Discrete spectrum of the quantum harmonic oscillator

Step 1 (ladder operators). Define $\hat a=\sqrt{\tfrac{m\omega}{2\hbar}}\left(\hat X+\tfrac{i}{m\omega}\hat P\right)$ and $\hat a^\dagger=\sqrt{\tfrac{m\omega}{2\hbar}}\left(\hat X-\tfrac{i}{m\omega}\hat P\right)$. Using $[\hat X,\hat P]=i\hbar$, a direct computation gives $[\hat a,\hat a^\dagger]=\tfrac{m\omega}{2\hbar}\left(\tfrac{-i}{m\omega}[\hat X,\hat P]+\tfrac{i}{m\omega}[\hat P,\hat X]\right)=\tfrac{1}{2\hbar}\left(\hbar+\hbar\right)=1$.

Step 2 (rewrite the Hamiltonian). Inverting, $\hat X=\sqrt{\tfrac{\hbar}{2m\omega}}(\hat a+\hat a^\dagger)$ and $\hat P=i\sqrt{\tfrac{m\omega\hbar}{2}}(\hat a^\dagger-\hat a)$; substituting into $\hat H=\tfrac{\hat P^2}{2m}+\tfrac12 m\omega^2\hat X^2$ and using $[\hat a,\hat a^\dagger]=1$ to reorder terms gives $\hat H=\hbar\omega\left(\hat a^\dagger\hat a+\tfrac12\right)$. Define the number operator $\hat N=\hat a^\dagger\hat a$; it is self-adjoint, and for any state $\phi$, $\langle\phi|\hat N|\phi\rangle=\|\hat a\phi\|^2\ge0$, so every eigenvalue $n$ of $\hat N$ satisfies $n\ge0$.

Step 3 (ladder relations). From $[\hat a,\hat a^\dagger]=1$ one gets $[\hat N,\hat a]=-\hat a$ and $[\hat N,\hat a^\dagger]=\hat a^\dagger$. So if $\hat N|n\rangle=n|n\rangle$, then $\hat N\hat a=\hat a\hat N-\hat a=\hat a(\hat N-1)$, giving $\hat N(\hat a|n\rangle)=(n-1)(\hat a|n\rangle)$: $\hat a$ lowers the eigenvalue by $1$ (or annihilates the state), and $\hat a^\dagger$ raises it by $1$.

Step 4 (termination and the spectrum). Because $\hat N\ge0$, repeatedly applying $\hat a$ to any eigenstate cannot produce eigenvalues below $0$; the descending chain $n,n-1,n-2,\dots$ must terminate at $n=0$ (a non-integer starting value would force a negative eigenvalue after finitely many steps, a contradiction), so every eigenvalue of $\hat N$ is a nonnegative integer, and the ground state $|0\rangle$ satisfies $\hat a|0\rangle=0$. Applying $\hat a^\dagger$ repeatedly to $|0\rangle$ produces normalizable eigenstates $|n\rangle\propto(\hat a^\dagger)^n|0\rangle$ for every $n=0,1,2,\dots$, and no others exist. Substituting $\hat N|n\rangle=n|n\rangle$ into $\hat H=\hbar\omega(\hat N+\tfrac12)$ gives $E_n = \hbar\omega\left(n+\tfrac12\right)$.