Hidden States in Quantum Harmonic Oscillator

Question

I was just revising my notes of Quantum harmonic Oscillator when an idea struck my head. To build up to the question, we start with the creation and annhilation operators. $$a=\sqrt{\frac{m\omega}{2\hbar}}X+i\frac{1}{\sqrt{2m\hbar\omega}}P$$ which is the annhilation operator and the creation operator is: $$a^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}X-i\frac{1}{\sqrt{2m\hbar\omega}}P$$ My question is that: We say that $$a|n\rangle=C_n|n-1\rangle$$ for some constant $C_n$ which is a complex number. My question is how can we be sure that these operators actually just reduce the states by one level only. why can't be two levels? Basically what I want to prove that, for an arbitrary case: $$a|n\rangle=C_m|m\rangle$$ for some eigenstate $|m\rangle$ of the hamaltonian. What I want to prove that is- there is no other eigenstate of H(Hamaltonian) between $|n\rangle$ and $|m\rangle$ (Between the states means that a state with eigenvalue greater than state $|m\rangle$ but less than $|n\rangle$).
Thanks in advance.

Answer 1

The level is by definition the eigenvalue of the state w.r.t. the number operator $$ N = a^\dagger a . $$ By definition $$ N | n \rangle = n | n \rangle. $$ Using the commutator $[a,a^\dagger]=1$, we find $$ [ N , a ] = - a . $$ It follows that \begin{align} N ( a |n\rangle ) &= [ N , a ] |n\rangle + a N | n \rangle = ( n - 1 ) ( a |n\rangle ) \end{align} It follows that the state $a|n\rangle$ has level $n-1$ so $$ a|n\rangle= C_n |n-1\rangle. $$