the integral can be evaluated [@knp86; @kno79ajp]. The result is $$I = \left[\frac{\pi}{\cosh\eta}\right] \exp{( 2rs \tanh\eta)}
  \exp{\left(\frac{2rr'}{\cosh\eta}\right) }.$$ We can now expand this expression and choose the coefficients of $r^{n+k}, s^{k}, r'^{n}$ for\
$H_{(n + k)}(x), H_{n}(y)$, and $H_{n}(z')$ respectively. The result is $$A_{n;k} = \left(\frac{1}{\cosh\eta}\right)^{(n+1)}
     \left[\frac{(n+k)!}{n!k!}\right]^{1/2} (\tanh\eta)^{k}.$$ Thus, the series becomes $$\label{seri55}
\chi_{n}(x') \chi_{0}(y') = \left(\frac{1}{\cosh\eta}\right)^{(n+1)}
 \sum_{k} \left[\frac{(n+k)!}{n!k!}\right]^{1/2}
       (\tanh\eta)^{k} \chi_{k+n}(x)\chi_{k}(y) .$$ If $n = 0$, it is the squeezed ground state, and this expression becomes the entangled state of Eq.(\[gau03\]).

![Sear transformation of the Gaussian form given in Eq.(\[gau00\]).[]{data-label="shear00"}](shear00b.eps)

E(2)-sheared States {#shear}
===================

Let us next consider the effect of shear on the Gaussian form. From Fig. \[shear00\] and Fig. \[sqz00\], it is clear that the sheared state is a rotated squeezed state.

In order to understand this transformation, let us note that the squeeze and rotation are generated by the two-by-two matrices $$K = \pmatrix{0 & i \cr i & 0}, \qquad J = \pmatrix{0 & -i \cr i & 0}, \qquad$$ which generate the squeeze and rotation matrices of the form $$\begin{aligned}
&{}& \exp{\left(-i\eta K\right)} =
\pmatrix{\cosh\eta & \sinh\eta \cr \sinh\eta & \cosh\eta },   \nonumber\\[1ex]
&{}& \exp{\left(-i \theta J\right)} = \pmatrix{\cos\theta & -\sin\theta \cr
     \sin\theta &  \cos\theta} 