\,dt\no\\ &&=\tint_0^T[\s p(t)\s{\dot q}(t)-H(p(t),q(t))\s]\,dt\;,\en namely the classical action functional enhanced by the fact that $\hbar>0$ still. The Hamiltonian connection, called the {\it Weak Correspondence Principle} \cite{wcp}, is given, assuming that $P$ and $Q$ are irreducible, by \bn H(p,q)\hskip-1.3em&&\equiv\

\no\\
&&=\<0|\s \H(P+p\one,Q+q\one)\s|0\>\no\\
&&= \H(p,q)+{\cal O}(\hbar;p,q)\;. \en
While it is clear, in our approach, that the relation of the classical variables, $(p,q)$, to the quantum variables, $(P,Q)$, is entirely different from the usual procedure, {\it it is of primary importance to appreciate that the new procedures have---at this point---led to the same result as conventional canonical quantization, specifically, the same Hamiltonian operator, modulo terms of order $\hbar$, that is used in the standard approach.}.
Although there is much more to this alternative version of the classical/quantum connection, including canonical coordinate transformations and Cartesian coordinates, etc., (see \cite{eQ}), the few equations presented here illustrate the alternative procedure that we shall employ in analyzing the vector and matrix models.
\section{Rotationally-Symmetric Vector Models}
Let us consider the quantization of the vector model with a classical Hamiltonian given by
\bn H_c( \pa,\qa)=\half\s[\s\pa^2+m_0^2\s\qa^2\s]+\l\s(\qa^2)^2 \;;\en
a definition of the symbols appears below Eq.~(\ref{vm}).
As a consequence of rotational invariance, every classical solution is equivalent to a solution for $N=1$ if $\pa\s\|\qa$
at time $t=0$, or to a solution for $N=2$ if $\pa\!\not\!\|\s \qa$ at time $t=0$. Moreover, solutions for $N=\infty$ may be derived from those for $N<\infty$ by the limit $N\ra\infty$, provided we maintain $(\pa^2+\qa^2)<\infty$.
A conventional canonical quantization begins with $\pa\ra\Pa$, $\qa\ra\Qa$, which are irreducible operators that obey $[Q_l,P_n]=i\hbar\delta_{ln}\one$ as the only non-vanishing commutation relation. For a free model, with mass $m$ and $\l=0$, the quantum Hamiltonian
$\H_0=\half
:( \Pa^2+m^2\s\Qa^2):$, where $:(\cdot):$ denotes normal ordering. It has the feature that the Hamiltonian operator for $N=\infty$ is obtained as the limit of those for which $N<\infty$.
Moreover, with the normalized ground state $|0\>$ of the Hamiltonian operator chosen as the fiducial vector for canonical coherent states, i.e., \bn |\pa,\qa\>=\exp[-i\qa\cdot\Pa/\hbar]\s\exp[i\pa\cdot\Qa/\hbar]\s|0\>\;,\en
it follows that
\bn H_0(\pa,\qa)\hskip-1.3em&&=\<\pa,\qa|\s\half:(\Pa^2+m^2\s\Qa^2):|\pa,\qa\>\no\\
&&=\half\s(\pa^2+m^2\s\qa^2)\en
as desired, for all $N\le\infty$, provided that $(\op^2+\oq^2)<\infty$.
For comparison with later relations, we recall the characteristic function (i.e., the Fourier transform) of the ground-state distribution for the free vector model with mass $m$ given by
\bn C_0(\of)\hskip-1.3em&&=M^2_0\int e^{\t i\of\cdot\ox/\hbar}\,e^{\t -m\s\ox^2/\hbar}\,\Pi_{n=1}^N\s dx_n\no\\
&&=e^{\t-\of^2/4\s m\s\hbar}\;, \en
where $M_0$ is a normalization factor, and this result formally holds for $N\le\infty$.
However, canonical quantization of the interacting vector models with $\l>0$ leads to trivial results for $N=\infty$. To show this, we assume that the Schr\"odinger representation of the ground state of an interacting model is real, unique, and rotationally invariant. As a consequence, the characteristic function of the ground-state distribution has the form (note: $\of^2\equiv|f|^2\equiv\Sigma_{n=1}^N\s f_n^2$ and $r^2\equiv\Sigma_{n=1}^N\s x_n^2$)
\bn C_N(\overrightarrow{f})\hskip-1.3em&&=\int e^{\t i\Sigma_{n=1}^N f_n\s x_n/\hbar}\,\Psi_0(r)^2\,\Pi_{n=1}^N\s dx_n\no\\
&&=\int e^{\t i|f|\s r\s\cos(\theta)/\hbar}\,\Psi_0(r)^2\,r^{N-1}\, dr\s\sin(\theta)^{N-2}\, d\theta\,d\Omega_{N-2}\no\\
&&\simeq M'\int e^{\t-\of^2\s r^2/2(N-2)\hbar^2}\,\Psi_0(r)^2\,r^{N-1}\, dr\s d\Omega_{N-2}\no\\
&&\ra\int_0^\infty e^{\t-b\s \of^2/\hbar}\,w(b)\,db \en
assuming convergence, where a steepest descent integral has been performed for $\theta$, and in the last line we have taken the limit $N\ra\infty$; additionally,
$w(b)\ge0$, and $\int_0^\infty w(b)\s db=1$. This is the result based on symmetry. Uniqueness of the ground state then ensures that $w(b)=\delta(b-1/4{\widetilde m})$, for some $\widetilde{m}>0$, implying that the quantum theory is that of a free theory, i.e., the quantum theory is trivial! In addition, the classical limit of the resultant quantum theory is a free classical theory, which differs from the original, nonlinear classical theory.
This kind of result signals an unsuccessful quantization.
\subsection{Nontrivial quantization of vector models}
The way around this unsatisfactory result is to let the representations of $\Pa$ and $\Qa$ be {\it reducible}. The weak correspondence principle, namely
$ H(\pa,\qa)\equiv \<\pa,\qa|\s\H\s|\pa,\qa\>$, ensures that the enhanced classical Hamiltonian depends only on the proper variables, but, unlike conventional quantization procedures, there is no rule forbidding the Hamiltonian from being a function of other, non-trivial operators that commute with $\oP$ and $\oQ$, thus making the usual operators reducible. A detailed study \cite{RS} of the proper reducible representation, still in accord with the argument above that limits the ground-state functional form to be a Gaussian, leads to the following formulation. Let $\Ra$ and $\Sa$ represent a new set of canonical operators, independent of the operators $\Pa$ and $\Qa$, and which obey the commutation relation $[S_l,R_n]=\i\hbar\delta_{ln}\one$. We introduce two sets of compatible annihilation operators that annihilate the unit vector $|0,0;\zeta\>$, namely
\bn &&
[\s m(\s\Qa+\zeta\s \Sa)+i\Pa\s]\s|0,0;\zeta\>=0\;,\no\\
&& [\s m(\s \Sa+\zeta\s \Qa)+i\Ra\s]\s|0,0;\zeta\>=0\;. \en
If we introduce eigenvectors $|\ox,\oy\>$ such that $\oQ|\ox,\oy\>=\ox|\ox,\oy\>$ and $\oS|\ox,\oy\>=\oy|\ox,\oy\>$, it follows that
\bn \<\ox,\oy|0,0;\zeta\>=M\s \exp[-(\ox^2+2\s\zeta\ox\cdot\oy+\oy^2)/2\hbar]\;\;, \label{gs}\en
where $M$ is a normalization factor, and the condition $0<\zeta<1$ ensures normalizability. These two annihilation operators lead directly to
two related free Hamiltonian operators,
\bn &&\H_{0\,PQ}\equiv\half:(\Pa^2+m^2(\Qa+\zeta\Sa)^2\s):\;,\no\\
&&\H_{0\,RS}\equiv\half:(\Ra^2+m^2(\Sa+\zeta\Qa)^2\s):\;,\en
for which (\ref{gs}) is a common, unique, Gaussian ground state.
Let new coherent states, which span the Hilbert space of interest, be defined with this ground state as the fiducial vector, as given by
\bn|\pa,\qa\>\equiv \exp[-i\qa\cdot\Pa/\hbar]\s\exp[i\pa\cdot\Qa/\hbar]\s|0,0;\zeta\>\;,\en
and it follows that
\bn H(\pa,\qa)\hskip-1.3em&&=\<\pa,\qa|\s\{\s \H_{0\,PQ}+\H_{0\,RS}+4v\s:\H_{0\,RS}^2:\s\}\s|\pa,\qa\>\no\\
&&=\half[\pa^2+m^2(1+\zeta^2)\s\qa^2]+v\s\zeta^4\s m^4\s (\qa^2)^2\no\\
&&\equiv \half[\s\pa^2+m_0^2\s \qa^2\s]+\l\s(\qa^2)^2\en
as required; and this solution is valid for {\it all} $N$, $1\le N\le\infty$,
provided that $(\pa^2+\qa^2)<\infty$.
\section{Rotationally-Symmetric Matrix Models}
As discussed in Sec.~1, we now turn our attention to the quantization of classical systems that have a classical Hamiltonian given by
\bn H_c(p,q)=\half\s[ \tr(p^2)+m_0^2\s \tr(q^2)\s]+\l\s\tr(q^4)\;,\en
where the variables $p$ and $q$ are $N\times N$ real, symmetric matrices, as explained after Eq.~(\ref{mm}).
Such models are invariant under matrix transformations $p\ra O\s p\s\s O^T$ and $q\ra O\s q \s O^T$, where
$O\in {\bf O}(N)$. The free model, with mass $m$ and $\l=0$, is readily quantized by promoting $p\ra P$ and $q\ra Q$, which are Hermitian, symmetric (in their indices) matrix operators with the property that the only nonvanishing commutator is
\bn [\s Q_{ab}, P_{cd}\s] =i\hbar \s\half(\s \delta_{ac}\delta_{bd}+\delta_{ad}\delta_{bc}\s)\s\one\;.\en
The free Hamiltonian operator is given by
\bn \H_0\equiv\half\s:\s[\s\tr(P^2)+m^2\s\tr(Q^2)\s]:\;, \en
and the normalized ground state $|0\>$ of this Hamiltonian is unique and rotationally invariant, and is given by
\bn \ \no\\
&& =\<0|\, \half\s:\s[\s\tr((P+p\one)^2)+m^2\s\tr((Q+q\one)^2)\s]:\s|0\>\no\\
&&=\half\s[\s\tr(p^2)+m^2\s\tr(q^2)\s]\;, \en
as desired.
Again, for comparison purposes, we compute the characteristic function for the ground-state distribution of the free model, which is given by
\bn C_0(f)\hskip-1.3em&&=M'^2\int\s e^{\t i\tr(f\s x)/\hbar}\,e^{\t-m\tr(x^2)/\hbar}\,\Pi_{a\le b=1}^{N,N}\s dx_{ab}\no\\
&&=e^{\t-\tr(f^2)/4m\hbar}\;, \en
where now $f$ denotes a real, symmetric $N\times N$ matrix, and this relation formally holds for all $N\le\infty$.
When $\l>0$, conventional quantization procedures would require a rescaling of the quartic interaction term by replacing $\l$ by $\l/N$, as dictated by a perturbation analysis. However, under reasonable assumptions, just as in the vector case discussed in Sec.~2, we are able to show that when $\l>0$ and $N=\infty$ the quantum theory for the matrix models is trivial ($=$ free) just as was the case for the vector models. To show this, we again appeal to the ground state which we assume to be real, unique, and rotationally invariant for the models under consideration. While in the vector models rotational invariance meant the ground state was a function of only one variable, namely $\ox^2$, that is not the case for the matrix models. For example, the ground state could be a function of $\tr(x^2)$, $\tr(x^3)$, $\tr(x^4)$, $\det(x)$, etc. We recognize the fact of many invariant forms, but for simplicity we shall just display only two of them, namely, $\tr(x^2)$ and $\tr(x^4)$.
As before, we consider the characteristic function of the ground-state distribution in the Schr\"odinger representation given by
\bn C_N(f)\hskip-1.3em&&=\int e^{\t i\tr(f\s x)/\hbar}\,\Psi[\tr(x^2),\tr(x^4)]^2\,\Pi_{a\le b=1}^{N,N} dx_{ab}\;, \label{int}\label{ee}\en
where, again, $f$ is a real, symmetric $N\times N$ matrix. As defined, $C_N(f)$ is clearly invariant under rotations such as $f\ra O\s f\s O^T$. As a real, symmetric matrix, we can imagine choosing $O\in{\bf O}(N)$ so as to diagonalize $f$, namely the matrix is now of the form where $f_{ab}=\delta_{ab}\s f_a$, with $f_a$ being the diagonal elements. This means that the expression
\bn \tr(f\s x)={\ts\sum}_{a=1}^N f_a\s x_{aa}\;, \en
and thus only the $N$ {\it diagonal elements} of the real, symmetric $N\times N$ matrix $x$
enter into the exponent of (\ref{ee}).
Let us introduce a suitable form of spherical coordinates in place of the $N^*\equiv N(N+1)/2$ integration variables $x_{ab}, \,a\le b$. We choose the radius variable $r$ so that $r^2\equiv\tr(x^2)= \Sigma_{a,b=1}^{N,N}\s x_{ab}^2$, and the first $N$ angles are identified with the diagonal elements of $x$ leading to
$x_{11} = r \cos(\theta_1)$, and for $2\le a\le N$, we set
$x_{aa} = r [\s\Pi_{l=1}^{a-1}\sin(\theta_l)\s] \cos(\theta_{a})$; we also extend the latter notation to include $x_{11}$. The remaining $N(N-1)/2$ variables $x_{ab}$ for $a**$, $0<\zeta<1$, and require that
\bn &&[\s m\s(Q+\zeta S)+i\s P\s]\s|0,0;\zeta\>=0\;,\no\\
&&[\s m\s(S+\zeta Q)+i\s R\s]\s|0,0;\zeta\> =0\;. \en
In terms of eigenvectors $|x,y\>$ for both $Q$ and $S$, respectively, it follows that
\bn \ =M\,\exp\{- m\s[\s\tr(x^2+2\s\zeta\s x y+y^2)\s]/2\hbar\s\}\;,\en
a Gaussian state in accord with the conclusion of (\ref{yes2}).
There are two related, Hermitian Hamiltonian expressions of interest, given by
\bn \H_{1\,PQ}\hskip-1.3em&&=\half\s\tr\{[\s m\s(Q+\zeta S)-i\s P\s]\,[\s m\s(Q+\zeta S)+i\s P\s]\}\no\\
&&=\half\s:\s[\tr(P^2)+m^2\s\tr((Q+\zeta\s S)^2)\s]: \;,\en
and
\bn \H_{1\,RS}\hskip-1.3em&&=\half\s\tr\{[\s m\s(S+\zeta Q)-i\s R\s]\,[\s m\s(S+\zeta Q)+i\s R\s]\}\no\\
&&=\half\s:[\s\tr(R^2)+m^2\s\tr((S+\zeta\s Q)^2)\s]: \;,\en
and $|0,0;\zeta\>$ is the unique ground state for both of them. For coherent states, we choose
\bn |p,q\>=\exp[-i\tr(q P)/\hbar]\,\exp[i\tr(p\s\s Q/\hbar)\s]\,|0,0;\zeta\>\;, \en
and, for the final, total Hamiltonian, it follows that
\bn H(p,q)\hskip-1.3em&&=\**

\no\\
&&=\half\s[\s \tr(p^2)+m^2(1+\zeta^2)\s\tr(q^2)\s]+ v\s m^4 \s \zeta^4 [\s\tr(q^2)\s]^2\no\\
&&\equiv\half\s[\tr(p^2)+m^2_0\s\tr(q^2)\s]+\l\s[\s\tr(q^2)\s]^2\;,\label{yes3}\en
as desired.
The solution associated with Eq.~(\ref{mm}) is different from that just presented, but, again, according to the analysis that led to (\ref{yes2}), we are still obliged to look for a solution based on a Gaussian ground state.
Once again we start with the canonical matrix operator pair $P$ and $Q$, as well as the canonical matrix operator pair $R$ and $S$ that we used in the solution of the model (\ref{yes3}). However, we now need yet another matrix operator pair, namely, the independent canonical matrix pair $T$ and $U$. This time we start with a ground state denoted by the unit vector $|0,0,0;\xi\>$ which is annihilated by three operators, namely
\bn &&\hskip-2.75em[\s m\s(Q+\xi(S+U))+iP\s]\s|0 , 0,0;\xi\>=0\;,\no\\
&&[\s m\s(S+\xi\s Q)+iR\s]\s|0,0,0;\xi\>=0\;,\no\\
&&[\s m\s(U+\xi\s Q)+iT\s]\s|0,0,0;\xi\>=0\;. \en
If we introduce eigenvectors $|x,y,z\>$ for the three operators $Q, S$, and $U$, respectively, then it follows that in the Schr\"odinger representation, the ground state is given by
\bn \=M''\exp\{- m\s[\s\tr(x^2+y^2+z^2+2\s\xi(x\s y+ x z))\s]/2\hbar\s\}\;,\en
and the condition $0<\xi<1/\sqrt{2}$ ensures normalizability.
There are now {\it four} Hamiltonian-like expressions of interest, the first three of which have the vector $|0,0,0;\xi\>$ as their common, unique ground state, namely,
\bn \H_{2\,PQ}\hskip-1.3em&&=\half\s\tr\s\{\s[\s m\s(Q+\xi(S+U))-iP\s]\,[\s m\s(Q+\xi(S+U))+iP\s]\s\}\no\\
&&=\half\s:\{\tr\s(P^2)+m^2\s\tr[(Q+\xi\s(S+U))^2]\s\}:\;,\en
\bn H_{2\,RS}\hskip-1.3em&&=\half\s\tr\s\{\s[\s m\s(S+\xi\s Q)-iR\s]\,[\s m\s(S+\xi\s Q)+iR\s]\s\}\no\\
&&=\half\s:\{\tr\s(R^2)+m^2\s\tr[(S+\xi\s Q)^2]\s\}:\;,\en
\bn H_{2\,TU}\hskip-1.3em&&=\half\s\tr\s\{\s[\s m\s(U+\xi\s Q)-iT\s]\,[\s m\s(U+\xi\s Q)+iT\s]\s\}\no\\
&&=\half\s:\{\tr\s(T^2)+m^2\s\tr[(U+\xi\s Q)^2]\s\}:\;,\en
and the fourth Hamiltonian expression, which also has $|0,0,0;\xi\>$ as a ground state, is given by
\bn\H_{2\,RSTU}\hskip-1.3em&&=\tr\{[\s m\s(S+\xi\s Q)-iR\s]\,[\s m\s(S+\xi\s Q)+iR\s]\s\no\\
&&\hskip1.9em\times\s[\s m\s(U+\xi\s Q)-iT\s]\,[\s m\s(U+\xi\s Q)+iT\s]\s\} \no\\
&&=\tr\s\{\s:\s[\s R^2+m^2\s(S+\xi\s Q)^2]\s:\,:\s[\s T^2+m^2\s( U+\xi Q)^2]:\}\;.\en
For coherent states, we choose
\bn |p,q\>\equiv \exp[-i\tr(qP)/\hbar]\,\exp[i\tr( p\s\s Q)/\hbar]\,|0,0,0;\xi\>\;,\en
and it follows, for the final, total Hamiltonian, that
\bn H(p,q)\hskip-1.3em&&=\ \no\\
&&=\half\s[\s\tr(p^2)+m^2(1+2\s\xi^2)\s\tr(q^2)\s]+v\s m^4\s\xi^4\s\tr(q^4)\no\\
&&\equiv\half\s[\s\tr(p^2)+m^2_0\s\tr(q^2)\s]+\l\s\tr(q^4)\;,\label{yes9}\en
as desired! This expression is valid for {\it all} $N\le\infty$.
\subsubsection*{Critical commentary}
It is worthwhile to examine a natural candidate that was {\it not} chosen for the last model, and to see why we did not choose that natural ``solution''. We start be asking why was it necessary to employ {\it three} sets of canonical pairs when it seems that {\it two} canonical pairs should be enough. Before focussing on just two operators, however, let us first assume
that a new unit vector, $|0,0;\xi\>$---hoping for two pairs and not three---is annihilated by three operators, namely
\bn &&\hskip.03em A\s\s|0,0;\xi\>\equiv(1/\sqrt{2m\hbar})\s [\s m(Q+\xi(S+U))+i\s P\s]\s|0,0;\xi\>=0\;,\no\\
&& B\s\s|0,0;\xi\>\equiv(1/\sqrt{2m\hbar})\s [\s m(S+\xi Q)+i\s R\s]\s|0,0;\xi\>=0\;,\no\\
&& C\s\s|0,0;\xi\>\equiv (1/\sqrt{2m\hbar})\s[\s m(U+\xi Q)+i\s T\s]\s|0,0;\xi\>=0\;. \en
The matrix annihilation operators $A$, $B$, and $C$ are associated with matrix creation operators, $A^\dag$, $B^\dag$, and $C^\dag$, and
\bn [\s A_{ab},A^\dag_{cd}\s]=[\s B_{ab},B^\dag_{cd}\s]=[\s C_{ab},C^\dag_{cd\s}]=\half(\delta_{ad}\delta_{cb}+\delta_{da}\delta_{cb})\one\;. \en
For simplicity hereafter, we sometimes set $m=\hbar=1$. Thus
$\H_{2\,PQ}=\tr(A^\dag\s A)$, $\H_{2\,RS}=\tr (B^\dag\s B)$, and $\H_{2\,TU}=\s\tr(C^\dag C)$.
For a fourth expression, let
us introduce ${\cal F}\equiv \tr(B^\dag\s B^\dag\s B\s B)$. With coherent states given (with $\hbar$) by
\bn|p,q\>=\exp[-i\tr(qP)/\hbar]\s \exp[i\tr(p\s\s Q)/\hbar]\s|0,0;\xi\>\;,\en
it follows (restoring $m$ part way through) that
\bn W(p,q)\hskip-1.3em&&\equiv \ \no\\
% &&\<\xi|\half\tr((A^\dag-ip+mq) (A_+ip+mq))+\tr((B^\dag+m\xi q) (B+m\xi q))+4v\s\tr(B^\dag % % (B^\dag+m\xi q)) B B)\s||0,0;\xi\>\no\\
&&=\half\s[\s\tr(p^2)+m^2(1+\xi^2)\s\tr(q^2)\s]+v\s m^4\s\xi^4\s\tr(q^4)\no\\
&&\equiv\half\s[\s\tr(p^2)+m^2_0\s\tr(q^2)\s]+\l\s\tr(q^4)\s\;,\en
as desired---or so it would seem.
The evaluation carried out so far is insufficient to determine whether ${\cal F}\equiv
\tr(B^\dag B^\dag B B)$ is a genuine Hermitian {\it operator} or, instead, merely a {\it form} requiring restrictions
on both kets {\it and} bras. For ${\cal F}$ to be an acceptable operator, it is necessary that
\bn \<\phi|\s {\cal F}^\dag\s {\cal F}\s|\phi\><\infty \en
for a dense set of vectors $|\phi\>$. If we are able to bring the expression ${\cal F}^\dag {\cal F}$ into normal order, without potential divergences, then we can, for example, use coherent states to establish that $
{\cal F}$ is an acceptable operator. After a lengthy calculation, it follows that
\bn {\cal F}^\dag {\cal F}\hskip-1.3em&& \equiv \tr(B^\dag B^\dag B B)\, \tr(B^\dag B^\dag B B)\no\\
&& = \;: \tr(B^\dag B^\dag B B) \,\tr(B^\dag B^\dag B B) : \no\\
&&\hskip2em + : \tr(B^\dag B^\dag B B B B^\dag ) :
+ : \tr(B^\dag B^\dag B B B^\dag B) :\no\\
&&\hskip2em +\,:\tr(B^\dag B^\dag B B^\dag B B) : +\s \tr(B^\dag B^\dag B^\dag B B B) \no\\
&&\hskip2em +\,\s(\fiveoverfour+\half\s N\s)\, \tr(B^\dag B^\dag B B) +\, \quarter\,\tr(B^\dag B^\dag) \tr(B B) \;. \label{bad}\en
This calculation was performed for $N\times N$ matrices, and the last line of (\ref{bad}) has a coefficient $N$ which means as $N\ra\infty$ this term would diverge and cause suitable states with two or more excitations to have an infinite expectation value for the supposed operator ${\cal F}=\tr(B^\dag B^\dag B B)$. Hence, ${\cal F}$ is a {\it form} rather than an acceptable operator, and this fact rules out this proposed quantum solution of the model represented by (\ref{mm}). A similar computation shows that $\tr(B^\dag B B^\dag B)$ is also a form for similar reasons, and so is $\tr(B^\dag C^\dag C B)$.
On the other hand, we now show that $\H\equiv \tr(B^\dag B C^\dag C)\,[\s=\tr(C^\dag C B^\dag B)\s] $ is a genuine operator for all $ N\le\infty\s$!
To do so, let us bring the expression
$\H^\dag \H$ into normal ordered form. It follows that
\bn \H^\dag\H\hskip-1.3em&&\equiv \tr( C^\dag C B^\dag B)\,\tr( B^\dag B C^\dag C ) \no\\
&&=\;:\tr(C^\dag C B^\dag B)\,\tr( B^\dag B C^\dag C): \no\\
&&\hskip2em +\,\half\s:\tr(C^\dag C B^\dag B C^\dag C): +\,\half\s:\tr(C^\dag C B^\dag C C^\dag B):\no\\
&&\hskip2em +\,\half\s:\s\tr(C^\dag B B^\dag C B^\dag B):+\,\half\s :\tr( C^\dag C B^\dag B B^\dag B):\no\\
&&\hskip2em+\,\quarter\s:\tr(C^\dag C B^\dag B):+\quarter\,:\tr(C^\dag B B^\dag C):\no\\
&&\hskip2em+\,\quarter\s:\tr(C^\dag B)\,\tr(B^\dag C):+\,\quarter\s:\tr(C^\dag C )\,\tr(B^\dag B):\;,\en
{\it with no factor of $N$}. Thus, for all $N\le\infty$, the expression for $\H$ is an acceptable operator and not merely a form.
And that is why we have chosen to use that operator to build our interaction in (\ref{yes9}).
%\subsubsection*{Extension to unitary invariance}
It should be observed that all the models which were successfully treated have total Hamiltonians that are well defined and do not exhibit infinities even though they deal with nonlinear models and an infinite number of degrees of freedom when $N=\infty$. The spectrum of these total Hamiltonians can be computed, and that exercise has already been partially carried out for the vector model in the second article listed in \cite{RS}.
\section*{Acknowledgements}
It is a pleasure to thank C.B.~Thorn and E.~Onofri for helpful discussions and/or correspondence regarding traditional $1/N$-expansion techniques. In addition, thanks are extended to J.~Ben Geloun for earlier discussions that centered on alternative matrix models \cite{JBG}, which have led the author to examine the matrix models addressed in the present paper.
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\medskip of independent-value matrix models'', arXiv:1306.4403.
\end{thebibliography}
\end{document}
T4 = Tr(a^d a^d a a)
that led to T4^d T4 given by
T4^d T4 = Tr(a^d a^d a a) Tr(a^d a^d a a)
= : Tr(a^d a^d a a) Tr(a^d a^d a a) : + : Tr(a^d a^d a a^d a a) : + : Tr(a^d a^d a a a^d a) :
+ : Tr(a^d a^d a a a a^d) : + Tr(a^d a^d a^d a a a) + Tr(a^d a^d) Tr(a a) + Tr(a^d a^d a a) N
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E. Brezin, C. Itzykson, G. Parisi, J.B. Zuber (Saclay). Nov 1977. 20 pp.
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Loop Equations and Nonperturbative Effects in Two-dimensional Quantum
Gravity
F. David (Saclay, SPhT). Mar 19, 1990. 14 pp.
Published in Mod.Phys.Lett. A5 (1990) 1019-1030
SACLAY-SPHT-90-093, SACLAY-SPHT-90-043
DOI: 10.1142/S0217732390001141
Detailed record - Cited by 182 records
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A Nonperturbative Treatment of Two-dimensional Quantum Gravity
David J. Gross, Alexander A. Migdal (Princeton U.). Dec 1989. 48 pp.
Published in Nucl.Phys. B340 (1990) 333-365
PUPT-1159
DOI: 10.1016/0550-3213(90)90450-R
Detailed record - Cited by 395 records
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Nonperturbative Solution of the Ising Model on a Random Surface
David J. Gross, Alexander A. Migdal (Princeton U.). Dec 1989. 12 pp.
Published in Phys.Rev.Lett. 64 (1990) 717
PUPT-1156
DOI: 10.1103/PhysRevLett.64.717
Detailed record - Cited by 262 records
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Nonperturbative Two-Dimensional Quantum Gravity
David J. Gross, Alexander A. Migdal (Princeton U.). Oct 1989. 13 pp.
Published in Phys.Rev.Lett. 64 (1990) 127
PUPT-1148
DOI: 10.1103/PhysRevLett.64.127
Detailed record - Cited by 878 records
MAYBE The standard $1/N$ expansion stories are related with ${\bf U}(N)$-invariant Hamiltonians for large $N$ \cite{1overN} that focus of special subspaces of the traditional Fock-Hilbert space, and these subspaces depend explicitly on the parameter $N$. As $N\ra\infty$, these subspaces detach themselves from the traditional Hilbert space within which our analysis takes place. This disjoint separation also means that the operators allowed in the detached Hilbert space are generally {\it not} allowed in the tradu=itional Hilbert space, and vice versa. Consequently, the standard large $N$ programs (which, e.g., are inspired by planar diagrams in QCD) is disconnected from the large $N$ problems discussed in the present paper. Specifically, our Hamiltonians are endowed with ${\bf O}(N)$ symmetry when $N\le\infty$. We seek to quantize such models within a natural Hilbert space with the aid of suitable coherent states, and we seek a quantization procedure in which, as much as possible, the parameter $N$ is used {\it only} to specify the number of vector or matrix components, just as the case in the classical theory. For our models, it is always important to keep a suitable connection between the classical and quantum theories. MAYBE
Models of the kind introduced above have many components and, for large $N$, very large symmetry groups. Among other methods to study such models, are the familiar perturbation expansions in $1/N$ based on the change $\l\ra \l/N$ as dictated by a traditional perturbation analysis; e.g., see \cite{1overN}. Such procedures arise from the natural assumption that conventional quantization techniques are appropriate, but
we will be led in other directions and conclude that standard $1/N$ expansions are not relevant for these problems. % probably
\bn \H^\dag\H\hskip-1.3em&&\equiv \tr( C^\dag C B^\dag B)\,\tr( B^\dag B C^\dag C ) \no\\
&&=\;:\tr(C^\dag C B^\dag B)\,\tr( B^\dag B C^\dag C) :+\,:\tr( C^\dag C B^\dag B B^\dag B):\no\\
&&\hskip2em +\,:\tr(C^\dag C B^\dag B C^\dag C): +\,:\tr(C^\dag C )\,\tr(B^\dag B):\;,\en
\bn \tr({\cal F}^\dag {\cal F})\hskip-1.3em&& \equiv \tr(B^\dag B^\dag B B)\, \tr(B^\dag B^\dag B B)\no\\
&& = \;: \tr(B^\dag B^\dag B B) \,\tr(B^\dag B^\dag B B) : +
: \tr(B^\dag B^\dag B B^\dag B B) :\no\\ &&\hskip2em + : \tr(B^\dag B^\dag B B B^\dag B) :
+ : \tr(B^\dag B^\dag B B B B^\dag) :\no\\
&&\hskip2em +\s \tr(B^\dag B^\dag B^\dag B B B) + \tr(B^\dag B^\dag) \tr(B B)\no\\
&&\hskip2em +\s \tr(B^\dag B^\dag B B)\, N \;. \label{bad}\en
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