Does the initial position of a particle matter in QM? Will the "exact" position of a particle at time $t=0$ determines its position at $t=t_p$?
For example, if the initial position of particle in a 1-D potential box [-1,1] is at (0,1), how long does it takes for the particle to be distributed as $\psi^*\psi$ in the box? 
Sorry if the question is too beginner-ish
 A: There is no initial position, any more than any other position. There's an initial wave function, $\psi(x, y, z) = \psi(x, y, z, 0)$. Or more generally an initial state vector $|\psi(0)>$. Yes, the initial wave function matters. And it determines -- unless there's a "measurement" that happens, and even then that depends on how you interpret QM -- the evolution thereafter at succeeding times $t_p$, as that's what Schrodinger equation tells you.
EDIT: You could set up a "wavefunction" with exact position but it's nonsense physically, it's a Dirac delta. Think about this: Dirac delta doesn't belong to the Hilbert space of complex valued square-integrable functions of 3 real variables since it's not a function at all (well not that kind of function anyways). So it's not a physical state of the system as that is the space of physical states of the system and it's not in there. It's kinda like $\pm\infty$ on the (extended) real number line, these are not real numbers they're more like "limits" of the real numbers. In like wise the Dirac delta is like some kind of "limit" of the wavespace. An idealization but not reality.
A: The particle has a definite position at t=0 means that initially the wavefunction is a position eigenstate. If initial position in an 1-D box is $x_0$, the wavefunction at t=0 is 
$$\psi(x,t=0) = \delta(x-x_0)$$
Now, with time the initial wavefunction will evolve according to Schrodinger equation. For a system with Hamiltonian $H$, the state at time $t=t_p$ will be 
$$\psi(x,t=t_p) = exp(\frac{-iHt_p}{\hbar})\psi(x,t=0)\\=exp(\frac{-iHt_p}{\hbar})\delta(x-x_0)$$ 
One way to evaluate this equation is to expand $\delta(x-x_0)$ in terms of the eigenfunctions of the Hamiltonian and evolve each eigenfunction individually. In general, the wavefunction will not remain a position eigenstate at $t>0$. It will spread out instantly.  
A: In Quantum Mechanics you cannot specify the position of a particle with infinite accuracy as it violates the position-momentum uncertainty relation.  Here you are specifying that the particle is at a specific location at $t=0$.
Infinite accuracy in position ($ \Delta x = 0 $)  results in infinite uncertainty in momentum ($ \Delta p = \infty $)
You can only say that at $t=0$ the particle has a definite probability to be found at a specific position
