Momentum uncertainty of free particle I've read several Q&A's regarding free particles and the associated wave packet in this website, but found the answer to my question nowhere. It's OK to attribute a Gaussian wave packet to the free particle in order to normalize it. The question is now what the uncertainty  in momentum is. The start point to find the wave function is the equation:
$$H \psi = E \psi$$
we find an exponential function which is not a proper solution, so we add up infinite number of these waves and make a wave packet that also has a width in momentum space. 
But we know that writing the above equation means we have measured energy of the particle. So, it has already gone to an eigenvector of energy and its energy and momentum are definite, not uncertain. If we want to decrease the momentum uncertainty to zero, we have to choose a Dirac delta function as the distribution function in momentum space. What happens is that the wave packet breaks into a simple exponential function, what we had at the beginning. What am I missing?
 A: First of all I think I should clarify the confusion of the terminology "uncertainty in $x$". What we mean by uncertainty is simply the standard deviation of the observable defined by:
$$\Delta x= \sqrt{ \left< x^2 \right> - \left< x \right>^2}$$
where $<A>$ denotes the expectation value of the operator $A$ is some state $\psi$. You should be aware that I won't put any hats (^) on the operators, so when I write $x$, what I really mean is $\hat x$. Furthermore I'll drop all the $x$ (the variable not the operator) dependencies e.g. I'll write $\psi$ to indicate $\psi(x)$.
Note that all of the proper states ie square integrable functions, that solve the Schrödinger Equation, must satisfy the uncertainty relationship, given by
$$\Delta x \Delta p \ge \frac \hbar 2$$
so saying that the state is an "eigenvector of energy and its momentum is definite, not uncertain" is not true in general, though it is correct for this specific case. 

Returning to the question of Gaussian wave packages, you should note that the equation
$$H\psi = E \psi$$
does not collapse the wave function. Take for example the state $\psi = a_1 \phi_1 + a_2\phi_2$, where $\phi_i$ are the energy eigenfunctions of the Hamiltonian $H$ with energy eigenvalue $E_i$ and $a_i \in \mathbb C$ satisfy $\sum_i |a_i|^2 = 1$. Now applying the Hamiltonian gives:
$$H\psi = a_1 H \phi_1 +  a_2 H \phi_2 = a_1E_1\phi_1+ a_2 E_2\phi_2$$
Note that this describes a state having probability $|a_i|^2$ of having energy $E_i$ and it is certainly not an eigenstate of energy neither did we collapse the wave function by acting on it by the Hamiltonian. In particular applying the Hamiltonian to a wave function $\psi$ gives the probability density of the state $\psi$ having energy $E_i$.
The Gaussian wave package is just a particular superposition of energy eigenstates and everything I did above applies. However you should note that a Gaussian package is a superposition of infinitely many energy eigenfunctions, whereas I only took a superposition of 2 states in my example.
