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Contribution to the Proceedings of the 21st Max Born Symposium, Wroc\l aw,
Poland June 26--28, 2006.
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\title{ On the Spectral Zeta Function for the Non-commutative
Harmonic Oscillator }
\author{ Takashi Ichinose
\thanks{$\,\,$ Supported in part by JSPS Grant-in-Aid
for Scientific Research (B) No. 16340038} \\
Department of Mathematics, Faculty of Science,\\
Kanazawa University, Kanazawa, 920-1192, Japan \\
e-mail: ichinose@kenroku.kanazawa-u.ac.jp \\[2ex]
Masato Wakayama
\thanks{$\,\,$ Supported in part by JSPS Grant-in-Aid
for Scientific Research (B) No. 15340012}
\\ Faculty of Mathematics, Kyushu University,\\
Hakozaki, Fukuoka, 812-8581, Japan \\
e-mail: wakayama@math.kyushu-u.ac.jp }
\begin{document}
\maketitle
\begin{abstract}
The spectral zeta function for the so-called non-commutative
harmonic oscillator
is able to be meromorphically extended to the whole complex
plane, having only one simple pole at the same point $s=1$ where
Riemann's zeta function $\zeta(s)$ has,
and possesses a trivial zero at each non-positive even integer.
The essential part of its proof is sketched. A new result is also given
on the lower and upper bounds of the eigenvalues of the non-commutative
harmonic oscillator.
\end{abstract}
\noindent
{\bf Keywords:} spectral zeta functions; Riemann's zeta function;
harmonic oscillator; non-commutative harmonic oscillators; Weyl's law;
Bernoulli's number.
\section{Introduction and Results}
The aim of this note is to give a survey of our recent results in
\cite{paperIW1} (cf. \cite{paperIW2})
on the spectral zeta function for ---
the zeta function associated with the spectrum of ---
the non-commutative harmonic oscillator $Q := Q(x, D_x)$,
improving a previous result there
on the lower and upper bounds of the eigenvalues of $Q$, and also
to mention some recent results.
This $Q$ was introduced by A.~Parmeggiani and the second author
(\cite{paperPW1, paperPW2}) as a differential operator
\begin{eqnarray}
Q(x, D_x) &\!:=\!&
A\Big(-\frac{\partial_x^2}2+\frac{x^2}2\Big) +
J\Big( x\partial_x +\frac12\Big) \nonumber\\
&\!=\!& \left(\begin{array}{cc}
\alpha\Big(-\frac{\partial_x^2}2+\frac{x^2}2\Big)
& -\Big( x\partial_x +\frac12\Big)\\
x\partial_x +\frac12
&\beta\Big(-\frac{\partial_x^2}2+\frac{x^2}2\Big)
\end{array} \right)\,,
\qquad x\in {\bf R}\,, \label{def-of-NCHO}
\end{eqnarray}
with $D_x := -i\partial_x$ and $\partial_x:=\frac{d}{dx}$,
acting on the Hilbert space
$L^2({\bf R},{\bf C}^2) = L^2({\bf R})\otimes {\bf C}^2$,
which is dependent on two positive parameters $\alpha$ and $\beta$
with $\alpha\beta>1$ through
\begin{equation}
A=\left(\begin{array}{cc} \alpha &0\\0&\beta \end{array}\right),
\qquad
J=\left(\begin{array}{cc} 0 &-1\\1&0 \end{array}\right).
\end{equation}
It may not be an operator being a Hamiltonian coming from
some existing system in quantum mechanics in the present stage.
In the papers \cite{paperPW1} and \cite{paperPW2}, among others,
the following basic facts for $Q=Q(x,D_x)$ were already observed:
$Q$ is a selfadjoint operator in $L^2({\bf R},{\bf C}^2)$, and
essentially selfadjoint on ${\cal S}({\bf R}, {\bf C}^2)$.
It has only discrete spectrum consisting of the eigenvalues
$\{\lambda_n\}_{n=1}^{\infty}$ such that
$ 0 < \lambda_1 \leq \lambda_2 \leq \lambda_3
\leq \cdots \quad \uparrow \infty,
$
with uniformly bounded multiplicity. As a result, the first eigenvalue
$\lambda_1$ is positive, so that $Q$ has a bounded inverse
$Q^{-1}$. However,
the value of any eigenvalue $\lambda_n$ is hardly computed
explicitly (A numerical study of the spectrum was carried out
in \cite{paperNNW}).
The two relations of non-commutativity, i.e. the one for matrices
$[A,\,J]=(\beta-\alpha)\left(\begin{array}{cc} 0 &1\\1&0
\end{array}\right)\not=0$ when $\alpha\not=\beta$ and
the canonical commutation relation $[\partial_x,\,x]=1$,
make highly non-trivial the spectrum
of the non-commutative harmonic oscillator, which thus in general is
a non-trivial pair of the harmonic oscillators.
We define the spectral zeta function associated with $Q$ as
\begin{equation}
\zeta_Q(s) := \sum_{n=1}^{\infty} \frac1{{\lambda_n}^s}\,.
\end{equation}
Then it was shown in \cite{paperIW2}
that the series on the right-hand side converges
absolutely in the half plane $\hbox{\rm Re}\, s >1$.
$\zeta_Q(s)$ has dependence on the two parameters
$\alpha$ and $\beta$ (essentially, on
the ratio $\alpha/\beta$), since each eigenvalue $\lambda_n$ does.
Notice here that this spectral zeta function is giving
a deformation of the Riemann zeta function $\zeta(s)$,
which in turn, we may think,
is essentially associated with the spectrum of the harmonic
oscillator $-\frac{\partial_x^2}2+\frac{x^2}2$.
Indeed, when $\alpha=\beta$, the
non-commutative harmonic oscillator becomes (equivalent with)
a couple of two usual
harmonic oscillators whose eigenvalues are given by
$(n+\frac12) \sqrt{\alpha^2-1}\; (n=0,1,2,\ldots)$
with multiplicity two (see \cite{paperPW2}).
Although no eigenvalues of $Q$ can be calculated explicitly,
we would expect to find with this zeta function $\zeta_Q(s)$ another way
to investigate the spectral properties of $Q$ and,
for instance, some knowledge of its $n$-th eigenvalue to be gained
through the analytic property of the spectral zeta function.
We have shown in \cite{paperIW1} the following theorem and
corollary on $\zeta_Q(s)$.
\begin{theorem}{Theorem} \label{thMM1}
There exist constants $C_{Q,j} \;(j=1,2,\ldots)$ such that
$\zeta_Q(s)$ is, for every positive integer $n$, represented as
\begin{equation}
\zeta_Q(s) = \frac{1}{\Gamma(s)}\Bigg[
\frac{\alpha+\beta}{\sqrt{\alpha\beta(\alpha\beta -1)}}\,
\frac{1}{s-1}
+ \sum_{j=1}^n \frac{C_{Q,j}}{s+2j-1}
+ H_{Q,n}(s)\Bigg],
\end{equation}
where $\Gamma (s)$ is the gamma function and
$H_{Q,n}(s)$ is a holomorphic function in $\hbox{\rm Re}\,s>-2n$.
Consequently, the spectral zeta function $\zeta_Q(s)$ is meromorphic
in the whole complex plain {\bf C} with a simple pole at $s=1$
and has zeros for $s = 0, -2, -4, \dots$ (non-positive even integers).
\end{theorem}
Here, note that $\frac1{\Gamma (s)}$ is holomorphic in the whole complex
plane ${\bf C}$ and has zeros at $s= 0, -1, -2, \cdots.$ In this sense,
one says that the non-positive even integers are the trivial
zeros of $\zeta_Q(s)$.
\begin{corollary}{Corollary} \label{coRR1}
\hbox{\rm (Weyl's law)}
\begin{equation}
\sum_{\lambda_n 1,
$$
so that
\begin{equation} \label{zeta-by-Mellin}
\zeta_Q(s) = \hbox{\rm Tr}\,Q^{-s}
= \frac1{\Gamma (s)} \int_0^{\infty} dt\, t^{s-1}
\hbox{\rm Tr}\, K(t).
\end{equation}
\medskip
As mentioned before, we cannot obtain any of the eigenvalues $\lambda_n$
explicitly, but
estimate their lower and upper bounds. The following theorem
gives an improvement of the result in \cite{paperIW1}, where we
treated only the case for the first eigenvalue $\lambda_1$
by use of a Lieb--Thirring inequality \cite{paperLT}.
\begin{theorem}{Theorem} \label{thMM2}
Let $\lambda_{2j-1},\,\, \lambda_{2j}, \,\, j= 1, 2, \dots,$
be the $(2j-1)$-th and $2j$-th eigenvalues of $Q$.
Then
\begin{equation} \label{bound-of-first-e.v.}
\big(j- \frac12\big)
\min\{\alpha,\beta\}\sqrt{1-\frac{1}{\alpha\beta}}
\leq \lambda_{2j-1} \leq \lambda_{2j}
\leq \big(j-\frac12\big)
\max\{\alpha,\beta\}\sqrt{1-\frac{1}{\alpha\beta}}.
\end{equation}
\end{theorem}
Note that the two closed intervals made
of the bounds (\ref{bound-of-first-e.v.})
for two pairs of the eigenvalues
$\{\lambda_{2j-1},\, \lambda_{2j}\}$
and $\{\lambda_{2k-1},\, \lambda_{2k}\}$ with $j1$,
of which the series on the right
converges absolutely in $\hbox{\rm Re}\, s >1$.
This $\zeta (s)$ has only one simple pole at $s=1$, and
has also zeros at $s= -2, -4, \dots$ (\hbox{negative even integers}),
and satisfies the functional equation
\begin{equation}\label{FE}
\pi^{-\frac{s}{2}}\Gamma\Big(\frac{s}{2}\Big)\zeta(s)
= \pi^{-\frac{1-s}{2}}\Gamma\Big(\frac{1-s}{2}\Big)\zeta(1-s),
\end{equation}
and the Euler product expression
\begin{equation}\label{EP}
\zeta(s) = \prod_{p: \hbox{\rm prime}}\frac1{1-p^{-s}}
\qquad (\hbox{\rm Re}\, s >1).
\end{equation}
Note that $\alpha =\beta$ if and only if $J$ and $A$ commute, and hence
if and only if $Q$ (or $Q^{-1}$) and $A$ commute
if and only if $Q$ (or $Q^{-1}$) and $J$ commute.
In this case, $Q$ is unitarily equivalent to
a couple of the ordinary harmonic oscillators
$\sqrt{\alpha^2-1}Q_0$, where
$Q_0 \equiv \Big(-\frac{\partial_x^2}2+\frac{x^2}2\Big)\otimes I$
with $I$ being the $2\times 2$ identity matrix.
In particular, when
$\alpha =\beta = \sqrt{2}$, $Q$ and $Q_0$
are unitarily equivalent to each other, whence it follows that
\begin{equation} \label{expression-of-zetaQ_0}
\zeta_{Q}(s)=\zeta_{{Q_0}}(s) = 2\sum_{n=0}^{\infty} \frac1{(n+\frac{1}{2})^s}
= 2(2^s-1)\,\zeta (s).
\end{equation}
Note here also that
when $\alpha =\beta$, the inequalities (\ref{bound-of-first-e.v.})
become equalities; it implies that
$\lambda_{2j-1} = \lambda_{2j}
= \big(j-\frac12 \big) \sqrt{{\alpha}^2-1}$ in this case.
But it is hard to expect $\zeta_Q(s)$ to have some functional
equation and/or Euler product expression.
There are many spectral zeta functions in geometry,
dynamical systems,.., but
not many seem to be known to be meromorphically extended to the
whole complex plain ${\bf C}$. Among those studies,
D.~Robert \cite{paperR} considered, for a general class of elliptic
systems of pseudo-differential operators $P$, their power $P^{-s}$
through the Dunford integral (only) to do a meromorphic continuation
of $\hbox{\rm Tr}\, P^{-s}$ to the whole complex plane ${\bf C}$.
\medskip
In Section 2 we give an outline of the proof of Theorem~\ref{thMM1} and
a proof of Theorem~\ref{thMM2},
and in Section 3 some remarks and open questions.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5
\section{Proofs of Theorem~\ref{thMM1} and Theorem~\ref{thMM2}}
{\it Outline of Proof of Theorem 1}.
\medskip
The actual proof is a little lengthy because of non-commutativity.
We have denoted by $K(t) = K(t,x,y)$
$:= e^{-tQ}(x,y)$ the heat kernel for $Q$.
From the expression (\ref{zeta-by-Mellin}) of $\zeta_Q(s)$ through
the Mellin transform, we write
\begin{equation}
\zeta_Q(s) = \frac1{\Gamma (s)}
\Big[\int_0^1 +\int_1^{\infty}\Big]dt\, t^{s-1}
\hbox{\rm Tr}\, K(t)
=: Z_0(s) + Z_{\infty}(s).
\end{equation}
Then it is easy to see the second term
$Z_{\infty}(s)$ in the last member
is holomorphic in ${\bf C}$.
So we must study $Z_0(s)$. To do so, we need an asymptotic
expansion for small $t>0$ like
$$
\hbox{\rm Tr}\, K(t) \sim c_{-1}t^{-1}
+ c_0 + c_1t + c_2 t^2 + c_3 t^3 + \cdots,
\quad 0< t << 1.
$$
\medskip
Define the operator $K_1(t)$ for
$f \in {\cal S}({\bf R}, {\bf C}^2)
= {\cal S}({\bf R}) \otimes {\bf C}^2$,
by the pseudo-differential operator
\begin{eqnarray}
K_1(t)f(x) &\!=\!& \int K_1(t,x,y) f(y) dy \nonumber\\
&\!=\!& \frac1{2\pi}\int\int e^{i(x-y)\xi}
\exp\Big[-t\Big(A\frac{\xi^2 +y^2}{2} + Jyi\xi\Big)\Big]f(y) dy d\xi.
\end{eqnarray}
Put
\begin{equation}
R_2(t) := K(t) - K_1(t),
\end{equation}
so that
$
K(t) = K_1(t) + R_2(t).
$
\begin{proposition}{Proposition}\label{prPP1}
\begin{eqnarray*}
Z_0(s)&\!=\!& \frac1{\Gamma(s)}\int_0^1 dt\, t^{s-1}\hbox{\rm Tr}\,K(t)\\
&\!=\!& \frac1{\Gamma(s)}\int_0^1dt\, t^{s-1}\hbox{\rm Tr}\, K_1(t)
+ \frac1{\Gamma(s)}\int_0^1 dt\,t^{s-1}\hbox{\rm Tr}\, R_2(t)\\
&\!=\!& \frac{\alpha +\beta}{\sqrt{\alpha\beta(\alpha\beta -1)}}
\frac1{\Gamma (s)}\frac1{s-1}
+\frac1{\Gamma(s)}\int_0^1 dt\,t^{s-1}\hbox{\rm Tr}\, R_2(t).
\end{eqnarray*}
\end{proposition}
\medskip
It is hence sufficient to treat only the second term
$\frac1{\Gamma(s)}\int_0^1 dt\,t^{s-1}\hbox{\rm Tr}\, R_2(t)$
in the last member of the equation in Proposition~\ref{prPP1}.
Since by (\ref{heat-eq.-of-Q})
$$
0 = [\partial_t + Q]K(t,x,y)
= [\partial_t + Q]K_1(t,x,y) + [\partial_t + Q]R_2 (t,x,y),
$$
we have
\begin{equation} \label{eq-of-R_2}
[\partial_t + Q]R_2(t,x,y) = -[\partial_t + Q]K_1(t,x,y)
=: F(t,x,y).
\end{equation}
\noindent
By Duhamel's principle, we can solve (\ref{eq-of-R_2}) to get
\begin{eqnarray*}
R_2(t) &\!=\!& \int_0^{t} e^{-(t-u)Q} F(u) du\\
&\!=\!& \int_0^{t} du \int K_1(t-u,x,z)F(u,z,y)dz
+ \int_0^{t} du \int R_2(t-u,x,z)F(u,z,y)dz\\
&\!=:\!& K_2(t) + R_3(t),
\end{eqnarray*}
so that
$
K(t) = K_1(t) + K_2(t) + R_3(t).
$
\medskip
Repeating this iteration yields
\begin{eqnarray}
&& K(t) = K_1(t) + K_2(t)+ \cdots + K_n(t) + R_{n+1}(t),\\
&& K_m(t) =
\int_0^tdu_1 \int_0^{t-u_1} du_2\cdots
\int_0^{t-u_1-u_2 \cdots - u_{m-2}} du_{m-1} \nonumber\\
&&\qquad\quad\times K_1(t-u_1 - \cdots - u_{m-1})
F(u_{m-1}) \cdots F(u_2)F(u_1), \quad 2\leq m \leq n,\\
&& R_{n+1}(t) =
\int_0^tdu_1 \int_0^{t-u_1}du_2 \cdots
\int_0^{t-u_1-u_2 \cdots - u_{m-1}} \nonumber\\
&&\qquad\quad \times K(t-u_1 - \cdots - u_{m})
F(u_m)F(u_{m-1}) \cdots F(u_2)F(u_1)du_m.
\end{eqnarray}
So we have to study the traces of $K_m(t)$ and $R_{n+1}(t)$
for small $t >0$.
\begin{lemma}{Lemma}\label{leMM1}
For every small $\varepsilon >0$ one has
\begin{eqnarray*}
&& |\hbox{\rm Tr}\,R_2(t)|
\leq C(\varepsilon) t^{-\varepsilon},\\
&& |\hbox{\rm Tr}\,R_{n+1}(t)|
\leq C^n\frac{\Gamma(1/2)^n}{\Gamma(1+n/2)}t^{n/2}, \quad n \geq 2,
\end{eqnarray*}
where $ C(\varepsilon)$ is a positive constant independent of
$t$ but dependent on $\varepsilon$, and $C$
a positive constant independent of $t$ and $n$.
\end{lemma}
This lemma implies that the Mellin transform of $\hbox{\rm Tr}\,R_2(t)$
is holomorphic in $\hbox{\rm Re}\,s > \varepsilon$ and hence
in $\hbox{\rm Re}\,s > 0$ because $\varepsilon$ is arbitrary,
and that of $\hbox{\rm Tr}\,R_{n+1}(t)$ $(n\geq 2)$ in
$\hbox{\rm Re}\,s > -\frac{n}{2}$.
A little patient, lengthy calculation yields the following key
asymptotics of $K_m(t)$ as $t \downarrow 0$.
\begin{lemma}{Lemma}\label{leMM2}
For $m= 2,3, \dots$, one has for $t \downarrow 0$,
$$
\hbox{\rm Tr}\,K_m(t) \sim \sum_{j=0}^{\infty} c_{m,j}t^{j}\,,
$$
with $c_{m,j} =0$ for $0\leq j < m-2$ and $j=2\ell$ being positive
even integers.
\end{lemma}
Hence we have by Proposition~\ref{prPP1}, Lemma~\ref{leMM1}
and Lemma~\ref{leMM2}
%%%%%%
\begin{eqnarray*}
\hbox{\rm Tr}K_1(t) &\!=\!&
\frac{\alpha+\beta}{\sqrt{\alpha\beta(\alpha\beta-1)}}t^{-1}\,,\\
\hbox{\rm Tr}K_2(t) &\!\sim\!&
c_{2,1}t+c_{2,3}t^3+c_{2,5}t^5+c_{2,7}t^7+c_{2,9}t^9+\cdots\,,\\
\hbox{\rm Tr}K_3(t) &\!\sim\!&
c_{3,1}t+c_{3,3}t^3+c_{3,5}t^5+c_{3,7}t^7+c_{3,9}t^9+\cdots\,,\\
\hbox{\rm Tr}K_4(t) &\!\sim\!&
\qquad\quad c_{4,3}t^3+c_{4,5}t^5+c_{4,7}t^7+c_{4,9}t^9
+\cdots\,,\\
\hbox{\rm Tr}K_5(t) &\!\sim\!&
\qquad\quad c_{5,3}t^3+c_{5,5}t^5+c_{5,7}t^7+c_{5,9}t^9
+\cdots\,,\\
\hbox{\rm Tr}K_6(t) &\!\sim\!&
\qquad\qquad\quad\quad\,\,\,
c_{6,5}t^5+c_{6,7}t^7+c_{6,9}t^9+\cdots\,,\\
\hbox{\rm Tr}K_7(t) &\!\sim\!& \qquad\quad\quad\qquad\;
\qquad \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdot\cdot\,,\\
\cdots &\!\sim\!& \qquad\qquad\quad\quad\qquad\;\;
\qquad \cdots\cdots\cdots\cdots\cdots\cdots\,, \\
\hbox{\rm Tr}R_{n+1}(t)&\!=\!& \,\, O(t^{n/2})\,.
\end{eqnarray*}
%%%%%
Thus with the Mellin transform we can conclude
\begin{eqnarray*}
\zeta_Q(s) &\!=\!&
\frac1{\Gamma (s)}
\frac{\alpha+\beta}{\sqrt{\alpha\beta(\alpha\beta-1)}}\frac1{s-1}
+(c_{2,1}+c_{3,1})\frac1{\Gamma (s)}\frac1{s+1}\\
&&\, + (c_{2,3}+c_{3,3}+c_{4,3}+c_{5,3})\frac1{\Gamma (s)}\frac1{s+3}\\
&&\,\,
+ (c_{2,5}+c_{3,5}+c_{4,5}+c_{5,5}+ c_{6,5})\frac1{\Gamma (s)}\frac1{s+5}
+ \cdots\cdots + \frac1{\Gamma (s)} H_{Q,n}(s),
\end{eqnarray*}
showing Theorem~\ref{thMM1}.
\bigskip
{\it Proof of Theorem~\ref{thMM2}}.
\medskip
We develop a method used in \cite{paperIW1}, {\it Remark 3}, p.714,
to give an alternative proof of the assertion for
$\lambda_1$, not appealing to a Lieb--Thirring inequality.
Though we cannot solve
(\ref{heat-eq.-of-Q}) explicitly, we can solve the modified equation
\begin{equation}
[\partial_t + Q'(x,D_x)]K'(t,x,y) =0\,,
\end{equation}
where
\begin{eqnarray}
Q' &\!:=\!& A^{-1/2}QA^{-1/2} = \frac12(-\partial_x^2+x^2)
+ \gamma J(x\partial_x + \frac12), \nonumber \\
&\!=\!& \frac12 (-i\partial_x +i\gamma Jx)^2
+ \frac{1-\gamma^2}{2}x^2,
\qquad\qquad\qquad\qquad
\gamma := \frac1{\sqrt{\alpha\beta}}\,.
\end{eqnarray}
In fact, we can show
\begin{eqnarray}
K'(t,x,y) &\!=\!& e^{-tQ'}=(1-\gamma^2)^{1/4}\exp[\frac12 \gamma(x^2-y^2)J]
\nonumber\\
&&\qquad\qquad \times
p((1-\gamma^2)^{1/2}t, (1-\gamma^2)^{1/2}x, (1-\gamma^2)^{1/2}y),
\end{eqnarray}
where
$p(t,x,y) := \exp[-\frac12(-\partial_x^2+x^2)](x,y)$
is the heat kernel of the ordinary harmonic oscillator.
Indeed, it was shown in \cite{paperPW2}
that $Q'$ is unitarily equivalent to
$\frac{\gamma}{2}
[-\partial_x^2 +(\alpha\beta -1)x^2] \otimes I$, and also
\begin{equation}
\zeta_{Q'}(s) = \hbox{\rm Tr}\, {Q'}^{-s}
=(1-\gamma^2)^{-s/2}\zeta_{Q_0}(s)
= (1-\gamma^2)^{-s/2}2(2^s-1)\zeta(s),
\end{equation}
with $\zeta_{Q_0}(s)$ in (\ref{expression-of-zetaQ_0}).
Hence we know that every eigenvalue $\lambda'_{n}$
of $Q'$ has multiplicity two,
so that its $(2j-1)$-th and $2j$-th eigenvalues
$\lambda'_{2j-1}$ and $\lambda'_{2j}$
coincide and are equal to
\begin{equation}
\lambda'_{2j-1}=\lambda'_{2j}
= \big(j-\frac12\big)\sqrt{1-1/(\alpha\beta)}.
\end{equation}
Note that $Q=A^{1/2}Q'A^{1/2}$.
Using this fact and the min-max principle (e.g. \cite{bookRS}),
we have with $n:= 2j-1$ or $n:=2j$, $j=1, 2, \dots$,
\begin{eqnarray*}
\lambda_n
&\!=\!& \sup_{u_1, \cdots, u_{n-1}}\,\,
\inf_{u\not= 0;\,
u \perp [u_1, \cdots, u_{n-1}]}
\frac{(u,Qu)}{\|u\|^2}\\
&\!=\!& \sup_{u_1, \cdots, u_{n-1}}\,\,
\inf_{u\not= 0;\,
u \perp [u_1, \cdots, u_{n-1}]}
\frac{(A^{1/2}u,Q'A^{1/2}u)}{\|u\|^2} \\
&\!=\!& \sup_{u_1, \cdots, u_{n-1}}\,\,
\inf_{u\not= 0;\,
u \perp [A^{-1/2}u_1, \cdots, A^{-1/2}u_{n-1}]}
\frac{(A^{1/2}u,Q'A^{1/2}u)}{\|u\|^2}\\
&\!=\!& \sup_{u_1, \cdots, u_{n-1}}\,\,
\inf_{u\not= 0;\,
A^{1/2}u \perp [u_1, \cdots, u_{n-1}]}
\frac{(A^{1/2}u,Q'A^{1/2}u)}{\|u\|^2}\\
&\!=\!& \sup_{u_1, \cdots, u_{n-1}}\,\,
\inf_{v \not= 0;\,
v \perp [u_1, \cdots, u_{n-1}]}
\frac{(v,Q'v)}{\|A^{-1/2}v\|^2} \qquad (v :=A^{1/2}u )\\
&\!=\!& \sup_{u_1, \cdots, u_{n-1}}\,\,
\inf_{v \not= 0;\,
v \perp [u_1, \cdots, u_{n-1}]}
\frac{(v,Q'v)}{\|v\|^2}\frac{\|v\|^2}{\|A^{-1/2}v\|^2}.
\end{eqnarray*}
Here we have denoted by $[u_1, \dots, u_{n-1}]$
the subspace spanned by the vectors $u_1, \dots, u_{n-1}$
in the domain of $Q$; note that $Q$ and $Q'$ have the same domain.
It is crucial above that the third equality holds,
because if the vectors $\{u_1, \dots, u_{n-1}\}$ are linearly independent,
so are the vectors $\{A^{-1/2}u_1, \dots, A^{-1/2}u_{n-1}\}$,
since $A$ is one-to-one and has a bounded inverse $A^{-1}$, and
because a vector $w$ belongs to the domain of $Q$ or $Q'$ if and only if
$A^{-1/2}w$ does. We note that with
$m:= \min\{\alpha,\beta\},\,\, M:= \max\{\alpha,\beta\}$,
$$
\frac{1}{M}\|v\|^2 \leq \|A^{-1/2}v\|^2=(v, A^{-1}v)
\leq \frac{1}{m}\|v\|^2,
\quad v \in L^2({\bf R},{\bf C}^2),
$$
whence $m \leq {\|v\|^2}/{\|A^{-1/2}v\|^2} \leq M$ for $v \not= 0$.
It follows that
\begin{eqnarray*}
\lambda_{2j-1} \leq \lambda_{2j}
&\! \leq\!& M \sup_{u_1, \cdots, u_{2j-1}}\,\,
\inf_{v \not= 0;\,
v \perp [u_1, \cdots, u_{2j-1}]}
\frac{(v,Q'v)}{\|v\|^2}
= M\lambda'_{2j}= M\lambda'_{2j-1}\,,\\
\lambda_{2j} \geq \lambda_{2j-1}
& \geq \!& m \sup_{u_1, \cdots, u_{(2j-1)-1}}\,\,
\inf_{v \not= 0;\,
v \perp [u_1, \cdots, u_{(2j-1)-1}]}
\frac{(v,Q'v)}{\|v\|^2}
= m \lambda'_{2j-1}\,.
\end{eqnarray*}
Thus we have shown the lower and upper bounds of $\lambda_n$
as in Theorem~\ref{thMM2}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some Remarks and Open Questions}
We have shown one of the basic analytic properties of
the spectral zeta function $\zeta_Q(s)$ for
the non-commutative harmonic oscillator $Q =Q(x,D_x)$
in (\ref{def-of-NCHO}).
There still remain a number of open questions.
\medskip
1. Can one take the limit $n\to \infty$ in the Theorem
\ref{thMM1} ? In other words,
does the series $\sum_{j=1}^n\frac{C_{Q,j}}{s+2j-1}$
converge when $n\to \infty$ ? If $\alpha=\beta$, then one knows that this
is true. For the precise situation, see the last discussion given
in \cite{paperIW1}, pp.735--738.
\medskip
2. In view of the Euler product expression (\ref{EP}), one recognizes
easily that
Riemann's zeta function $\zeta(s)$ has no zeros in $\hbox{\rm Re}\, s >1$.
We have proved that $\zeta_Q(s)$ has no zeros in
$\hbox{\rm Re}\,s>c$ for sufficiently large $c$, though.
Can one determine the zero-free region of $\zeta_Q(s)$
in the half plane $\hbox{\rm Re}\, s >1$
where the defining series of $\zeta_Q(s)$ converges absolutely ?
\medskip
3. $\zeta(s)$ has an infinite number of
nontrivial zeros on the line $\hbox{\rm Re}\, s= \frac12$.
Is there any such a critical line or curve for $\zeta_Q(s)$ ?
If so, what is it ? Also, does
$\zeta_Q(s)$ possess any functional equation or Euler product
expression such as $\zeta (s)$ does (though it is hardly expected) ?
See also the very final remark in \cite{paperIW1}.
\medskip
4. Compute the values of $\zeta_Q(s)$ for $s= n$ being
positive integers greater than $1$.
For Riemann's zeta function $\zeta(s)$, the
Bernouille numbers $B_{2m}$ $(m \in {\bf N})$ describe such special
values: $\zeta(2m)=\frac{(-1)^{m-1}}{2}\frac{(2\pi)^{2m}}{(2m) !} B_{2m}$.
\medskip
In \cite{paperIW2}, we have computed
$\zeta_Q(2)$ and $\zeta_Q(3)$ and discussed a connection of these values
with (singly confluent) Heun's differential equations.
Here a Heun differential equation is a Fuchsian ordinary equation
with four regular singular points in a complex domain.
Actually, $\zeta_Q(2)$ is represented as a contour integral for a
holomorphic solution in the unit disk of this equation, and
also $\zeta_Q(3)$ as one for the holomorphic solution
of the same equation but with an extra inhomogeneous term.
Based on these results, H.~Ochiai \cite{paperO3} has
obtained a beautiful explicit
expression of $\zeta_Q(2)$ in terms of the complete
elliptic integral of the first kind. In fact,
by suitable change of the variable,
he could show that the singly confluent Heun equation in question
turns to be a Gaussian hypergeometric differential equation.
Exploiting a similar idea, in \cite{paperKW1},
K.~Kimoto and the second author have obtained an analogous explicit
expression for the value $\zeta_Q(3)$ by using a Gauss hypergeometric
function and solving the aforementioned inhomogeneous equation,
and shown that the coefficients of the Taylor expansion of
the holomorphic solutions of these Heun equations
possess some remarkable congruence relations such as
the Ap\'ery numbers have. We note that
the Ap\'ery numbers are known to be used
for proving the irrationality of $\zeta(2)=\frac{\pi^2}6$
and $\zeta(3)$ (see \cite{paperVP}). Moreover, in \cite{paperKW2},
one finds that the above Heun equation (for describing the value $\zeta_Q(2)$)
is the Picard--Fuchs equation for the universal family of elliptic curves
equipped with rational 4-torsion and the parameter for a family of such
curves can be regarded as a modular function of the congruent subgroup
$\Gamma_0(4)$.
According to the result \cite{paperO1} (see also \cite{paperO2}),
the spectral problem of the non-commutative harmonic oscillator is
known to be equivalent to solving a certain Heun's equation (but
not confluent).
\medskip
5. Understand clearly the situation or intrinsic reason why our special values
are so related with the confluent Heun equations as in view of the
results in \cite{paperO1}.
Explore, moreover, any relation between the values of
$\zeta_Q(s)$ at
$s=-2m+1$ and those at $s=2m$ via Heun's equation to compensate for
a missing state of such a functional equation as
$\zeta(s)$ has in (\ref{FE}).
\medskip
6. Calculate the `characteristic polynomial' of $Q$. To be more precise,
let us recall
the determinant of an operator
$A={\rm diag}(a_1,a_2,a_3, \ldots)$ defined by
\begin{equation}
\det A:
=\exp\Big(-{\rm Res}_{s=0}\frac{\zeta(s,A)}{s^2}\Big).
\end{equation}
Here $\zeta(s,A)$ is a Dirichlet series defined by
$\zeta(s,A)=\sum_{n=1}^\infty {a_n}^{-s}\;(\hbox{\rm Re}\,s\gg1)$
and assumed to be meromorphic around
$s=0$ (see \cite{paperIllies}, \cite{paperKW0} for
this definition of the zeta
regularized determinants and products).
A similar technique we developed in \cite{paperIW1} and \cite{paperIW2}
allows us to state a conjecture that
the Hurwitz type spectral zeta function defined by
\begin{equation}
\zeta_Q(s,z):=\sum_{n=1}^\infty (\lambda_n-z)^{-s}
\quad \qquad (z\notin \{\lambda_n\}_{n=1,2,3 \ldots})
\end{equation}
can be meromorphically extended to the whole complex plane.
If this is true, the determinant
$\det (Q-zI):=\exp\big(-{\rm Res}_{s=0}\frac{\zeta_Q(s,z)}{s^2}\big)$
can be defined.
For instance, for the particular operator $Q_0$, i.e. the $Q$
with $\alpha=\beta=\sqrt2$, mentioned in (\ref{expression-of-zetaQ_0}),
one actually shows from the Lerch formula \cite{paperL} that
$$
\det(Q_0-zI)= 2\pi \Gamma\Big(\frac12-z\Big)^{-2}.
$$
\medskip
We close the present note
by asking the following fundamental but probably a
difficult question.
\medskip
7. For $Q$, is there a pair of {\it annihilation} and
{\it creation operators} ?
\bigskip\noindent
{\it Acknowledgement}.
One of the authors (T.I.) is grateful to Professor Witold Karwowski and
Professor Robert Olkiewicz, the organizers of the 21st Max Born Symposium,
for the invitation and the hospitality at Institute of Theoretical Physics,
Wroc\l aw University.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{document}
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