We want to find the spectrum of the Harmonic Oscillator Hamiltonian: $$H=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega ^2 \hat{x}^2$$ From what I have seen in many books the procedure is as follows: We can define a destruction operator $a$ and a creation operator $a^\dagger$, such as: $$H=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega ^2 \hat{x}^2=\hbar \omega \left(a^\dagger a +\frac{1}{2}\right)$$ We then call: $$N := a^\dagger a$$ Now: if we find the spectrum of $N$ we have automatically found the spectrum of $H$, obviously. So let's start digging for the spectrum of $N$: we can easily prove that: $$[a^\dagger , a]=-1 \ \ ; \ \ [N , a]=-a \ \ ; \ \ [N , a^\dagger]=a^\dagger$$ And now, using these commutators, we can show the following: calling an eigenstate of $N$: $|n\rangle$ (so: $|n\rangle \ | \ N|n\rangle = \lambda _n |n\rangle$) $$Na^\dagger|n\rangle=(a^\dagger N + [N , a^\dagger])|n\rangle=(\lambda _n +1)a^\dagger | n \rangle$$ $$Na|n\rangle=(a N + [N , a])|n\rangle=(\lambda _n -1)a | n \rangle$$ So now, with this, we have managed to prove two things:
- The operators $a^\dagger , a$ by acting on an eigenstate of $N$ allow us to find another eigenstate of $N$ (Fact 1)
- The newly found eigenstate has eigenvalue one unit higher or lower, with respect to $\lambda _n$, depending on which of the two operators we chose to use (Fact 2)
Wonderful. Another thing that we can prove is that it must exist an eigenstate $|0\rangle$ of $N$ such that: $$a|0\rangle=0 \ \ \ (Fact \ 3)$$ this can be easily proven: the eigenvalues of the Harmonic Oscillator Hamiltonian must be positive (we omit the proof of this), but the operator $a$ always lowers the value of $\lambda$ by one, so the only option is that, at some point; applying $a$ on an eigenstate of $N$ returns the null vector.
Now here is my problem: In all the sources that I have found the author then goes on to state that fact 1, 2 and 3 allow us to say that the complete collection of all the eigenstates of the Harmonic Oscillator Hamiltonian is
$$|n\rangle = K_n (a^\dagger)^n|0\rangle \ \ \ \ \ \ n \in \mathbb{N} \tag{1}$$ with $K_n$ some number. So essentially all the eigenstates can be found by applying the creation operator $a^\dagger$ to the lowest state $|0\rangle$. My problem with this is: surely all the vectors that we can obtain with (1) are eigenstates of $N$ but there could be other eigenstates hiding somewhere! We have never managed to prove, in all this, that all the eigenstates of $H$ are findable by the creation and destruction operators, there could be other eigenstates that we cannot reach by applying $a,a^\dagger$.
How can we show that this is not the case? How can we show that all the eigenstates are in (1)? Is there something that I am missing?