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?