Review of QHO

$H=\frac{P^2}{2m}+\frac{m{\omega }^2{X}^2}{2}$ where $X$ and $P$ are the position and momentum operators respectively.

Let us introduce the creation and annihilation operators such that, $[a,a^{†}]=1$ and we write ansatz for $X$ and $P$ in terms of the ladder operators as,

$X=(a+a^{†})C$

$P=i\hslash (a-a^{†})B$.

Then, $[X,P]=-2i\hslash BC$ and to recover the commutation to be $i\hslash$, we require $-2BC=1$.

Further, we can fix the Hamiltonian to have the form,

$H=(\frac{1}{2m}(-B^2{\hslash }^2)(a^2+(a^{†}{)}^2-\{a,a^{†}\}))\hslash +\frac{m{\omega }^2}{2}(C^2)(a^2+(a^{†}{)}^2+\{a,a^{†}\})\hslash$

$=E(\{a,a^{†}\})=E(a{a}^{†}+\frac{1}{2})\hslash$.

Hence, using these two equations, we can solve for B and C obtaining,

$X=\sqrt{\frac{\hslash }{2m\omega }}(a+a^{†})$

$P=-i\sqrt{\frac{\hslash m\omega }{2}}(a-a^{†})$.

We also obtain a relation E = ℏω, i.e.,

$H=\hslash \omega (a{a}^{†}+\frac{1}{2})$.

Hence, we interpret these ladder operators as creating and destroying quanta since,

$[H,a^{†}]=\hslash \omega a^{†}$

$[H,a]=-\hslash \omega a$.

The action of $a^{†}(a)$ on a Hamiltonian eigenstate is to increase (decrease) the energy by $\omega$. Given the positive definite spectrum, there must be a state $|0⟩$ such that $a|0⟩=0$ and the whole spectrum is spanned by the Hilbert space: $\oplus (a^{†}{)}^n|0⟩$ which has the following features:

• Energy is $\sim \sqrt{{\omega }^2}$ and we have a positive spectrum.
• Creation and annihilation operators offer a simple picture of quanta.