Introduction
We, physicists, work with our operators very differently from mathematicians. We are unencumbered by rigor and our mathematics proceeds with the motivation to forge ahead to gain insight as quickly as possible. Only later when our results are meaningful that we bother to step back and ask ourselves if our steps in reaching our conclusions can be rigorously justified.
Tidying our maths up is not an unproductive exercise. It is in fact essential and can give us more in depth insight into our problems at hand. Retracing our steps back and discovering the conditions under which our manipulations are justified can lead us to uncover the necessary physical conditions that can lead to the phenomenon that our lackadaisical maths have led us to.
Our operators are mostly formal. That is we define them by simply specifying how they act, and from their definitions we derive their properties without asking whether the operations involved in the derivations are valid or not. Take the position $\mathbf{x}$ and momentum $\mathbf{p}$ operators as examples. We define them by their action on functions $\psi(x)$ by $\mathbf{x}\psi(x)=x \psi(x)$ and $\mathbf{p}\psi(x)=-i \hbar \psi'(x)$, without being specific on the properties of $\psi(x)$. From these definitions we obtain the well-known canonical commutation relation $(\mathbf{x p}\psi(x)-\mathbf{px}\psi(x))=i\hbar \psi(x)$, which we write simply as $[\mathbf{x},\mathbf{p}]=i\hbar \mathbf{I}$. And from this relation we deduce further properties of $\mathbf{x}$ and $\mathbf{p}$.
But there are many ways that can go wrong if do not bother to be rigorous on the definition of the position and momentum operators in the relevant Hilbert space. For a quantum particle in the entire real line, the position and momentum operators are both unbounded self-adjoint operators, with spectra spanning the real line. In the half line, the position operator is self-adjoint and semi-bounded and the momentum operator is not self-adjoint. In a bounded segment, the position operator is self-adjoint and bounded, and there are infinitely many possible unbounded momentum operators corresponding on the chosen boundary condition.
Clearly we can easily go astray if we fail to recognize that the properties of the position and momentum operators depend on the Hilbert space and the required boundary conditions. One erroneous claim that has been perpetuated for a long time that arose from lack of mathematical rigor is the claim that the canonical commutation relation $[\mathbf{x},\mathbf{p}]=i\hbar \mathbf{I}$ implies that any pair of operators $\mathbf{Q}$ and $\mathbf{P}$ satisfying the same relation $[\mathbf{Q},\mathbf{P}]=i\hbar \mathbf{I}$ share the same properties as the position and momentum operators in the real line, in particular, any such pair have spectra spanning all real numbers. This erroneous claim has proved to be prejudicial to the development of the quantum theory of time.
In the paper under review, Joseph and I address the question on the status of the Bender-Dunne basis operators, $\mathbf{T}_{m,n}$, as Hilbert space operators. These operators arose in the work of Bender and Dunne in solving the Heisenberg equation of motion $i \hbar \dot{\mathbf{q}}= [\mathbf{q},\mathbf{H}]$, $i\hbar\dot{\mathbf{p}}=[\mathbf{p},\mathbf{H}]$ for the quantum mechanical Hamiltonian $\mathbf{H}=H(\mathbf{q},\mathbf{p})$ in one degree of freedom, where $\mathbf{q}$ and $\mathbf{p}$ are generalized position and momentum operators, respectively, satisfying the well-known canonical commutation relation $[\mathbf{q},\mathbf{p}]=i\hbar \mathbf{I}$. They obtained an implicit solution to the problem by solving the operator equation $[F(\mathbf{q},\mathbf{p}),H(\mathbf{q},\mathbf{p})]=i\hbar\mathbf{I}$ for $F(\mathbf{q},\mathbf{p})$. Their solution constitutes expanding $H(\mathbf{q},\mathbf{p})$ and $F(\mathbf{q},\mathbf{p})$ in terms of the basis operators, $\mathbf{T}_{m,n}$, i.e. $H(\mathbf{q},\mathbf{p})=\sum_{m,n} h_{m,n} \mathbf{T}_{m,n}$ and $F(\mathbf{q},\mathbf{p})=\sum_{m,n}\alpha_{m,n}\mathbf{T}_{m,n}$, where the coefficients $\alpha_{m,n}$’s are determined by the $h_{m,n}$’s.
Our interest in the Bender-Dunne operators stem from the observation that their problem of solving $[F(\mathbf{q},\mathbf{p}),H(\mathbf{q},\mathbf{p})]=i\hbar\mathbf{I}$ can be taken as solving for a time operator $\mathbf{T}$ for the system described by the given Hamiltonian $\mathbf{H}$, i.e. finding $\mathbf{T}$ that satisfies $[\mathbf{T},\mathbf{H}]=i\hbar \mathbf{I}$. Time operators, of course, is our staple object of research. The Bender-Dunne operators appeared earlier in our construction of time of arrival (TOA) operators in the non-interacting and interacting cases. The TOA-operators for spatially confined particles turned out to be bounded, self-adjoint operators. Moreover, they are compact and hence with spectrum consisting of eigenvalues only. This is a consequence of the fact that the relevant $\mathbf{T}_{mn,n}$ operators are bounded and self-adjoint in the Hilbert space of a spatially confined particle.
We wish now to go beyond the confined case and extend our analysis to the quantum particle in the entire real line. This requires inquiring into the status of the Bender-Dunne basis operators as operators in the Hilbert space $L^2(\mathbb{R})$, the Hilbert space of square integrable, complex-valued functions in the real line. It is our hope that this will give us more insight into the nature of time in the quantum domain. The status of the TOA-operators in $\mathbb{R}$ remain formal at the moment; investigating their properties as Hilbert space operator may lead us to their spectral decompositions that are necessary in extracting falsifiable time of arrival distributions. What we have accomplished in the paper is just a small part of the whole we wish to accomplish: there still remains studying the Hilbert space properties of the TOA-operators, and hopefully, our clarification of the Hilbert space status of the $\mathbf{T}_{m,n}$’s will help towards this goal.
But there is more value to the results of the paper than its relevance to the quantum time problem. Bender and Dunne recognized the formality of the progress they reported because the operator solutions $F(\mathbf{q},\mathbf{p})$ could be “extremely singular” as they contained arbitrary inverse powers of $\mathbf{q}$ and $\mathbf{p}$ coming from the $\mathbf{T}_{m,n}$’s. With that they, too, recognized the need to clarify the actions of the operators $\mathbf{T}_{m,n}$ in the Hilbert space $L^2(\mathbb{R})$. Doing so could lift the formality of the solutions $F(\mathbf{q},\mathbf{p})$, which could pave a way to solve spectral problems of Schrodinger operators. That is possible if their solutions $F(\mathbf{q},\mathbf{p})$ can be consistently interpreted as generators of energy translations in the relevant Hilbert space. We hope that our work is relevant towards the realization of Bender and Dunne idea to use $F(\mathbf{q},\mathbf{p})$ in investigating Schrodinger operators.
In this post I wish to make the paper more accessible to a wider audience by providing the necessary mathematical preliminaries that have been summarily assumed in the paper. Emphasis will be given on how to define a Hilbert space operator.
Hilbert space operators
An operator $\mathbf{A}$ on a Hilbert space $\mathcal{H}$ is always a pair—a subset $D(A)$ of $\mathcal{H}$, called the domain of $\mathbf{A}$, and a rule on how $\mathbf{A}$ maps $D_A$ in $\mathcal{H}$. The domain $D(A)$ of $\mathbf{A}$ maybe the entire $\mathcal{H}$ or a proper subset of $\mathcal{H}$. An important requirement for $\mathbf{A}$ to be an operator in $\mathcal{H}$ is that for every $\psi$ in $D$ the vector $\mathbf{A}\psi$ must belong to the Hilbert space $\mathcal{H}$. When the Hilbert space is finite dimensional, the domain of every operator is the entire Hilbert space. However, when the Hilbert space is infinite dimensional the domain can be the entire Hilbert space or just a proper subset of the Hilbert space. In general the former holds when the operator is bounded, and the later when the operator is unbounded.
For an operator to be a meaningful quantum operator, its domain must necessarily be dense in the entire Hilbert space. An operator whose domain is dense is called densely defined. Many useful properties of an operator derives from the denseness of its domain. The domain is necessarily dense if the operator has to have a unique adjoint. And only when the adjoint exists that we can inquire whether a given operator is self-adjoint or at least maximally symmetric. Quantum observables are postulated to be maximally symmetric operator, self-adjoint operators included, they being special cases of maximally symmetric operators. For this reason, it is important to establish that the Bender-Dunne operators are densely defined if there is any chance that they represent quantum observables. It is not unreasonable to expect that they are quantum observables because they are a particular quantization of classical observables.
Now a subspace $D$ of a Hilbert space $\mathcal{H}$ is called dense in $\mathcal{H}$ if for every vector $\psi$ in $\mathcal{H}$ and every $\epsilon>0$ there exists a vector $\phi$ in $D$ such that $||\psi-\phi||<\epsilon$. That is $D$ is dense if every vector of $\mathcal{H}$ can be arbitrarily approximated by a vector in $D$. A given subspace $D$ can be established to be dense using any of the following equivalent statements:
- $D$ is dense in $\mathcal{H}$.
- For every vector $\psi$ in $\mathcal{H}$ there exists a Cauchy sequence $\{\psi_1,\psi_2,\psi_3,\dots\}$ in $D$, i.e. $\psi_k$ belongs to $D$ for all $k$, such that $\lim_{k\rightarrow\infty}||\psi-\psi_k||=0$.
- If $\left<\left.\psi\right|\phi\right>=0$ for all $\phi$ in $D$, then $\psi=0$, i.e. the vector that is orthogonal to all vectors of $D$ is only the zero vector.
- $D$ consists of a basis for $\mathcal{H}$.
The paper utilized the fourth statement. To show that a sequence of linearly independent vectors $\{\varphi_1,\varphi_2,\varphi_3,\dots\}$ forms a basis for a Hilbert space $\mathcal{H}$, it is sufficient to establish that the only vector in $\mathcal{H}$ that is orthogonal to all vectors of the sequence is the zero vector, i.e., $\left<\left.\psi\right|\varphi_k\right>=0$ for all $k$ implies that $\psi=0$.
The Bender-Dunne operators
For $m,n\geq 0$ the $\mathbf{T}_{m,n}$’s are defined as the Weyl-ordered form of the classical function $q^n p^m$,
\begin{equation}\label{exp1}
\mathbf{T}_{m,n}=\frac{1}{2^n}\sum_{k=0}^{n} {n \choose k} \mathbf{q}^k \mathbf{p}^m \mathbf{q}^{n-k}, \;\; m,n=0,1,2,\dots.
\end{equation} Using the canonical commutation relation, $[\mathbf{q},\mathbf{p}]=i\hbar I$, expression \eqref{exp1} can be reordered to assume the equivalent form
\begin{equation}\label{exp2}
\mathbf{T}_{m,n}=\frac{1}{2^m} \sum_{j=0}^{m} { m \choose j} \mathbf{p}^j \mathbf{q}^n \mathbf{p}^{m-j}, \;\; m,n=0,1,2,\dots.
\end{equation}
Equations \eqref{exp1} and \eqref{exp2} can be used to define the $\mathbf{T}_{m,n}$’s when one of the indices is negative and the other positive. Equation \eqref{exp1} is used to define $\mathbf{T}_{m,n}$ when $m<0$ and $n>0$; on the other hand, equation \eqref{exp2} is used when $m>0$ and $n<$. Explicitly we have \begin{equation}\label{exp3} \mathbf{T}_{-|m|,n}=\frac{1}{2^n}\sum_{k=0}^{n} {n \choose k} \mathbf{q}^k \mathbf{p}^{-|m|} \mathbf{q}^{n-k}, \;\; m=-1,-2, \dots, \; n=0,1,2,\dots. \end{equation} \begin{equation}\label{exp4} \mathbf{T}_{m,-|n|}=\frac{1}{2^m} \sum_{j=0}^{m} { m \choose j} \mathbf{p}^j \mathbf{q}^{-|n|} \mathbf{p}^{m-j}, \;\; m=0,1,2,\dots,\ ; n=-1,-2,\dots. \end{equation} In these forms $\mathbf{T}_{-|m|,n}$ and $\mathbf{T}_{m,-|n|}$ are consistent with the interpretation that they are the respective Weyl quantizations of the classical functions $p^{-|m|} q^n$ and $p^m q^{-|m|}$. Equations \eqref{exp1} and \eqref{exp2} can be further extended to cover both negative indices. The appropriate extension is obtained by replacing the binomial coefficient with gamma functions and extending the summation from negative to positive infinity. The resulting coefficients are obtained by an appropriate limiting procedure. As an example of the utility of the Bender-Dunne operators, a time of arrival operator $\mathbf{T}$ can be constructed by solving the formal canonical commutation relation $[\mathbf{H},\mathbf{T}]=i\hbar \mathbf{I}$ for a given Hamiltonian $\mathcal{H}$. The harmonic oscillator, whose Hamiltonian is given by \begin{equation} \mathbf{H}_{HO}=\frac{1}{2m}\mathbf{p}^2 + \frac{1}{2} m \omega^2 \mathbf{x}^2, \end{equation} has the corresponding time of arrival operator \begin{equation} \mathbf{T}_{HO}=-\sum_{k=0}^{\infty} (-1)^k \frac{m^{2k+1} \omega^{2k}}{2k+1} \; \mathbf{T}_{-2k-1,2k+1}.\label{toahere} \end{equation}The generalized position $\mathbf{q}$ and momentum $\mathbf{p}$ operators have been replaced by the one-dimensional position $\mathbf{x}$ and momentum $\mathbf{p}$ operators, which will be the case from here on. Notice that $\mathbf{T}$ involves only negative powers of $\mathbf{p}$ and positive powers of $\mathbf{x}$. This is true in general for analytic potentials. It is for this reason that the paper considered only the case $\mathbf{T}_{-m,n}$ for positive integers $m$ and $n$.
The operators $\mathbf{T}_{-m,n}$ as integral operators
Now the operators $\mathbf{T}_{-m,n}$ are formal and they do not take on specific meaning until it is decided on what space of functions they act on. If we take them as meaningful quantum mechanical operators, then they must be meaningful Hilbert space operators. However, that is still not enough. It is assumed that $\mathbf{x}$ and $\mathbf{p}$ are the position and momentum operators of a quantum particle in one-dimension. So we must still decide whether the particle has access to the entire real line or just a portion of it. In the paper currently under review, we considered the case when the configuration space is the entire real line. The relevant Hilbert space is then $L^2(\mathbb{R})$, the space of square integrable complex valued functions in $\mathbb{R}$ with the inner product $\left<\psi\left|\phi\right>\right.=\int_{-\infty}^{\infty} \psi^*(x) \phi(x) \mbox{d}x$
Next we must decide on how they explicitly act on the vectors in their respective domains. There are two possibilities: Either we define their actions in position or in momentum representation. For the case that we wish to consider it seems that it is more transparent to work in the momentum representation and affect the following replacements
\begin{eqnarray}
\mathbf{x} \;\; &\mapsto & i\hbar \;\; \frac{\mbox{d}}{\mbox{d}p},\;\;\;\;
\mathbf{p} \;\; &\mapsto & p .
\end{eqnarray} The problem with this is that the solutions involve an infinite summation in both indices; see, for example, equation \eqref{toahere}. This leads to a differential operator in $p$ that is infinite in order with essential singularity at $p=0$ coming from the infinite inverse power of $p$.
In turns out that both problems can be solved, at least formally, by defining the operators in the position space. This leads them acting as integrals operators,
\begin{equation}
\left(\mathbf{T}_{-m,n}\varphi\right)\!(x)=\int_{-\infty}^{\infty} \left<\left. x\right|\right.\mathbf{T}_{-m,n}\left|\left. y\right>\right. \varphi(y)\, \mbox{d}y
\end{equation}The kernel is obtained using the the fact that $\left|\left.x\right>\right.$ is a generalized eigenvector of the position operator $\mathbf{x}\left|\left.x\right>\right.=x \left|\left.x\right>\right.$,
\begin{eqnarray}
\left<\left. x\right|\right.\mathbf{T}_{-m,n}\left|\left. y\right>\right. &=& \frac{1}{2^n}\sum_{k=0}^{n} {n \choose k} \left<\left. x\right|\right.\mathbf{x}^k \mathbf{p}^{-m} \mathbf{x}^{n-k}\left|\left. y\right>\right. \nonumber \\
&=& \frac{1}{2^n}\sum_{k=0}^{n} {n \choose k} x^{k} y^{n-k}\left<\left. x\right|\right.\mathbf{p}^{-m} \left|\left. y\right>\right. \nonumber \\
&=&\frac{(x+y)^n}{2^n} \left<\left. x\right|\right.\mathbf{p}^{-m} \left|\left. y\right>\right.\label{bok}
\end{eqnarray}The kernel $\left<\left. x\right|\right.\mathbf{p}^{-m} \left|\left. y\right>\right.$ is obtained by inserting the resolution of the identity $\mathbf{I}=\int_{-\infty}^{\infty} \left|\left. p\right>\right.\! \left<\left. p\right|\right. \mbox{d}p$, where $\left|\left. p\right>\right.$ is the generalized eigenvector of the momentum operator, $\mathbf{p}\left|\left. p\right>\right.=p\left|\left. p\right>\right.$. Then we have
\begin{eqnarray}
\left<\left. x\right|\right.\mathbf{p}^{-m} \left|\left. y\right>\right. &=& \int_{-\infty}^{\infty} \frac{\left<\left. x|\right. p\right> \left<\left. p|\right.y\right>}{p^m}\, \mbox{d}p\nonumber \\
&=&\frac{1}{2\pi \hbar}\int_{-\infty}^{\infty} p^{-m} \mbox{e}^{i (x-y)p/\hbar}\, \mbox{d}p, \label{kern}
\end{eqnarray}where we have used $\left<\left. x|\right.p\right>=\mbox{e}^{i x p/\hbar}/\sqrt{2\pi\hbar}$ to obtain the second line.
Now the singularity of $\mathbf{T}_{-m,n}$ shows up in the divergence of the integral in equation \eqref{kern}. The operator $\mathbf{T}_{-m,n}$ is not meaningful until we assign a value to the divergent integral. Here we chose to interpret the integral as a distributional integral, so that the integral is the Fourier transform of the generalized function $p^{-m}$. The Fourier transform of a distribution always exists. We use the well known result
\begin{equation}
\int_{-\infty}^{\infty} x^{-m} \mbox{e}^{i \sigma x} \mbox{d}x=\frac{\pi \, i^m}{(m-1)!} \sigma^{m-1} \mbox{sgn}\,\sigma,\;\;\; m=1, 2, \dots
\end{equation}where $\mbox{sgn}\,\sigma$ is the sign function. Applying this to equation \eqref{kern} and substituting the result back into equation \eqref{bok} give the kernel we seek
\begin{equation}
\left<\left. x\right|\right.\mathbf{T}_{-m,n}\left|\left. y\right>\right. = \frac{i (-1)^{(m-1)/2}}{2^{n+1} \hbar^m (m-1)!} (x+y)^n (x-y)^{m-1} \mbox{sgn}(x-y)\label{kernel} .
\end{equation}The operator $\mathbf{T}_{-m,n}$ is now defined as an integral operator whose kernel is given by equation \eqref{kernel}.
However, defining how $\mathbf{T}_{-m,n}$ acts is only half the problem. The minimum requirement for $\mathbf{T}_{-m,n}$ to be a meaningful quantum operator in $L^2(\mathbb{R})$ is the existence of a subset $D$ of functions in $L^2(\mathbb{R})$ with the property that every $\psi(x)$ in $D$ the function $(\mathbf{T}_{-m,n}\psi)(x)$ is again square integrable in the real line, and that $D$ is dense in $L^2(\mathbb{R})$. This subset $D$ may be assigned as the domain of $\mathbf{T}_{-m,n}$.
The paper under review proves the existence of a dense domain of the operator $\mathbf{T}_{-m,n}$ in the Hilbert space $L^2(\mathbb{R})$. I will not repeat the proof here, but I will, in the following section, reproduce the proof for the case $\mathbf{T}_{-1,1}$. The proof for the general case is an extension of the proof for $\mathbf{T}_{-1,1}$ and is sufficiently detailed in the paper.
The operator $\mathbf{T}_{-1,1}$
Here let us consider the formal operator $\mathbf{T}_{-1,1}$. This operator is proportional to the first time of arrival operator which has the subject of numerous investigations. For simplicity’s sake we just consider the operator $T_{-1,1}$ which is proportional to $\mathbf{T}_{-1,1}$ and it is given by the integral operator
\begin{equation}\label{toa}
\left(T_{-1,1} \varphi \right)\!(x) = \int_{-\infty}^{\infty}(x+y)\mbox{sgn}(x-y)\varphi (y) dy .
\end{equation}We wish now to show that it has a non-trivial dense domain in the Hilbert space $L^2(\mathbb{R})$. Let us assume that $T_{-1,1}$ has a non-trivial domain, $D(T_{-1,1})\subseteq L^2(\mathbb{R})$. It is now our objective to show that $D(T_{-1,1})$ exists and is dense.
Let us first establish that $T_{-1,1}$ has a non-trivial domain which is a proper subset of the Hilbert space. If $\varphi(x)$ belongs to the domain, $D(T_{-1,1})$, of $T_{-1,1}$, then $(T_{-1,1}\varphi)(x)$ is necessarily square integrable,
\begin{equation}\label{sqr}
\int_{-\infty}^{\infty}|(T_{-1,1} \varphi )(x)|^{2}\mbox{d}x<\infty.
\end{equation}The domain is non-trivial because the vector $\psi_1(x)=(1-2x)e^{-x^2}\in L^2(\mathbb{R})$ is mapped by $T_{-1,1}$ to a vector in $L^2(\mathbb{R})$, in particular,
\begin{equation}
(T_{-1,1}\psi_1)\!(x)=\int_{-\infty}^{\infty}(1-2y^2)e^{-y^2}(x+y)\mbox{sgn}(x-y)\mbox{d}y=(1+4x^2)e^{-x^2},
\end{equation} which is clearly square integrable. However, its domain cannot be the entire Hilbert space because $T_{-1,1}$ acting on $\psi_2(x)=e^{-x^2}\in L^2({\mathbb{R}})$ yields the function
\begin{equation}
(T_{-1,1}\psi_x)\!(x)=\int_{-\infty}^{\infty} (x+y) \mbox{sgn}(x-y) \mbox{e}^{-y^2}\mbox{d}y = \left( x\sqrt {\pi }{\rm erf} \left(x\right){{\rm e}^{{x}^{2}}}-1
\right) {{\rm e}^{-{x}^{2}}}\label{be}
\end{equation} where $\mbox{erf}(x)$ is the error function. With $\mbox{erf}(\infty)=1$, equation \eqref{be} has the asymptotic behavior $\psi_2(x)\sim \sqrt{\pi} \, x$ as $|x|\rightarrow\infty$; thus equation \eqref{be} is not square integrable in the real line. The function $\mbox{e}^{-x^2}$ which belongs to the Hilbert space is mapped outside of the Hilbert space so that it does not belong to the domain of $T_{-1,1}$.
To identify the elements of the domain of $T_{-1,1}$, let us determine the necessary and sufficient conditions for the square integrability condition \eqref{sqr} to be satisfied. To proceed we expand equation \eqref{toa} using the identity $\mbox{sgn}(x)=H(x)-H(-x)$, where $H(x)$ is the Heaviside step function. The expansion yields
\begin{eqnarray}\label{expandtoa}
\left(T_{-1,1} \varphi \right)\!\!(x) = x\left(\int_{-\infty}^{x}\!\!\!\!\varphi (y) \mbox{d}y\!\! -\!\! \int_{x}^{\infty}\!\!\!\!\!\!\varphi (y) dy\right) \!\!+\! \! \left(\int_{-\infty}^{x}\!\!\!\!\!y \varphi (y) \mbox{d}y \!\!-\!\! \int_{x}^{\infty}\!\!\!\!\! y \varphi(y) \mbox{d}y\right) .
\end{eqnarray} A necessary but not sufficient condition for $(T_{-1,1}\varphi)\!(x)$ to be square integrable is that the right hand side of equation \eqref{expandtoa} vanishes as $|x| \rightarrow \infty$. This implies the following necessary integral conditions
\begin{equation}\label{zerocond}
\int_{-\infty}^{\infty}\varphi (y) \mbox{d}y = 0, \;\;\;\;\; \int_{-\infty}^{\infty} y\varphi (y) dy = 0 .
\end{equation}The function $\psi_1(x)$ satisfies both integral conditions; while, the function $\psi_2(x)$ does not.
Now assuming that $\varphi(x)$ satisfies these integral conditions, a sufficient condition to ensure that $\varphi(x)$ belongs to the domain is that each term in the right hand side of \eqref{expandtoa} is square integrable at infinity. Let us consider the term $x\int_{-\infty}^x \varphi(y) \mbox{d}y$. This term must vanish faster than $|x|^{-1/2}$ or $o(|x|^{-1/2})$ to be square integrable. This requires that the integral involved has the behavior $|\int_{-\infty}^x \varphi(y) \mbox{d}y|=O(|x|^{-\sigma})$ with $\sigma>3/2$ or $|\int_{-\infty}^x \varphi(y) \mbox{d}y|=o(|x|^{-3/2})$ at infinity. Carrying out the same analysis to the rest of the terms, we arrive at the following large $x$ behaviors of the integrals,
\begin{equation}\label{asymcond}
\left|\int_{-|x|}^{\infty}\varphi (\pm y) \mbox{d}y\right| = o(|x|^{-3/2}), \;\;\;\;\; \left|\int_{-|x|}^{\infty} y\varphi(\pm y) \mbox{d}y\right| = o(|x|^{-1/2}), |x|\rightarrow\infty .
\end{equation}These conditions are sharper than that appeared in the published version of the paper under review.
Let me now show how elements of the domain of the $T_{-1,1}$ can be constructed. Since the conditions are a pair, we need at least a pair of linearly independent vectors and their linear sum to ensure that the integral conditions \eqref{zerocond} are satisfied. Let $\varphi(x) = \alpha \varphi_{1}(x) + \beta \varphi_{2}(x)$, where $\varphi_{1}(q)$ and $\varphi_{2}(x)$ are linearly independent functions, and $\alpha$ and $\beta$ are constants such that conditions \eqref{zerocond} are satisfied. Substituting $\varphi(x)$ to equations \eqref{zerocond}, we obtain the following matrix expression for $\alpha$ and $\beta$,
\begin{equation} \left[ \begin{array}{cc}
\int_{-\infty}^{\infty}\varphi_{1}(x)\mbox{d}x & \int_{-\infty}^{\infty}\varphi_{2}(x)\mbox{d}x\\
\int_{-\infty}^{\infty}y\varphi_{1}(x)\mbox{d}x & \int_{-\infty}^{\infty}x\varphi_{2}(x)\mbox{d}x
\end{array} \right]
\left[ \begin{array}{c}
\alpha \\
\beta
\end{array} \right] = 0
\end{equation} A solution for $\alpha$ and $\beta$ exist if the determinant of the matrix of the coefficients vanishes. This gives the following condition for the integrals of $\varphi_{1}(x)$ and $\varphi_{2}(x)$,
\begin{equation}\label{constcond}
\left[\int_{-\infty}^{\infty}\varphi_{1}(x)dx\right]\left[\int_{-\infty}^{\infty}x\varphi_{2}(x)dx\right] = \left[\int_{-\infty}^{\infty}\varphi_{2}(x)dq\right]\left[\int_{-\infty}^{\infty}x\varphi_{1}(x)dx\right]
\end{equation}
Equation \eqref{constcond} can be satisfied by letting $\varphi_{1}(x)$ and $\varphi_{2}(x)$ have definite parities. That is, either both of them are even or both are odd. With these conditions, we can write
\begin{equation}
\varphi^{(1)}(x) = \left(\int_{-\infty}^{\infty}\varphi_{1}(y)dy\right)\varphi_{2}(x) – \left(\int_{-\infty}^{\infty}\varphi_{2}(y)dy\right)\varphi_{1}(x),\label{kaw1}
\end{equation}for even $\varphi_{1}(x)$ and $\varphi_{2}(x)$, and
\begin{equation}
\varphi^{(2)}(x) = \left(\int_{-\infty}^{\infty}y\varphi_{1}(y)dy\right)\varphi_{2}(x) – \left(\int_{-\infty}^{\infty}y\varphi_{2}(y)dy\right)\varphi_{1}(x),\label{kaw2}
\end{equation} for odd $\varphi_{1}(x)$ and $\varphi_{2}(x)$. By direct substitution, one can see that they satisfy equation (\eqref{zerocond}) provided that the integrals exist. The vectors $\varphi^{(1)}(x)$ and $\varphi^{(2)}(x)$ must satisfy the minimum asymptotic conditions \eqref{asymcond}.
The domain of $D(T_{-1,1})$ is certainly not trivial. We now show that it is dense by showing that a complete and orthogonal basis exists in the domain of $T_{-1,1}$. Let us denote this basis set by $S(T_{-1,1})$. Elements of $S(T_{-1,1})$ are mapped by $T_{-1,1}$ into $L^2(\mathbb{R})$, so that they necessarily satisfy the integral conditions \eqref{zerocond}.
We construct $\mathcal{S}(T_{-1,1})$ from a complete set of vectors in the Hilbert space $L^2(\mathbb{R})$ with definite parities; the vectors in the chosen basis do not necessarily belong to $D(T_{-1,1})$. The set of vectors $\{\psi_{k}(x) = (2^{k/2}\sqrt{k!}\sqrt[4]{\pi})^{-1}H_{k}(x)\mbox{e}^{-x^{2}/2}, k=0,1,\dots\}$, where the $H_{k}(x)$’s are the Hermite polynomials, is complete in $L^2(\mathbb{R})$. Each $\psi_k(x)$ has a definite parity, even (odd) for even (odd) $k$. Using equations \eqref{kaw1} and \eqref{kaw2}, we define the following set of vector elements of $\mathcal{S}(T_{-1,1})$,
\begin{eqnarray}\label{basicvec}
\psi_{m,n}^{(1)}(x) &=& A_{n}^{(1)}\psi_{2m}(x)-A_{m}^{(1)}\psi_{2n}(x), \\
\psi_{m,n}^{(2)}(x) &=& A_{n}^{(2)}\psi_{2m+1}(x)-A_{m}^{(2)}\psi_{2n+1}(x),
\end{eqnarray}for all natural numbers $m$ and $n$ with $m \neq n$. The coefficients are given by
\begin{eqnarray}
A_{n}^{(1)} &=& \int_{-\infty}^{\infty}\psi_{2n}(y)dy = \frac{\sqrt{2}\sqrt[4]{\pi}\sqrt{(2n)!}}{2^{n}n!},\\
A_{n}^{(2)} &=& \int_{-\infty}^{\infty}y\psi_{2n+1}(y)dy = \frac{2\sqrt[4]{\pi}\sqrt{(2n+1)!}}{2^{n}n!}.
\end{eqnarray} These vectors satisfy the required the integral conditions \eqref{zerocond}. Moreover, it can be shown that the vectors $(T_{-1,1}\psi_{m,n}^{(1)})(x)$ and $(T_{-1,1}\psi_{m,n}^{(2)})(x)$ exponentially decay as $|x|\rightarrow \infty$ so that equations \eqref{asymcond} are satisfied.
We now show that $S(T_{-1,1})$ is a complete set (in fact over complete). A sequence of vectors is complete if the only vector orthogonal to every element of the sequence is the zero vector. Let $n$ be fixed and consider the sequence of vectors $\{\psi_{m,n}^{(1)},\psi_{m,n}^{(2)}, m=0,1,\dots\}$. This sequence is complete if $<\psi_{m,n}^{(k)}|\phi>=0$ for all $m=0,1,2,\dots$ and $k=1,2$ implies that $\phi=0$. Since the vectors $\psi_{k}(x)$ form a complete set of orthonormal vectors, we can write $\phi(x) = \sum_{k=0}^{\infty}\phi_{k}\psi_{k}(x)$. We must now show that $\phi_{k}=0$ for all $k$ so that $\phi(x)=0$. The orthonormality of the $\psi_k(q)$’s and the conditions $<\psi_{m,n}^{(1)}|\phi> = 0$ and $<\psi_{m,n}^{(2)}|\phi> = 0$ for all $m$ imply
\begin{eqnarray}\label{coeffbehaviour}
\phi_{2m} &=& \left(\frac{\sqrt{(2n)!}}{2^{n}n!}\right)^{-1}\left(\frac{\sqrt{(2m)!}}{2^{m}m!}\right)\phi_{2n}, \\
\phi_{2m+1} &=& \left(\frac{\sqrt{(2n+1)!}}{2^{n}n!}\right)^{-1}\left(\frac{\sqrt{(2m+1)!}}{2^{m}m!}\right)\phi_{2n+1} .
\end{eqnarray}
Let us consider the behavior of the coefficients for $m>n$ as $m\rightarrow\infty$. Using Stirling’s formula: $m! \sim \sqrt{2\pi}m^{m+1/2}\mbox{e}^{-m}$ as $m\rightarrow\infty$, we obtain the following asymptotic behaviors for the coefficients:
\begin{eqnarray}
\phi_{2m} &\sim& \left(\frac{\sqrt{(2n)!}}{2^{n}n!}\right)^{-1}\frac{1}{m^{1/4}}\phi_{2n}, \label{asym1}\\
\phi_{2m+1} &\sim& \left(\frac{\sqrt{(2n+1)!}}{2^{n}n!}\right)^{-1}m^{1/4}\phi_{2n+1}\label{asym2}
\end{eqnarray}
Since $\phi(x)$ should belong to $L^2(\mathbb{R})$, it is necessary that $\sum_{k=0}^{\infty}|\phi_{k}|^{2}<\infty$. We see immediately that from equation \eqref{asym2}, $\phi_{2m+1}$ cannot satisfy this for non-zero $\phi_{2n+1}$ since $\phi_{2m+1}$ algebraically diverges as $m\rightarrow\infty$. It is also not possible for the coefficients $\phi_{2m}$ in \eqref{asym1}, since they behave as $\phi_{2m} \sim \mathcal{O}(m^{-1/4})$ as $m\rightarrow\infty$ for non-zero $\phi_{2n}$. It is then necessary to have $\phi_{2n+1}=0$ and $\phi_{2n}=0$. From equation \eqref{coeffbehaviour}, we see that $\phi_{2m+1}=0$ and $\phi_{2m}=0$ for any $m \neq n$. We then have $\phi_{k}=0$ for any $k$. That is, the only vector orthogonal to the sequence for a given $n$ is the zero vector. Hence the sequence $\{\psi_{m,n}^{(1)},\psi_{m,n}^{(2)}, m=0,1,\dots\}$ is complete in $L^2(\mathbb{R})$. Also since $n$ is arbitrarily chosen, our conclusion should hold for any $n$, so that for each natural number $n$ we have a complete set. Then the set $S(T_{-1,1})$ is (over) complete. We can now assign the linear span of $S(T_{-1,1})$ to be the domain $D(T_{-1,1})$ of $T_{-1,1}$. Since $S(T_{-1,1})$ is a complete set, its linear span $D(T_{-1,1})$ is necessarily dense in $L^2(\mathbb{R})$. Hence the operator $T_{-1,1}:D(T_{-1,1})\subset L^2(\mathbb{R})\mapsto L^2(\mathbb{R})$ is a densely defined operator. The same method is applied in proving the general case for $\mathbf{T}_{-m,n}$. The Appendix of the paper gives a detailed account of the proof.
Conclusion
The paper under review has shown that the operators $\mathbf{T}_{-m,n}$ for positive integers $m$ and $n$ are densely defined operators in the Hilbert space $L^2(\mathbb{R})$. However, the progress reported is just the first step into fully understanding the Hilbert space properties of the Bender-Dunne basis operators. Many questions about them remain to be answered. Are they essentially self-adjoint in their domains? If not are they maximally symmetric? These questions are important in our attempt to understand the properties of time operators, and in obtaining spectral decompositions of them that yield definite predictions on the distributions of quantum time measurements. Also their explicit properties may help us investigate the property of time operators or the Bender-Dunne solutions as generators of energy translations. This is essential in implementing Bender and Dunne’s original idea of using their solutions in investigating the spectral properties of Schrodinger operators.