Interacting Bose gas on the Lattice
Matthew Egginton
0901849
Supervisor: Daniel Ueltschi
May 22, 2014
MEN S T A T
A G I MOLEM
SI
S
UN
IV
ER
S
I TAS WARWI C
EN
Mathematics Institute
Abstract
A Lower bound for the density of a Bose gas in Zd in the grand canonical
ensemble is calculated for general dimension d, and for l1 (Zd ) potential interactions. The rigorous formalism of the measure on continuous time random walks,
analogous to the Wiener measure, is briefly explained, along with justification of
the Feynman-Kac formula using this. These are then used to bound the density,
although with a flawed lemma. Throughout, attempts to further understand the
discrete Gaussian are made.
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Contents
1 Introduction
2
2 Introductory Material
2.1 Discrete Gaussian . . . . . . . . . . . . . . .
2.2 Discrete Gaussian and Fourier Transform .
2.3 Measure on continuous time Random walks
2.4 Feynman-Kac formula . . . . . . . . . . . .
2.5 Statistical Mechanics . . . . . . . . . . . . .
2.6 Properties of Functions . . . . . . . . . . .
3 Lower Bound on the Density of the
3.1 Bounding the “A” term . . . . . .
3.2 Bounding the “B” term . . . . . .
3.3 Lower Bound Result . . . . . . . .
3.4 Bounding the Integral Kernel . . .
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Bose gas
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4 Machinations on the Discrete Gaussian
30
4.1 Attempted proofs of some product rule . . . . . . . . . . . . . . . . . . . 30
4.2 Modifications of the Random Walk . . . . . . . . . . . . . . . . . . . . . 31
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M. Egginton
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Introduction
In 1995 the first Bose-Einstein condensate was produced by Eric Cornell and Carl
Wieman, using a gas of Rubidium atoms. A Bose-Einstein condensate is a state
of matter occurring in Bosonic gases with temperatures near absolute zero in which
quantum effects become apparent on a macroscopic scale. This state, predicted by
Einstein and Bose in 1925, is characterised, upon cooling, by the atoms falling into the
lowest energy state. This happens at a critical temperature, Tc , below which all the
particles are in the lowest energy state, and above, none are.
In this work I consider a bosonic gas that is able to move on the integer lattice
Zd . The interactions are given by the Bose-Hubbard model, which is a simplistic
model, but nonetheless useful. It can be used to describe motion on an optical lattice.
These are typically created as an interference pattern of laser beams, giving a spatially
periodic pattern. This potential field then traps the atoms, and resembles the atom
distribution of a crystalline solid.
Atoms near absolute zero in an optical lattice are considered as a standard realisation of the Bose-Hubbard model, and this model can also be used in quantum
computing.
One can relate the critical temperature to a critical fugacity zc . Then it suffices to
find a lower bound on the density of the system in order to find bounds on the critical
fugacity.
This work finds a lower bound on this density as follows:
Theorem 1.0.1 For a Bose gas in Λ ⊂ Zd in the grand canonical ensemble with
Hamiltonian HΛ,N = ∆D + U with U ∈ l1 (Zd ) the density is bounded below by
2
(0)a0 β − a0 βz 2 [3q0 (z)q1 (z) + q02 (z) + 2q1 (z)M + 3q0 (z)M ]
ρ(z) ≥ ρ(0) (z) − 2z 2 g2β
where
1
M=
1
2π
R
[−π,π]d
1−
4β
d
(1
−
cos
k
)
eik(x−y) dk
j
j=1
Pd
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M. Egginton
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Introductory Material
In this section we introduce the necessary tools to be used throughout the next section.
2.1
Discrete Gaussian
For background work on random walks and Markov chains, the reader should see
Lawler [4] or Grimmett and Stirzaker [3]. We have a Markov chain with increment
1
distribution p : Zd → R where p(e) = 2d
, for e a unit vector in Zd , and p(x) = 0 for
x not a unit vector. In other words we assign the probability of 1/2d to move to a
neighbouring point and probability of 0 for anywhere else. We define p(x, y) := p(y−x).
We then have, for x, y ∈ Zd x 6= y and ∆t small, that
P(St+∆t = y|St = x) := p(x, y)∆t + o(∆t)
(2.1.1)
by definition. In other words, we have that the probability that the state space at time
t + ∆t is y given that the state space was x at time t is linear in the time period ∆t
up to small perturbations. One also has, following immediately from the above, that
X
P(St+∆t = x|St = x) = 1 −
p(x, y)∆t + o(∆t)
(2.1.2)
y6=x
Henceforth, we denote by gt (x) the discrete Gaussian in space variable x and time
t. This is the transition probability for the simplest type of random walk, the simple
random walk. Observe here that one could combine the above into one line with a
definition
P(St+∆t = y|St = x) = δx,y + ∆tp̃(x, y) + O(∆t2 )
although here p̃ is defined slightly differently to the above p.
Definition 2.1.1 Suppose that P is the probability as above. Then we define the discrete Gaussian as
gt2 −t1 (x2 − x1 ) := P(ω(t2 ) = x2 |ω(t1 ) = x1 )
(2.1.3)
By computing the above probability explicitly we get.
Theorem 2.1.2 The discrete Gaussian in Zd , for x1 , x2 ∈ Zd and t1 , t2 ∈ T , two
“times” is given by
gt2 −t1 (x2 − x1 ) = e−(t2 −t1 )
X (t2 − t1 )N
(2d)N N !
N ≥0
X
1y1 =x1 1yN =x2
y1 ,...,yN
N
Y
1||yi −yi−1 ||=1
i=2
While it is not used in this work, it should be noted that it is expected that, for
fixed x, gt (x) ≈ πt (x) =
1
d
(2πt) 2
e−
|x|2
2t
for large times t.
Lemma 2.1.3 For all x ∈ Zd .
gt (x) ≤ gt (0)
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M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Proof
The Fourier transform of the discrete Gaussian (see subsection 2.2) is
gbt (k) = e−
Pd
j=1 (1−cos kj )t
and then using this we get
gt (x) = |gt (x)|
Z
1
=
2π
e− d
1
Pd
1
Pd
j=1 (1−cos kj )t
eikx dk
[−π,π]d
≤
Z
1
2π
e− d
j=1 (1−cos kj )t
dk
[−π,π]d
= gt (0)
as the exponential term is real and positive so we can remove the modulus.
Q.E.D.
Lemma 2.1.4
X
gt (x) = 1
x∈Zd
Lemma 2.1.5
gt+s (x) =
X
gs (y)gt (x − y)
y∈Zd
The proofs of the previous two lemmas are just an exercise in rearranging sums.
Definition 2.1.6 I define the discrete Laplacian to be
X
∆D f (x) :=
(f (y) − f (x))
yx̃
2.2
Discrete Gaussian and Fourier Transform
For our random walk, the generator is the discrete Laplacian, up to a constant, as seen
below.
It was mentioned to me that it would be possible to find a nicer formula for the
discrete Gaussian using Fourier analysis. To this end I introduce the Fourier transform
and do some work with this.
Definition 2.2.1 The discrete Fourier transform of a function φ : Zd → C is defined
to be a function F(φ) : [−π, π]d → C such that
X
φ̂(k) = F(φ)(k) :=
φ(x)e−ikx
x∈Zd
In analogy with the Fourier transform on [−π, π] where the Fourier coefficients are
sums over Z, our inverse Fourier transform will be
Z
1
−1
F f (k) =
f (k)eikx dk
(2.2.1)
2π
[−π,π]d
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Lemma 2.2.2
ID: 0901849
d
X
1
− ∆D = F −1 (
(1 − cos ki ))F
2
i=1
Proof
The statement above is equivalent to − 12 F∆D =
d
P
(1 − cos ki )F and we proceed
i=1
to prove this statement.
1
1 X
− F∆D φ(k) = − F
(φ(y) − φ(x))(k)
2
2 y∼x
1 X −ikx X
(φ(y) − φ(x))
e
=−
2
d
y∼x
x∈Z
1 X X −ikx
=−
e
(φ(x + h) − φ(x))
2
d
x∈Z
h
1 XX
φ(x)(e−ik(x−h) − e−ikx )
=−
2
d
x∈Z
=
=
=
=
1X
2
1
2
(1 − eikh )
h
d
X
X
φ(x)e−ikx
x∈Zd
h
1X
2
h
(1 − eikh )Fφ(k)
(2 − (eikj + e−ikj ))Fφ(k)
j=1
d
X
(1 − cos kj )Fφ(k)
j=1
By summing over all h, we mean summing over all possible unit vectors in Zd in both
positive and negative directions. Then kh for each h gives the element kj of k if h is
the jth unit vector of Zd .
Q.E.D.
We now introduce the following:
Definition 2.2.3 This is from [4]. We define the difference operators for a function
f : Zd → R by
∇x f (y) = f (y + x) − f (y)
1
1
∇2x f (y) = f (y + x) + f (y − x) − f (y)
2
2
We define the generator of a random walk, defined by increment distribution p : Zd →
R to be
X
Lp f (y) =
p(x)∇x f (y)
x∈Zd
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M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
One should observe that the transition probabilities, or in the simple random walk
the discrete Gaussian, for the random walk satisfies a type of “ Heat equation”,
d
gt (x) = Lp gt (x)
dt
In our case we have Lp =
(2.2.2)
1
2d ∆D ,
and if one Fourier transforms this equation, one gets:
d
1
F
gt (x) (k) = F ∆D gt (x)
dt
2d
and so, denoting Gt (k) = F(gt (x))(k) = ĝt (k) one obtains the differential equation
d
d
1X
Gt (k) = −
(1 − cos kj )Gt (k)
dt
d
j=1
and this has the solution
1
Pd
Gt (k) = Gt (0)e− d j=1 (1−cos kj )t
P
where Gt (0) = F(gt (x))(0) = Zd gt (x) = 1. Then we get that
Z
1 Pd
1
gt (x) =
e− d j=1 (1−cos kj )t eikx dk
2π
(2.2.3)
[−π,π]d
Recall the convolution theorem for Fourier transform, namely
Lemma 2.2.4

F

X
g(x − y)f (y) (k) = F(g)(k)F(f )(k)
y∈Zd
2.3
Measure on continuous time Random walks
We first give a brief method on the construction of the analogue to the Wiener measure
on continuous time random walks. This is done in [2, pp. 331-335] for the Wiener
measure.
d
Q We dhave a space of functions ω : [0, t] → Z , although we can think of this space as
[0,t] Z and we have, for θ = (t1 , ..., tN ) a projection hθ (ω) = (ω(t1 ), ..., ω(tN )). We
then define a process by the finite distributions
dµt1 ,...,tN (x1 , ..., xN ) = gt1 (x1 )gt2 −t1 (x2 − x1 )...gtN −tN −1 (xN − xN −1 )gt−tN (y − xN )
and then by a celebrated theorem of Kolmogorov there is a unique measure on the
space of functions that agree with these distributions, given certain conditions on the
distributions, that are satisfied. We have just explained existence of the following:
t the conditional discrete Wiener measure on the
Definition 2.3.1 We denote by Wx,y
space Ωx,y of paths from x to y in time t.
In direct analogy to this we can do the same, but without the restriction of the endpoint. This would give
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M. Egginton
MA469 - Interacting Bose gas on the Lattice
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Definition 2.3.2 We denote by Wxt the discrete Wiener measure on the space Ωx of
paths starting at x and finishing in time t.
There is another way in which to formulate the existence of such a measure, and
uses the Riesz-Markov representation theorem (the theorem found in [1, p.212], method
again in [2, pp. 331-335]). This alternative construction gives the formula that, for a
function f that depends on n points in time, we have
Z
t
f (ω(t1 ), ..., ω(tn ))dWx,y
(ω) =
(2.3.1)
X
f (x1 , ..., xn )gt1 (x1 )gt2 −t1 (x2 − x1 )...gtn −tn−1 (xn − xn−1 )gt−tn (y − xn )
=
x1 ,...,xn ∈Zd
For convenience we normalise as follows
Z
t
dWx,y
(ω) = gt (y − x)
(2.3.2)
The following lemma is stated in more generality than needed. We only need it for
f a constant function.
1
2 by
Lemma 2.3.3 We define ω t ω 0 , for ω ∈ Ωtx,z
and ω 0 ∈ Ωtz,y
(
ω(t) 0 ≤ t ≤ t1
ω t ω 0 (t) =
ω 0 (t) t1 ≤ t ≤ t1 + t2
t1 +t2 . Then we get
and so ω t ω 0 ∈ Ωx,y
Z
X ZZ
0
t
s
0
t+s
f (ω t ω )dWx,z (ω)dWz,y (ω ) = f (η)dWx,y
(η)
z∈Zd
for f a bounded function.
Proof We use the measure theory machine. Observe that this is trivially true in the
case with f a constant function. Suppose f (ω) = a. Then from the normalisation
above, we have
Z
X ZZ
X
t
s
t+s
adWx,z
(ω)dWz,y
(ω 0 ) =
agt (z − x)gs (y − z) = agt+s (y − x) = adWx,y
(η)
z∈Zd
z∈Zd
using properties of the discrete Gaussian.
Now suppose that f is a simple function, namely f (ω t ω 0 ) = F (ω t ω 0 (t1 ), ..., ω t
0
ω (tN )) where F : ZdN → R. Suppose t1 , ..., ti ≤ t and t ≤ ti+1 , ..., tN ≤ t + s. We get
X ZZ
t
s
F (ω t ω 0 (t1 ), ..., ω t ω 0 (tN ))dWx,z
(ω)dWz,y
(ω 0 ) =
z∈Zd
X
=
gt1 (x1 − x)...gti −ti−1 (xi − xi−1 )gt−ti (z − xi )gti+1 −t (xi+1 − z)×
x1 ,...,xN ,z∈Zd
...gtN −tN −1 (xN − xN −1 )gt+s−tN (y − xN )F (x1 , ..., xN ))
X
=
x1 ,...,xN
Z
=
gt1 (x1 − x)...gtN −tN −1 (xN − xN −1 )gt+s−tN (y − xN )F (x1 , ..., xN ))
∈Zd
t+s
F (η(t1 ), ..., η(tN ))dWx,y
(η)
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M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
As f is bounded, the dominated convergence theorem gives the result for general
f , by taking an approximating sequence of simple functions to f .
Q.E.D.
Lemma 2.3.4 For 0 ≤ t1 < ... < tN ≤ t we have
Z
Z
t
F (ω(t1 ), ..., ω(tN ))dW0,x (ω) = F (ω(t1 ), ..., ω(tN )1ω(t)=x dW0t (ω)
Proof
Z
F (ω(t1 ), ..., ω(tN ))1{ω(t)=x} dW0t (ω) :=
X
=
gt1 (x1 )gt2 −t1 (x2 − x1 )...gt−tN (xN +1 − xN )F (ω(t1 ), ..., ω(tN ))1{ω(t)=x}
x1 ,...,xN +1
X
=
gt1 (x1 )gt2 −t1 (x2 − x1 )...gt−tN (x − xN )F (ω(t1 ), ..., ω(tN ))
x1 ,...,xN
Z
=:
t
F (ω(t1 ), ..., ω(tN ))dW0,x
(ω)
Q.E.D.
Lemma 2.3.5 The following formula holds for f a function on paths from 0 in time t.
The discrete Wiener measure W0a,t represents the usual measure, except the a denotes
the variance of the paths. By ω + ω 0 we mean pointwise summation of the paths. Then
Z
ZZ
1,t 0
1,t
0
f (ω + ω )dW0 (ω)dW0 (ω ) = f (η)dW02,t (η)
However, the corresponding formula for the conditional discrete Wiener measure doesn’t
hold, namely
ZZ
Z
gt (x)gt (y)
1,t 0
1,t
2,t
(ω ) =
(ω)dW0,y
f (ω + ω 0 )dW0,x
(η)
f (η)dW0,x+y
g2t (y + x)
is not true.
Proof
For the first formula, observe that it suffices to consider f of the form
F (x1 , ..., xN ) = e−i
PN
j=1
kj xj
as any such f can be approximated by the simple F depending on N points, and the
simple F can be approximated by objects of the form of the right hand side. Then,
for N = 1 we get
ZZ
X
e−ik1 x1 −ik1 y1 gt1 (x1 )gt1 (y1 )
F (ω(t1 ) + ω 0 (t1 ))dW01,t (ω)dW01,t (ω 0 ) =
x1 ,y1
= ĝt1 (k1 )ĝt1 (k1 )
= ĝ2t1 (k1 )
X
=
e−ik1 z1 g2t1 (z1 )
z
Z1
=
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F (η)dW02,t (η)
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Now for N = 2 we get
ZZ
F (ω(t1 ) + ω 0 (t1 ), ω(t2 ) + ω 0 (t2 ))dW01,t (ω)dW01,t (ω 0 ) =
X
e−ik1 x1 −ik1 y1 −ik2 x2 −ik2 y2 gt1 (x1 )gt1 (y1 )gt2 −t1 (x2 − x1 )gt2 −t1 (y2 − y1 )
=
x1 ,y1 ,x2 ,y2
=
X
e−ik1 x1 −ik1 y1 gt1 (x1 )gt1 (y1 )ĝt2 −t1 (k2 )eik2 (x1 +y1 ) ĝt2 −t1 (k2 )
x1 ,y1
= (ĝt2 −t1 (k2 ))2 (ĝt1 (k1 − k2 ))2
= ĝ2t2 −2t1 (k2 )ĝ2t1 (k1 − k2 )
and then performing operations as in N = 1 case we get back to
X
e−ik1 z1 −ik2 z2 g2t2 −2t1 (z2 − z1 )g2t1 (z1 )
z1 ,z2
which is the value we want.
This suggests the general formula for this specific F of the form
ZZ
F (ω(t1 ) + ω 0 (t1 ), ..., ω(tN ) + ω 0 (tN ))dW01,t (ω)dW01,t (ω 0 )


N
X
= ĝtN −tN −1 (kN )ĝtN −1 −tN −2 (kN −1 − kN ) . . . ĝt2 −t1  (−1)j kj  ×
j=2


N
X
× ĝt1  (−1)j+1 kj 
j=1
and if true, then using the convolution properties of the Fourier transform we get the
desired result. We prove this
P by induction. Suppose it is true for N = L and all
functions F of the form e− mj xj for some constants mj . Then
ZZ
F (ω(t1 ) + ω 0 (t1 ), ..., ω(tL+1 ) + ω 0 (tL+1 ))dW01,t (ω)dW01,t (ω 0 )
PL+1
X
=
e−i j=1 kj (xj +yj ) gt1 (x1 )...gtL+1 −tL (xL+1 − xL )×
x1 ,...,xL+1
y1 ,...,yL+1
× gt1 (y1 )...gtL+1 −tL (yL+1 − yL )
= ĝtL+1−tL (kL+1 )
X
eikL+1 (xL +yL ) e−i
PL
j=1
kj (xj +yj )
gt1 (x1 )...
x1 ,...,xL
y1 ,...,yL
× gt−tL−1 (xL − xL−1 )gt1 (y1 )...gtL −tL−1 (yL − yL−1 )
Then by invoking the induction hypothesis, we get thePdesired result. Note that in
L
doing this we take as F the function eikL+1 (xL +yL ) e−i j=1 kj (xj +yj ) which gives the
required changes of sign needed in the arguments of the Fourier transformed discrete
Gaussians.
For the disproving of the second statement, take y = −x. Then x + y = 0. Also
take f (ω) = 1{η:η(s)=0 ∀s} (ω) and t = 1. This function is measurable with respect
to the discrete Wiener measures involved. Then on the right hand side, the set of
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M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
all
that stay always at zero has strictly positive measure, and so the value of
R paths
1 > 0. On the left hand side however, the set of paths ω + ω 0 that stay at zero
f dW0,0
has zero measure, as all the jumps ω makes must happen at exactly the same time the
jumps in ω 0 happen. Thus the left hand side is zero. This is thus a counter example
of the statement.
Q.E.D.
The latter being false here is serious for this work, see lemma 2.4.2.
2.4
Feynman-Kac formula
Now suppose we want to consider the motion of M particles in a subset Λ of the integer
lattice. This motion is governed by a Hamiltonian of the form
1
HΛ,M = − ∆D + U (x)
2
where ∆D f (x) =
(2.4.1)
P
(f (y) − f (x)) is the discrete Laplacian and U is a potential inyx̃
teraction function, typically bounded and a same site type interaction. The type of
interaction we consider uses U ≥ 0 and so is repulsive. This Hamiltonian represents
an interaction where two particles interact when they are on neighbouring sites ( the
discrete Laplacian) with a same site external potential. This can be though of as
Bose-Hubbard style interaction (see [5]).
The operator e−βH where β is inverse temperature is important statistical mechanically (see the next subsection), and one aims to find a representation of this in the
form
X
L(x, y)f (y)
(2.4.2)
e−βH f (x) =
y∈Zd
Proceeding similarly to Feynman’s path integral formulation, and using the Trotter
product formula (found in [8]), one can find that (Ginibre [2, pp. 344-345] performs
the exact method),
Z
Rβ
β
L(x, y) =
e− 0 U (ω(s))ds dWx,y
(ω)
(2.4.3)
Ωx,y
We set K(x, y) to be the integral kernel of the operator e2β∆D − eβ(2∆D −U ) , i.e.
Z
R
1 4β
4β
1 − e− 4 0 U (ω(s))ds dWx,y
K(x, y) =
(ω)
(2.4.4)
Ωx,y
Definition 2.4.1 [6, p.5] We introduce the notation, for ω, ω 0 : [0, 2β] → Zd ,
Ū (ω, ω 0 ) =
1
2
2β
Z
U (ω(s) − ω 0 (s))ds
(2.4.5)
0
X
V (ω) =
Ū (ωi , ωj )
(2.4.6)
Ū (ωl , ωl00 )
(2.4.7)
0≤l<m≤k−1
0
0
V (ω, ω ) =
k−1 kX
−1
X
l=0 l0 =0
If the paths are enumerated ωi , i ∈ I then we write Vij = V (ωi , ωj ).
10 of 33
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
While I believe the following lemma to be false, although initially both myself and
my supervisor believed it to be true (see the last lemma in the previous section), this
development came too close to the deadline to enable me to change most of the results
which use this. I thus leave it in as it is frequently referred to. I leave the proof in to
show how the result seems so feasible.
Lemma 2.4.2 [6, p.5] For any x, y, x0 , y 0 ∈ Zd we have that
ZZ
0
2β
0
0 0
0
0
1 − e−Ū (ω,ω ) dWx,y
(ω)dWx2β
0 ,y 0 (ω ) = H(x, y, x , y )K(x − x , y − y )
where H(x, y, x0 , y 0 ) =
g2β (y−x)g2β (y 0 −x0 )
g4β (y−y 0 −x+x0 )
Proof The difference ω − ω 0 of two Brownian bridges is again a Brownian bridge, but
with double variance. Then
ZZ
0
1 − e−Ū (ω,ω )
2β
0
dWx,y
(ω)dWx2β
0 ,y 0 (ω ) =
g2β (y − x)g2β (y 0 − x0 )
R
1 2β
Z
1 − e− 2 0 U (ω(2s))ds)
4β
dWx−x
=
0 ,y−y 0 (ω)
g4β (y − y 0 − x + x0 )
and then by noting that g2β (y − x)g2β (y 0 − x0 ) and g4β (y − y 0 − x + x0 ) are independent
of the paths in consideration, we get that
ZZ
0
0
2β
1 − e−Ū (ω,ω ) dWx,y
(ω)dWx2β
0 ,y 0 (ω ) =
Z
R
g2β (y − x)g2β (y 0 − x0 )
4β
− 12 04β U (ω(s))ds
=
1
−
e
dWx−x
0 ,y−y 0 (ω)
g4β (y − y 0 − x + x0 )
= H(y − x, y 0 − x0 )K(x − x0 , y − y 0 )
as required.
2.5
Q.E.D.
Statistical Mechanics
If the reader is unfamiliar with statistical mechanics, then I found [7] particularly
insightful on the basics.
Statistical mechanics is the study of systems where the particle numbers are too
high to consider the system by changes in momentum and position of all particles
independently. One instead considers a probability distribution of position and momentum. However, one strives to express this function using macroscopic properties
of the system, namely the energy of the system, number of particles and volume in
consideration. Of key significance is the entropy S, which is a function dependent on
these three things that is maximised in equilibrium.
For our situation, we consider a case called the Grand Canonical ensemble. This is
where we assume that our system is able to exchange both energy and particles with a
reservoir, and where we assume the reservoir is much larger than the system of interest.
We assume that the reservoir is much larger than the system in consideration so that
its properties are not significantly affected by relatively small changes in its energy or
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
particle number. The entropy of the reservoir is given by S = k ln ΩR (ER , NR , VR ),
and the probability distribution is given by
P (E, N, V ) =
Ω(E, N, V )ΩR (ET − E, NT − N, VR )
ΩT
i.e. is the chance of being in the specified state out of all states possible. Ω is the
function
ZZ
1
δ(E − H(q, p)) dq dp
Ω(E, V, N ) = 3N
h N!
One can find that the probability in terms of the position and momentum is given by
P (p, q) =
1 −βHN +βµN
Ωe
Z
(2.5.1)
where H is the Hamiltonian of the system and β = 1/T is inverse temperature. Z
is the normalising factor, called the Grand Canonical Partition function. I define the
fugacity to be z = eβµ . If we sum over eigenstates, we have
X X
X
Z=
Ωe−βHN +βµN =
z N Tre−βHN
(2.5.2)
N ≥0 H=E
N ≥0
Definition 2.5.1 We define the pressure to be
1
ln Z
β|Λ|
(2.5.3)
∂
(p(β, z))
∂z
(2.5.4)
p(β, z) =
and we define the density to be
ρ(z) = βz
The following is a summary of rewriting the partition function and the density
using the Feynman-Kac formula. Due to the fact that the partition function has a
trace involved, it makes sense to consider loops, i.e. functions f : [0, t] → Zd for which
f (0) = f (1).
We now consider winding loops, as these are important in order to rewrite the
density. Let Ωk be the set of continuous paths [0, 2βk] → Zd that are closed, i.e.
ω(0) = ω(2βk). We denote an element of Ωk by ω = (x, k, ω) where x ∈ Zd is the
starting point, k is the winding number and ω(0) = x = ω(2βk). For 0 ≤ l ≤ k − 1 we
let ω l denote the lth leg of ω. This is equal to
ω l (s) = ω(2βl + s)
with 0 ≤ s ≤ 2β. Then let Ω = ∪k≥1 Ωk .
We define the measure µk on Ωk by
P
1
−
2
zk
2βk
dµk (ω) = dxχΛ (ω)dWx,x
(ω)e 0≤l<m≤k−1
k
2β
R
0
U (ωl (s)−ωm (s))ds
(2.5.5)
One can extend this onto Ω = ∪k≥1 Ω as follows. While here I have underlined for
emphasis that ω and ω are different, henceforth I do not distinguish between the two.
12 of 33
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Definition 2.5.2
dµ(ω) =
X zk
k≥1
k
2βk
dxχΛ (ω)dWx,x
(ω)e−V (ω)
where
e
−V (ω)
−
=e
P
=e
0≤l<m≤k−1
P
−
Ū (ωl ,ωm )
0≤l<m≤k−1
1
2
2β
R
U (ωl (s)−ωm (s))ds
0
One can then rewrite the density using the Feynman-Kac formula to get
Theorem 2.5.3 [6]
P
V (ωi ,ωj )
X 1 Z −
e 1≤i<j≤n
dµ(ω1 )...dµ(ωn )
Z=
n!
n≥0
(2.5.6)
Ωn
P
Z
−
Vij
1 X
1
ρ(z) =
k1 e 1≤i<j≤l dµ(ω1 )...dµ(ωl )
|Λ|Z
(l − 1)! Ωl
(2.5.7)
l≥1
It can also be shown that ρ(z) ≤ ρ(0) (z) where the (0) signifies that the potential
function is everywhere zero. This is called the ideal gas. We can show that
X
ρ(0) (z) =
z n g2βn (0)
(2.5.8)
n≥1
One can calculate this using the above expression for the density, with U = 0, and
then noting that most of the terms in the numerator give the partition function, and
you are left with the formula above.
2.6
Properties of Functions
It is not clear as to why we study this function presently, but I hope the reader is
patient, as all will be clear in section 3.
Definition 2.6.1 For α ∈ R, I define the function
X
qα (z) =
nα z n g2βn (0)
n≥1
Proposition 2.6.2 (Properties of qα ) The following are true:
1. The function qα is well defined for 0 ≤ z < 1.
2. For any α, as z → 1 the series diverges.
Proof
1. Clear for z = 0 as we have the zero sequence. We now consider the ratio test:
(n + 1)α z n+1 g2β(n+1) (0)
n + 1 α g2β(n+1) (0)
an+1
=
=z
an
nα z n g2βn (0)
n
g2βn (0)
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
We have that (at least I expect it to) g2βn (x) → ∞ as n → ∞. Thus the ratio
g2β(n+1) (0)
tends
g2βn (0)
g
(0)
that z 2β(n+1)
g2βn (0)
to 1. Thus there is an N ∈ N such that for all n ≥ N we have
≤ 1 for m ∈ [0, 1). Similarly we have an M ∈ N such that for all
α g2β(n+1) (0)
n ≥ M we have that z n+1
n
g2βn (0) ≤ 1 for z ∈ [0, 1). Thus the ratio of our
sequence tends to a value less than 1, and so the series converges.
2. The series diverges at z = 1, as then all terms in the sequence are greater than
1, and so the sequence is not null.
Since qα (z) converges, we have nα z n g2βn (0) → 0 and so these must be dominated
by a constant, importantly independent of both n and z. Then the Dominated
convergence theorem gives
X
X
nα g2βn (0)
lim nα z n g2βn (0) =
lim qα (z) =
z→1
n≥0
z→1
n≥0
and the final term diverges.
Q.E.D.
Lemma 2.6.3 Suppose that f is a non negative function. Then
−
e
Pn
i=1
f (xi )
≥1−
n
X
(1 − e−f (xi ) )
i=1
Proof We show that the left hand side is greater than the right hand side initially,
and also that the derivative of the LHS is greater than that of the RHS. Thus the
LHS is greater than the RHS. We proceed by induction in n. For n = 1 we have
e−f (x) ≥ 1 − 1 + e−f (x) which is obvious.
Suppose true for all k ≤ n − 1. Then we have, for xn , that at xn = 0 then the
inequality holds by induction hypothesis. Now
∂ − Pni=1 f (xi )
∂f − Pni=1 f (xi )
e
=−
e
∂xn
∂xn
!
n
X
∂f −f (xn )
∂
−f (xi )
(1 − e
) =−
1−
e
∂xn
∂xn
i=1
but
−
∂f −f (xn )
∂f − Pni=1 f (xi )
e
≤−
e
∂xn
∂xn
and so the inequality holds.
Q.E.D.
Lemma 2.6.4
−
e−V (ω) = e
P
0≤l<m≤k−1
Ū (ωl ,ωm )
≥1−
X
(1 − e−Ū (ωl ,ωm ) )
0≤l<m≤k−1
Proof This is a direct application of lemma 2.6.3, taking f = Ū .
14 of 33
Q.E.D.
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Lemma 2.6.5 Suppose that a > 0, then
1 − e−a ≤ a
Proof The differential of the left hand side is e−a and right hand side is 1 with
respect to a and e−a ≤ 1 for all a, so the differential of the left hand side is less than
the differential of the right hand side. At a = 0, we have 0 ≤ 0 and so the left hand
side is always less than or equal to the right hand side.
Q.E.D.
15 of 33
M. Egginton
3
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Lower Bound on the Density of the Bose gas
The methods used in this section are similar to the paper by Ueltschi and Seiringer
[6, pp.6-7]. We now aim to find a lower bound on the density which can be expressed
in terms of the potential function U and the density of the ideal gas, with some other
position invariant terms as well. The majority of the work here still holds, but the last
line or two of each calculation uses lemma 2.4.2 and so the final results are false.
We now proceed to split up the density term into parts where we have only one path
interacting and where we have two paths interacting. In doing this, the interactions
for higher numbers of paths become the partition function Z, thus cancelling the Z in
the denominator. Once this is achieved, we also strive to remove the dependence on
|Λ| in the denominator, so that the results hold in the infinite volume limit.
We have from before that
P
Z
−
Vij
1 X
1
ρ(z) =
k1 e 1≤i<j≤l dµ(ω1 )...dµ(ωl )
(3.0.1)
|Λ|Z
(l − 1)!
l≥1
Ωl
We now bound the exponentials from below i.e. we use lemma 2.6.3 to find a lower
bound of this. This is essentially splitting the interactions of the first path with the
others, and only bounding those below.
Inputting this into (3.0.1) we get that


P
Z
l
−
Vij
X
X
1
1
ρ(z) ≥
k1 1 −
(1 − e−V1j ) e 2≤i<j≤l dµ(ω1 )...dµ(ωl )
|Λ|Z
(l − 1)! Ωl
j=2
l≥1
(3.0.2)
Then by splitting this term we get
P
Z
−
Vij
k1
1 X
ρ(z) ≥
e 2≤i<j≤l dµ(ω1 )...dµ(ωl )−
|Λ|Z
Ωl (l − 1)!
l≥1
Z
l
−
Vij
k1 X
1 X
(1 − e−V1j )e 2≤i<j≤l dµ(ω1 )...dµ(ωl )
−
|Λ|Z
Ωl (l − 1)!
P
l≥1
(3.0.3)
j=2
We now try to eliminate the partition function from these expressions, so as to get
a bound that is valid in the infinite volume case. Now using the definition of Z and
separating out the ω1 and ω2 terms in the second term we get, using lemma 2.3.3, that
Z
Z
XZ
1
1
1
−V12
ρ(z) ≥
k1 dµ(ω1 ) −
k1 (1 − e
)dµ(ω1 )dµ(ω2 )
×
|Λ| Ω
|Λ|Z Ω2
(l
−
1)!
Ωl−2
l≥2
×
l
X
(1 − e−V1j )e
−
P
2≤i<j≤l
Vij
dµ(ω3 )...dµ(ωl )
(3.0.4)
j=3
and so we have, since V is repulsive (positive), and thus −(1 − e−V ) ≥ −e0 = −1 and
so
Z
Z
1
1
ρ(z) ≥
k1 dµ(ω1 ) −
k1 (1 − e−V12 )dµ(ω1 )dµ(ω2 )×
|Λ| Ω
|Λ|Z Ω2
P
−
Vij
X 1 Z
1≤i<j≤n
×
e
dµ(ω1 )...dµ(ωn ) (3.0.5)
n! Ωn
n≥0
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
and noting that the part in the last sum is again the partition function, we get:
Z
Z
1
1
ρ(z) ≥
k1 dµ(ω1 ) −
k1 (1 − e−V12 )dµ(ω1 )dµ(ω2 )
(3.0.6)
|Λ| Ω
|Λ| Ω2
Then the first term above becomes, using lemma 2.6.4 (where we note for k = 0, 1 that
we have no terms),


Z
Z
X
X
X
1
1
2βk
k1 dµ(ω1 ) ≥
dWx,x
(ω) 1 −
(1 − e−Ū (ωl ,ωm ) )
zk
|Λ| Ω
|Λ|
x∈Λ
0≤l<m≤k−1
k≥1
Z
X
X
1
2βk
=
zk
dWx,x
(ω)−
|Λ|
x∈Λ
k≥1
Z
X
1 X kX
2βk
−
z
dWx,x
(ω)
(1 − e−Ū (ωl ,ωm ) )
|Λ|
x∈Λ
k≥2
0≤l<m≤k−1
X
X
1
g2βk (x − x)−
=
zk
|Λ|
x∈Λ
k≥1
Z
X
1 X kX
2βk
−
z
dWx,x
(ω)
(1 − e−Ū (ωl ,ωm ) )
|Λ|
x∈Λ
k≥2
0≤l<m≤k−1
{z
}
|
=:A
(0)
=ρ
(z) − A
(3.0.7)
We now note that the second term above is
1
|Λ|
k1 P
k2
P
Z
−
Ū (ω1,i ,ω2,j )
1
k1 (1 − e−V12 )dµ(ω1 )dµ(ω2 ) ≤
k1 (1 − e i=1 j=1
)dµ(ω1 )dµ(ω2 )
|Λ| Ω2
Ω2
Z
1
=
k1 (1 − e−k1 k2 Ū (ω1,1 ,ω2,1 ) )dµ(ω1 )dµ(ω2 )
|Λ| Ω2


kP
1 k2
Z
−
Ū
(ω
,ω
)
1,1 2,1
1
 dµ(ω1 )dµ(ω2 )
=
k1 1 − e i=1
|Λ| Ω2
!!
Z
kX
1 k2
1
k1 1 − 1 −
dµ(ω1 )dµ(ω2 )
≤
(1 − e−Ū (ω1,1 ,ω2,1 ) )
|Λ| Ω2
i=1
Z
1
≤
k 2 k2 (1 − e−Ū (ω1,1 ,ω2,1 ) )dµ(ω1 )dµ(ω2 )
|Λ| Ω2 1
=: B
(3.0.8)
Z
where we multiply by k1 k2 because we can reorder the points to get ω1,1 and ω2,1 to
be whichever leg we want.
We now bound A and B from above and summarise in the following two subsections.
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3.1
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Bounding the “A” term
We first proceed to split A up into different parts as follows:
Z
X
1 X kX
2βk
A=
z
dWx,x
(ω)
(1 − e−Ū (ωl ,ωm ) )
|Λ|
x∈Λ
k≥2
0≤l<m≤k−1
Z
2
X
X
z
4β
=
dWx,x
(ω)
(1 − e−Ū (ωl ,ωm ) )+
|Λ|
x∈Λ
0≤l<m≤1
Z
X
1 X kX
2βk
+
z
dWx,x
(ω)
(1 − e−Ū (ωl ,ωm ) )
|Λ|
x∈Λ
k≥3
0≤l<m≤k−1
Z
X
X
1
2βk
zk
dWx,x
(ω)×
= A1 +
|Λ|
x∈Λ
k≥3
X
(1 − e−Ū (ωl ,ωm ) ) 1{l=m+1} + 1{l6=m+1}
×
0≤l<m≤k−1
1 X kX
= A1 +
z
|Λ|
k≥3
Z
X
2βk
dWx,x
(ω)
x∈Λ
(1 − e−Ū (ωl ,ωm ) )1{l=m+1} +
0≤l<m≤k−1
1 X kX
+
z
|Λ|
k≥3
Z
2βk
dWx,x
(ω)
x∈Λ
X
(1 − e−Ū (ωl ,ωm ) )1{l6=m+1}
0≤l<m≤k−1
= A1 + A2 + A3
In other words we have A1 where k = 2, A2 where we have at least winding number
of 3 and the interactions are between neighbouring parts of the path, and A3 where
the winding number is at least 3 and the interactions are not between neighbouring
parts of the path. It should be clear that these completely describe all such paths. It
is not obvious why we split the terms up like this at first sight, but I hope on viewing
the simplicity of the resulting bounds that the reader will see the justification of this
method.
We bound each Ai separately.
Lemma 3.1.1
A1 ≤ z 2
2 (0) X
g2β
g4β (0)
K(x, −x)
x∈Zd
Proof We can write A1 as follows, since we have two legs in the path ω starting and
ending at x and so we have
Z
z2 X
4β
A1 =
dWx,x
(ω)(1 − e−Ū (ω1 ,ω2 ) )
(3.1.1)
|Λ|
x∈Λ
Then by splitting this into its two constituent parts, by introducing x2 as an intermediary point in between the two paths over time 2β and rewriting the interaction term
(i.e. we use lemma 2.3.3) we get
Z
z2 X
A1 =
dWx2β
(ω1 )dWx2β
(ω2 )(1 − e−Ū (ω1 ,ω2 ) )
(3.1.2)
1 ,x2
2 ,x1
|Λ|
x1 ,x2 ∈Λ
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Now by using lemma 2.4.2 we get
z2
|Λ|
A1 =
X
K(x1 − x2 , x2 − x1 )
x1 ,x2 ∈Λ
g2β (x1 − x2 )g2β (x2 − x1 )
g4β (x1 − x2 − x1 + x2 )
(3.1.3)
Now expanding the domain of summation to Z d ( the functions are all positive so
this only makes the value larger), and performing a change of variables, by setting
x = x1 − x2 and choosing y = x2 , we get
A1 ≤
g2β (x)g2β (−x)
z2 X
K(x, −x)
|Λ|
g4β (0)
d
(3.1.4)
x,y∈Z
Now using lemma 2.1.3 we get
A1 ≤ z 2
2 (0) X
g2β
g4β (0)
K(x, −x)
(3.1.5)
x∈Zd
as required.
Q.E.D.
Lemma 3.1.2
A2 ≤ z 2 [q1 (z) + 2q0 (z)]
X g2β (x)g2β (y)
K(x, y)
g4β (x − y)
d
x,y∈Z
where qα (z) =
P
n≥1 n
αzng
2βn (0).
Proof The second term is
Z
1 X kX
2βk
A2 =
z
dWx,x
(ω)
|Λ|
k≥3
x∈Λ
X
1{m=l+1} (1 − e−Ū (ωl ,ωm ) )
(3.1.6)
0≤l<m≤k−1
where the 1 is forcing the neighbouring segments of the loop to interact. Now, since
k ≥ 3 we can split the loop from x to x up into a loop from x1 = x to x2 , x2 to x3 then
x3 to x1 , taking 2β for the first two loops and 2β(k − 2) for the last. Then we have
1
A2 =
|Λ|
X
X
z
k
Z
dWx2β
(ω1 )dWx2β
(ω2 )dWx2β(k−2)
(ω3 )×
1 ,x2
2 ,x3
3 ,x1
x1 ,x2 ,x3 ∈Λ k≥3
X
×
1{m=l+1} (1 − e−Ū (ωl ,ωm ) )
0≤l<m≤k−1
1
=
|Λ|
X
Z
X
dWx2β
(ω1 )dWx2β
(ω2 )
1 ,x2
2 ,x3
x1 ,x2 ,x3 ∈Λ
z
n+2
Z
dWx2βn
(ω3 )×
3 ,x1
n≥1
×
X
1{m=l+1} (1 − e−Ū (ωl,1 ,ωm,1 ) )
(3.1.7)
0≤l<m≤n+1
Since the integrals around the loop are independent of the place we start, and because
we integrate over the starting point, we can assume we start at x1 always, so long as
we multiply by the number of segments in the loop, which is k = n + 2. We thus get
X Z
X
1
−Ū (ω1 ,ω2 )
2β
A2 =
(ω
)(1−e
(n+2)z n+2 g2βn (x1 −x3 )
dWx2β
(ω
)dW
1
2
,x
x
,x
1 2
2 3
|Λ|
n≥1
x1 ,x2 ,x3 Λ
(3.1.8)
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MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Then using lemma 2.1.3 and then lemma 2.4.2 we get
A2 ≤
1
|Λ|
X
x1 ,x2 ,x3 ∈Λ
X
g2β (x1 − x2 )g2β (x2 − x3 )
(n + 2)z n+2 g2βn (0)
K(x1 − x2 , x2 − x3 )
g4β (x1 − x2 − (x2 − x3 ))
n≥1
(3.1.9)
Then by changing variables by x = x1 − x2 and y = x2 − x3 and w = x3 we get
A2 ≤
X
1 X X g2β (x)g2β (y)
(n + 2)z n+2 g2βn (0)
K(x, y)
|Λ|
g
(x
−
y)
4β
d
(3.1.10)
n≥1
x,y∈Z w∈Λ
then by integrating out the w we get
X g2β (x)g2β (y)
X
(n + 2)z n+2 g2βn (0)
K(x, y)
g
(x
−
y)
4β
d
A2 ≤
(3.1.11)
n≥1
x,y∈Z
Then we first evaluate the sum over n to get


X
X
X
(n + 2)z n+2 g2βn (0) = z 2 
nz n g2βn (0) + 2
z n g2βn (0) = z 2 [q1 (z) + 2q0 (z)]
n≥1
n≥1
n≥1
(3.1.12)
and then using this gives
A2 ≤ z 2
X g2β (x)g2β (y)
K(x, y)[q1 (z) + 2q0 (z)]
g4β (x − y)
d
(3.1.13)
x,y∈Z
as required
Q.E.D.
I feel that it is important to note that here there is no simple way in which to remove
the quotient of Gaussians from the bound here, unless the function is bounded. We
leave this for discussion later though.
Lemma 3.1.3
A3 ≤ 2z 2 q0 (z)[q1 (z) + q0 (z)]
X
K(x, y)
x,y∈Zd
Proof We have that
A3 =
1 X kX
z
|Λ|
k≥3
x∈Λ
Z
2βk
dWx,x
(ω)
X
(1 − e−Ū (ωl ,ωm ) )1{l6=m−1}
(3.1.14)
0≤l<m≤k−1
We can rewrite this as follows. We suppose that the two legs to interact are between
x1 to x2 and x3 to x4 . We then have to vary the lengths of x2 to x3 and x4 to x1 . Let
the lengths of these two legs be k1 and k2 respectively. Then, since starting at any
point is the same as starting at x1 , we must multiply by k = k1 + k2 + 2. We thus get
Z
X
1 X k
A3 =
z 1{k1 +k2 +2=k}
dWx2β
(ω1 )×
1 ,x2
|Λ|
x1 ,x2 ,x3 ,x4 ∈Λ
k≥3
Z
Z
Z
2β
1
2
(ω4 )(1 − e−Ū (ω1 ,ω3 ) )k (3.1.15)
× dWx2βk
(ω
)
dW
(ω
)
dWx2βk
2
3
x3 ,x4
2 ,x3
4 ,x1
20 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
We can rewrite this by setting k1 the winding number of ω2 and k2 to be the winding
number of ω4 . Then the sum over k becomes a sum over k1 and k2 and we get
X Z
1
(ω3 )(1 − e−Ū (ω1 ,ω3 ) )×
A3 =
dWx2β
(ω1 )dWx2β
1 ,x2
3 ,x4
|Λ|
x1 ,...,x4 ∈Λ
Z
X
1
2
(ω2 )dWx2βk
(ω4 )(k1 + k2 + 2) (3.1.16)
×
z k1 +k2 +2 dWx2βk
2 ,x3
4 ,x1
k1 ,k2 ≥1
Then by evaluating the integrals over ω2 and ω4 we get
X Z
z2
A3 =
(ω3 )(1 − e−Ū (ω1 ,ω3 ) )×
dWx2β
(ω1 )dWx2β
1 ,x2
3 ,x4
|Λ|
x1 ,...,x4 ∈Λ
X
×
(k1 + k2 + 2)z k1 +k2 g2βk1 (x2 − x3 )g2βk2 (x4 − x1 ) (3.1.17)
k1 ,k2 ≥1
now using lemma 2.1.3 we get
X Z
z2
A3 ≤
(ω3 )(1 − e−Ū (ω1 ,ω3 ) )×
dWx2β
(ω1 )dWx2β
1 ,x2
3 ,x4
|Λ|
x1 ,...,x4 ∈Λ
X
×
(k1 + k2 + 2)z k1 +k2 g2βk1 (0)g2βk2 (0)
(3.1.18)
k1 ,k2 ≥1
then using lemma 2.4.2 we get
A3 ≤
z2 X
(k1 + k2 + 2)z k1 +k2 g2βk1 (0)g2βk2 (0)×
|Λ|
k1 ,k2 ≥1
X
×
x1 ,...,x4 ∈Λ
g2β (x1 − x2 )g2β (x3 − x4 )
K(x1 − x3 , x2 − x4 )
g4β ((x1 − x2 ) − (x3 − x4 ))
(3.1.19)
by extending the domain of summation and changing variables as follows x = x1 − x3 ,
y = x2 − x4 , w = x3 − x4 and another variable suitably chosen we get
A3 ≤
|Λ|z 2 X
(k1 + k2 + 2)z k1 +k2 g2βk1 (0)g2βk2 (0)×
|Λ|
k1 ,k2 ≥1
×
X
w,x,y∈Zd
21 of 33
g2β (x − y − w)g2β (w)
K(x, y) (3.1.20)
g4β (x − y)
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
and then using lemma 2.3.3 and then evaluating the sum over k1 , k2 as follows,
X
(k1 + k2 + 2)z k1 +k2 g2βk1 (0)g2βk2 (0)
k1 ,k2 ≥1
=
X
k1 z k1 +k2 g2βk1 (0)g2βk2 (0) + k1 z k1 +k2 g2βk1 (0)g2βk2 (0)
k1 ,k2 ≥1
+ 2z k1 +k2 g2βk1 (0)g2βk2 (0)



X
X
k1 z k1 g2βk1 (0) 
=
z k2 g2βk2 (0) +
k1 ≥1
k2 ≥1

+
X

k2 z k2 g2βk2 (0) 
k2 ≥1

X
z k1 g2βk1 (0)
k1 ≥1

+2
X

z k1 g2βk1 (0) 
k1 ≥1

X
z k2 g2βk2 (0)
k2 ≥1
= 2q0 (z)[q1 (z) + q0 (z)]
we get
A3 ≤ 2z 2 q0 (z)[q1 (z) + q0 (z)]
X g4β (x − y)
K(x, y)
g (x − y)
d 4β
(3.1.21)
x,y∈Z
and simplifying we get
A3 ≤ 2z 2 q0 (z)[q1 (z) + q0 (z)]
X
K(x, y)
(3.1.22)
x,y∈Zd
as required.
Q.E.D.
3.2
Bounding the “B” term
We now proceed to split up B as follows:
Z
1
B=
k 2 k2 (1 − e−Ū (ω1,1 ,ω2,1 ) )dµ(ω1 )dµ(ω2 )
|Λ| Ω2 1
X Z
1 X
1
=
k1 z k1 +k2
dWx2βk
(ω1 )×
1 ,x1
|Λ|
x1 ,x2 ∈Λ
k1 ,k2 ≥1
Z
2
× dWx2βk
(ω2 )e−V (ω1 )−V (ω2 ) (1 − e−Ū (ω1,1 ,ω2,1 ) )
2 ,x2
Z
Z
z2 X
2β
=
dWx1 ,x1 (ω1 ) dWx2β
(ω2 )(1 − e−Ū (ω1,1 ,ω2,1 ) )+
2 ,x2
|Λ|
x1 ,x2 ∈Λ
X Z
1 X
1
+
dWx2βk
(ω1 )×
k1 z k1 +k2
1 ,x1
|Λ|
x1 ,x2 ∈Λ
k1 ,k2 ≥2
Z
2
× dWx2βk
(ω2 )e−V (ω1 )−V (ω2 ) (1 − e−Ū (ω1,1 ,ω2,1 ) )
2 ,x2
22 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
X Z
1 X
k1 +1
1
= B1 +
k1 z
dWx2βk
(ω1 )×
1 ,x1
|Λ|
x1 ,x2 ∈Λ
k1 ≥2
Z
× dWx2β
(ω2 )e−V (ω1 ) (1 − e−Ū (ω1,1 ,ω2,1 ) )+
2 ,x2
X Z
1 X
k1 +k2
1
+
k1 z
dWx2βk
(ω1 )×
1 ,x1
|Λ|
x1 ,x2 ∈Λ
k1 ,k2 ≥2
Z
2
× dWx2βk
(ω2 )e−V (ω1 )−V (ω2 ) (1 − e−Ū (ω1,1 ,ω2,1 ) )
2 ,x2
= B1 + B2 + B3
Here we have split B up into parts with both paths having winding number 1, with
the second path having winding number 1 and then both having winding number at
least 2.
We now bound each Bi separately.
Lemma 3.2.1
B1 ≤ z
g 2 (0)
2 2β
g4β (0)
X
K(x, x)
x∈Zd
Proof We have that B1 is equal to the following, as k1 = 1 = k2 .
Z
z2 X
B1 =
dWx2β
(ω1 )dWx2β
(ω2 )(1 − e−Ū (ω1 ,ω2 ) )
1 ,x1
2 ,x2
|Λ|
(3.2.1)
x1 ,x2 ∈Λ
Now using lemma 2.4.2 we get that
B1 =
z2
|Λ|
X
x1 ,x2 ∈Λ
g2β (x1 − x1 )g2β (x2 − x2 )
K(x1 − x2 , x1 − x2 )
g4β (x1 − x2 − x1 + x2 )
(3.2.2)
Then from lemma 2.1.3, and a change of variables, we get
B1 ≤
z 2 g2β (0)g2β (0) X X
K(x, x)
|Λ|
g4β (0)
d
(3.2.3)
x∈Z y∈Λ
and integrating with respect to y gives
B1 ≤ z 2
2 (0) X
g2β
g4β (0)
K(x, x)
(3.2.4)
x∈Zd
as required.
Q.E.D.
Lemma 3.2.2
B2 ≤ z 2 (q1 (z) + q0 (z))
X g2β (0)g2β (x − y)
K(x, y)
g4β (x + y)
d
x,y∈Z
23 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Proof Suppose that k2 = 1 and that k1 is the loop that varies over larger winding
numbers. Then we have
Z
X
1 X
B2 =
zdWx2β
(ω2 )e−V (ω2 )
kz k dWx2βk
(ω1 )(1 − e−Ū (ω1,1 ,ω2 ) )e−V (ω1 )
2 ,x2
1 ,x1
|Λ|
x1 ,x2 ∈Λ
k≥2
(3.2.5)
We can bound the self interaction terms by one to get
Z
X
1 X
B2 ≤
zdWx2β
(ω
)
kz k dWx2βk
(ω1 )(1 − e−Ū (ω1,1 ,ω2 ) )
2
2 ,x2
1 ,x1
|Λ|
x1 ,x2 ∈Λ
(3.2.6)
k≥2
We observe that the interaction term depends only on the first leg of ω1 and so we
isolate that from the rest of ω1 . We use lemma 2.3.3 so we get
Z
X
XZ
1 X
k
B2 ≤
zdWx2β
(ω
)
kz
dWx2β
(ω1,1 )×
2
2 ,x2
1 ,x3
|Λ|
x1 ,x2 ∈Λ
x3 ∈Λ
k≥2
Z
× dWx2β(k−1)
(ω3 )(1 − e−Ū (ω1,1 ,ω2 ) )
(3.2.7)
3 ,x1
Rearranging gives
1
B2 ≤
|Λ|
Z
X
(ω1,1 )(1 − e−Ū (ω1,1 ,ω2 ) )×
zdWx2β
(ω2 )dWx2β
2 ,x2
1 ,x3
x1 ,x2 ,x3 ∈Λ
×
X
kz k
Z
dWx2β(k−1)
(ω3 )
3 ,x1
(3.2.8)
k≥2
Using the normalisation of Wiener integrals we have chosen and reordering the sum
gives
X Z
1
dWx2β
(ω2 )dWx2β
(ω1,1 )(1 − e−Ū (ω1,1 ,ω2 ) )×
B2 ≤
2 ,x2
1 ,x3
|Λ|
x1 ,x2 ,x3 ∈Λ
X
×
(k + 1)z k+2 g2βk (x1 − x3 ) (3.2.9)
k≥1
Then by using lemma 2.1.3 we get that
X Z
X
1
2β
−Ū (ω1,1 ,ω2 )
dWx2β
(ω
)dW
(ω
)(1
−
e
)
(k + 1)z k+2 g2βk (0)
B2 ≤
2
1,1
,x
x
,x
2 2
1 3
|Λ|
x1 ,x2 ,x3 ∈Λ
k≥1
(3.2.10)
Using lemma 2.4.2 we get that
B2 ≤
1
|Λ|
X
X
(k + 1)z k+2 g2βk (0)
x1 ,x2 ,x3 ∈Λ k≥1
g2β (0)g2β (x3 − x1 )
K(x2 − x1 , x2 − x3 )
g4β (x2 − x1 − x2 + x3 )
(3.2.11)
Now by setting x = x2 − x1 and y = x2 − x3 and w = x3 we get
B2 ≤ z 2
X X
x,y∈Zd k≥1
(k + 1)z k g2βk (0)
g2β (0)g2β (y − x)
K(x, y)
g4β (x − y)
24 of 33
(3.2.12)
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Thus evaluating the sum over k we get
X g2β (0)g2β (x − y)
K(x, y)
g4β (x + y)
d
B2 ≤ z 2 [q1 (z) + q0 (z)]
(3.2.13)
x,y∈Z
as required.
Q.E.D.
Lemma 3.2.3
X
B3 ≤ z 2 q0 (z)[q0 (z) + q1 (z)]
K(x, y)
x,y∈Zd
Proof
XZ
1 X
k1
1
B3 =
k1 z
dWx2βk
(ω1 )e−V (ω1 ) ×
1 ,x1
|Λ|
x1 ∈Λ
k1 ≥2
X
XZ
k2
2
×
z
dWx2βk
(ω2 )e−V (ω2 ) (1 − e−Ū (ω1,1 ,ω2,1 ) )
2 ,x2
k2 ≥2
(3.2.14)
x2 ∈Λ
then by bounding the self interaction terms by 1 we get
XZ
X
XZ
1 X
k1
2βk1
k2
2
k1 z
dWx1 ,x1 (ω1 )
z
dWx2βk
B3 ≤
(ω2 )(1 − e−Ū (ω1,1 ,ω2,1 ) )
2 ,x2
|Λ|
k1 ≥2
x1 ∈Λ
k2 ≥2
x2 ∈Λ
(3.2.15)
If we take out integration over the first leg of each loop and reorder the sum over k1 , k2
to go from 1 instead of 2 (i.e. we use lemma 2.3.3) to get
X Z
1
dWx2β
(ω1,1 )dWx2β
B3 ≤
(ω2,1 )(1 − e−Ū (ω1,1 ,ω2,1 ) )×
1 ,x3
2 ,x4
|Λ|
x1 ,...,x4 ∈Λ
×
X
(k1 + 1)z
k1 +k2 +2
Z
1
2
dWx2βk
(ω3 )dWx2βk
(ω4 )
3 ,x1
4 ,x2
k1 ,k2 ≥1
(3.2.16)
Then by using the normalisation of the Wiener integral we get
X Z
1
dWx2β
(ω1,1 )dWx2β
(ω2,1 )(1 − e−Ū (ω1,1 ,ω2,1 ) )×
B3 ≤
1 ,x3
2 ,x4
|Λ|
x1 ,...,x4 ∈Λ
×
X
(k1 + 1)z k1 +k2 +2 g2βk1 (x3 − x1 )g2βk2 (x4 − x2 )
k1 ,k2 ≥1
(3.2.17)
and then by using lemma 2.1.3 we get
X Z
1
dWx2β
(ω1,1 )dWx2β
(ω2,1 )(1 − e−Ū (ω1,1 ,ω2,1 ) )×
B3 ≤
1 ,x3
2 ,x4
|Λ|
x1 ,...,x4 ∈Λ
×
X
(k1 + 1)z k1 +k2 +2 g2βk1 (0)g2βk2 (0)
k1 ,k2 ≥1
25 of 33
(3.2.18)
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Then by taking this sum out of the integral and using lemma 2.4.2 we get
1 X
(k1 + 1)z k1 +k2 +2 g2βk1 (0)g2βk2 (0)×
|Λ|
B3 ≤
k1 ,k2 ≥1
×
X
x1 ,...,x4 ∈Λ
g2β (x1 − x3 )g2β (x2 − x4 )
K(x1 − x2 , x3 − x4 )
g4β(x1 −x2 −x3 +x4 )
(3.2.19)
and then by changing variables to x = x1 − x2 , y = x3 − x4 , w = x2 − x4 and h = x4
we get
X
B3 ≤
X
(k1 + 1)z k1 +k2 +2 g2βk1 (0)g2βk2 (0)
k1 ,k2 ≥1
x,y,w∈Zd
g2β (x − y − w)g2β (w)
K(x, y)
g4β(x−y)
(3.2.20)
and then by integrating
out
by
h
and
w
from
one
of
the
properties
of
the
discrete
P
Gaussian, namely z∈Zd gt (x − z)gs (z − y) = gt+s (y − x), we get that
X
B3 ≤
(k1 + 1)z k1 +k2 +2 g2βk1 (0)g2βk2 (0)
k1 ,k2 ≥1
X g4β (x − y)
K(x, y)
g4β(x−y)
d
(3.2.21)
x,y∈Z
Then by rewriting the sum as follows
X
(k1 + 1)z k1 +k2 +2 g2βk1 (0)g2βk2 (0) =
k1 ,k2 ≥1
= z2
X X
k1 z k1 z k2 g2βk1 (0)g2βk2 (0) + z 2

X

k1 z k1 g2βk1 (0) 
k1 ≥1

X
z k2 g2βk2 (0) +
k2 ≥1

+ z2
X

z k1 g2βk1 (0) 
k1 ≥1
2
= z q0 (z)q1 (z) +
z k1 z k2 g2βk1 (0)g2βk2 (0)
k1 ≥1 k2 ≥1
k1 ≥1 k2 ≥1
= z2
X X

X
z k2 g2βk2 (0)
k2 ≥1
q02 (z)
we get that
B3 ≤ z 2 q0 (z)[q1 (z) + q0 (z)]
X g4β (x − y)
K(x, y)
g (x − y)
d 4β
(3.2.22)
x,y∈Z
as required.
3.3
Q.E.D.
Lower Bound Result
Collating all of lemmas 3.1.1-3.1.3 and 3.2.1-3.2.3 we get the following:
26 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Theorem 3.3.1 The grand canonical density is bounded below by
2 (0) X
g2β
ρ(z) ≥ ρ(0) (z) − z 2
g4β (0)
K(x, −x)
x∈Zd
X g2β (x)g2β (y)
K(x, y)
g4β (x − y)
x,y∈Zd
X
K(x, y)
− 2z 2 q0 (z)[q1 (z) + q0 (z)]
− z 2 [q1 (z) + 2q0 (z)]
x,y∈Zd
−z
g 2 (0)
2 2β
g4β (0)
X
K(x, x)
x∈Zd
X g2β (0)g2β (x − y)
K(x, y)
g4β (x − y)
d
x,y∈Z
X
2
K(x, y)
− z q0 (z)[q0 (z) + q1 (z)]
− z 2 [q1 (z) + 2q0 (z))
x,y∈Zd
≥ ρ(0) (z) − z 2
2 (0)
g2β

g4β (0)


X
K(x, −x) +
X
K(x, x)
x∈Zd
x∈Zd
− 3z 2 q0 (z)[q1 (z) + q0 (z)]
X
K(x, y)
x,y∈Zd
− z 2 [q1 (z) + 2q0 (z)]
X g2β (x)g2β (y)
K(x, y)
g4β (x − y)
d
x,y∈Z
− z 2 [q1 (z) + 2q0 (z)]
X g2β (0)g2β (y − x)
K(x, y)
g4β (x − y)
d
x,y∈Z
As one can see, two of these estimates, namely for A2 and B2 are not of the form that
we set out to obtain. Some methods to resolve this are considered in later subsections.
However, we continue blindly on and assume that the functions that appear are “nice
” in the sense that they do what we want. In other words I assume that
g2β (0)g2β (x − y)
= H(0, 0, x, y)
g4β (x + y)
g2β (x)g2β (y)
= H(0, x, 0, y)
g4β (x − y)
are position invariant, or simply bounded above by some value, so as to remove them
from the summation over Λ.
3.4
Bounding the Integral Kernel
P
We first consider a bound that uses a0 = x∈Zd U (x), but it is necessary to suppose
that U ∈ l1 (Zd ) and that U (x) < ∞ for all x ∈ Zd . We wish to evaluate
R
X
X Z 1 4β
4β
K(x, y) =
1 − e− 4 0 U (ω(s))ds dWx,y
(ω)
x,y∈Zd
x,y∈Zd
Theorem 3.4.1 The integral kernel K(x, y) is bounded as follows:
X
K(x, y) ≤ βa0
x,y∈Zd
27 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Proof Now using lemma 2.6.5 we get that
X Z − 41
1−e
R 4β
0
U (ω(s))ds
4β
dWx,y
(ω)
x,y∈Zd
Z Z 4β
1 X
4β
≤
U (ω(s))dsdWx,y
(ω)
4
0
d
x,y∈Z
Now since U (ω(s)) ≤ ||U ||∞ < ∞ and is positive we can apply Fubini to this to get
that
Z Z 4β
Z 4β Z
1 X
1 X
4β
4β
U (ω(s))dsdWx,y (ω) =
U (ω(s))dWx,y
(ω)ds
4
4
0
0
d
d
x,y∈Z
x,y∈Z
and then due to the integral on the ω only depending on the path at the point ω(s)
this becomes an integral over z with gaussian kernels added in: we have
Z 4β Z
Z 4β X
1 X
1 X
4β
gs (z − x)U (z)g4β−s (y − z)ds
U (ω(s))dWx,y
(ω)ds =
4
4
d 0
d 0
d
x,y∈Z
x,y∈Z
z∈Z
and then again noting that the integrand is bounded and positive so we can apply
Fubini to get that
Z 4β X
1 X
gs (z − x)U (z)g4β−s (y − z)ds =
4
0
z∈Zd
x,y∈Zd
Z
1 4β X
=
gs (z − x)U (z)g4β−s (y − z)ds
4 0
d
x,y,z∈Z
and then
Z
1 4β X
gs (z − x)U (z)g4β−s (y − z)ds =
4 0
x,y,z∈Zd
Z
X
X
1 4β X
=
U (z)
(gs (z − x))
(g4β−s (y − z)) ds
4 0
d
d
d
z∈Z
x∈Z
y∈Z
and then appealing to lemma 2.1.4 we get that this is
1
4
Z
4β
0
X
z∈Zd
U (z)
X
(gs (z − x))
x∈Zd
1
(g4β−s (y − z)) ds =
4
d
X
y∈Z
as required
Z
0
4β
X
U (z)ds
z∈Zd
Q.E.D.
Following the same procedure as the proof of the above theorem, one can bound
the following:
Lemma 3.4.2
X
K(x, x) ≤ βg4β (0)a0
x∈Zd
X
K(x, −x) ≤ βa0 g4β (0)
x∈Zd
28 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Using these bounds, we get
Theorem 3.4.3
2
ρ(z) ≥ ρ(0) (z) − 2z 2 g2β
(0)a0 β − a0 βz 2 [3q0 (z)q1 (z) + q02 (z) + 2q1 (z)M + 3q0 (z)M ]
where M is a bound on the Gaussian ratio that appears several times.
We now turn to bounding the ratio of Gaussians. We use the Fourier transform as
introduced in subsection 2.2.
g2β (x)g2β (y)
=
g4β (x − y)
1
2π
1
e− d
R
Pd
j=1 (1−cos kj )2β
[−π,π]d
1
2π
1
(2π)2
=
[−π,π]d
[−π,π]d
1
2π
1
(2π)2
≤
[−π,π]d
[−π,π]d
1
2π
eihy dh
eih(x−y) dh
j=1 (1−cos kj )2β
1
e− d
Pd
1
eikx e− d
j=1 (1−cos kj )4β
Pd
j=1 (1−cos hj )2β
eihy dkdh
eik(x−y) dk
[−π,π]d
1
R
j=1 (1−cos hj )2β
[−π,π]d
j=1 (1−cos hj )4β
Pd
R
e− d
R
Pd
[−π,π]d
1
R
1
e− d
R
Pd
1
e− d
R
e− d
R
1
eikx dk 2π
R
Pd
j=1 (1−cos kj )2β
1−
[−π,π]d
4β
d
1
eikx e− d
Pd
j=1 (1−cos hj )2β
eihy dkdh
(1
−
cos
k
)
eik(x−y) dk
j
j=1
Pd
1
≤
1
2π
R
[−π,π]d
1−
4β
d
(1
−
cos
k
)
eik(x−y) dk
j
j=1
Pd
Where we note that we have integrated a holomorphic function on a compact domain,
and so it has finite integral. This bound now concludes our bounding of the density,
barring the error from misusing lemma 2.4.2.
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M. Egginton
4
4.1
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
Machinations on the Discrete Gaussian
Attempted proofs of some product rule
It should be clear from the previous part that the difficulties in getting a lower bound
dependent only on some position invariant functions and on the integral kernel K(x, y)
were due to the lack of a nice formula for
gt (x)gt (y)
In the continuous case, there is an easy formula following directly from the parallelogram law, namely
πt (x)πs (y) = πt+s (x + y)πt+s (x − y)
Initially, I proceeded blindly into attempting to prove this, and I describe some of my
attempts below.
I initially tried working with the formula given for the gaussian, namely
gt2 −t1 (x2 − x1 ) = e−(t2 −t1 )
X (t2 − t1 )N
(2d)N N !
N ≥0
X
1
y1 ,...,yN
y1 =x1
yN =x2
||yn −yn−1 ||=1
and then for the product of two of these, one has a formula




 −t X
tN
gt (x)gs (y) = 
e

(2d)N N !

N ≥0

X
y1 ,...,yN
y1 =0
yN =x
||yn −yn−1 ||=1



  −s X
sM
e
1

(2d)M M !

M ≥0

X
z1 ,...,zM
z1 =0
zM =y
||zn −zn−1 ||=1
and then one can rearrange this sum to get





−(t+s)

=e
N
M
(2d) N ! (2d) M ! 

N ≥0 M ≥0

X X
sM
tN
X
y1 ,...,yN
y1 =0
yN =x
||yn −yn−1 ||=1
= e−(t+s)
X X
N ≥0 M ≥0
= e−(t+s)
tN sM
(2d)N +M N !M !
X X
N ≥0 M ≥0
X
y1 ,...,yN
y1 =0
yN =x
||yn −yn−1 ||=1




1



1

X
z1 ,...,zM
z1 =0
zM =y
||zn −zn−1 ||=1
X
z1 ,...,zM
z1 =0
zM =y
||zn −zn−1 ||=1
X
tN sM
1
(2d)N +M N !M ! y1 ,...,yN ,z1 ,...,zM
y1 =0
yN =x
||yn −yn−1 ||=1
z1 =0
zM =y
||zn −zn−1 ||=1
30 of 33
1



1






1



M. Egginton
MA469 - Interacting Bose gas on the Lattice
= e−(t+s)
X X
N ≥0 M ≥0
ID: 0901849
X
tN sM
1
(2d)N +M N !M ! y1 ,...,yN ,z1 ,...,zM
y1 =0
yN =x
||yn −yn−1 ||=1
z1 =x
zM =x+y
||zn −zn−1 ||=1
But I saw no way in which to rewrite the sum over numbers of paths, and so gave
up on that method. I next tried going back to the definition of the discrete gaussian.
Suppose we have the probability P as in section 1.1. Then by definition we have
gt2 −t1 (x2 − x1 ) = P({ω : ω(t1 ) = x1 }|{ω : ω(t2 ) = x2 })
I introduce the notation that Ωtx = {ω : ω(t) = x} and also Ωtx,y = {ω : ω(0) =
x, ω(t) = y} i.e. the former is the set of all paths that pass through x at time t and
the latter is the set of all paths that start at x and end at y. With this notation we
have that
gt (x)gs (y) = P(Ωt0,x )P(Ωs0,y )
and due to independence we have this equal to
= P ⊗ P(Ωt0,x × Ωs0,y )
At this stage I made a crazy guess, i.e. I tried to establish the result similarly to the
continuous case, by trying to link this with
gt+s (x + y)gt+s (x − y)
as this would be the natural formula I though. Doing the same as above, one gets that
t+s
gt+s (x + y)gt+s (x − y) = P ⊗ P(Ωt+s
0,x+y × Ω0,x−y )
and we want to relate the two sets in this product measure space. If we could find a
measurable isomorphism between the two then we would be done. However, there is no
bijection between the two, so we cannot. For example, if we define a map Ωt0,x ×Ωs0,y →
(
ω(h) 0 ≤ h ≤ t
t+s
Ωt+s
0,x+y × Ω0,x−y by (ω, σ) 7→ (ω ? σ, ω ? (−σ)) where ω ? σ(h) =
σ(h) t ≤ h ≤ t + s
t+s
t
s
then this maps Ω0,x × Ω0,y each element uniquely into the subset of Ωt+s
0,x+y × Ω0,x−y
that passes through the intermediary point x, x at time t. This clearly cannot be the
t+s
t+s
whole of Ω0,x+y
× Ω0,x−y
, and so there cannot be an isomorphism.
4.2
Modifications of the Random Walk
These being unsuccesful, Daniel Ueltschi suggested to me to completely change the
discrete Gaussian for another similar function, by going back to the definition and
using another function where I have specified the properties that I want. I shall now
explain my attempts.
One can observe in the previous section that it would suffice to have the following
properties in order to obtain the bounds in the form that we want:
1. gt (0) ≥ gt (x) for fixed t ≥ 0 and x ∈ Zd .
2. gt (x) ≤ gs (x) for fixed x ∈ Zd and t ≤ s.
31 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
3. gt (x)gt (y) = g2t (x + y)g2t (x − y) for x, y ∈ Zd and t ≥ 0.
I summarise the properties that these give on the increment distribution in the
following lemma
We have that P(ω(t) = x|ω(0) = 0) = δx,0 + tp(x) + O(t2 ) for small t. Thus we
have that
gt (x)gt (y) = (δx,0 + tp(x) + O(t2 ))(δy,0 + tp(y) + O(t2 ))
= δx,0 δy,0 + tδx,0 p(y) + tδy,0 p(x) + O(t2 )
and
g2t (x + y)g2t (x − y) = δx+y,0 + tp(x + y) + O(4t2 )
δx−y,0 + tp(x − y) + O(4t2 )
= δx+,0 δx−y,0 + 2tδx+y,0 p(x − y) + 2tδx−y,0 p(x + y) + O(t2 )
and thus we want these two to be equal, i.e.
δx,0 δy,0 + tδx,0 p(y) + tδy,0 p(x) = δx+,0 δx−y,0 + 2tδx+y,0 p(x − y) + 2tδx−y,0 p(x + y)
but setting x = y = 0 gives
1 + tp(0) + tp(0) = 1 + 2tp(0) + 2t(p(0)
which is true only if p(0) = 0 which is very undesirable, as we have no reason for the
paths to always jump. We thus have that no such useful Gaussian with this formula
exists, and so we had to stick with this and work with it.
32 of 33
M. Egginton
MA469 - Interacting Bose gas on the Lattice
ID: 0901849
References
[1] G.B. Folland. Real Analysis: Modern Techniques and Their Applications. John
Wiley and Sons, Inc, USA, 1999.
[2] J Ginibre. Some applications of functional integration in statistical mechanics.
Mécanique statistique et théorie quantique des champs, 1970.
[3] G. Grimmett and D. Stirzaker. Probability and Random Processes. Oxford University Press, Oxford, 2010.
[4] G.F. Lawler and V. Limic. Random Walk: A Modern Introduction. Cambridge
University Press, Cambridge, 2010.
[5] S Sachdev. Quantum Phase Transitions. Cambridge University Press, Cambridge,
2011.
[6] R Seiringer and D Ueltschi. Rigorous upper bound on the critical temperature of
dilute bose gases. Physical Review B, 2009.
[7] R.H Swendsen. An Introduction to Statistical Mechanics and Thermodynamics.
Oxford University Press, 2012.
[8] Hale F Trotter. On the product of semi-groups of operators. Proceedings of the
American Mathematical Society, 10(4):545–551, 1959.
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