What is the difference between an eigenfunction and a wavefunction? This question is an additional point of clarification to my previous question about adding position and momentum eigenstates.
For simplicity, suppose I had a particle in an eigenstate of momentum, $|p\rangle$, with eigenvalue $p$ (I know this is non-normalizable, but bear with me here - the problem I pose exists for superpositions of these states over $\mathbb{C}$ as well). I want to find its wavefunction in the position basis. This is most commonly given as:
$$f_p(x)=\langle x|p\rangle=\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}$$
But this isn't quite the wavefunction, since it doesn't have the right dimensions of $\mathsf{L}^{-1/2}$; instead, it has dimension $\mathsf{M}^{-1/2}\mathsf{L}^{-1}\mathsf{T}^{1/2}$. The same problem exists with position eigenstates:
$$g_y(x)=\delta(x-y)$$
These eigenstates have dimension $\mathsf{L}^{-1}$.
However, I recall that in my studies of the energy eigenstates of various scenarios, like the infinite square well and free particle, the eigenstates all had the right dimensions to be wavefunctions.
In the answers to my previous question, it was said that the eigenstates $|x\rangle$ and $|p\rangle$ are "rigged vectors" that carry dimensions, instead of state vectors, which are dimensionless.
So, what are these "rigged vectors"? Is there something special about energy that makes its eigenstates have the "right" dimensions? More generally, what are the differences between an eigenstate $|q\rangle$ of an observable $\hat{Q}$ and a wavefunction $\Psi(x)$?
 A: For a discrete basis $|n\rangle$, the orthogonolity condition $\langle n |m\rangle = \delta_{nm}$ imposes that the wave-functions satisfy :
$$\int \text d^dx \psi_n^*(x) \psi_m(x) = \delta_{nm} \tag{1}$$
This imposes that the wave-functions have dimension $\mathsf L^{-d/2}$.
For the momentum eigenstates, the orthogonality condition is $\langle p|q\rangle = \delta^{(d)}(p-q)$ imposes on the wave-functions :
$$\int \text d^dx \psi_p^*(x)\psi_q(x) = \delta^{(d)}(p-q) \tag{2}$$
Since the delta-function on the RHS has dimension $\mathsf M^{-d}\mathsf L^{-d} \mathsf T^d$, the wave functions must have dimension $\mathsf M^{-d/2} \mathsf L^{-d} \mathsf T^{d/2}$.
Likewise, for position eigenstates, we find that the normalisation condition $\langle x |y \rangle = \delta^{(d)}(x-y)$ imposes that  the wavefunctions have dimension $\mathsf L^{-d}$.
More generally, for a given hermitian operator, when the spectrum is discrete, the orthonormality condition gives Kronecker deltas like $(1)$, so that the natural way to normalize the wavefunctions gives them dimension $\mathsf L^{-d/2}$. When the spectrum is continuous, the wavefunctions are not normalizable (but one can still make sense of them rigorously as "rigged vectors"). The orthogonality condition involves a delta-function which is dimensionful, so the natural normalization for the wavefunctions gives them dimension different from $\mathsf L^{-d/2}$.
This only concerns the natural normalization though. If you are willing to introduce some dimensionful consants, then you can always decide to have your wave functions with dimension $\mathsf L^{-d/2}$ (or really any dimension you like). For example, if $a$ is an arbitrary length scales, you can define position eigenstates $|\xi\rangle$ by (say $d=1$ for simplicity):
$$\hat x|\xi\rangle  =a\xi|\xi\rangle$$
where $\xi$ is dimensionless, along with the normalisation condition :
$$\langle \xi|\xi'\rangle = \delta(\xi - \xi')$$
The wavefunctions are then $\psi_\xi(x) = \sqrt{a}\delta(x-a\xi)$, which indeed have dimension $\mathsf L^{-1/2}$
