# Eigenstates of the creation operator

We know that coherent states $$\vert\alpha\rangle$$ are eigenvectors of the annihilation operator $$\hat{a}$$, i.e. $$\hat{a} \vert\alpha\rangle = \alpha \vert\alpha\rangle$$ while the creation operator $$\hat{a}^\dagger$$ has no eigenvector.

Now, I have few questions:

1. Would it be correct to say that $$\langle\alpha\vert$$ is (left) eigenvector of $$\hat{a}^\dagger$$ ? Can we use this formalism, and does it have some (physical) meaning?
2. In view of the above, I would say that the naive argument that is often found: "a coherent state resembles a classical state because if you annihilate excitations it does not change", is rather wrong. In fact the converse should also be true, which is only the case if $$\vert\alpha\rangle$$ is eigenstate of $$\hat{a}^\dagger$$ as well.
• $a^\dagger$ does not have right eigenvectors, but it has a left eigenvector, $\langle\alpha|$, precisely, as suggested in the question. This fact is routinely used in derivation of the path integral formulation in terms of coherent states. This is really a homework question. Commented Jan 20, 2021 at 12:49