$\psi_{0} \in D^{\infty}(a^{\dagger}) \equiv \cap_{k=0}^{\infty} D( (a^{\dagger})^{k})$,

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

\
\
Then, we may define a sequence $\{ \psi_{n} \}$ in ${\cal H}$ by $$\begin{aligned}
\psi_{n} 
&\equiv& \frac{1}{\sqrt{n!}} \; (a^{\dagger})^{n} \psi_{0}, \;\;\; n= 0,1, \cdots \nonumber \\
&=& \frac{1}{\sqrt{n}} \; a^{\dagger} \psi_{n-1}, \;\;\; n =1,2, \cdots . \nonumber \end{aligned}$$ And we put an operator $N^{\dagger} \equiv a^{\dagger} b^{\dagger}$. Furthermore we have the following\

[*Proposition 4.2.*]{}

*The following statements hold.*

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

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

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

\

[*Proof.*]{} These proofs follow from Ref. [@h-t2].\
\
The above Assumption 1 and Assumption 2 coincide with the assumptions of Ref. [@h-t2]. We weaken the assumption of Ref. [@h-t2] to the next assumption in order to construct semi-regular biorthogonal pairs.\

[*Assumption 3.*]{}

