It seems that ladder operators exist for nearly all operators. We have the ladder operators for the hamiltonian, for spin, and even for many particle systems where the ladder operators act as creation and annihilation operators.
Are there ladder operators for every observable? Is there a general way to determine what form the ladder operators would take?
Thank you.
Provided that the spectrum of an operator is ordered: $$ \hat{A}|n\rangle=a_n|n\rangle $$ we can always construct operator $$ b=\sum_k |k\rangle\langle k+1|,b^\dagger=\sum_k|k+1\rangle\langle k| $$ so that $$ b|n\rangle = |n-1\rangle,b^\dagger|n\rangle=|n-1\rangle $$ (Some care is needed when $n$ is close to the boundaries of the spectrum - e.g., for a harmonic oscillator $n$ cannot be less than $0$, so that $b|0\rangle=0$.)
Now, we can see that $b\neq b^\dagger$, i.e., the ladder operator is not Hermitian, and as such cannot correspond to an observable quantity.
A ladder operator is nothing else than a shift operator with respect to the eigenvalues of another operator in the end. Roger V. explains in his answer how to construct the said ladder operator in the case of a discrete spectrum. When the latter is continuous, then it is given simply by $\hat{L} = e^{a\partial_x}$, hence $\hat{L}|x\rangle = |x+a\rangle$, where $|x\rangle$ denotes the eigenvector of an operator $\hat{X}$.
