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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.

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  • $\begingroup$ I think you are failing to distinguish between the right and the left eigenstates. These are not the same for a non-hermitian operator. $\endgroup$ Dec 9 '20 at 8:52
  • $\begingroup$ Thanks, thats where i was unsure really, because technically the bra is the eigenstate in what i did, not the ket $\endgroup$
    – Dk65
    Dec 9 '20 at 8:56
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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 \begin{equation} | \alpha \rangle = \sum_{n=0}^\infty C_n (a^\dagger)^n | 0 \rangle. \end{equation} Using the defining property of an eigenstate we get \begin{equation} \sum_{n=0}^\infty C_n (a^\dagger)^{n+1} | 0 \rangle=\alpha\sum_{n=0}^\infty C_n (a^\dagger)^{n} | 0 \rangle, \end{equation} which can be rewritten as \begin{equation} \sum_{n=1}^\infty \left ( C_{n-1}-\alpha C_n \right ) (a^\dagger)^{n} | 0 \rangle=\alpha C_0 | 0 \rangle, \end{equation} that in turn cannot be satisfied, since all $(a^{\dagger})^n | 0 \rangle$ are linearly independent.

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  • $\begingroup$ okay thanks that makes sense, I was wondering if my proof would work if i had apriori knowledge that eigenstates for the annihilation operator exist. Then could I use my argument? $\endgroup$
    – Dk65
    Dec 9 '20 at 8:54
  • $\begingroup$ 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. $\endgroup$
    – nwolijin
    Dec 9 '20 at 8:57

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