Is there a simple way of finding the eigenstates of the creation and annihilation operator in QM?

Question

How can I find the eigenstates of creation and annihilation operator in QM?

My attempt:

Such eigenstate will obey: $$ a^{\dagger} |\psi \rangle = \alpha |\psi \rangle. $$

We can expand $|\psi \rangle$ in terms of the quantum SHM eigenstates: $|\psi \rangle = \sum_{n=0}^{\infty} c_n |n\rangle $.

Knowing the action of the creation operation of quantum SHM eigenmodes ($a^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle)$:

$$ a^{\dagger} |\psi \rangle = a^{\dagger} \sum_{n=0}^{\infty} c_n |n\rangle = \sum_{n=0}^{\infty} c_n \sqrt{n+1}|n+1\rangle$$

from which the state $|0\rangle$ is now missing, so it will never be equal to the RHS of the first expression, $\alpha |\psi\rangle = \sum_{n=0}^{\infty} \alpha c_n |n\rangle $.

Answer 1

Write an arbitrary state as

$$|\Psi\rangle = \sum_{n=0}^{\infty} c_n |n\rangle \,.$$

Now apply the raising operator

$$ \begin{align} a^\dagger |\Psi\rangle &= a^\dagger \sum_{n=0}^{\infty} c_n |n\rangle \\ &= \sum_{n=0}^{\infty} c_n \sqrt{n+1} |n+1\rangle \\ &= \sum_{n=1}^{\infty} c_{n-1} \sqrt{n} |n\rangle \end{align} $$

If $|\Psi\rangle$ is an eigenstate of $a^\dagger$ with eigenvalue $\alpha$ then we have

$$\sum_{n=0}^{\infty} \alpha c_n|n\rangle = \sum_{n=1}^\infty c_{n-1} \sqrt{n}|n \rangle \, .$$

You already got this far. Indeed, the only solution to this equation is $c_n=0$ for all $n$. Therefore, there is no eigenstate of $a^\dagger$.

The eigenstates of $a$, which are called "coherent states" are given by

$$ |\alpha \rangle = e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}|n\rangle \, . $$ You can check easily by applying $a$ to $|\alpha \rangle$ that $|\alpha \rangle$ is an eigenstate of $a$.