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$\newcommand{\dydxt}[2]{\frac{d#1}{d#2}}
\newcommand{\dydx}[2]{\frac{\partial#1}{\partial#2}}$
For simplicity we consider the case of a single particle moving in a potential.
Suppose we are given a wave function
\begin{eqnarray*} \psi(x)=\left(\frac{\alpha}{\pi}\right)^{\frac{1}{4}}
\exp(-\alpha x^2/2)\end{eqnarray*}
as the sate of the particle at some fixed time. What do we conclude? Is it the ground state of a particle bound to the point $x=0$ by a harmonic force or is it the gaussian wave-packet of a free particle?
The point to understand here is that we can say nothing about a particle by just looking at its wave function. We must know the Hamiltonian of the system. We have to see how does the wave function change with time in order to see whether it belongs to a free particle or an interacting particle. We must know the {\em time dependence} of the wave function. And not just of one wave function but that of a complete basis of wave functions.
For the above example if we know $\psi(x,t)$ for at least a small time interval, then we can calculate
\begin{eqnarray*} i\hbar\dydxt{\psi(x,t)}{t}+\frac{\hbar^2}{2m}\dydx{^2\psi(x,t)}{x^2}. \end{eqnarray*}
If it is zero (for all $x$), even then we can say it is a wave function of a free particle. But whether the system is actually free or not would require knowing the Hamiltonian as an operator.
