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?

  • 1
    $\begingroup$ Note that Cosmos does adress the loophole at the end of the answer you refer to. $\endgroup$
    – NDewolf
    Jan 19 at 21:06
  • $\begingroup$ @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. $\endgroup$
    – SMA.D
    Jan 19 at 21:21
  • $\begingroup$ 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. $\endgroup$ Jan 19 at 21:28
  • 2
    $\begingroup$ 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... $\endgroup$ Jan 20 at 1:22
  • 1
    $\begingroup$ @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. $\endgroup$
    – SMA.D
    Jan 20 at 4:28

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