## Momentum Eigenstates

For simplicity I consider only the one-dimensional case.
Why are momentum eigenstates as follows?

When we ask the same question about position eigenstates |x> we have no doubts:

but it’s not so obvious why momentum eigenstates are in the form of fig.1.
Leaving out the constant portion of them (fig. 1), we want to understand why they must be in the form:

So we have got a momentum eigenvalue p and we want to deduce its associated eigenstate form in the |x> base (as a function of x).

First of all: can its magnitude depend on position? No! Because otherwise, it would mean that, chosen p, some positions are more special than others. But momentum is logically independent of position.

So it must have constant magnitude over all positions, from – infinity to + infinity.

Could it be real, or equivalently, could its phase be constant? If so, we wouldn’t have any way to distinguish momentum eigenstates!
Can phase be a function of x different from that expressed in fig. 3 (phase = kx)?

First of all, the eigenstate function must be periodic, otherwise we wouldn’t have any information (independent of position ) to associate to momentum p: phase periodicity is the only possible information that is not dependent on position.

The simplest function with this characteristic is that of fig.3.

But you could answer back: why not a different periodic function, not regular within its period as that of fig. 3? in fact it seems the only requirement should be that the opposite momentum eigenstate phase should satisfy relation (to be specular):