# Why are eigenstates of position (or momentum) not well defined?

For a free particle, there is a corresponding free wavefunction freely developing through spacetime. A wavefunction can be broken up in Hilbert space just as a vector in ordinary space can be broken up into its components. The base vectors of the Hilbert space are the eigenfunctions (eigenstates) of an operator acting on the wavefunction.
When a measurement is made the wavefunction is projected onto one of the eigenstates (with the corresponding eigenvalue, the result of the measurement). When we make a position measurement, though, the wavefunction ends up being localized in a small volume. There is not just one value of position, but a continuous small range.
It is, in theory, possible to make a precise measurement, giving us a precise value of the position. The corresponding "eigenfunctions" are said to be Dirac delta functions. Or, better, Dirac delta distributions.
Now I've read many times that these "eigenfunctions" (Dirac deltas) are not really quantum mechanical states. For example, @ACuriousMind writes in a comment to this question:

It might be worth it to add a caveat about |x⟩ and |p⟩ not actually being states

While @LubosMotl writes:

Yes, they [the Dirac deltas] are the (not normalizable) eigenstates of the operators. And some things are not stated somewhere because not every book lists all true statements.

I'm not sure if these two comments contradict each other. An eigenstate has to be normalizable. Motl writes that this is not the case. Is this what ACuriousMind means when he says that eigenstates of position and momentum don't exist? Is the fact that the Dirac delta is a distribution (and not a function) of importance? Does this mean that the wavefunction can't be written as a superposition of states (regarding the position or momentum operator)? Somehow infinity is involved, I guess.
It could also be that it's just a matter of words. A particle has to be in some "state".

• To be short (hence as a comment): yes, it being a distribution makes it impossible to be a state. States should be square-integrable functions, i.e. elements of $L^2(\mathbb{R}^n)$. Through convolutions one can define states in terms of ''infinite'' superpositions of such distributional ''eigenstates''. Mar 24, 2021 at 8:54

I believe it's because these eigenstates aren't normalizable and so don't actually belong to the Hilbert space we are interested in. You can however use them to construct a well-defined state by a linear combination of them. If position eigenstates were your basis, then simply the coefficients of the linear combination would give you a function of $$x$$: $$\begin{equation} f(x)=\int dy f(y)\delta(x-y) \end{equation}$$
While for the momentum eigenstates you would get a Fourier transform: $$\begin{equation} \tilde{f}(x)=\int dy f(y)e^{ixy} \end{equation}$$
The Hilbert space is the set of all physical states, which means that the are all normalized so that the produce the correct statistics in calculations. If you want to conform to strict mathematical practices, then the basis for the Hilbert space must be a subset of the elements in the Hilbert space. Hence, all the elements of the basis would be normalized states. However, that would mean that these elements cannot obey an orthogonality condition of the form $$\langle a|b\rangle = \delta(a-b) ,$$ because that would imply $$\|a\|^2 = \langle a|a\rangle \neq 1 .$$ So it turns out that such a (mathematically preferred) basis have to be a discrete basis with an orthogonality condition expressed in terms of a Kronecker delta instead of a Dirac delta $$\langle m|n\rangle = \delta_{m,n} ,$$