Regarding the eigenstate of creation operator $\hat{a}^\dagger$, the answer to this question shows that the eigenstate does not exist.
However, it is stated in another answer, that the proof has loophole and the state $$|\psi\rangle=\delta(a^\dagger - \beta) |0\rangle =\frac{1}{2\pi} \int \!\! dk ~e^{ik(a^\dagger -\beta)} |0\rangle $$
satisfies $$a^\dagger | \psi\rangle=\beta| \psi\rangle,$$
I couldn't find the loophole in the proof. Is $|\psi\rangle$ really an eigenstate of $\hat{a}^\dagger$? How can we obtain it? What is the loophole in the proof?
