Algebraic solution

The time independent Schrödinger equation for this first order potential approximation is:

$$-\frac{\hbar}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2x^2\psi = E \psi$$

$$\frac{1}{2m}\left(\left(\frac{\hbar}{i}\frac{d}{dx}\right)^2 + (m\omega x)^2\right)\psi = E \psi$$

$$\frac{1}{2m}\left(\left(\frac{h}{i}\frac{d}{dx}-im\omega x\right)... ...frac{h}{i}\frac{d}{dx}+im\omega x\right)-\frac{\hbar\omega}{2}\right)\psi=E\psi$$

$$\left(a_-a_+ - \frac{\hbar \omega}{2}\right)\psi = E \psi$$ (4.1)

Here we have:

$$a_{\pm}=\frac{1}{\sqrt{2m}}\left(\frac{h}{i}\frac{d}{dx}\pm im\omega x\right)$$ (4.2)

It is easy to show that $a_-a_+ - a_+a_- = \hbar \omega$, and as well, we have:

$$\left(a_+a_- + \frac{\hbar \omega}{2}\right)\psi = E \psi$$ (4.3)

We now introduce the following chain of logic:

$$\left(a_+a_- + \frac{\hbar\omega}{2}\right)(a_+\psi)=(a_+a_-a_+ + \frac{\hbar\omega a_+}{2})\psi$$

$$=a_+\left((a_-a_+ - \frac{\hbar \omega}{2})\psi+ \hbar\omega\psi\right) =(E+\hbar\omega)(a_+\psi)$$

$$\therefore \left(a_+a_- + \frac{\hbar\omega}{2}\right)(a_+\psi)=(E+\hbar\omega)(a_+\psi)$$

$$\ \ \left(a_-a_+-\frac{\hbar \omega}{2}\right)(a_-\psi)=(E-\hbar\omega)(a_-\psi)$$ (4.4)

It is now obvious to see why $a_+$ is called the raising operator and $a_-$ is called the lowering operator. We do of course require a minimum lowering, a ground level. This requirement puts upon us that $\exists \ \psi_o \ \ni \ a_-\psi_o = 0 \Longrightarrow \psi_o(x)=A_oe^{-\frac{m\omega}{2\hbar}x^2}$. We finally conclude that $E_o = \hbar\omega/2$, and

$$\psi_n(x)=A_n(a_+)^ne^{-\frac{m\omega}{2\hbar}x^2}, \ \ E_n=(n+1/2)\hbar\omega$$ (4.5)