applicable to $\chi_{n}(y)$: $$b = \frac{1}{\sqrt{2}}\left( y + \frac{\partial}{\partial y}\right),
            \quad\mbox{and}\quad
b^{\dag} = \frac{1}{\sqrt{2}} \left(y - \frac{\partial}{\partial y}\right) .$$ Thus $$\begin{aligned}
&{}& \left(a^{\dag}\right)^{n} \chi^{0}(x) = \sqrt{n!}~ \chi_{n}(x), \nonumber \\[1ex].
&{}& \left(b^{\dag}\right)^{n} \chi^{0}(y) = \sqrt{n!} \chi_{n}(y) ,\end{aligned}$$ and $$a~\chi_{0}(x) = b~\chi_{0}(y) = 0 .$$ In terms of these variables, the transformation leading the Gaussian function of Eq.(\[gau00\]) to its squeezed form of Eq.(\[gau03\]) can be written as $$\label{exp01}
\exp{\left\{\frac{\eta}{2}\left( a^{\dag}b^{\dag} - a~b \right)\right\}} ,$$ which can also be written as $$\label{exp02}
\exp{\left\{-\eta\left(x\frac{\partial}{\partial y} +
         y\frac{\partial}{\partial x}\right)\right\}} .$$

Next, we can consider the exponential form $$\label{exp05}
  \exp{\left\{(\tanh\eta) a^{\dag} b^{\dag}) \right\}} ,$$ which can be expanded as $$\label{seri11}
    \sum_{n} \frac{1}{n!} (\tanh\eta)^{n} \left(a^{\dag} b^{\dag}\right)^{n}.$$ If this operator is applied to the ground state of Eq.(\[gau00\]), the result is $$\label{seri12}
    \sum_{n} (\tanh\eta)^{n} \chi_{n}(x) \chi_{n}(y) .$$ This form is not normalized, while the series of Eq.(\[seri00\]) is. What is the origin of this difference?

There is a similar problem with the one-photon coherent state [@klaud73; @sal07]. There, the series comes from the expansion of the exponential form $$\label{exp22}
\exp{\left\{\alpha a^{\dag}\right\}} ,$$ which can be expanded to $$\sum_{n} \frac{1}{n!} \alpha^{n} \left(a^{\dag}\right)^{n} .$$ However, this operator is not unitary. In order to make this series unitary, we consider 