= Solution
Generalized <position eigenstates> satisfy $\langle x|x'\rangle=\delta(x-x')$ and $\int|x\rangle\langle x|dx=I$. Define the position-space <wavefunction> $\psi(x)=\langle x|\psi\rangle$. Completeness then gives $\langle\psi|\psi\rangle=\int|\psi(x)|^2dx$. Position acts by multiplication, $\langle x|\widehat x|\psi\rangle=x\psi(x)$. Using $\langle x|p\rangle=(2\pi\hbar)^{-1/2}e^{ipx/\hbar}$ and <momentum> completeness, differentiating the Fourier integral gives $\langle x|\widehat p|\psi\rangle=-i\hbar\psi'(x)$ on the <momentum operator>'s domain.
For the unit-mass, unit-frequency oscillator define
$$
a=\frac{\widehat x+i\widehat p}{\sqrt{2\hbar}},\quad a^\dagger=\frac{\widehat x-i\widehat p}{\sqrt{2\hbar}},\quad[a,a^\dagger]=1,\quad H=\hbar(a^\dagger a+1/2).
$$
Since $[H,a]=-\hbar a$, a simultaneous <energy eigenstate> and $a$-eigenstate would satisfy $[H,a]|\psi_\alpha\rangle=0=-\hbar\alpha|\psi_\alpha\rangle$, forcing $\alpha=0$. For $\alpha=0$, the energy is $\hbar/2$, the smallest possible because $\langle a^\dagger a\rangle=\|a\psi\|^2\geq0$.
The position-space equation is $(x+\hbar\partial_x)\psi_\alpha=\sqrt{2\hbar}\alpha\psi_\alpha$. Solving and normalizing, with $\alpha=u+iv$, gives up to a constant phase
$$
\boxed{\psi_\alpha(x)=(\pi\hbar)^{-1/4}\exp\left[-\frac{(x-\sqrt{2\hbar}\,u)^2}{2\hbar}+i\sqrt{\frac2\hbar}\,v x\right].}
$$
Its squared modulus is a normalized Gaussian, so every complex $\alpha$ gives a normalizable <coherent state>.
For a creation eigenstate, $(x-\hbar\partial_x)\psi=\sqrt{2\hbar}\beta\psi$ instead gives $\psi=C\exp[x^2/(2\hbar)-\sqrt{2/\hbar}\,\beta x]$. Every nonzero such function grows quadratically exponentially at infinity and fails square integrability. Hence \b[there are no nonzero normalizable eigenstates of $a^\dagger$]. This last symbol is $a^\dagger$ in the original PDF, although the converted TeX drops its dagger.
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