Condition for discrete spectrum and ladder operator In my first course in quantum mechanics we have seen three operators with discrete spectrum: the Hamiltonian of an harmonic oscillator $\hat{H}$, the square of the angular momentum $\hat{L^{2}}$ and $\hat{L_{z}}$. I followed the algebraic method for each case. It's seems to me that this method depends on these two condition:

*

*the spectrum of the operator is bounded (from below for $\hat{H}$, from below and above for $\hat{L_{z}}$ for each eigenvalue of $\hat{L^{2}}$)

*there are raising and lowering ladder operator

Are this two condition sufficient to state that the spectrum of a general self-adjoint operator is discrete? If it is true, there are arguments to state the uniqueness of the ladder operator's step?
 A: The existence of ladder operators is trivially equivalent to the discreteness of the spectrum (of the operator whose spectrum is being traversed by the said ladder operators).
Let's consider a generic discrete Hermitian operator $\hat{O} = \sum_no_n\vert n\rangle\langle n\vert$. Clearly, one can construct the operators $\hat{a}^\dagger\equiv  \sum_nc_n\vert n+1\rangle\langle n\vert$ and $\hat{a}\equiv\sum_nd_n\vert n-1\rangle\langle n\vert$. These operators will be the creator and annihilation operators respectively. One would need to adjust the summation ranges according to the boundedness but that is more or less irrelevant.
Similarly, if you have the creation operators (or annihilation operators) then, by definition, it means that they map a given eigenstate of the operator to the next eigenstate (or previous eigenstate) of the operator. This very definition implies that we are talking about an operator with a discrete spectrum because otherwise the talk of next or previous eigenstate does not make any sense.
