数理解析研究所講究録
第 1921 巻 2014 年 122-135

122

Renormalization Group Analysis of$2D0(N)$ Spin Models
K. R. Ito*
Institute for Fundamental Sciences
Setsunan University
Neyagawa, Osaka 572-8058, Japan

January 12, 2014

Abstract

The classical $O(N)$ spin models in two dimensions have been believed free fromany phase transitions if $N$ is larger than or equal to 3. We show that if $N$ is large,
then the block-spin-type transformations can be applied through Fourier (duality)transformation. This enables us to prove the result claimed in the title of this paper.
PACS Numbers $05.50+q$ , 11. $15Ha$ , 64. $60-i$

1Introduction

Though quark confinement in 4 dimensional (4D) non-Abelian lattice gauge theories andspontaneous mass generations in $2D$ non-Abelian sigma models are widely believed [1],we still do not have a rigorous proof. These models exhibit no phase transitions in thehierarchical model approximation of Wilson-Dyson type or Migdal-Kadanov type [12].
In ref. [14], we considered a transformation of random walk (RW) which appears in
the $O(N)$ spin models [3, 4]. This was extended by the cluster expansion [5, 11, 19, 20],
and we showed in the $2DO(N)$ sigma model that :$\frac{\beta_{c}}{N}\geq$

const $\log N$(1.1)

In this paper, we apply a block-spin transformation to the functional integral of thesystem, and establish the following theorem:*email:ito@mpg.setsunan.ac.jp, also ito@kurims.kyoto-u.ac.jp

123
Main Theorem. There exists no phase transition in two-dimensional $0(N)$ invariant
Heisenberg model for all if $N$ is large enough.
To appeal to the $1/N$ expansion [17], we scale the inverse temperature by $N.$
is denoted simply orin [14] and in our bound (1.1).) The dimensional $0(N)$ spin
(Heisenberg) model at the inverse temperature
is defined by the Gibbs expectation
values(1.2)$\beta$$(N\beta$$\beta$$v$$\beta_{c}$$\beta$$N\beta$$\langle f\rangle\equiv\frac{1}{Z_{\Lambda}(\beta)}\int f(\phi)\exp[-H_{\Lambda}(\phi)]\prod_{i}\delta(\phi_{i}^{2}-N\beta)d\phi_{i}$Here$\Lambda=\Lambda_{0}=[-(L/2)^{M}, (L/2)^{M})^{v}\subset Z^{\nu}$is the large square with center at the origin, where is chosen odd $(e.g. L=3)$ and$M$ is a large integer.
Moreover $\phi(x)=(\phi(x)^{(1)}, \cdots, \phi(x)^{(N)})$ is the vector valued spinis theatis the partition function defined so that $<1>=1$ . Moreover
Hamiltonian given by$L$$x\in\Lambda,$$H_{\Lambda}$$Z_{\Lambda}$$H_{\Lambda} \equiv-\frac{1}{2}\sum_{|x-y|_{1}=1}\phi(x)\phi(y)$

,

(1.3)

where
First substitute the identity $\delta(\phi^{2}-N\beta)=\int\exp[-ia(\phi^{2}-N\beta)]da/2\pi$ into eq.(1.2). We setwith the condition [3, 4] that$|x|_{1}= \sum_{i=1}^{\nu}|x_{i}|.$${\rm Im} a_{i}<-\nu$${\rm Im} a_{i}=-( v+m^{2}/2) , {\rm Re} a_{i}=\frac{1}{\sqrt{N}}\psi_{i}$(1.4)

where $m^{2}>$ will be determined soon. Thus we have
$Z_{\Lambda} = c^{|\Lambda|} \int\cdots\int\exp[-W_{0}(\phi, \psi)]\prod\frac{d\phi_{j}d\psi_{j}}{2\pi}$$= c^{|\Lambda|} \det(m^{2}-\triangle)^{-N/2}\int\cdots\int F(\psi)\prod\frac{d\psi_{j}}{2\pi}$

(1.5)

where
$W_{0}( \phi, \psi) = \frac{1}{2}\langle\phi, (m^{2}-\triangle+\frac{2i}{\sqrt{N}}\psi)\phi\rangle-\sum_{j}i\sqrt{N}\beta\psi_{j}$$F( \psi) =\det^{-N/2}(1+i\alpha G\psi)\exp[i\sqrt{N}\beta\sum_{j}\psi_{j}]$(1.6a)(1.6b)(1.6c)$\alpha = 2/\sqrt{N}$Here ’s are constants being diﬀerent on lines,
is the lattice Laplacian, $G=(m^{2}-\triangle)^{-1}$ is the covariant matrix. The two point functions are given by$c$$\triangle_{ij}=-2v\delta_{ij}+\delta_{|i-j|,1}$$\langle\phi_{0}\phi_{x}\rangle = \frac{1}{\tilde{Z}}\int\cdots\int(m^{2}-\triangle+i\alpha\psi)_{0x}^{-1}F(\psi)\prod\frac{d\psi_{j}}{2\pi}$

(1.7)

124
where is the obvious normalization constant. Choose the mass parameter $m=m_{0}>0$
so that $G(O)=\beta$ , where$\tilde{Z}$(1.8)$G(x) = \int\frac{e^{ipx}}{m_{0}^{2}+2\sum(1-\cos p_{i})}\prod_{i=1}^{\nu}\frac{dp_{i}}{2\pi}$, and we find that
This is possible for any if and onlyasfor $\nu=2$ , which is consistent with the renormalizaiton group analysis, see e.g. [6]. Thus
we can rewrite$m^{2}\sim 32e^{-4\pi\beta}$$\nu\leq 2$$\beta$$\betaarrow\infty$(1.9)$F(\psi) = \det_{3}^{-N/2}(1+i\alpha G\psi)\exp[-\langle\psi, G^{02}\psi\rangle]$for $\nu\leq 2$ , where $\det_{3}(1+A)=\det[(1+A)e^{-A+A^{2}/2}]$ and $G^{02}(x, y)=G(x, y)^{2}$ so that
$R(G\psi)^{2}=\langle\psi,$. Moreover $F(\psi)$ is integrable if and only if $N>2$ , and thus
and $N>2$ are required.
If $m$ is so chosen, the determinant $\det_{3}(1+i\alpha G\psi)^{-N/2}$ may be regarded as a small
perturbation to the Gaussian measure. This is the case if $N$ isvery large or if is very small $(e.g. N\log N>\beta)$ , in which caseand we can$\det_{3}^{-N/2}(1+i\alpha G\psi)$
disregard
and the model is exactly solvable in this limit. Thus we have$G^{02}\psi\rangle$$\nu\leq 2$$\sim\exp[-\langle\psi, G^{02}\psi\rangle]\prod d\psi$$\Vert|\alpha G||\ll 1$$\beta$$\langle\phi_{0}\phi_{x}\rangle = \frac{1}{Z}\int(m_{0}^{2}-\triangle+i\alpha\psi)_{0x}^{-1}\exp[-R(G\psi)^{2}]\prod d\psi$(1.10)$\leq (m_{0}^{2}-\triangle)_{0x}^{-1}\leq c\exp(-m_{0}|x|)$

But this argument fails for large since
is of long-range and the expansion of thedeterminant is not justified at all.
On the other hand, this argument can be justified if the main part of the integral
consists ofsuch that
O. In this case, the expansion of thedeterminant is justified. Our main argument in this paper is to justify this argument.
The renormalization group (RG) method is the method to integrate the functional
integration recursively introducing block spin operators and $C’$ defined by$G$$\beta$$\psi$$|\psi|<N^{\epsilon}\beta^{-1/2}$$\sum_{x}\psi_{x}\sim$$C$$\phi_{1}(x) = (C\phi)(x)$$\equiv$$\frac{1}{L^{2}}\sum_{\zeta\in\Delta_{0}}f(Lx+\zeta)$(l.lla)$\psi_{1}(x) = (C’f)(x)$$\equiv$$L^{2}(Cf)(x)$(l.llb)where
andis the square of size $L\cross L(L\geq 2)$ center at the origin. $C$ and$C’$ consist of
averaging over the spins in the blocks and the scaling of the coordinates,i.e.,. We integrate out the remaining degrees of freedom which we call$x\in\Lambda\cap L\Lambda$$\Lambda=\Lambda_{0}arrow\Lambda_{1}$$\triangle_{0}$125fluctuation fields ( and ) and continue these steps,
and $\Lambda_{n}arrow\Lambda_{n+1}arrow\cdots(n=0,1,2, \cdots)$ . We repeat this process by finding matrices
andsuch that$\tilde{\psi}$$\phi_{n}arrow\phi_{n+1}arrow\cdots,$ $\psi_{n}arrow\psi_{n+1}arrow\cdots$$\xi$$A_{n}$$\tilde{A}_{n}$(1.12a)(1.12b)$\phi_{n} = A_{n+1}\phi_{n+1}+Q\xi_{n}$$\psi_{n} = \tilde{A}_{n+1}\phi_{n+1}+Q\tilde{\psi}_{n}$and(1.13a)$\langle\phi_{n}, G_{n}^{-1}\phi_{n}\rangle = \langle\phi_{n+1}, G_{n+1}^{-1}\phi_{n+1}\rangle+\langle\xi_{n}, \Gamma_{n}^{-1}\xi_{n}\rangle$(1.13b)$\langle\psi_{n}, H_{n}^{-1}\psi_{n}\rangle = \langle\psi_{n+1}, \hat{H}_{n+1}^{-1}\psi_{n+1}\rangle+\langle\tilde{\psi}_{n}, Q^{+}H_{n}^{-1}Q\tilde{\psi}_{n}\rangle$

where$G_{n}^{-1}$and$H_{n}^{-1}$

are the main Gaussian parts in$W_{n}$, and(1.14a)$G_{n} = CG_{n-1}C^{+}=C^{n}G_{0}(C^{+})^{n}$$(Q\xi)(x)$$=$(1.14b)$\{\begin{array}{ll}\xi(x) if x\in\Lambda_{n}’-\sum_{\zeta\in\Delta(x),\zeta\neq x}\xi(\zeta) if x\not\in\Lambda_{n}’\end{array}$(1.14c)$\Lambda_{n}’ = \Lambda_{n}\backslash L\Lambda_{n}$is the square of size
. Namely
center at$(n=0,1,2, \cdots)$ is the operator to make zero-average fluctuations
from

where

$L\cross L$$\triangle(x)$$Q:R^{\Lambda_{n}’}arrow R^{\Lambda_{n}}$$x(\in\Lambda_{n}\cap L\Lambda_{n})$$Q\xi_{n}$$\{\xi_{n}(x)$: $x\in$$\Lambda_{n}’\}.$

In our case, we start with
$G_{0} = (-\triangle+m_{0})^{-1}(x, y)$$\sim \beta-\frac{1}{2\pi}\log|x-y|$$H_{0} = \frac{\mathring{1}}{G^{2}}(x, y)$$\sim \frac{1}{|x-y|^{4}}$where$H_{0}^{-1}$

is derived from the formal $Narrow\infty$ limit of $F(\psi)$ . Thus we see that
$G_{1}(x, y) = (CG_{0}C^{+})(x, y) \sim\frac{1}{L^{4}}\sum_{\zeta,\xi\in\triangle 0}\log(Lx-Ly+\zeta-\xi)$$\sim G_{0}(x, y)$$H_{1}(x, y) = (C’H_{0}C^{\prime+})(x, y) \sim\sum_{\zeta,\xi\in\triangle 0}(Lx-Ly+\zeta-\xi)^{-4}$$\sim H_{0}(x, y)$

as $|x-y|\gg 1$ . This means that the main Gaussian terms are left invariant by(self-similarity).

$C$

and $C’$126
Define(1.15a)(1.15b)(1.15c)(1.15d)(1.15e)(1.15f)$\mathcal{A}_{n} = A_{1}A_{2}\cdots A_{n}$$\tilde{\mathcal{A}}_{n} = \tilde{A}_{1}\tilde{A}_{2}\cdots\tilde{A}_{n}$$\varphi_{n} =\mathcal{A}_{n}\phi_{n}$$z_{n} = \mathcal{A}_{n}Q\xi_{n}$$\mathcal{G}_{n} = \mathcal{A}_{m}G_{n}\mathcal{A}_{n}^{+}$$\mathcal{T}_{n} = \mathcal{A}_{n}Q\Gamma_{n}Q^{+}\mathcal{A}_{n}^{+}$

so that(1.16a)(1.16b)$\varphi_{n} = \varphi_{n+1}+z_{n}$$\mathcal{G}_{n} = \mathcal{G}_{n+1}+\mathcal{T}_{n}$(1.16c)$G_{0} = \sum \mathcal{T}_{n}$(1.16d)$\mathcal{G}_{\mathring{0}^{2}} = \sum_{n}(\mathcal{G}_{\mathring{n}}^{2}-\mathcal{G}_{n+1}^{02})$$= \sum_{n}(\mathcal{T}_{n}^{02}+2\mathcal{G}_{n+1}\circ \mathcal{T}_{n})$Since$R(G\psi)^{2}=\langle\psi,$ $G^{02}\psi\rangle$(1.16e)in (1.9), we will see that

$H_{n}^{-1}\sim \mathcal{T}_{n}^{02}+2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}\sim2\beta_{n+1}\mathcal{T}_{n}$(1.17)

Here we use the following notation (Hadamard product)$(A oB)(x, y)=A(x, y)B(x, y) , T^{02}=ToT$

2

Hierarchical Model Revisited

Before beginning our BST, we study some remarkable features in this model by thehierarchical approximation of Dyson-Wilson type [13] in which the Gaussian part$\exp[-(1/2)\langle\phi_{n}, (-\triangle)\phi_{n}\rangle]$

is replaced by the hierarchical one:$\exp[-(1/2)\langle\phi_{n+1}, (-\Delta)_{hd}\phi_{n+1}\rangle-(1/2)\langle\xi_{n}, \xi_{n} n=0, 1, \cdots$

Put $g_{0}(\phi)=\delta(\phi^{2}-N\beta)$ . Choosing a box of size
spins $\phi+and\phi_{-}$ (two ’s in the box), we put$\phi_{n}$$\sqrt{2}\cross\sqrt{2}$$\phi\pm\equiv\phi\pm\xi$

at the nth step including two, where $\phi=\phi_{n+1}$ and$\xi=\xi_{n}.$127
and put $\phi=(\varphi, 0)\in R_{+}\cross R^{N-1},$ $\xi=(s, u)\in R\cross R^{N-1}$ and$f(x)=g_{n}(x)e^{-x/4}$ . Then putting $x=\phi^{2}$ , we haveThen$2\xi^{2}=\phi_{+}^{2}+\phi_{-}^{2}-2\phi^{2}$$g_{n+1}(x) = e^{x/2} \int f((\phi+\xi)^{2})f((\phi-\xi)^{2})dsd^{N-1}u$$= e^{x/2} \int f((\varphi+s)^{2}+u^{2})f((\varphi-s)^{2}+u^{2})dsd^{N-1}u$$= \frac{e^{x/2}}{\sqrt{x}}\int_{\mathcal{D}}f(p)f(q)\mu(p, q, x)^{(N-3)/2}dpdq$$\mu(p, q_{)}x) = \frac{p+q}{2}-x-\frac{(p-q)^{2}}{16x}$

where$\mathcal{D}\subset[0, N\beta]^{\cross 2}$

is defined so that $\mu(p, q, x)\geq 0$ and(2.1)$\frac{(p-q)^{2}}{16x}=\frac{(\phi_{+}^{2}-\phi_{-}^{2})^{2}}{16\phi^{2}}=\frac{\langle\phi,\xi\rangle^{2}}{\phi^{2}}$form the block spin such
This is a part of the probability that two spins$\phi^{2}=x$$p=N\beta,$$\exp[x/2+(1/2)(N-3)\log(p-x)]$ has a peak
. If $f(p)$ has a peak atthatat $x=N(\beta-1+O(N^{-1}))$ .
What we learn from this model is the following which will appear in the real system:$\emptyset\pm\equiv\phi\pm\xi$$\phi$

1. The curvature of $V_{n}=-\log g_{n}$ at its bottom $x=N\beta_{n}$ isdeviation of$x=\phi_{n}^{2}$

from$N\beta_{n}$is$N^{-1}$, and then the$N^{1/2}.$

2. $\beta_{n}\sim\beta-O(n)$is given by the Gaussian variables $u\in R^{N-1}$ of
3. The deviation$|\phi_{n,+}\phi_{n,-}-N\beta_{n}|=|\phi_{n+1}^{2}-N\beta_{n+1}+:u^{2}:_{1}|\simN^{1/2}$
short correlation. In fact$|\phi_{n}(x)\phi_{n}(y)-N\beta_{n}|$The factor
4. One block spin transformation yields the factor
relevant but logarithmic in the action. Thus its eﬀects are negligible.
$x^{-1/2}\sim\beta_{n}^{-1/2}$

5. $g_{n+1}(x)$ in analytic in $0<x<N\beta(N\geq 3)$ if so is$g_{n}(x)$.

$x^{-1/2}$

is$(g_{1}=(e^{x/2}/\sqrt{x})(N\beta-$$x)^{(N-3/2)})$

6. The probability such that $x=\phi^{2}>N\beta_{n_{0}}$ tends to zero rapidly as $(n_{0}<)narrow\infty,$. This is the mass generation in the hierarchical model.
and$g_{n}(x)arrow\delta(x)$Though this model is very much simplified, it is very surprising that this model contain almost all properties and problems which the real system has. The property (3) iscan
expansion since this means thatimportant and related to the.
replaced by$N^{-1}$$\mathcal{G}_{n}(x, y)$$\varphi_{n}(x)\varphi_{n}(y)/N$

128
One serious problem is that the factor $(x)^{-1/2}=\exp[-\log(\phi^{2})]$ andis relevantin the terminology of renormalization group analysis, i.e., the coeﬃcient may grow exponentially fast as $narrow\infty$ . To control this, we introduce an artificial relevant potential. We note that $(\phi_{0}^{2}-N\beta)^{2}=0$ by thewhich absorb the eﬀects of. Thus one of the main tasks in this paper is to show that
initial condition
are uniformly bounded in$g_{n}(\phi_{n}^{2}-N\beta_{n})^{2}$$\log(\phi^{2})$$\log(\varphi^{2})$$\delta(\phi_{0}^{2}-N\beta)$$g_{n}$$n.$3

RG Flow of the Real System

We combine two types of block transformations towhich is the dimensional
pure
with
imaginary
type
coupling.
boson model ofIn this approach, weinteraction
can expect all coeﬃcients are bounded and small through the block spin transformations.
Thus perturbative calculations are useful. We have two types of block spin transformations. One is the block spin transformation of the $N$ component boson model of mass
, and the other is the block spin transformation of the auxiliary field . The twodimensional boson field is dimensionless and the auxiliary field has the dimension$1engh^{-2}$ , and they have diﬀerent scalings. The
field keeps
on the surface of the$N$ dimensional ball of radius. We will see that by one step of the BSTs of and, the radius is shrinked to, where $\beta_{1}=\beta-O(1)$ .
We turn to our model and sketch our main ideas and procedures. Our method ofanalysis depends on . For $n<\log\beta$ we can forget the term
, but for $n>\log\beta$ thisterm is rather large and we cannot disregard. Assume $n>\log\beta$ and assume that
the Gibbs factor at the step is given by$W_{0}(\phi, \psi)$$\nu$$\phi^{2}\psi$$m_{0}^{2}$$\psi$$\psi$$\phi$$\phi_{0}=\phi$$\psi$$(N\beta)^{1/2}$$\phi$$(N\beta_{1})^{1/2}$$\psi$$\log\phi^{2}$$n$$V_{n}^{(1)}$

$n$

(3.1)

$\exp[-W_{n}(\varphi_{n}, \psi_{n})-\sum_{X}\delta W_{n}(X;\varphi_{n}, \psi_{n})]$where
is the main term which controls the system andarepolymers whose supports spread over paved set
are very small
but analytic domain of
may be small.for large $X$ . Our basic induction assumption isthat the main partis given by$\delta W_{n}(X;\varphi_{n}, \psi_{n})$$W_{n}(\varphi_{n}, \psi_{n})$$X\subset\Lambda.$$\delta W_{n}(X;\varphi_{n}, \psi_{n})$$\varphi_{n}$$W_{n}(\phi_{n}, \psi_{n})$$W_{n}( \phi_{n}, \psi_{n}) = \frac{1}{2}\langle\phi_{n}, G_{n}^{-1}\phi_{n}\rangle_{\Lambda_{n}}+\frac{i}{\sqrt{N}}\langle(:\phi_{n}^{2_{:_{G_{n}}}}, \psi_{n}\rangle_{\Lambda_{n}}+\langle\psi_{n}, H_{n}^{-1}\psi_{n}\rangle_{\Lambda_{n}}$(3.2a)$+V_{n}^{(1)}+V_{n}^{(2)}$(3.2b)$V_{n}^{(1)} = \frac{1}{2N}\langle:\phi_{n}^{2}:_{G_{n}}, g_{n}:\phi_{n}^{2}:c_{\mathfrak{n}}\rangle_{\Lambda_{\mathfrak{n}}}$$V_{n}^{(2)}$$=$$\frac{\gamma_{n}}{2}$$\langle$:$\phi_{n}^{2}:_{G_{n}},$$\tilde{A}_{n-1}E^{\perp}G_{n-1}^{-1}E^{\perp}\tilde{A}_{n-1}^{+}$:$\phi_{n}^{2}:c_{n}\rangle_{\Lambda_{n}}$(3.2c)

whereis a constant matrix discussed later,is the projection operator to the setof block-wise zerxaverage functions, i.e. $\mathcal{N}(C)=\{f\in R^{\Lambda} : (Cf)(x)=0, \forall x\in\Lambda_{1}\}$ , and$\tilde{A}_{n}$$E^{\perp}$

129:$\phi_{n}^{2}:c_{n}$is the Wick product of$\phi_{n}^{2}$

with respect to$G_{n}$. Furthermore we use the notation$\langle f, g\rangle=\sum_{\Lambda x\in}f(x)_{9}(x) , \langle f, g\rangle_{\Lambda_{n}}=\sum_{\Lambda_{n}x\in}f(xg(x))$. ThusThe point is thatacts as a diﬀerential operator andcorresponds to $(p-q)^{2}/16x$ and is irrelevant.
. The term
contains
The relevant terms
is a dummy and is not necessary in principle since :is artificially inserted to control
at the beginning. The term
) stay(defined on
This is relevant, but we can show that the coupling constantsorbounded. In the case of hierarchical model, we do not need any information offorsince the hierarchical Laplacian is local and (then) we have some a priori
bound for which are locally defined. But in the present model, however, it seems to be
to control
convenient to have the termandis absorbed by the parameters
We show that the change of the action. Here$E^{\perp}$$E^{\perp}(-\triangle)E^{\perp}$$G_{n}^{-1}\sim-\triangle$$V_{n}^{(2)}$$\prod_{i=1}^{4}\nabla_{\mu_{i}}$$V_{n}^{(1)}$$\varphi_{0}^{2}:_{G_{0}}$$\langle$$V_{n}^{(1)}$$\}:\varphi_{0}^{2}:c_{0}\rangle=0$$\log\phi^{2}.$$\Lambda_{n}$$g_{n}$$W_{n}$$g_{n}$$\phi_{n}^{2}<N\beta_{n}$$g_{n}$$V_{n}^{(1)}$$\log\varphi_{n}^{2}.$$W_{n}$$\beta_{n},$$g_{n}$$\gamma_{n}$$\beta_{n}$$g_{n}$$=$$=$(3.3a)(3.3b)(3.3c)-const. $n+o(n)$0(1)$\beta$$\gamma_{n} = O((\beta_{n}N)^{-1})$$H_{0}^{-1}=0,$$\gamma_{0}=0$

and$\beta_{0}=\beta$

and we discarded irrelevant terms.

Remark 1 It is noteworthy that we can put$\frac{1}{L^{2n}}\sum_{x\in D\cap\Lambda}f(x)_{9}(x)\equiv\int_{D/L^{n}}f(\chi)g(\chi)d^{2}\chi,$and which are diﬀerentiable on the finer lattice spacevariable $x/L^{n}$ on

for

$f$

$g$

$L^{-n}\Lambda$. Here$\chi$

is the new$L^{-n}\Lambda.$

4

Outline of the Proof

We here sketch our proof which consists of several steps:[step 1]and letbe the nth block spin $(\phi_{n+1}=C\phi_{n})$ : Set $\phi_{n}=A_{n+1}\phi_{n+1}+$
Let
is
and
, whereare the fluctuation field living on
are O.
the zero-average matrix so that the block averages of$\Lambda_{n}=L^{-n}\Lambda\cap Z^{2}$$Q\xi_{n}$$\xi_{n}(x)$$\phi_{n}$$\Lambda_{n}’=\Lambda_{n}\backslash LZ^{2}$$Q\xi$$\langle\phi_{n}, G_{n}^{-1}\phi_{n}\rangle_{\Lambda_{n}}=\langle\phi_{n+1}, G_{n+1}^{-1}\phi_{n+1}\rangle_{\Lambda_{n+1}}+\langle\xi_{n}, \Gamma_{n}^{-1}\xi_{n}\rangle_{\Lambda_{n}’}$$Q:R^{\Lambda’}arrow R^{\Lambda}$

130
where

$G_{n+1}^{-1}=A_{n+1}^{+}G_{n}^{-1}A_{n+1}$

and$Q^{+}G_{n}^{-1}Q=\Gamma_{n}^{-1}$. Namely $A_{n+1}=G_{n}C^{+}G_{n+1}^{-1}.$[step 2]
We have a relevant term, and then it is convenient to consider the Gaussian integral by. Define: (not by z) since:$q(z)\equiv 2\varphi_{n}z_{n}+:z_{n}^{2}$$\varphi_{n}^{2}:c_{\mathfrak{n}}=:\varphi_{n+1}^{2}:c_{n+1}+q(z)$$P(p) = \int\exp[i\langle\lambda, (p-q)\rangle]d\mu(\xi)\prod d\lambda$$z_{n} = \mathcal{A}_{n}Q\Gamma_{n}^{1/2}\xi$$d \mu(\xi) = \exp[-\frac{1}{2}\langle\xi, \xi\rangle_{\Lambda_{n}’}]\prod\frac{d\xi}{\sqrt{2\pi}}$

Then we have$P(p)$$=$$\int\exp[i\langle\lambda,p\rangle]\exp[-i\langle\lambda,$$(2\varphi_{n+1}(\mathcal{A}_{n}Q\Gamma_{n}^{1/2}\xi)+:(\mathcal{A}_{n}Q\Gamma_{n}^{1/2}\xi)^{2}$$d \mu(\xi)\prod d\lambda$$= \int\exp[-2i\langle\xi, \Gamma_{n}^{1/2}Q^{+}\mathcal{A}_{n}^{+}(\lambda\varphi_{n+1})\rangle-\frac{1}{2}\langle\xi, [1+2i\Gamma_{n}^{1/2}Q^{+}\mathcal{A}_{n}^{+}\lambda \mathcal{A}_{n}Q\Gamma_{n}^{1/2}]\xi\rangle]$$\cross\exp[i\langle\lambda,p\rangle+iN\langle\lambda, \mathcal{T}_{n}\rangle]\prod\frac{d\xi_{x}d\lambda(x)}{\sqrt{2\pi}}$

namely$P(p)$$=$$\int\exp[i\langle\lambda,p\rangle+iN\langle\lambda, \mathcal{T}_{n}\rangle]\det^{-N/2}(1+2i\mathcal{T}_{n}\lambda)$$\cross\exp[-2\langle\lambda, (\varphi_{n+1}\varphi_{n+1})\circ(\mathcal{A}_{n}Q\frac{1}{\Gamma_{\overline{n}^{1}}+2iQ^{+}\mathcal{A}_{n}^{+}\lambda \mathcal{A}_{m}Q}Q^{+}\mathcal{A}_{n}^{+})\lambda\rangle]\prod d\lambda(x)$(4.1)

We assume that we are outside of the domain wall region
definedbyand large field region$D_{w}(\varphi_{n})$$D(\varphi_{n})$(1)$D_{w}(\varphi_{n})=$paved set such that[$|\varphi_{n}(x)\varphi_{n}(y)-N\mathcal{G}_{n}(x, y)|\geq k_{0}N^{1/2+\epsilon}\exp$ $\frac{c}{10L^{n}}|x-y$(2)$D(\varphi_{n})=minima1$$|$:$\forall x\in D_{w},$ $\exists y\in D_{w}$paved set such that$\varphi_{n}^{2}(x):_{G_{n}}|\leq k_{0}N^{1/2+\epsilon}\exp[\frac{c}{10L^{n}}|x-y$$\forall x\in D(\varphi)$,$\forall y\in D(\varphi)^{c}$where $0<\epsilon<1/2$ and paved set is a collection of squares
each of which consists$L\crossL$(in). The powerof squaresof sizeis related to the central limit
theorem applied to the sum of $N$ independent Gaussian variables::. To imagine
why, consider spins
located on the bottom ofand put
Thus the parallel component of the fluctuation
is suppressed and only the orthogonal$\{\square \}$$\triangle\subset\Lambda$$N^{1/2}$$\Lambda_{n}$$\sum_{i=1}^{N}$$(\varphi_{n}^{2}-N\beta_{n})^{2}$$\varphi_{n}(x)$$z_{n}$

fluctuations occur.

$\xi_{i}^{2}$$\varphi_{n}=\varphi_{n+1}+z_{n}.$131

Figure 1: Two wine bottles and the propagation of spin waves which are orthogonal or
perpendicular. It costs energy to change the radius of the bottles but it is easy to fluctuate
perpendicular to the radius. This is caused by , the massive Gaussian field of $N-1$
degrees of freedom.
$\xi$

perpendicular to
Fluctuations
have $N-1$ degrees of freedom of Gaussianbyand expand thefields. See also the figure caption. We replace
determinant up to the second order:$\phi_{n}(x)$$\xi_{n}(x)$$\varphi_{n+1}\varphi_{n+1}$

(4.1)

$=$

$N\mathcal{G}_{n+1}$$\int\exp[i\langle\lambda,p\rangle-N\langle\lambda, (\mathcal{T}_{n}^{02}+2\mathcal{G}_{n+1}\circ \mathcal{T}_{n})\lambda\rangle]$$\cross\det_{3}^{-N/2}(1+2i\Gamma_{n}^{1/2}Q^{+}\mathcal{A}_{n}^{+}\lambda \mathcal{A}_{n}Q\Gamma_{n}^{1/2})$$\cross\exp[-2\langle\lambda, (:\varphi_{n+1}\varphi_{n+1} :)\circ \mathcal{T}_{n})\lambda\rangle+$( higher order terms)]$\sim \exp[-\frac{1}{4N}\langle p, \frac{1}{2\mathcal{G}_{n+\mathring{1}}\mathcal{T}_{n}+\mathcal{T}_{n^{02}}}p\rangle]$

The terms::,:
Putting$\varphi_{n+1}^{2}$$g_{n}$$\prod d\lambda(x)$(4.2): are treated by polymer expansion and yield relevant terms :.
which are fractions ofwith $p_{1}=C^{n+1}p$ and $C^{n+1}A=1$ , we see that $P(p)$ is given by$\langle$$\varphi_{n+1}\varphi_{n+1}$$\varphi_{n+1}^{2}$$\log(\varphi_{n}^{2})$$p=Ap_{1}+\tilde{Q}\tilde{p}$$\exp[-\frac{1}{4N}\langle p_{1},$$\frac{1}{C^{n}[2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}+\mathcal{T}_{n}^{02}](C^{+})^{n}}p_{1}\rangle_{\Lambda_{n+1}}-\frac{1}{4N}\langle\tilde{Q}\tilde{p},$$\frac{1}{2\mathcal{G}_{n+\mathring{1}}\mathcal{T}_{n}+\mathcal{T}_{n^{02}}}\tilde{Q}\tilde{p}\rangle]$(4.3)

Here we remind the reader that$C^{n+1}\mathcal{T}_{n}(C^{+})^{n+1} = 0$$C^{n+1}\mathcal{T}^{02}n_{\mathcal{G}_{n+1}\circ \mathcal{T}_{n}}(C^{+})^{n+1} \sim\sim \beta_{n+1}\mathcal{T}_{n}1$132
decays much faster than . Thismeans that the blockwise constant part of remains and the zero-average fluctuation
partof is almost absent since it approximately holds thatsince$\mathcal{T}_{n}=\mathcal{A}_{n}Q\Gamma_{n}Q^{+}\mathcal{A}_{n}^{+},$$C^{n}\mathcal{A}_{n}=1,$$CQ=0$ and$p_{1}$$\mathcal{T}_{n}$$\mathcal{G}_{n}$$p$$\tilde{Q}\tilde{p}$$p$$(2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}+\mathcal{T}_{n}^{02})\lceil_{\tilde{Q}\xi}=O(L^{2n})\beta_{n+1}$

though this has to be taken with a grain of salt. (We need a suble discussion on thispoint, however.)[step 3]

We calculate the next order Gibbs factor by multiplying the distribution function$P(p)=Prob(p=q(\xi))$ to the previous Gibbs factor. This idea perhaps goes back to [22].
In the present case, however,can be large $(\sim L^{2} on A_{n})$ and then we choose which
minimizes$p$$g_{n}$$F(p)$$=$$\frac{1}{4N}\langle p,$$\frac{1}{2\mathcal{G}_{n+\mathring{1}}\mathcal{T}_{n}+\mathcal{T}_{n^{02}}}p\rangle+\frac{1}{4N}\langle(:\varphi_{n+1}^{2}:_{G_{n+1}}+p)$,$g_{n}(:\varphi_{n+1}^{2}:c_{n+1}+p)\rangle$(4.4)$= \langle p, \frac{1}{D}p\rangle+\frac{1}{N}\langle(:\varphi_{n+1}^{2}:c_{n+1}, g_{n}p\rangle+\frac{1}{2N}\langle(:\varphi_{n+1}^{2}:_{G_{n+1}g_{n}:\varphi_{n+1}^{2}:c_{n+1}\rangle}$

(4.5)

where
$\frac{1}{D}=\frac{1}{4N}\frac{1}{2\mathcal{G}_{n+\mathring{1}}\mathcal{T}_{n}+\mathcal{T}_{n}^{02}}+\frac{1}{2N}g_{n}$To diagonalize this, we again set$p=\mathcal{A}p_{1}+\tilde{Q}\tilde{p}$

(4.6)

where

$\mathcal{A}=D(C^{+})^{n+1}[C^{n+1}D(C^{+})^{n+1}]^{-1}, C^{n+1}\tilde{Q}=0$

(4.7)

and

(4.8a)$F(p) = F_{1}(p)+F_{2}(p)$$F_{1} = \langle p_{1}, \frac{1}{C^{n+1}D(C^{+})^{n+1}}p_{1}\rangle_{\Lambda_{n+1}}+\frac{1}{N}\langle(:\varphi_{n+1}^{2}:c_{n+1}, g_{n}p\rangle$$+ \frac{1}{2N}\langle(E:\varphi_{n+1}^{2}:c_{n+1}, g_{n}E:\varphi_{n+1}^{2}:c_{n+1}\rangle$(4.8b)$F_{2} = \langle\tilde{Q}\tilde{p}, \frac{1}{D}\tilde{Q}\tilde{p}\rangle+\frac{1}{N}\langle(E^{\perp_{:\varphi_{n+1}^{2}:_{G_{n+1}}}}, g_{n}\tilde{Q}\tilde{p}\rangle$$+ \frac{1}{2N}\langle(E^{\perp}:\varphi_{n+1}^{2}:c_{n+1}, g_{n}E^{\perp}:\varphi_{n+1}^{2}:c_{n+1}\rangle$(4.8c)

where $E$ is the projection to block-wise constant functions (block of size $L^{n+1}\cross L^{n+1}$ )and $E^{\perp}=1-E$ . We moreover assume thatis a constant diagonal matrix. Then$g_{n}$$F_{1}$

133
and$F_{2}$

take their minima at the following points:$p_{1} = - \frac{1}{N}C^{n}Dg_{n}:\varphi_{n+1}^{2}:_{G_{n+1}}$(4.9)$= [-1+ \frac{1}{L^{2n}g_{n}}\frac{1}{C^{n}[2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}+\mathcal{T}_{n^{02}}](C^{+})^{n}}]C^{n}:\varphi_{n+1}^{2}:c_{n+1}$$\tilde{Q}\tilde{p}= -\frac{1}{2N}E^{\perp}Dg_{n}:\varphi_{n+1}^{2}:_{G_{n+1}}$(4.10)$= [-1+ \frac{1}{9n}\frac{1}{2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}+\mathcal{T}_{n^{02}}}]E^{\perp}:\varphi_{n+1}^{2}:c_{n+1}$

Thus$\min F_{1} = \frac{k}{4N}\langle C^{n+1}:\varphi_{n+1}^{2}:, \frac{1}{C^{n+1}[2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}+\mathcal{T}_{n^{02}}](C^{+})^{n+1}}C^{n+1}:\varphi_{n+1}^{2}:\rangle.$$\min F_{2}$$=$$\frac{1}{4N}\langle E^{\perp}:\varphi_{n+1}^{2}$:,$\frac{1}{2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}+\mathcal{T}_{n^{02}}}E^{\perp}:\varphi_{n+1}^{2}:\rangle$

We integrate overandaround the points (4.9) and (4.10) (steepest descentmethod) and we get some small terms coming from the integrations overand
The term $\min F_{1}$ means that the $9n$ term disappears and the coeﬃcient of the relevantterm:
can be regarded as a constant for $n>\log\beta$ since:(field on) and). This also implies that(on$\tilde{p}$$p_{1}$$E^{\perp}\tilde{p}.$$p_{1}$$C^{n+1}$$(: \varphi_{n+1}^{2}:)^{2}$$A_{n}$$C^{n+1}[2\mathcal{G}_{n+1}\circ \mathcal{T}_{n}+\mathcal{T}_{n}^{02}](C^{+})^{n+1}\sim 1$$\langle$:$\varphi_{n+1}^{2}:_{G_{n+1}}+p,$$\psi_{n}\rangle$$\frac{1}{L^{2n}}$$arrow$$\langle$which is consistent with our choice of the scaling of
tiallywhich is irrelevant. We remark that theis absorbed byand the Hamiltonian part of:$\log$$\mathcal{F}_{n}$$V_{n}^{(1)}$$\Lambda_{n+1}$$\varphi_{n+1}^{2}:c_{n+1},$$\psi$$\phi_{n+1}$$:\sim:\phi_{n+1}^{2}$$\varphi_{n+1}^{2}$(4.11)$E\psi_{n}\rangle$. The term $\min F_{2}$ is essenand:term is expanded and:through$\tilde{A}_{n}$$\varphi_{n+1}\varphi_{n+1}$$2 :\varphi_{n+1}(x)\varphi_{n+1}(y) \varphi_{n+1}^{2}(x):+:\varphi_{n+1}^{2}(y):-:(\varphi_{n+1}(x)-\varphi_{n+1}(y))^{2}$

The shifts of the variables$p_{1}$

and$\tilde{Q}\tilde{p}$are in the admissible deviations of$\varphi_{n+1}$:and$q_{n}.$[step 4]Thus we can iterate these steps. The most important point is that: –::obeys the Gaussian distribution uniformly in (CLT) and the coupling constant is kept
as a constant on the shell:near which the functional integrals have supports.
This ensures our scenario.
$q=:\varphi_{n}^{2}$$n$$\phi_{n}^{2}:_{G_{n}}=0$5

Remaining Problems

The following problems remain:$\varphi_{n+1}^{2}$$g_{n}$

134
1. Prove this for small $N.$
2. Prove this for quantum spins.

3. Solve the Millennium problem of quark confinement.
The present author hopes that the reader is ambitious enough to attack these problems.

Acknowledgements. This work was partially supported by the Grant-in-Aid for Scientific
Research, No.23540257, No. 26400153, the Ministry of Education, Science and Culture,
Japanese Government. Part of this work was done while the author was visiting INS
Planck Inst. for Physics (Muenchen) and UBC (Vancouver.) He would like to
Lyon,
thank K.Gawedzki, E.Seiler and D.Brydges for useful discussions and kind hospitalities
extended to him. Last but not least, he thanks T. Hara and H.Tamura for stimulating
discussions and encouragements.
${\rm Max}$

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