I begin with a definition of the $f$ representation and $q$ representation of a wave function [1]. Next, in Proposition 1, I use the definition to show that the use of Dirac-delta distributions as eigenfunctions is self consistent with the definition of wave function in both the $f$ representation and the $q$ representation. To be clear, in Proposition 1, the Dirac-delta distributions are not shown to be solutions of any particular eigen-value equation. Finally, I show that Dirac-delta distributions are the eigenfunctions of the position operator since they yield the correct value of the average position.
Definition [Wave function in the $f$ representation and in the $q$ representation]
$\,$
(1) The function $a(f)$ is called a wave function in the $f$ representation.
- $a(f) = \int \Psi(q)\,\Psi_f^*(q)\,dq$
with normalization $\int\Psi_{f}^*(q^\prime)\, \Psi_f(q)\,df = \delta(q^\prime - q)$
is an expansion of $a$ in terms of the functions
$\Psi_f^*$ along with the expansion coefficient
$\Psi(q)= \int a(f) \, \Psi_f (q)\,df$.
- $\left|a(f)\right|^2$ determines the probability for the physical quantity $f$ to lie in a given interval $df$.
- The functions $\Psi^*_f$ are the eigenfunctions of the coordinate $q$ in the $f$ representation.
(2) The function $\Psi(q)$ is called a wave function in the $q$ representation.
- $\Psi(q) = \int a_f\,\Psi_f(q)\,df$.
with normaliation $\int\Psi_{f^\prime}(q)\, \Psi_f^*(q)\,dq = \delta(f^\prime - f)$
is an expansion of $\Psi$ in terms of the functions
$\Psi_f$ along with the expansion coefficient
$a_f= \int\Psi(q)\,\Psi_f^*(q)\,dq$.
- $\left|\Psi(q)\right|^2$ determines the probability for the system to have coordinates lying in a given interval $dq$.
- The functions $\Psi_f(q)$ are the eigenfunctions of the quantify $f$ in the $q$ representation.
Now, I show that wave functions constructed with Dirac-delta distributions are consistent with the definition above.
Proposition 1
Let $q_f\in\mathbb{R}$ and Let $\theta\in [-\pi,\pi]$. Suppose
the functions $\Psi^*_{q_f} = e^{-i\theta}\,\delta(q-q_f)$ are the eigenfunctions of the coordinate $q$ in the $f$ representation.
Then:
(1) The wave function in the $f$ representation is an expansion of $a$ in terms of the functions
$\Psi_{q_f}^*$ where
$$a(q_f) = \int_{-\infty}^\infty \Psi(q)\,e^{-i\theta}\,\delta(q-q_f)\,dq = \Psi(q_f)\,e^{-i\theta}, $$
and with normalization
$$\int e^{-i\theta}\,\delta(q^\prime-q_f)\, e^{+i\theta}\,\delta(q-q_f)\,dq_f = \delta(q^\prime - q) = \delta(q - q^\prime)\tag{2}$$
and with expansion coefficient
$$\Psi(q)= \int_{-\infty}^\infty \,
\Psi(q_f)\,e^{-i\theta}\, e^{+i\theta}\,\delta(q-q_f)\,dq_f = \Psi(q).$$
Note that
$\left|a(q_f)\right|^2= \left|\Psi(q_f)\right|^2$ determines the probability for the physical quantity $q$ to lie in the interval $dq_f$.
(2) The wave function in the $q$ representation is
an expansion of $\Psi$ in terms of the functions
$\Psi_f$ along with the expansion coefficient $a_f$ where
$$
a_{q_f}= \int\Psi(q)\,e^{-i\theta}\,\delta(q-q_f)\,dq = \Psi(q_f)\,e^{-i\theta}
,
$$
and with normalization
$$
\int_{-\infty}^\infty e^{ i\theta}\,\delta(q-q_f^\prime)\, e^{-i\theta}\,\delta(q-q_f)\,dq =
\delta(q^\prime-q_f) = \delta(q_f^\prime - q_f)= \delta(q_f - q_f^\prime).
\tag{4}$$
Note that
$$
\left|\Psi(q)\right|^2
=
\left|a_{q_f}\right|^2
$$ determines the probability for the system to have coordinates lying in a given interval $dq$. Also note that
$$\Psi(q)
=
\int \Psi(q_f)\,e^{-i\theta}\,e^{+i\theta}\,\delta(q-q_f) \,dq_f
=
\Psi(q ).
$$
Now, Landau explains [1] that we define the integral operator $\hat{q}$ in such a way that the integral product of $\hat{q}\Psi$ and the complex conjugate function $\Psi$ is equal to the mean value $\overline{q}$:
$$
\overline{q} = \int_{-\infty}^\infty \Psi^*(q) \left(\hat{q}\Psi \right)(q)\,dq. \tag{6}
$$
He further states that the form of the operators for various physical quantities can be determined from direct physical considerations. Well, the physical consideration offered is that
$$
\overline{q} = \int_{-\infty}^\infty \Psi^*\!(q)\,q\,\Psi \!\left(q\right)\,dq. \tag{8}
$$
Comparing (6) and (8), we deduce that
$$
\hat{q} = q.
$$
Now, we speak here of a physical quantity $q$, which is the coordinate. To this physical quantity there corresponds its operator $\hat{q}$. The eigenfunctions $\Psi_{q_f}$ corresponds to states in which the physical quantity $q$ has a definite value. In other words,
the eigenfunctions $\Psi_{q_f}$ corresponds to a definite eigenvalue $q_f$. Suppose that $\Psi$ is in the definite state $\Psi_{q_f}$. Then I expect that \begin{align*}
\overline{q} &= q_f \,\delta(0)\tag{10}
\end{align*}
To understand how come a distribution is given on the right-hand side of (10), the interested reader should see note [2] below.
All that is left to show is that when the operator is applied to a wave function in a definite state that the associated eigenfunction produces the anticipated mean value of the position. In other words that $\Psi_{q_f} = e^{i\theta}$ produces $q_f\,\delta(0)$. In fact,
\begin{align*}
\overline{q} & = \int \Psi_{q_f}^*(q) \left(\hat{q}\Psi_{q_f}\right)(q)\,dq.
\\
& = \int \Psi_{q_f}^*(q) \,q \,\Psi_{q_f}\!\left(q\right)\,dq.
\\
& = \int e^{ -i\theta}\,\delta(q-q_f^\prime) \,q \,e^{ i\theta}\,\delta(q-q_f^\prime) \,dq.
\\
& = \int \delta(q-q_f^\prime) \,q \, \delta(q-q_f^\prime) \,dq.
\\
& = q_f^\prime \,\delta(q_f^\prime-q_f^\prime).
\\
& = q_f \,\delta(0) .
\end{align*}
We have shown what we wish to show: the usage of Dirac-distributions are consistent with representations of wave functions, and when applied to a wave function in a definite state, the associated eigenfunction produces the anticipated mean value of the position.
Bibliography
[1] Landau and Lifshitz, Coure on Theoretical Physics, Volume 3, pp. 10-11, 18.
[2] What I propose in the body of my answer may seem counterintuitive. For, in the case of a discrete spectrum of eigenvalues one might justifiable propose that
\begin{align*}
\overline{q} &= q_f .\tag{20}
\end{align*}
Many students of quantum physics seem to use (20) as well for the continuous spectrum $q$ as well. Yet, for a continuous spectrum of eigenvalues, as given in (2) and (4) we have that
$$\int_{-\infty}^\infty\Psi_{q_f^\prime}(q)\,\Psi_{q_f }(q)\,dq =\delta(q_f^\prime-q_f) $$
and that
$$
\int_{-\infty}^\infty\Psi_{q_f}(q^\prime) \Psi_{q_f }(q) =\delta(q_f^\prime-q_f)\,dq_f = \delta (q^\prime - q).
$$
Therefore to propose (20) appears unjustifiable to this author. Rather, what appears justifiable is more like
\begin{align*}
\overline{q}
&=
\int_{-\infty}^\infty\Psi_{q_f }^*(q)\,\left(\hat{q} \Psi_{q_f }\right)(q) \,dq
\\
&=
\int_{-\infty}^\infty\Psi_{q_f }^*(q)\, q_f\, \Psi_{q_f }(q) \,dq
\\
&=
q_f\,\delta (q_f - q_f)
\\
&=
q_f\,\delta (0)
.
\end{align*}
