This question is a small step removed from two earlier questions about the uncertainty relation in the QM context. knzhou's answer touches on the point, and L. Motl's link-answer here is a little closer.
The passage from Wikipedia in relevant part is (italics added):
"When a state is measured, it is projected onto an eigenstate in the basis of the relevant observable. For example, if a particle's position is measured, then the state amounts to a position eigenstate. This means that the state is not a momentum eigenstate, however, but rather it can be represented as a sum of multiple momentum basis eigenstates. In other words, the momentum must be less precise. This precision may be quantified by the standard deviations..."
Well, the crucial step is missing I think. I understand the wave-mechanical explanation well, and I appreciate that different observables are associated with different eigenstates.
Can someone fill in these two points or suggest a reference for:
In what sense can a position eigenstate be represented as a sum of momentum eigenstates? Maybe a homely example would do here...
How does the math of a position eigenstate represented by multiple momentum eigenstates translate into the product of variances?
If the answer is too involved a reference would be great.