State of a well determined position particle

[bluesunday]

I must write the state of a free particle with well determined position (r₀) at t=0. I assume that, as it has well-determined position, it's an eigenstate of position operator, with eigenvalue r₀. So is this it, in the momentum basis?

\mid \psi (t_{0})\rangle = \mid {\bold{r}}_{0}\rangle=\frac{1}{(2\pi \hbar)^{3/2}}\int d^{3}p\mid \bold{p}\rangle e^{-i\bold{p}\cdot \bold{r_{0}}/\hbar}

However, this would be the state of ANY particle, not just a free particle...

I need it in order to apply the evolution operator to it:

[URL]http://upload.wikimedia.org/math/1/0/3/10317da44bf13fbd709cd642c5143b9f.png[/URL]

with:

[PLAIN]https://www.physicsforums.com/latex_images/25/2520739-3.png

Being the hamiltonian of a free particle:

[PLAIN]https://www.physicsforums.com/latex_images/13/1375576-2.png

I will obviously get the state on the p-representation. Is this correct?

I seem to have very serious trouble with the fundamentals of Dirac notation...

 

[G01]

See your other thread on this problem...

 

[bluesunday]

I read your hint in the other thread. Then what I've just put here is right, isn't it? I put the state straightforward in the momentum basis just by introducing the closure relation for the |p> basis into the |r0> state...