In a lecture on Quantum mechanics, the professor concluded that if $ a$ is a linear operator with $ [a, a^\dagger] = 1$ , where $ a^\dagger$ is the adjoint of $ a$ and $ [a, a^\dagger] = aa^\dagger – a^\dagger a$ , and if $ a^\dagger a$ has an eigenvalue, then vectors of form \[ | n \rangle = (a^\dagger)^n | 0 \rangle \quad ( n \in \mathbb Z_{ \ge 0 }) \] are all eigenvectors of $ a^\dagger a$ . ($ | \lambda \rangle$ is the eigenvector of $ a^\dagger a$ which belongs to the eigenvalue $ \lambda$ ).

But he gives no reason why its eigenvalue exists. So, I asked him if an eigenvalue exists, and he said ‘I cannot answer. Roughly speaking, it is some limit of the case of a finite dimensional complex vector space.’

My question is: Let $ H$ be an infinite dimensional Hilbert space, say $ \ell^2$ space, whose inner product is denoted by $ \langle \cdot, \cdot \rangle$ . It is well-known that for a bounded linear operator $ T \in \mathcal L(H)$ , there corresponds a (bounded) linear operator $ T^\dagger \in \mathcal L(H)$ such that $ \langle x, Ty \rangle = \langle T^\dagger x, y \rangle$ for all $ x, y \in H$ . Then, is it true that for any bounded linear operator $ T \in \mathcal L(H)$ with $ [T, T^\dagger] = 1$ , there exists a scalar $ \lambda \in \mathbb C$ and a nonzero vector $ x \in H$ such that $ T^\dagger T x = \lambda x$ ?

How could I deal with it? Is there a ‘characteristic polynomial’ of $ T^\dagger T$ ?

Please give me the answer, or some references.