shown respectively by the solid and dashed curves in Fig. \[potential\]. The minimum of the well is inclined towards the left side when $\rho>\kappa$. The PT potential is one of the exactly solvable quantum mechanical potentials. The corresponding Schrödinger equation can be solved to obtain the energy eigenvalues and eigenfunctions [@klauder] $$\begin{aligned}
E_{n}&=&\frac{\hbar^2\alpha^2}{2m} (2n+\rho+\kappa)^2,\quad
n=0,1,2,... ~~\nonumber\\
\psi_{n}(x)&=&N
\cos^{\rho}(\alpha x) \sin^{\kappa}(\alpha x) P_n^{\rho-1/2,\kappa-1/2} [\cos(2\alpha x)],\nonumber\\
\label{PTeigenfunction}\end{aligned}$$ where $P_n^{\rho-1/2,\kappa-1/2} [\cos(2\alpha x)]$ is the Jacobi polynomial and $N$ is the normalization constant: $$N=\left[\frac{2 \alpha
\Gamma{(n+1)}(2n+\rho+\kappa)\Gamma{(n+\kappa+\rho)}}{\Gamma{(n+\rho+1/2)}
\Gamma{(n+\kappa+1/2)}}\right]^\frac{1}{2}.$$ It is worth mentioning that the PT potential has an underlying $SU(1,1)$ dynamical symmetry algebra, which admits an infinite number of energy levels. Some of them can be excited to create a CS wave-packet by using an appropriate laser. These states have many classical features and tend to preserve their properties even for long evolution times. Here, we consider a displacement operator CS [@perelomov] of this potential. Construction of this CS makes use of the exponential form for the solution of hypergeometric differential equation and the correct choice of $SU(1,1)$ generators [@guru1]. The CS depends upon a parameter $\beta$ and can be written as [@utpal] $$\chi_{\beta}(x)\;=\;\sum_{n=0}^{\infty}\;\;d_{n}^{\beta}\;\psi_{n}(x),$$ where the amplitudes $d_{n}^{\beta}$ are given by $$d_{n}^{\beta}\;=\;(-\beta)^n
\left[\frac{\Gamma{(\rho+n+1/2)}\Gamma{(\kappa+n+1/2)}}{2 \alpha
\Gamma{(\kappa+\rho+n)}\Gamma{(n+1)}(2n+\kappa+\rho)}\right]^{1/2}.$$ 