A common question I have seen on problems sets and on exam papers is as follows (own words):
Show that $a^\dagger$ and $a$ satisfy the canonical commutation relations. Hence show write down the eigenvalues and eigenvectors of $a^\dagger a$.
The expected answer is as follows :
The eigenvalues are $n=0,1,\ldots$ and the eigenvectors are - spare a normalization: $$(a^\dagger)^n \left|0\right>$$ where $a \left|0\right>=0$.
It is clear from the commutation relations that if we have an eigenvalue $\lambda$ then we have a string of eigenvalues: $$\ldots, \lambda -2,\lambda -1, \lambda, \lambda+1,\lambda+2,\ldots$$ if $\lambda$ is a natural number then this string is bounded by below by $0$ and it is always bounded above by $\infty$. The expected answer I have given above has made two (in my opinion, fairly large) assumptions:
- There is only one string present. I.e. there is no eigenvalue $\lambda_2$ such that $\lambda_2 \ne \lambda+k$ for some $k \in \Bbb{Z}$.
- The string that is present has $\lambda \in \Bbb{N}$.
These assumptions hold for the case of the Harmonic oscillator - this can be shown by explicitly solving the Hamiltonian. But often there are no harmonic oscillators in sight - for example in the Fock space representation$^\star$.
Thus my question is: under what circumstances do we expect the above assumptions to be valid and why?
$\star$ Only an example - not the focus of my question.
