Is There an Eigenstate for the Creation Operator?

[George Jones]

I apologize for the harsh tone of post #25.

Maybe I'm being picky, but I haven't seen anywhere in this thread a demonstration that $a^\dagger$ has no eigenvalues and eigenvectors. Such a demonstration follows directly from

so thus, could i now conclude that the eigenvalues are continuously increasing when applying the creation operator, and hence the operator does not have an eigenstate

but, at this introductory level, I think the details need to be filled in.

Since $\left\{ |n> \right\}$ is a basis for state space, an arbitrary state vector $|\psi>$ can written

$$|\psi> = \sum_{n=0}^\infty c_n |n>.$$

Is it possible to find just the right values of the $c_n$'s (e.g., some equal to zero, some equal to -1, some equal to $1/\sqrt{2}$, some equal to 42, etc.) such that $|\psi>$ is an eigenvector of $a^\dagger$?

Vigorous hand waving not allowed.

 

[indigojoker]

Since $\left\{ |n> \right\}$ is a basis for state space, an arbitrary state vector $|\psi>$ can written

$$|\psi> = \sum_{n=0}^\infty c_n |n>.$$

Is it possible to find just the right values of the $c_n$'s (e.g., some equal to zero, some equal to -1, some equal to $1/\sqrt{2}$, some equal to 42, etc.) such that $|\psi>$ is an eigenvector of $a^\dagger$?

Vigorous hand waving not allowed.

I don't think so, but I'm not sure how to prove why.