# Eigen-States of Creation Operator

I know the annihilation operator has eigen-states

$$\hat{\alpha} |\alpha \rangle = {\alpha} |\alpha \rangle$$

I also know the creation operator $$\alpha^{\dagger}$$ has no eigenstates. Is the following a correct proof of this fact

The annihilation operator has eigen-states

$$\hat{\alpha} |\alpha \rangle = {\alpha} |\alpha \rangle$$

Let's suppose the creation operator did infact have eigen-states as follows

$$\hat{\alpha}^{\dagger} |\beta \rangle = {\beta} |\beta \rangle$$

Taking dagger on both sides I get

$$\langle \beta | \hat{\alpha} = {\beta^{*}} \langle \beta |$$

But this implies $$|\beta\rangle$$ is an eigen-state of $$\hat{\alpha}$$ with eigen-value $$\beta^{*}$$

Now since $$\hat{\alpha}$$ and $$\hat{\alpha}^{\dagger}$$ do not commute, hence they share no similar eigen-states, hence this contradicts the assumption that the $$\hat{\alpha}^{\dagger}$$ has eigen-states.

• I think you are failing to distinguish between the right and the left eigenstates. These are not the same for a non-hermitian operator. Dec 9 '20 at 8:52
• Thanks, thats where i was unsure really, because technically the bra is the eigenstate in what i did, not the ket
– Dk65
Dec 9 '20 at 8:56

What you have presented cannot be a proof, for it is symmetric wrt interchanging $$a$$ and $$a^\dagger$$, you haven't used the property that tells them apart, i.e. $$[a,a^\dagger]=1$$. You can argue like that, assume there is an eigenstate of $$a^\dagger$$. It can be written in the following form $$$$| \alpha \rangle = \sum_{n=0}^\infty C_n (a^\dagger)^n | 0 \rangle.$$$$ Using the defining property of an eigenstate we get $$$$\sum_{n=0}^\infty C_n (a^\dagger)^{n+1} | 0 \rangle=\alpha\sum_{n=0}^\infty C_n (a^\dagger)^{n} | 0 \rangle,$$$$ which can be rewritten as $$$$\sum_{n=1}^\infty \left ( C_{n-1}-\alpha C_n \right ) (a^\dagger)^{n} | 0 \rangle=\alpha C_0 | 0 \rangle,$$$$ that in turn cannot be satisfied, since all $$(a^{\dagger})^n | 0 \rangle$$ are linearly independent.
• As I said, no. Turn it around and assume that there exist an eigenstate of $a$, and follow the same steps as in you suggested proof. You would have to conclude that it does not exist. Dec 9 '20 at 8:57