Solving by the method of separation of variables $\Psi (x, t) = \psi(x) \phi(t)$
\[\begin{align}\phi(t) &= e^{-iEt/\hbar} \\ -\frac{h^2}{2m}\frac{d^2 \phi}{d x^2} + V\Phi &= E\Phi\end{align}\]Separable solutions have meaningfull
$\left| c_n \right|^2 $ is the probability that $E_n$ is measured
\[\begin{align}\sum_{n=1}^{\infty} \left|c_n\right|^2 &= 1 \\ \left< H \right> &= \sum_{n=1}^{\infty} \left| c_n \right|^2 E_n \end{align}\]Potential energy is $V(x) = \frac{1}{2}kx^2$. Any potential is approximately parabolic in the neighborhood of a local minimum.
Let’s define Ladder operator as,
\[\hat{a}_\pm = \frac{1}{\sqrt{2\hbar m \omega}} \left(\mp i\hat{p} + m\omega x \right)\]Commutate -
\[\begin{align}\left[ x, \hat{p} \right] &= i\hbar \\ \left[\hat{a}_{-} , \hat{a}_{+} \right] &= 1\end{align}\]The solution for this equation
\[\begin{align}\psi_0 (x) &= \left( \frac{m\omega}{\pi\hbar} \right)^{1/4} e^{-\frac{mw}{2\hbar}x^2} \\ \psi_n (x) &= \frac{1}{\sqrt{n!}} \left(\hat{a}_{+} \right)^n \psi_0 (x) \end{align}\]$V(x) = 0$, the general solution is
\[\begin{align}\Psi(x, t) &= \frac{1}{\sqrt{2 \pi}} \int_{\infty}^{\infty} \psi(k) e^{i \left(kx - \frac{\hbar k^2}{2m}t\right)}dk \\ \psi(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}\Psi(x, 0)e^{-ikx}dx\end{align}\]$E < V(-\infty) \& V(+\infty)$ Then bound state, otherwise scattering state
For given potential $V(x) = -\alpha \delta(x)$, $E<0$ (bound states)
\[\psi(x) = \frac{\sqrt{m\alpha}}{\hbar} e^{-m\alpha|x|/\hbar^2}, E=-\frac{m\alpha^2}{2\hbar^2}\]$E>0$ (scattering states)
\[\begin{align}\psi(x)&=Ae^{ikx}+Be^{-ikx}(x<0) \\ &=Fe^{ikx}(x>0)\end{align}\]with
\[B=\frac{i\beta}{1-i\beta}A, F=\frac{1}{1-i\beta}A, \beta=\frac{m\alpha}{\hbar^2 k}\]and
\[\begin{align}R&=\frac{|B|^2}{|A|^2} = \frac{1}{1+\left(2\hbar^2 E/m \alpha^2 \right)^2} \\ T&=\frac{|F|^2}{|A|^2} = \frac{1}{1+\left( m\alpha^2 / 2\hbar^2 E \right)} \end{align}\]$R$ is reflection coefficient, $T$ is transmission coefficient.