Ladder operators, in quantum mechanics, could be used to derive eigenkets of energy with different eigenvalues (i.e. the system's energy) by raising and lowering it.

However, is there a proof that it proves a COMPLETE set of eigenkets?

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    $\begingroup$ Related: physics.stackexchange.com/q/23028/2451 , physics.stackexchange.com/q/54691/2451 , physics.stackexchange.com/q/300829/2451 and links therein. $\endgroup$ – Qmechanic Dec 5 '17 at 7:19
  • $\begingroup$ @Qmechanic I'm a 2nd year undergrad who never have been exposed to Hilbert spaces -- is there any other way to prove it? (my course in Oxford is based on the book "The Physics of Quantum Mechanics" which didn't seem to have proven it) $\endgroup$ – user148792 Dec 5 '17 at 7:22
  • $\begingroup$ If you don’t know anything besides the ladder operators, completeness is just an assumption you put in. $\endgroup$ – knzhou Dec 5 '17 at 9:55
  • $\begingroup$ Any good book on functional analysis will prove the spectral theorem for self-adjoint operators on a Hilbert space; ladder operators are related to lie algebras. $\endgroup$ – Mozibur Ullah Dec 5 '17 at 12:33

Relying only on commutation relations of $a$ and $a^\dagger$ it is not possible to prove completeness of the produced basis by the action of $a^\dagger$ on $\psi_0$, since counterexamples can be constructed easily. If you instead refer to the whole theory, assuming that the Hilbert space is $L^2(\mathbb R, dx)$ and $a$ and $a^\dagger$ are constructed as you know out of $X$ and $P$, you can prove completeness observing that (1) $\psi_0$ is the first Hermite function and (2) the recurrence formula of $\psi_n$, $\sqrt{n+1}\psi_{n+1} = a^\dagger \psi_n$ is the same as for the known Hilbert basis of Hermite functions. This proves completeness.

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