semi-regular biorthogonal pairs on the above commutation rule. We assume the following statements:\

[*Assumption 1.*]{}

*There exists a non-zero element $\phi_{0}$ of ${\cal H}$ such that*

\(i) $a \phi_{0}=0$,

\(ii) $\phi_{0} \in D^{\infty}(b) \equiv \cap_{k=0}^{\infty} D( b^{k})$,

\(iii) $b^{n} \phi_{0} \in D(a)$, $n=0,1, \cdots$.

\
\
Then, we may define a sequence $\{ \phi_{n} \}$ in ${\cal H}$ by $$\begin{aligned}
\phi_{n} 
&\equiv& \frac{1}{\sqrt{n!}} \; b^{n} \phi_{0}, \;\;\; n =0,1, \cdots \nonumber \\
&=& \frac{1}{\sqrt{n}} \; b \phi_{n-1}, \;\;\; n =1,2, \cdots . \nonumber \end{aligned}$$ Furthermore, we have the following\

[*Proposition 4.1.*]{}

*The following statements hold.*

\(1) $b^{n} \phi_{0} \in D(a^{m})$ and $$\begin{aligned}
a^{m} b^{n} \phi_{0}
&=& \left\{
\begin{array}{cl}
_{n}P_{m} b^{n-m} \phi_{0} \;\;\;\;\;\;\;\;\;\; &,m\leq n, \\
\nonumber \\
0 \;\;\; &,m > n.
\end{array}
\right. \nonumber \end{aligned}$$

\(2) $\phi_{n} \in D(N^{m})$ and $N^{m} \phi_{n}=n^{m} \phi_{n}$, $n,m=0,1, \cdots$. In particular, $N\phi_{n}=n \phi_{n}$, $n=0,1, \cdots$.

\(3) $$\begin{aligned}
a \phi_{n}
&=& \left\{
\begin{array}{cl}
0 \;\;\;\;\;\;\;\;\;\; &,n=0, \\
\nonumber \\
\sqrt{n} \phi_{n-1}, \;\;\; &,n=1,2, \cdots,
\end{array}
\right. \nonumber \\
\nonumber \\
b \phi_{n} &=& \sqrt{n+1} \phi_{n+1}  \;\;\;\; ,n=0,1, \cdots .\nonumber \end{aligned}$$

[*Proof.*]{} These proofs follow from Ref. [@h-t2].\

[*Assumption 2.*]{}

*There exists a non-zero element $\psi_{0}$ of ${\cal H}$ such that*

\(i) $b^{\dagger} \psi_{0}=0$,

\(ii) 