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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?

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    $\begingroup$ Your two conditions don't make much sense to me, because you didn't specify how the operator in question is related to your ladder operators. There are obviously operators with bounded continuous spectra. There also always exist ladder operators, just pick an arbitrary basis in the Hilbert space enumerated by integers and define your ladder operators in the same way as you do with oscillator ladder operators, only applied to this basis. For the oscillator, $H = a^{\dagger} a+1/2$ which is crucial for the proof of the discreteness of the spectrum of $H$, but you didn't give any such relation. $\endgroup$ Mar 19 at 12:45
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    $\begingroup$ Thanks you @Prof.Legolasov for your comment. I didn't specified the relation between the general self-adjoint operator and his ladder operator because it changes in the example i have seen (harmonic oscillator and angular momentum). Because i like to know if it is general, i didn't specified it. $\endgroup$
    – Luigi E.
    Mar 19 at 13:05
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    $\begingroup$ Well, your two claims obviously don't imply the discreteness of the spectrum in the general case where there isn't any relation between the operator in question and the ladder operators. $\endgroup$ Mar 19 at 13:07
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    $\begingroup$ Thanks you for your clarification @Prof.Legolasov. When I asked this question I suspected that the key was in the relation with the operator but I cannot see clear similarities between the two relation: $H = a^{\dagger}a + \frac{1}{2}$ and $L_{+}L_{-} = L^{2} - L^{2}_{z} + \hbar L_{z}$. Any suggestion where I can clarify this relation are welcomed. $\endgroup$
    – Luigi E.
    Mar 19 at 13:25
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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.

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  • $\begingroup$ "The existence of ladder operators is trivially equivalent to the discreteness of the spectrum" – the spectrum of what? $\endgroup$ Mar 19 at 12:39
  • $\begingroup$ @Prof.Legolasov I thought it was clear from the context but I added the clarification. I meant the spectrum of the operator whose spectrum is being traversed by the ladder operators. $\endgroup$
    – Dvij D.C.
    Mar 19 at 12:50
  • $\begingroup$ @Prof.Legolasov Just saw your comment on the question. Yes, the OP does not clarify the relation of the ladder operator to the operator in question. I am assuming that they meant a ladder operator who traverses the spectrum of the said operator. But given that then everything is trivial, maybe OP meant something else? I will modify my answer if/when OP responds to your comment :/ $\endgroup$
    – Dvij D.C.
    Mar 19 at 12:53
  • $\begingroup$ It is still unclear to me what it means for an operator "to have a spectrum that is traversed by the ladder operators". Can you please make a precise statement? What if only part of the spectrum is traversed by these operators, then your proof seems to not work? Or do you require that the entire spectrum is "traversed"? In the latter case, I think your claim is basically trivial: "if the spectrum is discrete, then it is discrete". $\endgroup$ Mar 19 at 12:54
  • $\begingroup$ @Prof.Legolasov Yes, it is the latter case and indeed I preface my answer by saying that it is a trivial claim ;) $\endgroup$
    – Dvij D.C.
    Mar 19 at 12:55

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