# Do eigenstates of the creation operator actually exist?

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?

• Note that Cosmos does adress the loophole at the end of the answer you refer to. Jan 19 at 21:06
• @NDewolf I didn't understand it. I thought any state should be represented by the basis states of the Hilbert space. Therefore the argument that prove that no eigenstate exists sounds valid to me. Jan 19 at 21:21
• It is an admittedly hyper-formal distribution loophole, and a seat-of-the-pants extension of the legitimate Hilbert space. But un-normalizable states, like $|x\rangle$, are part of the toolbox of the trade, and people utilize their formal algebraic features all the time. The "loophole" is not meant to violate principles of QM: it is meant to solve problems, if one were in the mood to. Much of the operator-valued distribution stunts of QFT may or may not have chapter-and verse constructive proofs. Jan 19 at 21:28
• Dirac, who had an electrical engineering degree, rarely glommed on the "existence" of entities or not. You might consider using these states as a basis of Hilbert space states, just as the nonexistent $|x\rangle$ can span bona-fide Hilbert space states. I've sat through so many lectures on the strict nonexistence of fish... Jan 20 at 1:22
• @CosmasZachos Thank you for your comment. Now I understand the loophole. The proof assumes that $c_n$ coefficients are complex numbers, while formally they can be derivatives of Dirac's delta function. Also the eigenstates of creation operator are not physical since they are not normalizable. Jan 20 at 4:28