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Quantum Physics in One Dimension:
Basic Training in Condensed Matter Physics
Erich Mueller
Lecture Notes: Spring 2018
Preface
About Basic Training
Basic Training in Condensed Matter physics is a modular team taught course
offered by the theorists in the Cornell Physics department. It is designed to
expose our graduate students to a broad range of topics. Each module runs
2-4 weeks, and require a range of preparations. This module, “Greens Function
Approach to Transport,” is designed for students who have completed a standard
one semester graduate solid state physics course. It assumes no prior exposure to
many-body Greens functions, but requires a knowledge of second quantization.
Prior Topics
2006 Random Matrix Theory (Piet Brouwer)
Quantized Hall Effect (Chris Henley)
Disordered Systems, Computational Complexity, and Information Theory
(James Sethna)
Asymptotic Methods (Veit Elser)
2007 Superfluidity in Bose and Fermi Systems (Erich Mueller)
Applications of Many-Body Theory (Tomas Arias)
Rigidity (James Sethna)
Asymptotic Analysis for Differential Equations (Veit Elser)
2008 Constrained Problems (Veit Elser)
Quantum Optics (Erich Mueller)
Quantum Antiferromagnets (Chris Henley)
Luttinger Liquids (Piet Brouwer)
2009 Continuum Theories of Crystal Defects (James Sethna)
Probes of Cold Atoms (Erich Mueller)
Competing Ferroic Orders: the Magnetoelectric Effect (Craig Fennie)
Quantum Criticality (Eun-Ah Kim)
i
2010 Equation of Motion Approach to Many-Body Physics (Erich Mueller)
Dynamics of Infectious Diseases (Chris Myers)
The Theory of Density Functional Theory: Electronic, Liquid, and Joint
(Tomas Arias)
Nonlinear Fits to Data: Sloppiness, Differential Geometry and Algorithms
(James Sethna)
2011 Practical Density Functional Theory (Tomas Arias)
Semiclassical Methods (Erich Mueller)
Ginzburg-Landau Theory of Superconductivity (Chris Henley)
Continuum Quantum Monte Carlo Methods in Chemistry and Physics
(Cyrus Umrigar)
2012 Quantum 14-15 puzzle: mechanics of a single hole in a spin-1/2 lattice
(Veit Elser)
The science of writing Science (Eun-Ah Kim)
Feynman diagrams and ab-initio computation of excited states in con-
densed matter (Tomas Arias)
Probes of Cuprate Superconductors (theorist’s view) (Chris Henley)
Physics of Life (Jane Wang)
2013 Many Body Field Theory (Erich Mueller)
The Theory of Density Functional Theory: Electronic, Liquid, and Joint
(Tomas Arias)
Conformal Symmetry (Andre Leclair)
Nonlinear Fits to Data: Sloppiness, Differential Geometry and Algorithms
(James Sethna)
2014 Rigidity (James Sethna)
Practical Density Functional Theory (Tomas Arias)
Quantum Monte-Carlo (Cyrus Umrigar)
Geometry in Quantum Mechanics (Erich Mueller)
2018 Greens Function approach to Transport (Erich Mueller)
Dynamical Systems that solve hard problems (Viet Elser)
Variational and Projector Monte Carlo Methods for Physics and Chem-
istry (Cyrus Umrigar)
Sloppy Models (Jim Sethna)
ii
Contents
Preface i
Contents iv
1 Introduction – Mar 13, 2019 1
A Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
B What is special about 1D . . . . . . . . . . . . . . . . . . . . . . 1
B.1 Fermi Liquid Theory . . . . . . . . . . . . . . . . . . . . . 1
C From “Coupled Masses” to “Classical Field Theory” . . . . . . . 4
D Quantum Coupled Masses . . . . . . . . . . . . . . . . . . . . . . 6
E HW 1 – Due March 20 . . . . . . . . . . . . . . . . . . . . . . . . 6
2 “Constructive” Bosonization – Mar 15, 2019 11
A Linearizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
B Density Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
C Klein factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
D Fermion Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 18
E Free Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
F Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3 Applications – Mar 20, 2019 24
A Occupation Number . . . . . . . . . . . . . . . . . . . . . . . . . 24
A.1 Other Correlation Function . . . . . . . . . . . . . . . . . 26
A.2 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
B HW 2 – Not Due . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4 Tonks Gas and the Leib-Lininger Model – Mar 27, 2019 33
.1 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . 33
.2 1D Hard Core Gas . . . . . . . . . . . . . . . . . . . . . . 34
A Spin Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
iii
5 Bethe Ansatz – Mar 29, 2019 38
A Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 38
B Three Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
C Thermodynamic Limit . . . . . . . . . . . . . . . . . . . . . . . . 46
iv
Chapter 1
Introduction – Mar 13,
2019
A. Reading
To learn more about Fermi liquid theory, and how it breaks down in 1D, see
Giamarchi, “Quantum Physics in One Dimension”, chapter 1. To see some
cartoon reasoning about “quasiparticles,” see Mattuck,“A Guide to Feynman
Diagrams in the Many Body Problem.”
B. What is special about 1D
This section takes some of the highlights from Giamarchi’s “Quantum Physics
in One Dimension,” chapter 1. Note, my treatment will be quite a bit simpler
than Giamarchi’s: Though if you have the background, Giamarchi tells a very
nice story.
There are at least two arguments for the importance of studying 1D systems.
The first, which is emphasized in Giamarchi’s chapter, is that 1D is special. The
traditional approaches from 3D break down in 1D. Understanding the break-
down sheds light on the original problem. The unusual physics of 1D also acts as
a template for other exotic phenomena. A second important part of 1D systems
is that we can actually calculate things in 1D. The approaches are technically
beautiful.
B.1. Fermi Liquid Theory
To appreciate the special properties of 1D systems, it is useful to learn a bit
about 3D systems. Giamarchi assumes that the reader already knows about
1
2
them, and does a fairly technical review of their properties. I will just briefly
go over the highlights.
The theory of typical 3D electron systems is “Fermi Liquid Theory.” This
approach was first put forth phenomenologically, and tested experimentally.
Later its theoretical foundations were established. These theoretical foundations
are one of the crowning achievements of many-body theory.
The central premise of Fermi Liquid Theory is that the excitations of a 3D
electron gas are in 1-to-1 correspondence with those of a non-interacting gas,
but these quasiparticles
• have a different effective mass
• have a finite lifetime (the lifetime scales as τ ∝ 1/E2, so low energy
excitations are long-lived)
• are weakly interacting (even if the interactions between the bare electrons
are strong.
The mental picture is that for a Fermi gas, the excitations are:
Imagine starting with these states and adiabatically turning on the interactions.
The resulting state is a ”quasiparticle” or ”quasihole.” One imagines that the
electron (or hole) becomes surrounded by a screening cloud.
3
Multiple representations are always useful, and one concrete picture, is to
consider the operator ck which annihilates a quasiparticle with momentum k.
This operator can be expressed in terms of the original fermion operators ak as
ck =√zak +
∑pq
fkpqak+p−1a†paq + · · · (1.1)
That is, the quasiparticle operator includes a linear combination of all terms
which have total momentum k and total particle number 1. The square of
the coefficient of the single-particle term, z, is known as the “quasi-particle
renormalization factor,” and physically corresponds to the square overlap of the
single particle state and the quasiparticle state. The central premise of Fermi
liquid theory is that z is non-zero.
Another useful picture to have in mind is the “single particle spectral den-
sity.” This is the quantity which is measured in photo-emission, or tunneling.
Imagine you try to inject an electron with momentum k and energy ω. In an
ideal gas, you will only succeed if ω = k2/2m. In an interacting gas, you in-
stead expect that your spectrum (the probability of success) will be bimodal (or
multimodal). You can add the particle, leading to a final state which contains
a single quasiparticle: this will give a Lorentzian contribution to the spectrum.
The width of the Lorentzian is related to the lifetime of the quasiparticle. The
area of the peak is related to the quasiparticle renormalization factor. Alter-
natively you can inject the electron and create a state with some number of
particle-hole excitations. This will give a broad background. Pictorially, the
graph on the left represents the spectral density of a noninteracting Fermi gas,
while the one on the right is a Fermi liquid:
The other story you will commonly see in the textbooks is the distribution
function nk, which represents the number of fermions with momentum k. In
a zero temperature non-interacting gas, nk = 1 for k < kf , and nk = 0 for
k > kf . In particular, there is a discontinuity at k = kf . In a Fermi liquid, the
discontinuity remains, and nk+f− nk−f = z:
4
In 1D we will find that z = 0, but there will still be a singularity at kf :
the occupations will have a kink there: nk ∼ n0 − (k − kf )α for k > kf , and
nk ∼ n0 + (kf − k)α for k < kf :
C. From “Coupled Masses” to “Classical Field Theory”
One of the most dramatic results that we will derive is that the excitations of a
1D fermi gas are exhausted by “phonons.” To understand what this means, it
is useful to remind ourselves where phonons come from. We will begin with a
1D model of masses connected by springs:
If we want to be able to take the continuum limit, then the masses and the
spring constants are related to the equilibrium separation by:
λ =M
d(1.2)
κ = kd. (1.3)
Here λ is the linear mass density, and κ is the stiffness.
The dynamical variables of this set of masses are the displacements from
5
equilibrium, si, and the momenta pi. The equations of motion read:
M∂2sj∂2t
= k(sj+1 + sj−1 − 2sj). (1.4)
This is solved by making the ansatz sj = ei(qdj−ωt, where xj = dj is the equilib-
rium position of the j′th mass. Substituting this into our equations of motion
gives
ω =
√k
M2 sin(qd/2) (1.5)
→ q
√kd
M/d= q
√κ
λ. (1.6)
We teach that argument in PHYS 2218 at Cornell, and it should feel pretty
familiar. One interesting way to think about things is to introduce the continu-
ous function s(x), where sj = s(xj). In the continuum limit this is a well-defined
procedure, and if you want to be pedantic there are formal ways of defining s(x).
We can relate s(x) to the local mass density. The separation between two neigh-
boring masses is
∆j = d+ sj + 1− sj = d
(1 +
∂s
∂x
). (1.7)
That is if s is uniform, then the separation is d. If ∂s/∂x > 0 then the separation
is greater than d. If you know something about elasticity, you would recognize
∂s/∂x as the 1D version of the strain tensor. Regardless, the mass density is
ρ =M
∆=M
d
1
1 + ∂s∂x
(1.8)
≈ λ
(1− ∂s
∂x
). (1.9)
Thus, we can identify −∂s/∂x as the fractional change in the density. Remem-
ber this – we will see this structure again. Note that s is independent of the
discretization.
The classical Lagrangian associated with this system is
L =∑j
M
2s2j −
k
2(sj+1 − sj)2 (1.10)
=∑j
dM
2ds2j − d
kd
2
(sj+1 − sj
d
)2
(1.11)
=
∫dx
[λ
2
(∂s
∂t
)2
− κ
2
(∂s
∂x
)2]. (1.12)
6
The conjugate variable to s(x) is the momentum density
π(x) =δL
δs(x)= λs(x). (1.13)
It has units of momentum per unit length. That is, the momentum of the j’th
mass is pj = dπ(xj).
The classical Hamiltonian is
H =
∫dx
1
2λπ2 +
κ
2(∂xs)
2. (1.14)
D. Quantum Coupled Masses
The quantum version of this problem is essentially identical to the classical prob-
lem. It is convenient to think about the operators sj and pj , with commutation
relations
[si, pj ] = i~δij . (1.15)
We can use our classical definitions of the fields
sj = s(xj) (1.16)
pj = dπ(xj). (1.17)
It is straightforward to argue that
[s(x), π(y)] = i~δ(x− y). (1.18)
We note that the density operator is ρ(x) = λ(1− s′(x)).
E. HW 1 – Due March 20
Problem 1.1. Number Phase Uncertainty: This problem makes concrete
some of the algebra that we will encounter in the lectures.
In your linear algebra class you may have learned about the polar-decomposition:
It is the statement that any matrix can be written as the product of a Hermitian
and Unitary matrix. The name comes from noting that a unitary matrix can
be written as the exponential of a Hermitian matrix, so it looks like the polar
decomposition of a complex number. In other words, the decomposition is:
M = eiθR (1.19)
where
R† = R (1.20)
θ† = θ. (1.21)
7
1.1.1. Two-level system: To make this very concrete, thing about a system
with two states, labeled |0〉 and |1〉. Typically we order these as |1〉, |0〉, but for
our purposes it will be better to write them as |0〉, |1〉. In this basis, with this
order, the “lowering” operator is
a =
(0 1
0 0
). (1.22)
Show that
{a, a†} = 1, (1.23)
and hence a can be thought of as a Fermion lowering operator (where we inter-
pret the two states as “full” and “empty”).
1.1.2. Show that
a =
(0 1
1 0
)(0 0
0 1
), (1.24)
and that the first matrix is Unitary, and the second Hermitian. Thus we write
eiθ =
(0 1
1 0
)(1.25)
R =
(0 0
0 1
). (1.26)
1.1.3. Calculate the anticommutator {eiθ, R†R}.
1.1.4. Find the matrix θ. Hint: Mathematica has a function MatrixLog. Alter-
natively you can write θ as a linear combination of the identity matrix and the
Pauli matrices, and solve for the coefficients. Note, because log is multivalued,
the answer is not unique.
Problem 1.2. Number-phase uncertainty for a bosonic mode:
The bosonic problem is a bit more complicated because the Hilbert space is
infinite. One strategy is to work in a truncated space, where you limit the
occupation to be 0, 1, 2, · · ·n, so the matrices are all finite. We can then take
n→∞ at the end.
1.2.1. Polar decomposition: In this finite space, the annihilation operator is
a =
0√
1 0
0 0√
2...
. . .
0 0√n− 1
0 . . . 0
. (1.27)
8
Show that
a =
0 1 0 0 · · ·0 0 1 0 · · ·...
0 . . . 1
1 . . . 0
0 0 0 0 · · ·0 1 0 0 · · ·0 0
√2 0 · · ·
...
0 . . .√n
. (1.28)
and that the first matrix is Unitary, and the second Hermitian.
1.2.2. Phase Matrix: The matrix θ is quite ugly here – so I won’t ask you to
calculate it. However it is straightforward to calculate
[eiθ, R†R]. (1.29)
A way to make the arithmetic simpler is to write
eiθ =
n∑j=1
|j − 1〉〈j|+ |n〉〈0| (1.30)
R†R =∑
j = 0n|j〉j〈j|. (1.31)
You will find
[eiθ, R†R] = eiθ +X. (1.32)
What is the matrix X?
1.2.3. In the limit n→∞, the matrix X involves a state with infinitely many
particles – and we typically argue that it can be neglected. If we do so, then
[eiθ, R†R] ≈ eiθ. Under this approximation, infer
[θ,R†R]. (1.33)
Problem 1.3. Coherent States
Consider the following two harmonic oscillator Hamiltonians, where we have
taken dimensionless units (~,m, ω all are unity):
H1 =p2
2+
1
2x2 (1.34)
H2 =p2
2+
1
2(x− x0)2. (1.35)
Let |ψ1〉 and |ψ2〉 be the ground states. Clearly
|ψ2〉 = eipx0 |ψ1〉, (1.36)
9
where exp(ipx0) is a translation operator.
We introduce the ladder operators:
a =1√2
(x+ ip) (1.37)
a† =1√2
(x− ip) . (1.38)
Show that |ψ2〉 is an eigenstate of a. Find the eigenvalue.
Hint: What is a|ψ1〉? What is [a, eipx0 ]?
Problem 1.4. Quantum Fluctuations in a 1D harmonic Chain
Consider the 1D chain described by Hamiltonian:
H =∑j
p2j
2m+
1
2k(xj+1 − xj)2, (1.39)
where [xi, pj ] = i~δij , and where j runs from 1 to N . For concreteness we will
use periodic boundary conditions (so xn+1 is identified with x1.
We can diagonalize this Hamiltonian by the transformation:
xj =∑q
eiqjuq(aq + a†−q) (1.40)
pj =∑q
e−iqjvqaq − a†−q
i.
The sums run over q = 2πn/N , with n running from −N/2 to N/2. It turns out
that we can choose uq and vq real. Furthermore, since xj and pj are Hermitian,
we need u−q = uq and v−q = vq. We want the operators aq to obey
[aq, a†q] = δq,q (1.41)
1.4.1. Show that if Eq. (1.41) is satisfied, then
[aq ± a†−q, aq′ ± a†−q′ ] = 0 (1.42)
for any q, q′. Hence the requirements [xi, xj ] = 0 and [pi, pj ] = 0 is automatically
satisfied by this ansatz
1.4.2. Show that if Eq. (1.41) is satisfied, then
[aq + a†−q,aq′ − a†−q′
i] = 2iδq,q′ . (1.43)
for any q, q′.
10
1.4.3. Use that last result to show that [xi, pj ] = i~δij if and only if
uq =
√~
2Nδq (1.44)
vq =
√~
2N
1
δq,
where δq is aribtrary (with δ−q = δq).
1.4.4. Substitute the ansatz, Eq. (1.40) into the Hamiltonian, using Eq. (1.44)
to write the Hamiltonian as a function of the raising/lowering operators and δq.
The resulting expression should be of the form
H =∑q
(aqa−q + a†−qa†q)~2
[Aqδ
2q −
Bqδ2q
](1.45)
+(aqa†q + a†−qa−q)
~2
[Aqδ
2q −
Bqδ2q
].
Find Bq and Aq. We then set
δ4q =
BqAq
, (1.46)
so that
H =∑q
(a†qaq +
1
2
)~ωq. (1.47)
Find ωq.
1.4.5. The ground state of this system will have aq|GS〉 = 0 for all aq. Use this
to write
〈(xi − xj)2〉 =∑q
u2q 4 sin2
(q(i− j)
2
)(1.48)
→ N
2π
∫dq u2
q 4 sin2
(q(i− j)
2
), (1.49)
where we converted the sum to an integral by noting that the separation between
q’s is 2π/N .
If |i − j| � 1, the sin function in Eq. (1.48) is rapidly oscillating, and one
might naively expect that one could replace it with 1/2. Show that if you do, the
integral is divergent. (Hint, the problem happens as q → 0). [You don’t need
to do it, but If you work a bit harder, you can even argue that the divergence
should be 〈(xi − xj)2〉 ∼ log |i− j|.]Regardless, his means that 〈(xi−xj)2〉 → ∞ as |i− j| → ∞. Physically, this
means that the quantum fluctuations disorder the chain. If you did the same
calculation in 3D, you would instead find the integral is convergent.
Chapter 2
“Constructive”
Bosonization – Mar 15,
2019
In this lecture we derive a quite remarkable result: The excitations of an in-
teracting 1D Fermi system can be put in a 1-to-1 correspondence with those
of a harmonic chain. I note that it is an obvious statement that among the
excitations of the electrons are “plasmons” – corresponding to density waves in
the electron fluid. This is true in any dimension. The special thing about 1D is
that all excitations can be written in terms of these phonons.
We will also first neglect spin, then add it later. Our treatment is similar
to Giamarchi chapter 2 – but maybe even closer to Delft and Schoeller’s article
”Bosonization for Beginners...” which is on the arXiv.
A. Linearizing
The first step will be to linearize about the Fermi surface. It turns out some
of our results do not depend on this approximation, but I think it is best to do
the simplest treatment first. The idea is that we imagine a setting where the
interactions are weak. In that case the ground state of the interacting system
will look a lot like the ground state of the non-interacting system, with the
exception of structure near the Fermi points. Thus it suffices to model physics
11
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201912
near the Fermi points. In other words we write the kinetic energy as
H = HR +HL +H ′ (2.1)
HR =∑
knear kf
vf (k − kf )c†kck (2.2)
HL =∑
knear−kf
vf (kf − k)c†kck (2.3)
H ′ =∑
other k
εkc†kck. (2.4)
We will come up with a model of HR and HL. We won’t worry about H ′, as it
will not be important to the physics we are after. Clearly if the interactions are
strong, this argument will break down and we will need to include H ′.
It makes sense to shift the origin of momenta, and write
ark = ck−kf (2.5)
a`k = ck+kf . (2.6)
when there is no possibility of confusion, we will leave off the r, ` indices. We
refer to these two “species” of particles as “right movers” and “left movers”.
I have been a little vague about what range of k we are summing over. In
principle it doesn’t matter – if all the action is happening at the Fermi surface,
I can include as many states as I want.
B. Density Operator
In this section we will see that the Fourier transform of the density acts as a
ladder operator. There are a couple tricky things here. The first is just keeping
track of L’s and 2π’s.
The best mnemonic that I have for the factors is to start with the electron
field operator – which is something like
ψ(x) =1√L
∑k
ckeikx, (2.7)
where k = 2πn/L. We can verify that I have the factors right by looking at the
density
ρ(x) = ψ†(x)ψ(x) =1
L
∑kq
ei(k−q)xc†qck. (2.8)
In particular, if we have a homogeneous gas 〈c†qck〉 is zero unless k = q and
k < kf , and we get
ρ(x) =N
L. (2.9)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201913
The Fourier transform of the density operator is
ρp =∑q
a†q+paq. (2.10)
That is
ρ(x) =1
L
∑p
eipxρp. (2.11)
Perhaps I should define ρp with a factor of 1/√L in it. In the end it doesn’t
matter – as long as you are consistent. This convention makes the following
arithmetic a bit easier.
Clearly if p� kf , we can write
ρp = ρRp + ρLp + ρotherp , (2.12)
where
ρRp =∑q
c†r,q+pcrq, (2.13)
and a similar expression for ρL. We will analyze the properties of ρRp .
Now we come to our first profound observation. Suppose |ψ〉 is an eigenstate
of HR:
H|ψ〉 = E|ψ〉. (2.14)
(Note I have left off the R.) Applying ρRp to |ψ〉 will give a superposition of terms
where various right-moving particle have had their momentum increased by p.
But the energy of a given particle is proportional to p – so we have increased
the energy by vfp. In other words
Hρp|ψ〉 = (E + vfp)ρp|ψ〉. (2.15)
Putting these two expressions together, we have Hρp|ψ〉 = ρp(H+vfp)|ψ〉. This
statement is true for all eignenstates, and hence
[ρp, H] = −vfpρp. (2.16)
This is exactly the requirement for a ladder operator. Actually – as you see
ρp appears on both the left and right hand side of that expression, so this is
actually the requirement to be a constant times a ladder operator. Below, we
will show that the correct normalization is (for p > 0),
b†p =2π
L
1
pρp (2.17)
bp =2π
L
1
pρ−p, (2.18)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201914
where these obey the standard relations [bp, b†q] = δp−q.
To establish this result, we need to understand what these operators do to
states. In particular we want to look at how ρpρq|ψ〉 differs from ρqρp|ψ〉. To
make the notation simpler, let us write momenta in units of 2π/~, which is the
spacing between momentum states.
Lets start with a Fermi sea |ψ〉 = | • • • • • : ◦ ◦ ◦ ◦ ◦〉. Applying ρ3 to this
gives three terms
ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = | • • • • ◦ : ◦ ◦ •◦ ◦◦〉 − | • • • ◦ • : ◦ •◦ ◦ ◦ ◦〉+ | • • ◦ • • : •◦ ◦ ◦ ◦◦〉
Following this with ρ2 gives a total of 8 terms:
ρ2ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = −| • • ◦ • •◦ : ◦ ◦ •◦ ◦◦〉+ | • • • ◦ ◦ : •◦ ◦ •◦ ◦◦〉+ | • • • • ◦ : ◦ ◦ ◦ ◦ •◦〉+| • ◦ • •◦ • : ◦ •◦ ◦ ◦ ◦〉 − | • • • ◦ • : ◦ ◦ ◦ •◦ ◦〉−| ◦ • •◦ • • : •◦ ◦ ◦ ◦◦〉 − | • • ◦ • ◦ : •◦ •◦ ◦ ◦ ◦〉+ | • • ◦ • • : ◦ ◦ •◦ ◦◦〉
For what it is worth, two of these cancel, leaving
ρ2ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = +| • • • ◦ ◦ : •◦ ◦ •◦ ◦◦〉+ | • • • • ◦ : ◦ ◦ ◦ ◦ •◦〉+| • ◦ • •◦ • : ◦ •◦ ◦ ◦ ◦〉 − | • • • ◦ • : ◦ ◦ ◦ •◦ ◦〉−| ◦ • •◦ • • : •◦ ◦ ◦ ◦◦〉 − | • • ◦ • ◦ : •◦ •◦ ◦ ◦ ◦〉
Remarkably, if you do these two operators in the opposite order ,you get
exactly the same terms, with exactly the same signs. Even more remarkably,
the same thing happens on any initial state. I will prove this below, but it is
worth playing with this to convince yourself. You will find, however, that if
p = −q, then they don’t commute.
ρ−3ρ3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = ρ−3 [| • • • • ◦ : ◦ ◦ •◦ ◦◦〉 − | • • • ◦ • : ◦ •◦ ◦ ◦ ◦〉+ | • • ◦ • • : •◦ ◦ ◦ ◦◦〉]= | • • • • •◦ : ◦ ◦ ◦ ◦◦〉+ | • • • •◦ • : ◦ ◦ ◦ ◦ ◦〉+ | • • •◦ • • : ◦ ◦ ◦ ◦◦〉= 3| • • • • • : ◦ ◦ ◦ ◦ ◦〉
but
ρ3ρ−3| • • • • • : ◦ ◦ ◦ ◦ ◦〉 = 0.
Having established plausibility, we now prove the general result [ρp, ρq] =
qδp+q – or in real units [ρp, ρq] = q(L/2π)δp+q. We start with
[ρp, ρq] =∑k,k′
[a†k+pak, a†k′+qa
′k]. (2.19)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201915
This commutator is zero unless k′+q = k or k+p = k′. We need to be somewhat
careful about the bounds on the sum, but for the moment lets not worry about
it,
[ρp, ρq] =∑k
a†k+pak−p −∑k′
a′k+qak′−p. (2.20)
We shift indices on the sum, and everything cancels – except the boundary
terms. In the case p 6= −q, we actually don’t need to worry about the boundary
terms, they are things like a†k+pak−p for k well inside the Fermi surface. Any
states we act on will have both those states occupied, and hence such terms do
nothing. When p = −q these expressions are a†k−qak−p, which is 1 if we are well
inside the Fermi surface, but −1 if we are well outside. A little thought should
convince you that [ρ−q, ρq] = q (or actually [ρ−q, ρq] = qL/2π). [When I have
a chance I’ll write it in more detail, but I already went through it in class, and
am running out of time in typing up the notes.]
We can therefore define ladder operators (for p¿0),
bp =
√2π
Lpρ−p (2.21)
b†p =
√2π
Lpρp. (2.22)
We further have
[bp, b†p] = δpp′ (2.23)
[bp, HR] = vfpbp
For left movers we would instead get (for p < 0)
bp,` =
√2π
L|p|ρ−p,` (2.24)
b†p,` =
√2π
L|p|ρp,`. (2.25)
We further have
[bp,`, b†p,`] = δpp′ (2.26)
[bp,`, HL] = vf |p|bp,`
Thus the Hamiltonian for the non-interacting system reads
H =∑p>0
vfp[b†pr bpr + b†p−`b−p`
]+ ∆ (2.27)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201916
where ∆ commutes with all of the b operators.
Claim: The only operators which commute with all of the b’s and conserve
the number of left and right movers are NL and NR – the total number of left
and right movers – and functions of these.
In other words, given any state, say
|ψ〉 = | • • ◦ • ◦ ◦ • ◦ ◦ ◦〉. (2.28)
The claim is that |ψ〉 can be expressed as a linear superposition of b’s acting on
|ψ0〉 = | • • • • ◦ ◦ ◦ ◦ ◦ ◦〉. (2.29)
This claim is equivalent to the statement that the number of ways of making
orthogonal states of energy E by acting on the ground state with b’s is the
same as the number of ways of making such a state by adding an equal number
of particles and holes. Both constructions make orthogonal states – hence the
Hilbert spaces will have the same dimension – and must be identical.
It is fun to do the counting for the first few states. For example, it is clear
that there is a unique state with energy 1. There are two states with energy 2:
| • • • ◦ : ◦ • ◦◦〉 and | • • ◦ • : • ◦ ◦◦〉. These are linear combinations of b†1b†1|ψ0〉
and b†2|ψ0〉.In general, the fact that these numbers are always the same is equivalent to
the “pentagonal number theorem” – which is at the core of the Jacobi triple
product formula, which can also be considered a Jacobi Theta function identity
(relating it to Wierstrauss functions). I won’t go through the proof of the Jacobi
identiy, but will show how to reduce this physics problem to that result.
First, the number of ways of using the boson operators to generate a state
of energy m is given by the number of partitions of the integer m. That
is the number of ways of choosing integers which add up to n. For exam-
ple, 2=1+1 or 2=2, has two ways. We write p(2) = 2. On the other hand
4=3+1=2+2+2+1+1=1+1+1+1 has 5 ways. We will define the generating
function
f(x) =∑m
p(n)xm. (2.30)
Standard generating function tricks give
f(x) =∏k
1
1− xk. (2.31)
[I’ll go through these in class.] This is a famous function in number theory.
Lets denote the number of ways of distributing an equal number of particles
and holes to produce energy m as s(m). We define
g(x) =∑m
s(m)xm. (2.32)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201917
The claim is that f(x) = g(x). Calculating g(x) is hard, so we introduce s(m,n)
as the number of ways of distributing particles and holes to produce energy m,
where the number of particles minus the number of holes is n. We define
G(x, y) =∑nm
s(m,n)xmyn. (2.33)
Clearly g(x) is the y0 term in the series. Formally we can write G as the product
of generating functions for each mode
G(x, y) =[(1 + xy)(1 + x2y)(1 + x3y) · · ·
]×[(1 + y−1)(1 + xy−1)(1 + x2y−1) · · ·
].
(2.34)
We then use the “Jacobi triple product” identity, which says
G(x, y) = f(x)∑n
xn(n−1)/2yn. (2.35)
Taking the y = 0 term of both side establishes the proof.
C. Klein factors
There is one other set of operators we will need, namely the ”Klein factors”
F †r and F †` , which are unitary operators which just shift the number of left and
right movers by 1. They are defined as the unique operators which commute
with the b’s, and shift the Fermi sea:
F †r | • • • • : ◦ ◦ ◦◦〉R = (−1)NL | • • • • : • ◦ ◦◦〉R (2.36)
Fr| • • • • : ◦ ◦ ◦◦〉R = (−1)NL | • • • ◦ : ◦ ◦ ◦◦〉R (2.37)
F †` | • • • • : ◦ ◦ ◦◦〉L = | • • • • : • ◦ ◦◦〉L (2.38)
F`| • • • • : ◦ ◦ ◦◦〉L = | • • • ◦ : ◦ ◦ ◦◦〉L (2.39)
On other states, it similarly shifts everything by one:
F †r | • • ◦ • : ◦ • ◦◦〉R = (−1)NL | • • • ◦ : • ◦ •◦〉R. (2.40)
It should be clear that they do indeed commute with the b’s, and
F †rFr = FrF†r = F †` F` = F`F
†` = 1 (2.41)
{Fr, F`} = 0. (2.42)
The factor of (−1)NL was chosen so that this last anticommutation relationship
would hold.
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201918
D. Fermion Operators
We can now write the fermion operators in terms of the phonons. In particular
ψ(x) =1√L
∑k
ckeikx (2.43)
=eikfx√L
∑q
arqeiqx +
e−ikfx√L
∑q
a`qeiqx + stuff away from Fermi surface.
In the obvious notation, we call the first two terms ψr and ψ`.
We will now construct ψr in terms of the b’s, by looking at the commutation
relations. Recall
ρr(p) =∑k
c†r k+pcr k (2.44)
Straightforward arithmetic gives
[ρ(p), cq] = −cq−p, (2.45)
and hence
[ρr(p), ψr(x)] = − 1√L
∑q
cr q−peiqx (2.46)
= −eipxψr(x). (2.47)
We now use that
[ρr(p), ρr(−p)] = −p L2π, (2.48)
to conclude that
ψr(x) = (· · · ) exp
(2π
Lpρr(−p)eipx
), (2.49)
where the term in parentheses commutes with ρr(p). The only expression con-
sistent with this (for all p), and which does the right thing to the number of
particles is
ψr(x) = AFr exp∑p
2π
Lpeipxρr(−p) (2.50)
= AFr exp∑p>0
√2π
Lp
[eipxbp − e−ipxb†p
], (2.51)
where A is a constant.
It is traditional to define
ϕr(x) = i2π
L
∑p>0
eipx
pρr(−p) (2.52)
= i
√2π
L
∑p>0
eipx√pbp, (2.53)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201919
and
φr(x) = ϕr(x) + ϕ†r(x) (2.54)
= i2π
L
∑p
eipx
pρr(−p). (2.55)
Then
ψr(x) = AFre−iφr(x) (2.56)
In addition to being a cute way of writing things, we can note an interesting
property of the field φr(x), in particular
∂xφr(x) = −2π
L
∑p
eipxρr(−p) (2.57)
= −2πρr(x). (2.58)
Thus φr has exactly the same meaning as the displacement field that we intro-
duced for the mass chain.
Now Giamarchi makes life more complicated by introducing two other fields
via:
φr(x) = φ(x)− θ(x) (2.59)
φ`(x) = −φ(x)− θ(x). (2.60)
The reason for doing this is that the physical density is the sum of the right-
moving and left-moving densities (plus stuff from other modes). Thus ∂xφ has
the meaning of the total density. The other field θ involves the mis-match of the
left-moving and right-moving densities. It is clearly related to the momentum
density, and will find
ρ(x) = − 1
π∂xφ(x) (2.61)
π(x) =1
π∂xθ(x). (2.62)
Indeed, it is a straightforward exercise, left as a homework exercise, that these
fields obey the right commutation relations.
E. Free Hamiltonian
Lets now consider how the Hamiltonian looks in terms of these operators. First,
as already argued, up to a constant, the free particle Hamiltonian is
H0 =∑p>0
vfp[b†pr bpr + b†−p`b−p`
]+ vf
2π
L(N2
R + N2L)− µ(NR + NL), (2.63)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201920
where we have added a chemical potential. This, and the field operators, are
more conveniently described by
bp>0 = brp (2.64)
bp<0 = b`p. (2.65)
The first term in Eq. (2.63) is then
Hb0 =
∑p
vfpb†pbp, (2.66)
where the superscript b reminds us that we have left off the N terms. Similarly
this makes the expression for θ and φ simpler,
φ(r) = −i2πL
∑p
(L|p|2π
)1/2e−ipx
p(b†p + b−p) (2.67)
θ(r) = −i2πL
∑p
(L|p|2π
)1/2e−ipx
|p|(b†p − b−p). (2.68)
For those who did Problem 1.4, you should recognize this as exactly the expres-
sions for
ρ(x) = − 1
π∂xφ(x) (2.69)
Π(x) =1
π∂xθ(x). (2.70)
In particular, straightforward arithmetic gives
Hb0 =
vf2π
∫dx[(∂xθ)
2 + (∂xφ)2]. (2.71)
F. Interactions
We now consider the role of interactions. A typical model for interactions is
Hint =
∫dx
∫dx′ V (x− x′)ψ†(x)ψ†(x′)ψ(x′)ψ(x). (2.72)
For electrons V could be a Coulomb potential, while for cold atoms it is some
sort of short range interaction. Note, that because of the anti-commutation
relations, one could also write this as
Hint =
∫dx
∫dx′ V (x− x′)ψ†(x)ψ(x)ψ†(x′)ψ(x′)− V (0)
∫dxψ†(x)ψ(x).
(2.73)
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201921
The latter term just shifts the chemical potential – possibly by an infinite
amount, depending on how V behaves at short distances. Thus some treat-
ments use
Hint =
∫dx
∫dx′ V (x− x′)ψ†(x)ψ(x)ψ†(x′)ψ(x′). (2.74)
You should keep in mind that the strictly correct expression is Eq. (2.72) or
(2.73), but as long as you don’t worry about a few infinities, it is fine to work
with Eq. (2.74).
We write Eq. (2.72) in momentum space as
Hint =1
L3
∑kpq
Vq ψ†k+qψ
†p−qψpψk. (2.75)
We think about this as a scattering process – where you have incoming particles
with momentum k and p, which exchange momentum q. We only care about
the cases where all of the momenta are near the Fermi surface. There are six
possibilities:
1. k, p, k + q, p− q are all near kf
2. k, p, k + q, p− q are all near −kf
3. k, k + p are all near kf while q and q − p are near −kf
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201922
4. k, k + p are all near −kf while q and q − p are near kf
5. k, q − p are near kf while q, k + p are near −kf
6. k, q − p are near −kf while q, k + p are near kf
The first two are referred to as “g4” processes – they involve just scattering
amongst right movers or amongst left movers. For these q ≈ 0, and one often
just replaces Vq with V0 for all these terms. The second two are referred to
as “g2” processes, and involve scattering between right and left movers. Again
q ≈ 0, and one often just replaces Vq with V0. Finally, the last terms are “g1”
CHAPTER 2. “CONSTRUCTIVE” BOSONIZATION – MAR 15, 201923
processes, which involve scattering from one Fermi surface to the other. Here
q ≈ 2kf , and one often just replaces Vq with V2kf .
If we make this approximation (replacing Vq with V0 in the g4 and g2 pro-
cesses, and replacing it with V2kf in the g1 processes – then both the g1 and g2
terms are of the form
H1/2 = (· · · )∫dx ρR(x)ρL(x). (2.76)
The g4 terms are of the form
H4 = (· · · )∫dx ρR(x)ρR(x) + ρL(x)ρL(x). (2.77)
Following Giamarchi, we write the interaction Hamiltonian as
Hint =
∫dx
g4
2(ρR(x)ρR(x) + ρL(x)ρL(x)) + g2ρR(x)ρL(x) (2.78)
=1
2π
∫dx
g4 − g2
2π(∂xθ)
2 +g4 + g2
2π(∂xθ)
2. (2.79)
Remarkably this is of exactly the same form as the free Hamiltonian, allowing
us to write
H =1
2π
∫dxu
[K(∂xθ)
2 +1
K(∂xφ)
], (2.80)
which is readily diagonalized to produce
H =∑k
ukb†k bk + const, (2.81)
and the ground state has bk|ψ〉 = 0.
The effects of interactions are parameterized by the coefficients u and K. In
particular, repulsive interactions correspond to K < 1, and attractive to K > 1.
The relationship between θ, φ and the b’s is the same as the relationship with
the b’s, except for a scaling. In particular, if we write θ =√Kθ and φ = φ/
√K,
these scaled fields have the same commutation relations as the original,
[φ(x), ∂xθ(x)] = [φ(x), ∂xθ(x)] = iπδ(x− y). (2.82)
But in these scaled variables, H looks like the free Hamiltonian,
H =1
2π
∫dxu
[(∂xθ)
2 + (∂xφ)], (2.83)
and hence
φ =φ√K
=
(−iπL
)∑p
(L
2π
)1/2e−ipx√|p| sign p
(b†p + b−p
)(2.84)
θ =√Kθ =
(iπ
L
)∑p
(L
2π
)1/2e−ipx√|p|
(b†p − b−p
). (2.85)
Chapter 3
Applications – Mar 20, 2019
In this lecture we will use the results from the previous lecture to calculate
various physical quantities. We will need the following facts. First, for an
eigenstate of any quadratic Hamiltonian
〈b1b2b3b4〉 = 〈b1b2〉〈b3b4〉+ 〈b1b3〉〈b2b4〉+ 〈b1b4〉〈b2b3〉, (3.1)
where each b is either a bq or a b†q. This fact, known as Wick’s theorem, is proven
in any standard book on many-body theory. It is a straightforward argument
which we skip here. A related fact is that for an eigenstate of a quadratic
Hamiltonian, and any operator X with 〈X〉 = 0, one has
〈e〈X〉〉 = e〈X2〉/2. (3.2)
Finally, we need that for the ground state
〈b1b2〉 = 0 (3.3)
unless there is some q such that b2 = b†q and b1 = bq
A. Occupation Number
The first thing that we want to calculate is the occupation number 〈nk〉 = 〈c†kck〉.If we choose k near kf , we can just consider
nkf+q = c†rqcrq (3.4)
=1
L
∫dx
∫dx′ ψ†r(x)ψr(x
′)eiq(x−x′). (3.5)
Taking the expectation value we have
〈nkf+q〉 =1
L
∫dx
∫dx′ 〈ψ†r(x)ψr(x
′)〉eiq(x−x′). (3.6)
24
CHAPTER 3. APPLICATIONS – MAR 20, 2019 25
We then note that the expectation value should be translationally invariant, so
〈nkf+q〉 =
∫dx 〈ψ†r(x)ψr(0)〉eiqx. (3.7)
We now use our bosonized expression
ψr(x) = AFre−i(φ−θ), (3.8)
which gives
〈nkf+q〉 = A2
∫dx 〈ei(φ(x)−θ(x))e−i(φ(0)−θ(0))〉eiqx. (3.9)
We can combine the two terms using “Baker-Campbell-Hausdorff,” which says
that if [A,B] = C, and C is a constant, then
eAeB = eA+B+(1/2)[A,B]. (3.10)
In the present case
[φ(x), θ(0)] =iπ
2sign(x), (3.11)
[θ(x), φ(0)] =iπ
2sign(x), (3.12)
and hence
〈nkf+q〉 = A2
∫dx 〈ei(φ(x)−φ(0))−i(θ(x)−θ(0))+iπ sign(x)e−i(−θ(0))〉eiqx(3.13)
= A2
∫dx 〈ei(φ(x)−φ(0))−i(θ(x)−θ(0))〉eiqx (3.14)
At this point we use Eq. 3.2, to get
〈nkf+q〉 = A2
∫dx e−
12 〈[φ(x)−φ(0))−i(θ(x)−θ(0)]2〉eiqx. (3.15)
Thus we see that calculating the momentum distribution reduces to calculating
the correlators of the fields φ and θ.
We start with
〈[φ(x)− φ(x′)]2〉 = K
(−iπL
)(L
2π
)∑pq
(e−ipx − e−ipx′√|p| sign(p)
)(e−iqx − e−iqx′√|q| sign(q)
)×〈(b†p + b−p)(b
†q + b−q)〉. (3.16)
The operator expectation value is 1 if q = −p and otherwise is 0, and hence
〈[φ(x)− φ(x′)]2〉 = K
π
L
∑p
sin2 p(x−x′)2
|p|. (3.17)
CHAPTER 3. APPLICATIONS – MAR 20, 2019 26
In the thermodynamic limit we replace the sum with an integral, and,
〈[φ(x)− φ(x′)]2〉 = K
∫ ∞0
dpsin2 p(x−x′)
2
p. (3.18)
Unfortunately that integral is infinite. In particular, the sin2 is rapidly oscillat-
ing, and (at least in the tails of the integral) can be replaced by 1/2. We then
get the area under the 1/p curve, which is infinite.
The standard argument to save us is to say that we are only considering p
near the Fermi surface, and hence we can introduce a cutoff. A better argument
is to use a finite range interaction to cut things off. Nonetheless, we introduce
a cutoff Λ, and replace the sin2 with 1/2 – this latter substitution only works if
k(x− x′) > 1, and hence we also introduce a lower cutoff 1/(x− x′). Thus
〈[φ(x)− φ(x′)]2〉 = K log[Λ(x− x′)]. (3.19)
Similarly we get
〈[θ(x)− θ(x′)]2〉 =1
Klog[Λ(x− x′)], (3.20)
and
〈[φ(x)− φ(x′)] [θ(x)− θ(x′)]〉 = 0. (3.21)
Plugging this into our expression for the occupation numbers yields
nkf+q ∼∫dx
eiqx
|x|(K+K−1)/2(3.22)
∼ q(K+K−1)/2−1. (3.23)
And hence, the occupation number has a power law singularity at the Fermi
surface.
In the non-interacting case, K = 1, instead we have that nk is flat, and we
have a discontinuity at the Fermi surface.
A.1. Other Correlation Function
We can do the same arithmetic with the density-density correlation function.
In particular,
ρ(x) = ρr + ρl + e2ikfxψ†rψl + e−2ikfxψ†l ψr (3.24)
= − 1
π∂xφ+A2e2ikfxe2iφ +A2e−2ikfxe−2iφ. (3.25)
CHAPTER 3. APPLICATIONS – MAR 20, 2019 27
The first part is smooth, while the latter show oscillations on the scale of 2kf .
Turning the crank yields
〈ρ(x)ρ(0)〉 = ρ2 − K
(2π)2
1
x2+ (· · · ) cos(2kfx)
(1
x
)2K
. (3.26)
We see that the larger K is, the faster the 2kf oscillations decay: Attractive
interactions make these oscillations decay faster, while repulsive makes them
decay slower. One way to interpret this is that repulsive interactions lead to a
fluctuating charge density wave.
A complementary variable is the “pairing operator,”
∆(x) = π†(x)ψ†(x+ a), (3.27)
where a is some small number. If a was zero, the ∆ involves taking two spinless
fermions from the same location in space, and hence ∆ would vanish. For small,
but finite a, ∆ is related to how likely it is that two fermions are close together.
Being sloppy about the arguments,
∆(x) ≈ e2ikfxψ†Rψ†R + e−2ikfxψ†Lψ
†L + ψ†Rψ
†L + ψ†Lψ
†R. (3.28)
The first two terms are typically small, and
∆(x) ∼ e2iθ + e−2iθ, (3.29)
is obviously complementary to ρ – which involved φ’s. Clearly
〈∆(x)∆†(0)〉 ∼ 1
x1/2K. (3.30)
Attractive interactions make the pair correlations stronger.
A.2. Spin
I just want to very quickly sketch out what happens when one adds spin. In the
absense of interactions the ↑ and ↓ fermions are independent, so everything we
did goes through with a φ↓, φ↑, θ↓ and θ↑. The Free Hamiltonian is
H0 =1
2π
∫dx vf
[(∂xθ↑)
2 + (∂xθ↓)2 + (∂xφ↑)
2 + (∂xφ↓)2]. (3.31)
The same interaction terms as before occur, but now we should be careful about
the spin. The g4 processes
CHAPTER 3. APPLICATIONS – MAR 20, 2019 28
can occur with both spins, and gives terms
H4 ∝∫dx [ρ2
r + ρ2l ] (3.32)
∝∫dx [∂x(φ↑ + θ↑ + φ↓ + θ↓)]
2 + [∂x(φ↑ − θ↑ + φ↓ − θ↓)]2
The g2 processes
give
H2 ∝∫dx [ρrρl] (3.33)
∝∫dx ∂x(φ↑ + θ↑ + φ↓ + θ↓)∂x(φ↑ − θ↑ + φ↓ − θ↓)
Finally, the g1 processes
separate into sets of terms. First if the spins are all the same we get
Ha1 ∝
∫dx ρr↑ρl↑ + ρr↓ρl↓ (3.34)
∝∫dx ∂x(φ↑ + θ↑)∂x(φ↑ − θ↑) + ∂x(φ↓ + θ↓)∂x(φ↓ − θ↓).
CHAPTER 3. APPLICATIONS – MAR 20, 2019 29
The terms where the spins are different have another structure
Hb1 ∝
∫dxψ†r↑ψ
†l↓ψr↓ψl↑ + ψ†r↓ψ
†l↑ψr↑ψl↓ (3.35)
∝∫dx cos(2i[φ↑ − φ↓]). (3.36)
These terms can be combined to give a Hamiltonian which separately involves
spin and charge degrees of freedom. In particular we define
φρ =1√2
(φ↑ + φ↓) (3.37)
φσ =1√2
(φ↑ − φ↓) (3.38)
θρ =1√2
(θ↑ + θ↓) (3.39)
θσ =1√2
(θ↑ − θ↓), (3.40)
in terms of which
H = Hρ +Hσ (3.41)
where
Hρ =
∫dx(· · · )(∂xθρ)2 + (· · · )(∂xφρ)2 (3.42)
Hσ =
∫dx(· · · )(∂xθσ)2 + (· · · )(∂xφσ)2 + (· · · ) cos(2
√2φσ). (3.43)
The charge degrees of freedom can be analyzed using the techniques we have
already discussed. The spin degrees of freedom are a bit more complicated: It
turns out there is a phase transition between a phase where φσ is frozen (which
it turns out corresponds to pairing) and one where it fluctuates. The phase
transition can be mapped onto the classical Kosterlitz-Thouless transition of a
2D XY model.
B. HW 2 – Not Due
These problems are just to help understanding. You neither need to do them
nor to hand them in.
Problem 3.1. Squeezed States
We can “solve” the Harmonic oscillator
H0 =p2
2+x2
2, (3.44)
CHAPTER 3. APPLICATIONS – MAR 20, 2019 30
with [x, p] = i, by the transformation:
a =1√2
(x+ ip) (3.45)
a† =1√2
(x− ip) , (3.46)
in terms of which
H0 = a†a+1
2. (3.47)
The operators a, a† obey the standard ladder operator commutators, and are
ladder operators for this Hamiltonian. The ground state is defined by
a|ψ0〉 = 0. (3.48)
Consider now the Hamiltonian
H = α
(λ2 p
2
2+
1
λ2
x2
2
), (3.49)
3.1.1. Show that x = x/λ and p = λp obey the same commutation relations as
x and p.
3.1.2. Use that result to argue that
H = α
(b†b+
1
2
), (3.50)
where b and b† obey the standard commutation relations. Express b and b† in
terms of x and p.
3.1.3. Express a and a† in terms of b and b†.
3.1.4. Consider the state |ψ〉 defined by
b|ψ〉 = 0. (3.51)
This is a “squeezed state” in the original variables. Find
〈ψ|a†a|ψ〉 (3.52)
〈ψ|aa†|ψ〉 (3.53)
〈ψ|aa|ψ〉 (3.54)
〈ψ|a†a†|ψ〉. (3.55)
The relevance to the Luttinger liquid is that the ground state of the interacting
system is exactly this sort of squeezed state.
CHAPTER 3. APPLICATIONS – MAR 20, 2019 31
3.1.5. A more tedious approach to the same problem is to consider the Hamil-
tonian
H = A(a†a+ aa†) +B(aa+ a†a†). (3.56)
Show that Eq. (3.56) is of the same form as Eq. (3.49), and relate α and λ to
A and B.
I am not going to make you go through it, but the “Bogoliubov” approach to
diagonalizing Eq. (3.56) is to make the general transformation:
a = ub+ vb† (3.57)
a† = ub† + vb, (3.58)
with u2 − v2 = 1. One can easily verify that the b’s obey the standard ladder
operator commutation relations. Substituting into Eq. (3.56) yields
H =[A(u2 + v2) +B(2uv)
](b†b+ bb†) +
[B(u2 + v2) +A(2uv)
](b†b† + bb).(3.59)
To diagonalize the Hamiltonian, one chooses u and v to obey
u2 − v2 = 1 (3.60)
B(u2 + v2) +A(2uv) = 0. (3.61)
One traditionally writes u = cosh(ξ) and v = sinh(ξ), in which case
u2 + v2 = cosh(2ξ) (3.62)
2uv = sinh(2ξ), (3.63)
and
tanh(2ξ) = −B/A. (3.64)
Clearly we require |A| ≥ |B|. Simple “trig” then gives
u2 + v2 =A√
A2 −B2(3.65)
2uv = − B√A2 −B2
, (3.66)
and
H =√A2 −B2(b†b+ bb†) (3.67)
which should agree with the less tedious “scaling” method.
Problem 3.2. This is mostly an exercise in book-keeping. Feel free to skip it if
it seems too dull. Nonetheless going through this once is useful – and if you end
CHAPTER 3. APPLICATIONS – MAR 20, 2019 32
up doing more 1D stuff, getting fast at it is useful, as you need to keep doing
this to find all your π’s and − signs.
3.2.1. In the main text we defined:
φr(x) = i2π
L
∑p
eipx
pρr(−p) (3.68)
φ`(x) = i2π
L
∑p
eipx
pρ`(−p), (3.69)
and we have
bp = θ(p)
√2π
L|p|ρr(−p) + θ(−p)
√2π
L|p|ρ`(p) (3.70)
or for p > 0
ρr(−p) =
√Lp
2πbp (3.71)
ρ`(p) =
√Lp
2πb−p (3.72)
ρr(p) =
√Lp
2πb†p (3.73)
ρ`(−p) =
√Lp
2πb†−p. (3.74)
Write φr and φ` in terms of the b’s.
3.2.2. Write φ = (φr − φ`)/2 and θ = −(φr + φ`)/2 in terms of the b’s.
3.2.3. As argued in the main text ρ = ∂xφ/π, and hence φ has the physical
meaning of a displacement field. Show that
[φ(x), ∂yθ(y)] = iπδ(x− y), (3.75)
and hence we can interpret Π(x) = ∂xθ(x)/π as the conjugate momentum to θ.
3.2.4. The Hamiltonian is
Hb0 =
∑p
vfpb†pbp. (3.76)
Show that
Hb0 =
vf2π
∫dx[(∂xθ)
2 + (∂xφ)2]. (3.77)
Chapter 4
Tonks Gas and the
Leib-Lininger Model – Mar
27, 2019
In the previous chapters we explored the “bosonization” techniques, largely in
the context of spin-free fermions. The remarkable result there was that the low
energy physics was describable in terms of a 1D elastic medium, and that phys-
ical quantities, such as the momentum occupation space occupation numbers,
could be expressed in terms of corelators of the fields describing this elastic
medium. Both a positive and negative feature of this bosonization approach is
that it is constructed in terms of operators. In the next two chapters we con-
sider a complementary point of view, and explicitly construct the wavefunction
of a 1D system. This is clearly more concrete, but it means that calculating
physical quantities is a little more difficult.
The approach, widely known as the “Bethe Ansatz” is applicable to a number
of systems with “point interactions.” It has been applied to bosons (the Lieb-
Lininger model), spin-1/2 fermions (the Gaudin-Yang model), spin-1/2 lattice
fermions (the Hubbard model), and spin chains (the XXZ model). We will
explicitly discuss only the simplest of these, namely the boson system.
.1. Permutations
Permutations play an important role in our arguments, so it is good to briefly
introduce some notation that we will use. Consider the permutations of n
objects. These form a group, Sn, consisting of n! elements, and known as the
“symmetric group”. We think about the permutations as functions which map
33
CHAPTER 4. TONKS GAS AND THE LEIB-LININGER MODEL – MAR 27, 201934
the integers 1, 2, · · · , n onto themselves. For example, with n = 5, we might
have σ(1) = 3, σ(2) = 2, σ(3) = 4, σ(4) = 1, σ(5) = 5. The composition of
two permutations is a permutation, and we just use standard “multiplication”
notation for composition. So if τ(1) = 5, τ(2) = 2, τ(3) = 1, τ(4) = 3, τ(5) = 4,
then we can define their product λ = στ by λ(1) = σ(τ(1)) = σ(5) = 5, and so
on. A permutation is uniquely defined by how it scrambles the ordered set, for
example 1, 2, 3, 4, 5. Thus we can write σ = [3, 2, 4, 1, 5] or τ = [5, 2, 1, 3, 4].
Another useful notation for permutations is to decompose them into disjoint
cycles. For example σ = (134)(2)(5), meaning that σ(1) = 3 and σ(3) = 4
and σ(4) = 1. Up to cyclical permutations of each cycle, this decomposition is
unique. We can safely ignore the 1-cycles (or unicycles as Veit suggested) as
they don’t do anything. Hence σ = (1, 2, 3) and τ = (1, 5, 4, 3).
A further refinement is that each cycle can be written as a product of
non-disjoint 2-cycles, also known as transpositions. For example, (1, 2, 3) =
(1, 2)(2, 3). The decomposition into transpositions is not unique, but the parity
of the number of transpositions is. The permutations consisting of even numbers
of transpositions forms a subgroup known as the “alternating group,” An.
Clearly the 2-cycles generate Sn. It turns out that a minimal set of generators
is just the “neighboring” transpositions (1, 2), (2, 3), · · · (n− 1, n). For example
(1, 3) = (1, 2)(2, 3)(1, 2). This decomposition is not unique.
.2. 1D Hard Core Gas
Let us now think about bosons with hard-core interactions. This is a bit of a
warm-up, that will introduce some of the basic concepts. The hard core Bose
gas is often known as the Tonks-Girardeau model, and sometimes it seems that
keeping track of the names of the various models is half the work. [For those
who care, Tonks actually studied the classical model, but nonetheless his name
is still associated with this system.]
As we know from solving the “particle in an infinite well,” infinitely strong
potentials can be replaced by a boundary condition on the wavefunction. By
the same argument, the hard core Bose gas can be described by a free-particle
Schrodinger equation,
−1
2
∑j
∂2jψ(x1, · · · , xn) = Eψ(x1, · · · , xn), (4.1)
with the boundary condition that ψ = 0 if any of the two x’s are equal. We
should also probably add the Bose symmetry requirement that ψ is unchanged
by permuting any two of the particles. Finally, if we want a finite particle
density, we should put this in a box of length L. Lets use periodic boundary
CHAPTER 4. TONKS GAS AND THE LEIB-LININGER MODEL – MAR 27, 201935
conditions, so ψ(x1 + L) = ψ(x1), where I have held all the other coordinates
fixed.
There is a wonderful trick for solving this problem. Namely we introduce
a function A(x1, x2, · · ·xn) which is always either 1 or −1. It equals 1 if x1 <
x2 < · · · < xN , or if the ordering of the particles is an even permutation. It is
−1 if it is an odd permutation. Given A, and a bosonic wavefunction ψB that
obeys our hard core boundary conditions, we can always generate a fermionic
wavefunction ψF = AψB . Conversely, given an arbitrary fermionic wavefunction
ψF we can generate a hard-core Bose wavefunction vie ψB = AψF . Clearly ψBobey EQ. (4.1) if and only if ψF also does. Thus we have established a one-to-
one mapping between hard core Bose wavefunctions, and non-interacting Fermi
wavefunctions in one dimension.
In particular, the fermionic wavefunction can always be expressed as a Slater
determinant
ψF =
∣∣∣∣∣∣∣∣eik1x1 eik1x2 · · · eik1xn
eik2x1 eik2x2 · · ·...
∣∣∣∣∣∣∣∣ (4.2)
=∑σ∈Sn
sig(σ)∏j
eikσ(j)xj (4.3)
where k1, k2, · · · kn are distinct wave-vectors, satisfying kj = nj(2π/L), and njare distinct integers. The energy is simply E =
∑j k
2j/2, and the ground state
consists of choosing all of the nj ’s to have as small as possible magnitudes – ie
forming a Fermi surface.
There is a minor subtlety worth mentioning. Namely, if there are an even
number of particles, and we want the bosonic wavefunction to have periodic
boundary conditions, then the fermions need to have antiperiodic boundary
conditions, ie kj = (nj + 1/2)(2π/L).
It is also useful to note that one can explicitly write down the ground state
wavefunction by using the fact that for numbers z1, z2, · · · zn, the Vandermond
determinant can be evaluated as∣∣∣∣∣∣∣∣∣∣1 1 1 · · ·z1 z2 z3 · · ·z2
1 z22 z2
3 · · ·...
∣∣∣∣∣∣∣∣∣∣=∏i<j
(zi − zj). (4.4)
This can be proven by noting that both sides vanish when zi = zj , and both
CHAPTER 4. TONKS GAS AND THE LEIB-LININGER MODEL – MAR 27, 201936
are n’th order homogeneous polynomials. We then write
ψf = e−inπL
∑j xj
∣∣∣∣∣∣∣∣∣∣1 1 · · · 1
e2πix1/L e2πix2/L · · · e2πixn/L
e4πix1/L e44πix2/L · · · e4πixn/L
...
∣∣∣∣∣∣∣∣∣∣(4.5)
= e−inπL
∑j xj∏i<j
(e2πixi/L − e2πixj/L( (4.6)
∝∏i<j
sinπ
L(xi − xj) (4.7)
The bosonic wavefunction is then
ψb =∏i<j
sinπ
L|xi − xj |. (4.8)
The remarkable thing is that all thermodynamic properties of the hard core
bosons and the free fermions are identical. Also the density, or the density-
density correlations functions should be identical. I should caution that other
properties are not the same. For example, the k’s in the determinant, referred to
as rapidities, do not correspond to the momenta of the bosons. The momentum
distribution is
nk =∝∫dx′1
∫dx1 dx2 · · · dxne−ik(x−x′)ψ∗(x′1, x2, · · ·xn)ψ(x1, x2, · · ·xn).
(4.9)
At small k, nk ∼ k−1/2, while for large k, nk ∼ k−4.
A. Spin Chains
To further illustrate the principle of mapping onto a Fermi system, we can
consider a chain of spin-1/2 objects,
H =∑j
Jx(Sxj Sxj+1 + Syj S
yj+1) + Jz(S
zj S
zj+1). (4.10)
This is referred to as the XXZ model, and represents some sort of anisotropic
magnetic interactions. The isotropic case Jx = Jz is the Heisenberg model,
while the extremes case Jx = 0 and Jz = 0 are the Ising model and XY models.
The operators Sx, Sy, Sz are Pauli operators. It is convenient to introduce
ladder operators S+ and S−, in terms of which
H =∑j
Jx2
(S+j S−j+1 + S−j S
+j+1) + Jz(S
zj S
zj+1). (4.11)
CHAPTER 4. TONKS GAS AND THE LEIB-LININGER MODEL – MAR 27, 201937
We map this onto a Fermi gas by identifying a down-spin site with an empty
site, and an up-spin site with an occupied site. One naturally expects something
like
S+j = f†j (4.12)
S−j = fj (4.13)
Szj = 2f†j fj − 1. (4.14)
One can readily verify that the {fj , f†j } = 1. Unfortunately the fj ’s on different
sites commute. We can fix this with what is known as a “Jordan-Wigner”
transform
a†j = (−1)∑k<j f
†kfkf†j (4.15)
aj = (−1)∑k<j f
†kfkfj , (4.16)
which gives
H =Jx2
∑i
(a†iai+1 + a†i+1ai
)+ Jz
∑i
(2ni − 1)(2ni+1 − 1). (4.17)
This is the Hubbard model. Note the special case Jz = 0 corresponds to free
Fermions, and we can trivially solve that case.
Chapter 5
Bethe Ansatz – Mar 29,
2019
The Bethe ansatz is a generalization of the approach in Sec. .2 – and is best
illustrated using a generalization of the model discussed their, namely a gas of
1D bosons with point interactions,
−1
2
∑j
∂2jψ + c
∑i<j
δ(xi − xj)ψ = Eψ. (5.1)
As before, we impose bosonic symmetry, and periodic boundary conditions in a
box of size L. The limit c→∞ corresponds to the Tonks gas, which we solved
by mapping onto free fermions. The limit c→ 0 corresponds to the free bosons,
with ψ = 1.
We will approach this problem, by first solving the two-body problem, then
the three-body problem, then generalizing to arbitrary particle number.
A. Two-Body Problem
The two-body problem is a classic undergraduate quantum mechanics problem.
In lecture I jumped to the answer, but in these notes I will go through all of the
details (since I never remember them, and have to do it from scratch anyways.
We begin with
−∂21 + ∂2
2
2ψ(x1, x2) + cδ(x1 − x2)ψ(x1, x2) = Eψ(x2, x2). (5.2)
38
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 39
We change variables to center-of-mass and relative:
y = x1 − x2 (5.3)
s =x1 + x2
2. (5.4)
We then use the chain rule to find:(∂ψ
∂x1
)x2
=
(∂ψ
∂y
)s
(y
∂x1
)x2
+
(∂ψ
∂s
)y
(s
∂x1
)x2
(5.5)
=
(∂ψ
∂y
)s
+1
2
(∂ψ
∂s
)y
. (5.6)
We then repeat to get
∂21 = ∂2
y + ∂y∂s +1
4∂2s . (5.7)
To get ∂22 , we just flip the sign of y, so
∂22 = ∂2
y − ∂y∂s +1
4∂2s . (5.8)
Hence, we can rewrite Eq. (5.9) as
−∂2yψ −
1
4∂2sψ + cδ(y)ψ = Eψ. (5.9)
We then integrate over y around y = 0 to yields
∂yψ|y=0+
y=0− = c ψ|y=0 . (5.10)
We now convert this back to our original variables to get
(∂1 − ∂2)ψ|x1=x2+εx1=x2−ε = 2c ψ|x1=x2
. (5.11)
The only tricky thing with that whole argument is getting the factors of 2 right.
We can think of Eq. (5.9) as a free Schrodinger equation with the boundary
condition in Eq. (5.11).
Lets now think about this as a scattering problem in the x1 − x2 plane. In
our elementary classes, we solve scattering problems by imagining an incoming
plane wave, and an outgoing plane wave, then choosing amplitudes and phases
so that the boundary conditions are satisfied. Thus we imagine an incoming
wave, ei(k1x1+k2x2), in the regime x1 > x2:
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 40
By Bose symmetry, one must also have an incoming wave ei(k1x2+k2x1) with
x1 < x2. One will also end up with reflected and transmitted rays from each
of these two incoming waves. The reflected ray from the x1 > x2 wave has the
same wave-vector as the transmitted ray from the x1 < x2 wave, so we only
need one coefficient R to describe both these processes. And hence we draw the
various rays as:
To be explicit, the wavefunction for x1 > x2 is
ψ12(x1, x2) = eik1x1+ik2x2 +Reik2x1+ik1x2 , (5.12)
and the wavefunction for x2 > x1 is
ψ21(x1, x2) = eik1x2+ik2x1 +Reik2x2+ik1x1 . (5.13)
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 41
We now implement the boundary condition in Eq. (5.11), and calculate
(∂1 − ∂2)ψ|x1=x2+ε = (∂1 − ∂2)ψ12|x1=x2(5.14)
= i(k1 − k2)(1−R)ei(k1+k2)x (5.15)
(∂1 − ∂2)ψ|x1=x2−ε = (∂1 − ∂2)ψ21|x1=x2(5.16)
= −i(k1 − k2)(1−R)ei(k1+k2)x (5.17)
ψx1=x2= (1 +R)ei(k1+k2)x. (5.18)
For the boundary condition to be satisfied, we need
i(1−R)(k1 − k2) = c(1 +R) (5.19)
which can be manipulated into
R = −eiφ (5.20)
= −(c− i(k1 − k2)
c+ i(k1 − k2)
), (5.21)
which defines the angle
φ = 2 arctank2 − k1
c. (5.22)
The first observation is that it makes sense for φ to simply be a phase: this
is consistent with conservation of probability. One may be curious about why I
included the minus sign in R = −eiφ. The reason is that with this definition, in
the hard core case φ = 0. It turns out that the expansion about this hard core
limit is better behaved than the expansion about the non-interacting limit, so
it is convenient to set up the notation in anticipation.
A second observation is that the boundary condition can be satisfied by any
k1 and k2. This is analogous to the observation in section .2, that the hard core
constraints and Bose symmetry could be satisfied by any Slater determinant. In
both that case, and this one, the quantization of the k’s comes from imposing
periodic boundary conditions:
ψ(x1 + L, x2) = ψ(x1, x2). (5.23)
If we take x1 < x2, this requires
ψ12(x1 + L, x2) = ψ21(x1, x2) (5.24)
which means
eik1Leik1x1+ik2x2 +Reik2Leik2x1+ik1x2 = eik1x2+ik2x1 +Reik2x2+ik1x1 .(5.25)
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 42
This must be true for all x1 and x2, so we need
Reik2L = 1 (5.26)
eik1L = R. (5.27)
If I multiply these two equations I find
ei(k1+k2)L = 1. (5.28)
This is just quantization of the center of mass momentum. If I divide the
equations, I get
ei(k1−k2) = R2 = e2iφ (5.29)
Taking the logarithm, and defining δ = k1 − k2, we have
δ = 2 logc− iδc+ iδ
+ 2πn (5.30)
= 2πn− 4 arctanδk
c. (5.31)
We can solve this graphically. The left hand side is a straight line, while the
right hand side is a series of steps. For smallish c, it looks like
The arctangent lines are labeled by n. The n = 0 case gives δ = 0, which
gives R = −1, and ψ = 0. Thus we should neglect the n = 0 line, which I
colored green. As c → 0, the remaining solutions approach δ = 2πn/L, n =
±0,±1,±2, · · · . They come in plus-minus pairs because permuting k1 and k2
yields the same wavefunction. For large c, the curves become flat:
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 43
Again as c → ∞ we have δ = 2πn/L, but this time n = ±1,±2, · · · . [Recall,
the green curve should be ignored.]
If c < 0 we also have complex solutions, which correspond to bound states.
We will not consider these here.
B. Three Particles
We now solve the three-particle problem. We simply follow the same approach
as the two-particle problem, and the only challenge is book-keeping. Instead
of dividing space into two sectors we divide it into six sectors: x1 < x2 < x3,
x2 < x1 < x3,· · · . I denote the wavefunction when x1 < x2 < x3 as ψ123, and
when x2 < x1 < x3 as ψ213, · · · . More formally, for a permutation σ we define
ψσ to describe the sector where xσ(1) < xσ(2) < xσ(3). We then postulate that
in each sector the wavefunction is a sum of plane waves, which we can think of
as all the rays which happen when one of the particles scatter off one-another,
ψ123(x1, x2, x3) = A123123e
i(k1x1+k2x2+k3x3) +A123213e
i(k2x1+k1x2+k3x3)(5.32)
+A123132e
i(k1x1+k3x2+k2x3) +A123231e
i(k2x1+k3x2+k1x3)
+A123312e
i(k3x1+k1x2+k2x3) +A123321e
i(k3x1+k2x2+k1x3).
That is A123τ is the coefficient of ei(kτ(1)x1+kτ(2)x2+kτ(3)x3). In other sectors one
would define
ψ213(x1, x2, x3) = A213123e
i(k1x2+k2x1+k3x3) +A213213e
i(k2x2+k1x1+k3x3)(5.33)
+A213132e
i(k1x2+k3x1+k2x3) +A213231e
i(k2x2+k3x1+k1x3)
+A213312e
i(k3x2+k1x1+k2x3) +A213321e
i(k3x2+k2x1+k1x3).
That is, in the sector where xσ(1) < xσ(2) < xσ(3), the factor Aστ is the coefficient
of ei(kτ(1)xσ(1)+kτ(2)xσ(2)+kτ(3)xσ(3)). The generalization to more particles should
be clear.
We now consider Bose symmetry, which says that ψ123(x, y, z) = ψ213(y, x, z).
Writing things out, we find A123τ = A213
τ . Clearly the generalization is that Aστis independent of σ. This generalization works for all particle numbers.
We now turn to the boundary conditions, for example, lets take x3 to be
bigger than x1 and x2, and consider the condition
(∂1 − ∂2)ψ|x1=x2+εx1=x2−ε = 2c ψ|x1=x2
. (5.34)
This can be expressed as
(∂1−∂2)ψ213(x, x, x3)−(∂1−∂2)ψ123(x, x, x3) = 2cψ213(x, x, x3) = 2cψ123(x, x, x3).
(5.35)
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 44
There are three sorts of terms, proportional to ei(k1+k2)x+k3x3 , ei(k2+k3)x+k1x3 ,
and ei(k1+k3)x+k2x3 . One can equate the coefficient of each of these. Lets look
at the terms proportional to ei(k1+k2)x+k3x3 :[i(k2 − k1)A213
123 + i(k1 − k2)A213213
]−[i(k1 − k2)A123
123 + i(k2 − k1)A213123
]= 2c(A123+A213)
(5.36)
We added the superscripts just to make it more obvious where each term came
from. We can then group terms, and find
A213
A123= R12 = −eiθ12 = −c+ i(k1 − k2)
c− i(k1 − k2). (5.37)
We can similarly define
Rij = −eiθij = −c+ i(ki − kj)c− i(ki − kj)
, (5.38)
which relatesAjikAijk
=AkjiAkij
= Rij . (5.39)
That is, because the boundary conditions relate permutations of neighboring
particles, we get relations between A’s where neighboring indices are exchanged.
Because the neighboring transpositions generate all the permutations, we can
find a general Aσ in terms of A123. We can further always choose A123 = 1.
That is
A123 = 1 (5.40)
A213 = R12 (5.41)
A132 = R23 (5.42)
A231 = R31R12 (5.43)
A312 = R13R23 (5.44)
A321 = R12R13R23. (5.45)
These are clearly consistent with one-another: There are two distinct ways to
reach A321 from A123, but they give the same R’s. The generalization to more
particles will also be consistent.
For any set of k’s, our ansatz solves Schrodinger’s equation (as long as no
two k’s are not identical – for otherwise the wavefunction vanishes). Adding
periodic boundary conditions will then restrict the allowed k’s. Take 0 <=
x1 < x2 < x3 < L, then
ψ(x1 + L, x2, x3) = ψ(x1, x2, x3). (5.46)
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 45
The wavefuction of the left is clearly ψ231, and the wavefunction on the right
is ψ123. There are several ways to do the arithmetic, but I like to use the Bose
symmetry so that these are both in the 123 sector,
ψ123(x2, x3, x1 + L) = ψ123(x1, x2, x3). (5.47)
Equating terms with the same x dependence yields
eik3LA123 = A312 = R13R23A123 (5.48)
and permutations. The way I like to think of it, is with periodic boundary
conditions you can go from 123 to 312 in two ways: You do two transpositions,
or you simply slip particle number 1 off the left. Equating these gives Eq. (5.48).
We consequently find three equations
R13R23 = eik3L (5.49)
R12R32 = eik2L (5.50)
R21R31 = eik1L (5.51)
Taking logs yields
k3L = (π + θ13) + (π + θ23) + 2πn3 (5.52)
k2L = (π + θ12) + (π + θ32) + 2πn2 (5.53)
k1L = (π + θ21) + (π + θ31) + 2πn1, (5.54)
where the nj are integers, and
θij = −i log
(c+ i(ki − kj)c− i(ki − kj)
)= 2 tan−1 ki − kj
c. (5.55)
If we define θii = 0, then the general many-particle result is clearly
kjL =∑i
θij + 2π(nj + ξ/2) (5.56)
where xi = 0 if there are an odd number of particles, but ξ = 1 for an even.
I claim that Eq. (5.56) has a unique solution once we specify the nj ’s. The
best way that I know to prove this uniqueness is to first show that Eq. (5.56) is
equivalent to extremizing the function
F =∑j
(k2j
2− 2π(nj + ξ/2)
Lkj
)− 1
2
∑ij
Λij (5.57)
with
Λij = − i
L[(c+ (ki − kj)) log (c+ (ki − kj)) + (c− (ki − kj)) log (c− (ki − kj))] .
(5.58)
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 46
We then prove that F is concave up. Thus we calculate the second derivatives
∂2F
∂k2i
= 1 +2
L
∑j 6=i
c
c2 + (ki − kj)2(5.59)
∂2F
∂ki∂kj= − 1
L
2c
c2 + (ki − kj)2. (5.60)
We now consider the quadratic form
G =∑ij
∂2F
∂ki∂kjxixj (5.61)
=∑i
x2i +
1
L
∑j
c
c2 + (ki − kj)2(xi − xj)2. (5.62)
Clearly, as the sum of positive quantities, G is positive – hence F is concave up.
C. Thermodynamic Limit
While numerically it is quite efficient to simply solve Eq. (5.56) for fixed N , it
is appealing to understand how to take the thermodynamic limit. To do so, we
consider an odd number of particles, and order the n’s such that n1 < n2 <
· · ·nN . The ground state will have
nj = −(N + 1)/2 + j. (5.63)
We then try to find the spacing between subsequent k’s:
(kj − kj−1)L =∑i
(θij − θij−1) + 2π. (5.64)
In the thermodynamic limit the spacing will be small, and we define the density
of k’s to be
Lf(kj) =1
kj − kj−1. (5.65)
We then divide Eq. (5.64) by (kj − kj−1)L to arrive at
1 =∑i
θij − θi,j−1
L(kj − kj−1)+ 2πf(kj). (5.66)
We then note that in the thermodynamic limit, the term in the sum is a deriva-
tive:
1 =∑i
1
L
∂θij∂kj
+ 2πf(kj). (5.67)
Finally we convert the sum to an integral,
1 =
∫dq f(q)∂kθ(q − k) + 2πf(k), (5.68)
CHAPTER 5. BETHE ANSATZ – MAR 29, 2019 47
where we have replaced kj with k and ki with q. We have also introduced the
function
θ(q) = −i logc+ iq
c− iq. (5.69)
Taking the derivative, we find the integral equation
2πf(k)− 1 =
∫dq
2c
c2 + (q − k)2f(q). (5.70)
This should be supplemented with the constraint∫f(k)dk =
N
L. (5.71)
The traditional approach to solving this integral equation involves first as-
suming that in the thermodynamic limit there is a maximum k, called k0. This
is certainly true in the hard core case, so it seems plausible. Thus f(k) = 0 for
all |k| > k0. We rescale by this maximum, writing x = k/k0. We also rescale
c, writing λ = c/k0. We also introduce γ = cL/N , which is a dimensionless
measure of the interaction strength. In the thermodynamic limit, one expects
that the interactions only enter in this combination. Finally to avoid confusion
we write g(x = k/k0) = f(k).
Equation (5.71) relates λ to g and the physical quantity γ via
λ = γ
∫ 1
−1
g(x)dx. (5.72)
The other equation becomes
2πg(y) = 1 + 2λ
∫ 1
−1
g(x)
λ2 + (x− y)2dx. (5.73)
One, simple approach to solving these equations is to begin with some trial
g0. one then uses Eq. (5.72) to calculate λ0. One substitutes this into Eq. (5.73)
to get a new g,
2πg1(y) = 1 + 2λ0
∫ 1
−1
g0(x)
λ2 + (x− y)2dx. (5.74)
one then iterates.