Informal proof. To show that possible eigenvectors of the position operator should necessarily be Dirac delta distributions, suppose that ψ ψ is an eigenstate of the position operator with eigenvalue x 0 . We write the eigenvalue equation in position coordinates, x ^ ^ ψ ψ ( x ) = x ψ ψ ( x ) = x 0 ψ ψ ( x ) recalling that x ^ ^ simply multiplies the wave-functions by the function x , in the position representation. Since the function x is variable while x 0 is a constant, ψ ψ must be zero everywhere except at the point x 0 . Clearly, no continuous function satisfies such properties, and we cannot simply define the wave-function to be a complex number at that point because its L 2 -norm would be 0 and not 1. This suggest the need of a "functional object" concentrated at the point x 0 and with integral different from 0: any multiple of the Dirac delta centered at x 0 . The normalized solution to the equation x ψ ψ = x 0 ψ ψ is ψ ψ ( x ) = δ δ ( x − − x 0 ) , or better ψ ψ = δ δ x 0 . Proof. Here we prove rigorously that x δ δ x 0 = x 0 δ δ x 0 . Indeed, recalling that the product of any function by the Dirac distribution centered at a point is the value of the function at that point times the Dirac distribution itself, we obtain immediately x δ δ x 0 = x ( x 0 ) δ δ x 0 = x 0 δ δ x 0 . Meaning of the Dirac delta wave. Although such Dirac states are physically unrealizable and, strictly speaking, they are not functions, Dirac distribution centered at x 0 can be thought of as an "ideal state" whose position is known exactly (any measurement of the position always returns the eigenvalue x 0 ). Hence, by the uncertainty principle, nothing is known about the momentum of such a state.