〉=Θβ |ψβ〉
c†αcβ
=∑αβ
Θβ 〈ψα|ψβ〉︸ ︷︷ ︸δαβ
c†αcβ
=∑α
Θαc†αcα
2.6.2 Two-Particle Operators
Two-body operators — such as the interaction potential — depend upon the coordinatesof a pair of particles. The first quantized form of the matrix elements of such an operatorare given by
Θlmpq =
∫dx1dx2ψ
∗l (x1)ψ∗m(x2)Θ(x1,x2)ψp(x2)ψq(x1)
Using the properly (anti) symmetrized wavefunctions for (fermions) bosons, the expectationvalues of Θ are
〈Θ〉 =
∫dx1dx2
1√2
(ψ∗l (x1)ψ∗m(x2)± ψ∗l (x2)ψ∗m(x1))
× (Θ(x1,x2) + Θ(x2,x1))1√2
(ψp(x2)ψq(x1)± ψp(x1)ψq(x2))
= (Θlmpq + Θmlqp)± (Θlmqp + Θmlpq)
This same expression can be recovered from the second quantized form
Θ =∑lmpq
Θlmpq c†l c†mcpcq
Taking the expectation with the second quantized form of the same states
〈Θ〉 = 〈0|clcmΘc†pc†q|0〉 =
∑abcd
Θabcd 〈0|clcmc†ac†b︸ ︷︷ ︸(δlbδam±δalδbm)
cccdc†pc†q|0〉︸ ︷︷ ︸
(δdpδcq±δdqδcp)
= (Θlmpq + Θmlqp)± (Θlmqp + Θmlpq)
30
Origin of minus signsThe minus signs here arise for fermions as terms anticommute if they have differentlabels
cccdc†pc†q|0〉 = −cdccc†pc†q|0〉 = cdc
†pccc
†q|0〉 = cdc
†p︸︷︷︸
when p=d⇒1−c†pcp
ccc†q︸︷︷︸
when c=q⇒1−c†c cc
|0〉
cccdc†pc†q|0〉 = −ccc†pcdc†q|0〉 = − ccc
†p︸︷︷︸
when p=c⇒1−c†pcp
cdc†q︸︷︷︸
when d=q⇒1−c†dcd
|0〉
2.7 Diagonalizing Quantum HamiltoniansThe aim in using second quantization techniques is to reduce the expectation values thatwe are interested in — as far as possible — to number operators. As we saw in the caseof the harmonic oscillator, it is then a simple matter to calculate the physical properties.Single particle operators can always be written in this form. Higher order many-particleoperators can be expanded in terms of single-particle operators using mean-field theoryand perturbative extensions (not part of this course).
2.7.1 Unitary transformations
Hamiltonians that conserve particle number can always be written
H =∑ij
Hij a†i aj.
For H to be Hermitian, Hij must be a Hermitian matrix, implying i. that it has realeigenvalues and ii. that it can be diagonalized by a unitary transformation to a new basis.Let α†i = Uij a
†j be a creation operator in such a basis ⇒ a†j = U †jiα
†i . Substituting into H,
H =∑ij
Hij a†i aj =
∑ijkl
HijUikα†kαlU
†lj =
∑kl
(∑ij
HijUikU†lj
)︸ ︷︷ ︸
(UTHU∗)kl−diagonal
α†kαl =∑l
εlα†l αl
Solving a problem defined by a particular H amounts to finding the unitary rotation thatdiagonalizes it. As we saw for the harmonic chain, often a Fourier transform forms onepart of this. In the following, we will see a number of example where there is a residualdiagonalization required - often of just a 2× 2 or 4× 4 matrix.
31
2.7.2 Bogoliubov Transformation
There are lots of physical systems for which a mean-field treatment leads to terms in theHamiltonian that are bilinear in creation or annihilation operators;
H = ε1c†1c1 + ε2c
†2c2 + λc†1c
†2 + λ∗c1c2.
These terms are present in Bose condensates (the subject of the next chapter), super-fluids, superconductors and anti-ferromagnets (a topic touched upon in Chapter 4). AHamiltonian with such terms does not preserve particle number — the groundstate is asuperposition of different numbers of particles, which occurs physically by the system ex-changing particles with the condensate or environment. A unitary transformation cannotbe used to diagonalize these Hamiltonians. Instead, one must use a Bogoliubov transfor-mation. This takes a slightly different form for bosons and fermions. The procedure is verysimilar in the two cases with some additional negative signs in the latter case.
Bosonic BogoliubovThe bosonic form of the Bogoliubov transformation is used for example for Bose condensatesand anti-ferromagnets. After a Fourier transform, the Hamiltonian takes the following form(usually with additional momentum dependence of each term that we suppress here):
H = ε(c†1c1 + c†2c2) + λ(c†1c†2 + c2c1).
This is diagonalized by a transformation of the form
c†1 = ud†1 + vd2, c†2 = ud†2 + vd1. (2.20)
There are various ways of going through the algebra to show this and to fix the coefficientsof u and v in terms of ε and λ. Here is the way that I always do it
• Introduce a matrix form
H =1
2
(c†1, c2, c
†2, c1
)ε λ 0 0λ ε 0 00 0 ε λ0 0 λ ε
c1
c†2c2
c†1
− ε (2.21)
The −ε term arises because the cc† terms are in the opposite order in half of theterms as written in this matrix form.
• The Bogoliubov transformation can be written in the formc1
c†2c2
c†1
︸ ︷︷ ︸
c
=
u v 0 0v u 0 00 0 u v0 0 v u
︸ ︷︷ ︸
M
d1
d†2d2
d†1
︸ ︷︷ ︸
d
(2.22)
The spinors c and d on the left hand right hand sides are referred to as Nambu spinors
32
• Preserving the commutation relation (which, remember, happens automatically withunitary transformations) requires
[c1, c†1] = [ud1 + vd†2, ud
†1 + +vd2]
= u2 [d1, d†1]︸ ︷︷ ︸
=1
+uv [d1, d2]︸ ︷︷ ︸=0
+uv [d†2, d†1]︸ ︷︷ ︸
=0
−v2 [d2, d†2]︸ ︷︷ ︸
=1
= u2 − v2
= 1
• Using u2 − v2 = 1 we can identify the inverse of M as
M−1 =
u −v 0 0−v u 0 00 0 u −v0 0 −v u
so that we can write the transformation, using the Nambu spinor notation, as
c = Md, c† = d†M ⇔ d = M−1c d† = c†M−1
• Next we insert resolutions of the identities MM−1 = 1 and M−1M = 1 between thespinors and the Hamiltonian matrix as follows:
H =1
2c†M−1︸ ︷︷ ︸
d†
M
ε λ 0 0λ ε 0 00 0 ε λ0 0 λ ε
M
︸ ︷︷ ︸H′
M−1c︸ ︷︷ ︸d
−ε
• The final task is to write out the components of H ′ and choose u and v so that theoff-diagonal elements are zero. We only need to focus on the 2× 2 block(
u vv u
)(ε λλ ε
)(u vv u
)=
(ε[u2 + v2] + 2λuv 2εuv + λ[u2 + v2]2εuv + λ[u2 + v2] ε[u2 + v2] + 2λuv
)• This can be solved by identifying u = cosh θ and v = sinh θ (so that u2 − v2 = 1)
and using the double angle formula for sinh θ and cosh θ to get u2 + v2 = cosh 2θ and2uv = sinh 2θ. Finally setting −λ/ε = tanh 2θ, we obtain
H = ε(d†1d1 + d†2d2
)− ε+ ε
ε =√ε2 − λ2
33
Fermionic BogoliubovThis is used for example for superconductivity and superfluidity of fermions (e.g. in 3He).The steps are very similar to those for bosons, but crucially, negative signs from the anti-commutation of fermionic operators changes things somewhat. The Hamiltonian takes thesame form as before, with the following Nambu spinor representation:
H =1
2
(c†1, c2, c
†2, c1
)ε λ 0 0λ −ε 0 00 0 ε −λ0 0 −λ −ε
c1
c†2c2
c†1
+ε
The opposite signs obtained for fermions are highlighted in red. These arise directly fromthe commutation relations, e.g. λc1c2 = −λc2c1 leads to the opposite sign for some of theλ terms. Similarly εc†1c1 = ε(1− c1c
†1) leads to a +ε rather than −ε at the end.
• This is diagonalised by a Bogoliubov transformation of the formc1
c†2c2
c†1
=
u v 0 0−v u 0 00 0 u −v0 0 v u
d1
d†2d2
d†1
(2.23)
The manipulations from here are rather similar
• The anti-commutation relations must be preserved by the fermionic Bogoliubov trans-formation. This implies that
c†1, c1 = ud†1 + vd2, ud1 + vd†2 = u2d†1, d+ v2d2, d†2 = u2 + v2 = 1,
which suggests taking u = cos θ and v = sin θ.
• Clearly M−1 = MT allowing us to diagonalize the Hamiltonian as follows:
H =1
2c†M︸︷︷︸d†
MT
ε λ 0 0λ −ε 0 00 0 ε −λ0 0 −λ −ε
M
︸ ︷︷ ︸H′
M−1c︸ ︷︷ ︸d
+ε
• As in the bosonic case, the final task is to write out the components of H ′ and chooseu and v so that the off-diagonal elements are zero. We only need to focus on the 2×2block(
u −vv u
)(ε λλ −ε
)(u v−v u
)=
(ε[u2−v2]−2λuv 2εuv + λ[u2−v2]2εuv + λ[u2−v2] −ε[u2−v2] + 2λuv
)which leads to the conditions tan 2θ = −λ/ε and ε = cos 2θε− sin 2θλ =
√ε2+λ2
• The resulting Hamiltonian is
H = ε(d†1d1 + d†2d2
)+ε−ε
ε =√ε2+λ2
34
Chapter 3
The Weakly-Interacting Bose Gas
Superfluidity was first observed in liquid helium below about 2.1K. Similar phenomena - atleast the phenomenon of Bose condensation - have since been observed in ultra-cold atomicgases and quantum magnets. After the rather formal interlude of the previous chapter, wenow poses the analytical tools to describe Bose condensation in detail.
3.1 The HamiltonianThe Hamiltonian for the weakly interacting Bose gas contains a kinetic term describing themotion of non-relativistic bosons in free space and a point repulsion between them (i.e theparticles excert no force upon one another until they are immediately adjacent and thenthey repel).
H = HKE +Hint =∑k
~2k2
2mc†kck +
U
2V
∑k,p,q
c†kc†pcqck+p−q
To understand the form of the repulsive interaction, we first note that it can be written infirst quantized form as
Hint =U
2
∑i 6=j
δ(xi − xj)
by which we imply that it is diagonal in the position basis and given by the product of thedensities at two different points. In second quantized form, in the position basis, then wehave
Hint =U
2
∫dxdy δ(x− y)n(x)n(y),
35
where n(x) = c†xcx is the number operator at the point x. Substituting Fourier expansionsfor the position-basis creation and annihilation operators
Hint =U
2
∫dxdy δ(x− y)
∫dk
(2π)ddp
(2π)ddq
(2π)ddl
(2π)deix.(k−p)+iy.(q−l)c†kcpc
†qcl
=U
2
∫dk
(2π)ddp
(2π)ddq
(2π)ddl
(2π)d
∫dxeix.(k−p+q−l)︸ ︷︷ ︸
=(2π)dδ(k−p+q−l)
c†kcpc†qcl
=U
2
∫dk
(2π)ddp
(2π)ddq
(2π)dc†kcpc
†qck−p+q
Notice that the momenta of the ingoing bosons that are annihilated matches that of theoutgoing, created bosons. This can be represented by a Feynman diagram:
p k q k + q p
p k q k + q p
p k q k + q p
p k q k + q p
3.2 Mean Field TheoryAt the end of Chapter 2, we learnt how to diagonalize a quadratic, second quantized Hamil-tonian. In the present case, however, the interaction is quartic in creation and annihilationoperators. Based upon our knowledge of the non-interacting Bose gas, we can develop aleading-order approximation that reduces the Hamiltonian to quadratic form.
Non-interacting Bose Gas: All particles in k = 0 state.
Interacting Bose Gas:Assume macroscopic occupation of the k = 0 state; 〈c†0c0〉 = N0. A number of steps followfrom this
• Replace c0 by a c-number (commuting number rather than operator) co →√N0.
• Expand H in powers of N0
Hint =UN2
0
2V+UN0
2V
∑k 6=0
(4c†kck + c†kc
†−k + ckc−k
)+O(1)
• The total number of particles N can be expressed as
N = N0 +∑k6=0
c†kck
36
• Substitute back and retain quadratic terms
Hint =UN2
2V+UN
2V
∑k 6=0
(2c†kck + c†kc
†−k + ckc−k
)• Adding to the non-interacting part of the Hamiltonian
H = VUρ2
2+∑k 6=0
[(~2k2
2m+ Uρ
)c†kck +
Uρ
2(c†kc
†−k + ckc−k)
](3.1)
• This has precisely the same form as the Hamiltonian considered in 2.7.2. It can bediagonalised with a Bogoliubov transformation resulting in a diagonal Hamiltonian
H =′∑
k 6=0
[εk(α†kαk + α†−kα−k) + εk −
~2k2
2m
],
εk =
[(~2k2
2m+ Uρ
)2
− (Uρ)2
]1/2
(3.2)
where∑′ indicates that the summation be carried out wit kx > 0 say and the point
k = 0 excluded.
3.3 Landau’s Critical Superfluid VelocityThe dispersion of excitation found above has different behaviour at large and small mo-menta;
εk ∼~2k2
2mlarge |k|
(such that
~2k2
2m>> Uρ
)∼
√Uρ
m|k| small |k|
The latter dependence at low |k| is responsible for the superfluid properties as can be seenby the following argument due to Landau:
Consider a superfluid of total mass M and velocity v. The only way that friction can ariseis by exciting some of the k 6= 0 modes that we have revealed above. Suppose producingone such mode reduces the bulk velocity to v −∆v.
Conservation of Momentum:
Mv = Mv −M∆v + ~k
Conservation of Energy:1
2Mv2 =
1
2M(v −∆v)2 + εk
Together which imply~k.v = εk
The quantity v.k takes different values depending upon the relative direction of v and kand is maximum for k parallel to v.
37
k
v.k
|k|
If |v| < εk/(~|k| for all k, then no bosons are excited and there is no friction. This impliesa critical velocity for superfluid flow
vc = mink
(εk~|k|
). (3.3)
From our previous results εk ≈√Uρ/m~k so that the critical velocity is given by
vc =
√Uρ
m.
The finite interaction crucially changes the properties of the Bose system allowing thepossibility of superfluidity.
3.4 Superfluid FractionThe Hamiltonian is diagonalized by a Bogoliubov transformation from the original Baseoperators ck to new ones αk with dispersion εk given by Eq.(3.2). Since the Hamiltonianfor the latter is diagonal, we can easily write an expression for the occupation of thesemodes at finite temperature in terms of the Bose distribution function
〈α†kαk〉 = nB(εk) =1
eβεk − 1.
This can be used to calculate thermodynamic properties of the Bose gas, such as thesuperfluid density;
ρ0 = ρ−∑k 6=0
〈c†kck〉. (3.4)
The trick is to use the form of the Bogoliubov transformation to write the expectationin Eq.(3.4) in terms of the α operators. The solution to this is the subject of one of theproblem sets.
38
Chapter 4
Quantum Magnets
In this chapter, we will use the language of second quantization to discuss properties ofquantum magnets — magnets in which atoms in a lattice have unpaired electron spins thatmay interact with the spins on other sites. We will focus upon insulating magnets in whichthe electrons that carry the spins are fixed in a give atomic orbital, as opposed to itinerant,where they would be free to hop around.
Despite the apparent simplicity of this set up, the collective quantum behaviour of suchsystems is incredibly rich. Just about every phenomenon of modern quantum physics isrevealed by these systems — they provide the first example of topology in quantum physics(through the Haldane conjecture that we will discuss later) and a magnetic model (due toKitaev) is the basis of some of the most widely studied quantum error correcting codes.The prototypical model of this behaviour is deceptively simple. The complications are alldown to emergent collective behaviour.
4.1 The Heisenberg modelThe Heisenberg model is the simplest model of quantum magnetic behaviour. Its Hamil-tonian is given by
H =∑〈ij〉
JijSi · Sj, (4.1)
where i, j are the sites of some lattice, it e.g. 2d sqare or triangular, 3d cubic etc. 〈i, j〉indicates a sum over neighbouring sites. Jij are exchange interactions between 2 spinson the sites i and j. Negative couplings favour parallel spins and ferromagnetic ordering,positive couplings favour anti-ferromagnetic couplings.
Commutation Relations: The spin operators obey commutation relations[Sµi , S
νj
]= i~εµνλSλi δij
The consequences of these are not trivial and render the sate-space and dynamics consider-ably more complicated than that of bosons. Classically, spin/angular momentum is a vectorof any length in any direction. Quantum mechanics restricts the length and direction:
39
Quantum Numbers: The total spin operator and its projection onto a particular axis (saythe z-axis) form a commuting set of operators
[S2, Sz
]= 0 with which we may associate
quantum numbers
S2|S,m〉 = ~2S(S + 1)|S,m〉Sz|S,m〉 = ~m|S,m〉 (4.2)
where m ∈ −S,−(S − 1), ...S i.e. a tower of 2S + 1 states. As you will recall fromprevious quantum mecahnics courses, this can be derived by constructing ladder operators
S± + Sx ± iSy (4.3)
that move states up and down this tower of states:
S±|S,m〉 = ~√
(S ∓m)(S ±+1)|S,m± 1〉S+|S, S〉 = 0
S−|S,−S〉 = 0 (4.4)
This algebra is actually very hard to deal with in general for systems of interacting spins.Notice, though, that the top and the bottom of the tower of states look rather like thebosonic state space. If the physics is such that these states dominate the behaviour, thenwe can make approximations that map our Hamiltonian based upon spin operators to onebased upon bosonic operators.
40
Origin of the Heisenberg ModelVarious different mechanisms collected together under the heading "exchange" con-tribute to the interaction between spins in the Heisenberg model
a) Direct Exchange:This leads to anti-ferromagnetic coupling and ultimately has the same origin asHund’s rul in atoms (though it has the opposite effect). Consider two neighbour-ing sites:
Symmetric spins⇒ antisymmetric spatial wavefunctiona ⇒ Node at x1 − x2 = 0 ⇒ Lower Coulomb
anti-symmetric spins⇒ symmetric spatial wavefunction ⇒ no node in wavefunction ⇒ Higher CoulombThis is also called potential exchange, because it relies upon the difference in potentialenergy between 2 configurations.
b) Super/Kinetic Exchange:This type of exchange is driven by virtual hopping between lattice sites. This spreadsthe electronic wavefunction and so lowers its kinetic energy.
Symmetric spins⇒ Pauli exclusion forbids double occupancy and there is no virtual hopping.
anti-symmetric spins⇒ Virtual hopping allowed ⇒ wavefunction spreads and lowers kinetic energy
[Anderson, PhyS. Rev. 79, 350 (1950)]aElectrons are fermions
4.2 Holstein-Primakoff transformationThe non-trivial commutation relations of the spin operators make them tricky to deal with.Bosonic operators are much easier and the Holstein-Primakoff transformation allows a wayto swap between the two.
Sz = S − b†b
S+ =√
2S
(1− b†b
2S
)1/2
b
S− =√
2Sb†
(1− b†b
2S
)1/2
(4.5)
41
One can easily check1 that the bosonic commutation relations [b, b†] = 1 reproduce thespin commutation relations in the form
[S+, S−
]= 2Sz. Identifying our boson operator
per site of the lattice, the fact that spin operators on different sites commute is readilyaccommodated by the similar commutation of bosonic operators on different sites.
Approximate Form:In the limit that the expectations of Sz on each site are near to S, there are very fewexcitations in the bosonic representation. In this case, the mapping may be reduced to
Sz = S − b†bS+ =
√2Sb
S− =√
2Sb† (4.6)
which amounts to turning the bosonic tower of states upside down and matching it withthe top of the tower of spin states. In physical situations in which only a few bosons areexcited, i.e. 〈b†b〉 1, the fact that the bosonic tower of states does not terminate does notshow up. We shall see now how this approximation allows us to calculate certain propertiesof magnetic systems.
4.3 The Heisenberg ferromagnetConsider an insultating magnet in which all of the nearest neighbour interactions are fer-romagnetic, i.e. in which J < 0.
4.3.1 The Groundstate
The groundstate is such that all of the spins are aligned — say in the z-directions — sothat Szx|S, S〉 = ~S|S, S〉 on each site and
|ψGroundstate〉 =⊗x
|S, S〉
This state is an eigenstate of the Heisenberg model. To see this, we write the Heisenbergmodel in the form
H = −J∑〈x,y〉
Sx · Sy = −J∑〈x,y〉
[SzxS
zy + (S+
x S−y + S−x S
+y )/2
]so that
H|ψGroundstate〉 = −J∑〈x,y〉
S2|ψGroundstate〉,
where we have used the fact that S+|S, S〉 = 0.1And you should!!
42
4.3.2 Excitations
Excitations about this groundstate can be found using the approximate form of the HolsteinPrimakoff transformation Eq(4.5), after which the Hamiltonian reduces to
H = −J∑〈x,y〉
S2 − JS∑〈x,y〉
(b†xby + b†ybx − b†xbx − b†yby
)and the groundstate
⊗x |S, S〉 becomes the vacumm state of the bosons
⊗ |0〉.Our task now is to diagonalize this Hamiltonian. Because the system is translationally
invariant, the first step towards this (actually the only step in this case) is to Fouriertransform. Let us do this for the simple cubic lattice, taking the Fourier transform relations
bx =1√N
∑k
e−ik·xbk bk =1√N
∑x
eik·xbx
where N is the total number of sites in the lattice, k = 2π(n,m, l)/L with n,m and lintegers and L the size of the lattice. With these definitions, we find
H = −1
2JS2Nz − JS
N
∑x,d
∑k,q
eix·(k−q)[eid·q − 1]b†kbq
= −1
2JS2Nz − JS
N
∑k
∑d
[eid·k − 1]b†kbk
where d is the displacement between adjacent lattice sites and z is the coordination numberof a site of the lattice. We may summarize this result in 3d as
H = − 1
2JS2Nz︸ ︷︷ ︸const
+∑k
εkb†kbk
εk = 2J [3− cos(akx)− cos(aky)− cos(akz)] .
We have reduce our description of the excitations to that of a bosonic, harmonic osicllator-like mode. Notice that since the ferromagnet is an eigenstate of the Hamiltonian, thereare no zero-point fluctuations. This is quite a special case. The bosonic modes have lowenergy dispersion
εk ∼ JSk2.
These modes are gapless — their energy goes to zero as |k| does. This is because theyare Goldstone modes; they arise because the spin rotational symmetry of the Hamiltonianis broken by the groundstate — it costs zero energy to rotate between these groundstatesand this coresponds to precisely the k = 0 limit of the excitations described here. The factthat the dispersion is quadratic (and not linear as for phonons) as |k| → 0 is because timereversal symmetry is broken (remember that angular momentum and hence spin reversesign under time-reversal).
43
4.3.3 Thermal Fluctuations
The breaking of rotational symmetry by the ferromagnetic state can be quantified by anorder parameter - the magnetization. The free energy of the system may in fact be ex-panded as a function of this order parameter. This powerful technique — known as aGinzburg-Landau expansion — can reveal key generic properties of thermodynamic phasesand transitions between them. Here we shall satisfy ourselves by considering the effect ofthermal fluctuations upon the magnetization of our system.
The magnetization per site (assuming polarization in the z-direction)is given by
M =1
N
∑x
〈Szx〉.
Using the Holstein-Primakoff transformation, Eq.(4.5) this can be written in terms of thebosonic operators as
M =1
N
∑x
〈S − b†xbx〉
= S − 1
N2
∑x
∑k,q
ei(k−q)·x〈b†kbq〉
= S − 1
N
∑k
〈b†kbk〉,
where we have also Fourier transformed as our Hamiltonian is diagonal in momentumspace. This latter fact, combined with the fact that there are not a fixed number of bosonsresults in the expectation 〈b†kbk〉 begin given by the Bose distribution. The magnetisationis therefore given by
∆M =1
N
∑k
1
eβεk − 1
Taking the continuum limit
=1
Ω
∫BZ
ddk1
eβεk − 1
Ω is the Brillouin zone volume, and εk ∼ SJk2
= T
∫ √T/SJ
0
kd−1dk
SJk2
∼
T[
1k
]√T/SJ
0d = 1
T [log[k]]
√T/SJ
0 d = 2
T [k]
√T/SJ
0 d = 3
The integral diverges at low k — called an IR (infra-red) divergence — in one and twodimensions. It converges in three dimensions so that ∆M ∼ T 3/2. The divergence in d = 1and 2 shows that fluctuations overwhelm antiferromagnetic order - an illustration of theMermin-Wagner theorem that a continuous symmetry cannot be broken in d ≤ 2 at finitetemperature.
44
4.4 The Heisenberg Anti-ferromagnetNext we shall consider the anti-ferromagnetic Heisenberg model on a bi-partite lattice
H = J∑〈x,y〉
Sx · Sy = J∑〈x,y〉
[SzxS
zy + (S+
x S−y + S−x S
+y )/2
]. (4.7)
A bi-partite lattice is one whose sites can be divided into two sets such that sites on oneset are only nearest neighbours in the other. The square, cubic and hypercubic lattice areexamples.
4.4.1 The Néel state
At the classical level, the Hamiltonian is minimised when the spins on the different sub-lattices are anti-parallel. This is known as the Néel state. On the square lattice it takesthe form
.
|ψNeél〉 =⊗x
|S,±S〉,
where the two different signs occur for the different sub-lattices. Unlike the ferromagnet,this is not an eigenstate of the Hamiltonian. This is demonstrated by the action of theS+x S−y terms. In the case of the ferromagnet, such terms always give zero as on one of the
sites S+|S, S〉 = 0. In the anti-ferromagnetic case, however, such terms are not zero —they generate quantum fluctuations away from the classical Néel state.
4.4.2 Holstein-Primakoff transformation
To study these quantum fluctuations away from the Néel state we use the Holstein-Primakofftransformation. The transformation must be applied slightly differently on the two sub-lattices.
A sub-lattice |S, S〉:
Sz = S − a†aS+ =
√2Sa
S− =√
2Sa†
45
B sub-lattice |S, S〉:
Sz = a†a− SS+ =
√2Sa†
S− =√
2Sa
This amounts to matching the bosonic tower of states to the spin tower of states from thebottom up in one case and from the top down in the other:
|S, Si
|S,Si
|0i |1i |2i|0i |1i |2i
|0i |1i |2i
|0i |1i |2i|0i |1i |2i
|0i |1i |2i
|S, Si
|S,Si
A sites B sitesA sites B sites
.
Substituting these transformations into the Heisenberg Hamiltonian, we find
H = −J∑〈x,y〉
S2 + JS∑〈x,y〉
[a†xax + a†yay + a†xa
†y + axay
].
The easiest way to proceed now is to introduce a coordinate z that runs over all of thelattice sites and a coordinate d that runs over nearest neighbour vectors. With these, theHamiltonian can be written
H = −JS2Nz
2+JS
2
∑z,d
[2a†zaz + a†za
†z+d + azaz+d
],
which after Fourier transform becomes
H = −JS2Nz
2+JSz
2
∑k
[a†kak + a†−ka−k + γk
(a†ka
†−k + aka−k
)],
where2 γk = 1z
∑d cos(k · d). This may be written in matrix form as
H = −JS(S + 1)Nz
2+JSz
2
∑k
(a†k, a−k
)( 1 γkγk 1
)(aka†−k
)
2Note that since d takes pairs of values on the hypercubic lattice, γk = 1z
∑d e
ik·d = 1z
∑d cos(k · d)
46
An alternativ approach — used in John Chalker’s notes, for example — is to rotatethe spins on the B sub-lattice
Sz → −Sz, Sx → −Sx, Sy → Sy
which preserves the commutation relations. The Hamiltonian then becomes
H = −J∑〈xy〉
[SzxS
zy + (S+
x S+y + S−x S
−y )/2
]and the Holstein-Primakov transformation can be used in the same form for eachsub-lattice.
4.4.3 Bogoliubov transformation
Evidently the Fourier transform alone is not enough to diagonalize the Hamiltonian. Theanomalous terms such as aka−k require us to use a Bogoliubov transfromation as developedin section 2.7.2. The result is.
H = −1
2JS(S + 1)Nz +
∑k
εk
(α†kαk + 1/2
)with
εk = JSz√
1− γ2k ∼ |k| at small k
and the Bogoliubov transformation given by(aka†−k
)=
(uk vkvk uk
)(αk
α†−k
)uk = cosh(θk)vk = − sinh(θk)
tanh(2θk) = −γk
Notice that the low-frequency dispersion is linear, unlike the ferromagnet. This is becausethe Néel state does not break time-reversal symmetry in a macroscopic sense (time-reversalsimply swaps the sublattices).
4.4.4 Fluctuations of the Antiferromagnet
As we have already noted, the classical Néel state is not an eigenstate of the anti-ferromagneticHeisenberg model. This has several consequences. Firstly, it leads to a zero-point contribu-tion to the energy. These zero-point fluctuations also reduce the sub-lattice magnetizationeven at zero temperature.
The average (staggered) magnetization per site is given by
S −∆S = S − 1
N
∑z
〈a†zaz〉 = S − 1
N
∑k
〈a†kak〉.
47
Since the Hamiltonian is diagonal in terms of α and α†, we must re-write the magnetizationin terms of them
S −∆S = S − 1
N
∑k
〈(ukα†k + vkα−k)(ukαk + vkα†−k)〉
= S − 1
N
∑k
u2k〈α†kαk〉+ v2
k〈α−kα†−k〉+ ukvk(〈α†kα†−k〉︸ ︷︷ ︸=0
+ 〈α−kαk〉︸ ︷︷ ︸=0
)
= S − 1
N
∑k
(u2k〈α†kαk〉+ v2
k〈α†−kα−k〉+ v2k
)At zero temperature, the number of bosons is zero. The constant, zero-point term is stillpresent, however;
∆S =1
N
∑k
v2k
=1
2Ω
∫BZ
dk
[1√
1− γ2k
− 1
]At small |k|, γk ∼ 1− |k|2/z
=1
4zΩ
∫dk kd−2
This integral is infra-red convergent for d ≥ 2, but logarithmically divergent in 1d (it alwaysconverges in the ultra-violet because of the cut-off at the Brillouin zone edge). Cutting theintegral explicitly with π/a at the Brillouin zone boundary and at π/L for a system of sizeL, we find
∆S ∼ 1
2π
∫ π/a
π/L
dk/k ∼ 1
2πlog(L/a).
The quantum fluctuations diverge as L increases such that the sublattice magnetization→ 0 when S = 1
2πlog(L/a). Rearranging, we may say that the lengthscale at which this
occurs is given by ξ ∼ ae2πS is approximately the correlation length in the one-dimensionalsystem. This result also demonstrates that the effects of quantum fluctuations are mostdramatic for small S.
4.4.5 Haldane’s Conjecture
Whilst the analysis given above is qualitatively correct for half-integer spins, it misses somekey physics for integer spins. The distinction was first realised by Duncan Haldane [Phys.Rev. Lett 50, 1153 (1983)] who, by considering topological effects of quantum spins knownas instantons, was able to argue that whilst 1/2 integer spin anti-ferromagnets are gapless,integer spin anti-ferromagnets have a gap to the first excitation above the groundstate.The following two sections consider briefly two different ways of studying the properties ofinteger and 1/2-integer spin chains. The details are not examinable per se (though one ofthe problems sets contains an example of the use of the Jordan-Wigner transformation).
48
4.4.6 Jordan-Wigner Transformation
A spin 1/2 has two states. Since fermionic creation and annihilation operators also havetwo states, it is tempting to try to construct a fermionic representation. It is natural toidentify
Szm = c†mcm − 1/2, with S+m ∝ c†m S−m ∝ cm.
However, although on a given site everything works fine (for example S+m, S
−m = cm, c†m =
1 ), there is a problem for pairs of spins on different sites. Spin operators at different sitescommute, but fermion operators on different sites anti-commute.
This is fixed by the Jordan-Wigner transformation:3
S+m = c†m
∏l<m
(1− 2nl)
S−m =∏l<m
(1− 2nl)cm
Szm = nm − 1/2 (4.8)
The strings,∏
l<m(1− 2nl), lead to additional negative factors that cancel those from theanti-commutation of fermionic operators:
c†m(1− 2nm) = −(1− 2nm)c†mcm(1− 2nm) = −(1− 2nm)cm
These expressions can be checked by evaluating the expectation between two fermionicFock states.
XY anti-ferromagnetLet us apply this method to the Hamiltonian
H = J∑n
[1
2(S+
n S−n+1 + S−mS
+n+1) + ∆SznS
zn+1
].
For ∆ = 1 this is the Heisenberg model, and for ∆ = 0 it is an XY model. It turns outthat we can solve the problem in the latter case. Using the Jordan-Wigner transformation,the Hamiltonian reduces to
H =J
2
∑n
[(c†ncn+1 + c†n+1cn)︸ ︷︷ ︸fermionic hopping
+ 2∆(c†ncn − 1/2)(c†n+1cn+1 − 1/2)︸ ︷︷ ︸interaction
]For ∆ = 0 this reduces to a model of non-interacting fermions hopping along the 1d line.A Fourier transform completes the diagonalisation of the problem, giving
H =∑k
J cos(ak)c†kck
3NB: one may also write this as S+m = c†m exp[iπ
∑l<m(1 − 2nl) and S+
m = exp[−iπ∑l<m(1 − 2nl)cmas can be verified by considering the action on fermionic Fock states.
49
We have reduced the problem to one of non-interacting fermions with dispersion ε(k) =cos(ak). Up to now, diagonalization has reduced our problems to non-interacting bosonswhose groundstate consists of the absence of bosons. In the non-interacting fermion case,the groundstate is given by a filled fermi sea — fermion states with ε(k) < 0 are occupiedand those with ε(k) > 0 are empty.
Brillouin Zone
States Occupied in groundstate
k
(k)
States Occupied in groundstate .
Spin Correlations:In the same way as the properties of a bosonic system or the fluctuations of the anti-ferromagnet after the Holstein-Primakoff transformation can be calculated by writing op-erators in terms of the bosons, so in the present case, we can calculate properties of theXY anti-ferromagnet by writing operators in terms of the fermion operators. Let us focusupon the correlations of the the z-component of spin in two different places4.
1
N
∑y
〈Szy Szy+x〉 =1
N
∑y
(〈c†y cy c†y+xcy+x〉 −
1
2〈c†y cy + c†y+xcy+x〉+
1
4
)
=
1
N2
∑k1,k2,k3,k4
〈c†k1 ck2 c†k3ck4〉eix(k4−k1)δ(k1 − k2 + k3 − k4)− 1
N
∑k
〈nk〉︸ ︷︷ ︸N/2
+1
4
There are two sets of contributions to this correlator: i. k1 = k2 and k3 = k4 with both
4Other correlators can be calculates, but involve strings between the points being correlated and so area little more complicated
50
electronic states occupied - cancels −1/4. ii. k1 = k4 occupied k2 = k3 unoccupied
1
N
∑y
〈Szy Szy+x〉 =1
N2
∑k1,k2
〈c†k1 ck2 c†k2ck1〉︸ ︷︷ ︸
〈c†k1 ck1 〉〈c†k2ck2 〉
eix(k2−k1)
=1
N2
∑k1,k2
nF (k1) (1− nF (k2)) eix(k2−k1)
=1
(2π)2
∫ π
−πdk1
∫ π
−πdk2nF (k1) (1− nF (k2)) eix(k2−k1)
= +1
(2π)2
∫ π
−πdk1
∫ π
−πdk2 (1− nF (k2)) eix(k2−k1)
− 1
(2π)2
∫ π
−πdk1
∫ π
−πdk2 (1− nF (k1)) (1− nF (k2)) eix(k2−k1)
= +1
(2π)2
∫ π
−πdk1e
−ixk1∫ π/2
−π/2dk2e
ixk2 − 1
(2π)2
∫ π/2
−π/2dk1e
−ixk1∫ π/2
−π/2dk2e
ixk2
= +1
(2π)2
∫ π
−πdk1e
−ixk1︸ ︷︷ ︸=2πwhen x→0
[2 sin(xπ/2)
x
]︸ ︷︷ ︸
≈2πδ(x)
− 1
(2π)2
[2 sin(xπ/2)
x
] [2 sin(xπ/2)
x
]
= δ(x)− 1
π2x2sin2(πx/2a)
where we have used the zero temperature limit in the last step so that n(k) = 1 forπ/2 < k < π and −π < k < −π/2 and zero otherwise. The resulting power-law decay 1/x2
is very different from the Néel state.
Excitations:The above calcuation shows that correlations in the groundstate of the 1d anti-ferromagnetare very different from the Néel state. The excitations are also very different from thespinwave excitations of the Néel state found in higher dimensions. The excitations at agiven wavevector q involve exciting a fermion from an occupied state to an unoccupiedstate — a particle-hole pair.
k
(k)
k
(k)
.
For a given wavevector q the energy can be distributed in a range of ways between theparticle and hole, giving a range of energies for each q.
51
kE
xci
tati
on
Ener
gy
.
This characteristic dispersion of energy is very different from the sharply defined disper-sion of spinwaves. It is frequently observed in neutron scattering experiments on one-dimensional magnets.
4.4.7 Integer Spin Case
In the above, we were able to discuss the properties of the spin 1/2 chain by consideringthe simplified, XY spin chain. Similarly, a simplified version of the spin 1 chain enables usto appreciate some of its properties. This is known as the AKLT (after Affleck, Kennedy,Lieb and Tasaki). The Hamiltonian is given by
H =∑n
[J1Sn · Sn+1 + J2(Sn · Sn+1)2
]and is solved by a cunning trick — the spin 1 on each site is considered to be constructedfrom the triplet sector of 2 spin 1/2. It turns out that the groundstate of this model canbe constructed by putting two of these fractional spins on neighbouring sites into a singletstate. This can be illustrated schematically as
" # " # " #S = 1 on site
| "#i| #"i
| "#i + | #"ip2
singlet correlation
| "#i | #"ip2
.
The gap to the first excitation above the groundstate can be understood as the energyrequired to break the singlet correlation between sites.
Notice that there are unpaired spins 1/2 at the ends of the chain. This is the first exampleof a topological edge state. The study of such states is an extremely active area of currentresearch, in part because of their possible use in quantum information processing.
52
Chapter 5
The Renormalization Group
The ideas of thermodynamics — developed by the Victorians intent upon optimising steamengines — provide some of the most fundamental notions of science. Indeed, it is not toomuch of a stretch to argue that they form a meta-theory of physics. All of the develop-ments of theoretical physics since this — quantum mechanics, special and general relativity,string theory, information theory etc — must obey the laws and constraints imposed bythermodynamics. The last piece of the story of thermodynamics was provided by a set ofideas known as the renormalisation group, first developed by Wilson and Fisher This isreally a collection of methods that embody the question,
" How does the behaviour of a system change when viewed on different length or timescales?"
Addressing this allowed a complete classification of classical phase transitions and wasessential in resolving the thorny issue of infinities in quantum field theory.
5.1 The 1d Classical Ising ModelIn order to give a feel for the renormalization group, we will consider a simple model thatcan be solved exactly — the 1d Ising model. This is model of classical spins σn = ±1 whoseHamiltonian is given by
H =∑n
(−Jσnσn+1 − hσn) . (5.1)
For reasons that will become apparent in a moment, we consider N spins, where N isan integer power of 2. The partition function for this model (from which all physicalobservables can be calculated) is given by
Z =∑σn
e−βH. (5.2)
The fundamental question of the renormalisation group is to ask how this model behaveson different length scales. In order to do this, as a first step, we will carry out the sum over
53
spins on every other site. Summing over spins on odd sites we find
Z =∑σn
eβ∑n(Jσnσn+1+hσn)
=∑σ2m
∑σ2m+1
eβh(σ0+σN )/2︸ ︷︷ ︸See note
exp
[β∑m
[J(σ2mσ2m+1 + σ2m+1σ2m+2) + hσ2m+1 + h(σ2m + σ2m−2)/2]
]
=∑σ2m
eβh(σ0+σN )/2 . . .
eβ[J(σ2m+σ2m+2)+h]
+e−β[J(σ2m+σ2m+2)+h]
eβh(σ2m+σ2m+2)/2 . . .
NB: the spins at the ends enter differently. We ignore this in what follows as we areinterested in the thermodynamic limit. It turns out after summing over the odd spins, thepartition function over the even spins takes the same form (up to an overall constant factorthat doesnt change the physics) with modified constants.
We can find the modified constants as follows:
• Compare
Ceβ′(J ′σ2mσ2m+2+h(σ2m+σ2m+2)/2) = eβ(h/2+J)(σ2m+σ2m+2)+βh + eβ(h/2−J)(σ2m+σ2m+2)−βh
• This must be true for all σ2m and σ2m+2. Comparing explicit expressions for differentvalues of σ2m and σ2m+2
σ2m = σ2m+2 = 1 Ceβ′(J ′+h′) = e2β(J+h) + e−2βJ (5.3)
σ2m = σ2m+2 = −1 Ceβ′(J ′−h′) = e−2βJ + e2β(J−h) (5.4)
σ2m = −σ2m+2 = 1 Ce−β′J ′ = eβh + e−βh (5.5)
• Solving for C, β′J ′ and β′h′
(5.3)/(5.4) e2β′h′ =e2β(J+h) + e−2βJ
e−2βJ + e2β(J−h)
(5.3)(5.4)/(5.5)2 e4β′h′ =
(e2β(J+h) + e−2βJ
) (e−2βJ + e2β(J−h)
)(eβh + e−βh)2
(5.3)(5.4)(5.5)2 C4 =(e2β(J+h) + e−2βJ
) (e−2βJ + e2β(J−h)
) (eβh + e−βh
)2
• Or alternatively, in terms of x = e−4βJ , y = e−2βh and z = C−4
x′ =x(1 + y)2
(x+ y)(1 + xy)
y′ =y(x+ y)
1 + xy
z′ =z2xy2
(x+ y)(1 + xy)(1 + y)2(5.6)
54
The equations (5.6) are known as the RG flow equations. They tell us how the parame-ters of the Hamiltonian change under our RG transformation (the particular transformationused here is known as decimation). Notice that in this case, the equations for x and y areindependent of z so we can study the renormalization in the xy-plane only.
Fixed Points:Fixed points are an important concept in using the renormalisation group. They representsets of parameters or Hamiltonians that do not change under renormalisation. The physicsis then said to be scale invariant and typically falls into classes of universal behaviour thatare independent of minor short-distance variations in the starting Hamiltonian. To findthe fixed points, we take x = x′ and y = y′. The solutions are
(x∗, y∗) = (1, 0 ≤ y∗ ≤ 1) line of fixed points with T →∞= (0, 1) fixed point with h = 0, T = 0
= (0, 0) fixed point with h =∞, T = 0 (5.7)
The flow to these fixed points can be represented diagrammatically as follows:
ferromagnetic fixed point
par
amag
net
icline
ferromagnetic fixed point
y x
y x
The results presented here were first derived in a seminal paper of Nelson and Fisher AnnPhys 91, 226(1975).
55
