I apologize for the late typed notes, but hopefully there will be a complete set by the next semester I teach again.
Let’s continue with the quantum harmonic oscillator, which is an important elementary model for many chemical applications. Here we will go over the ladder operator trick to obtain the eigenvalues and eigenvectors of the system, and also talk briefly about what we can do with them. I might be skipping some mathematical steps, but many detailed parts are carried out extensively in many sources online. Feel free to contact me if there is an error or confusion!
The hamiltonian of this system is $$ H = \frac{p^2}{2m} + \frac{m\omega ^2x^2}{2}$$
where $\omega \equiv \sqrt{k/m}$, the angular frequency. This should resemble some sort of spring constant to you.
To solve this hamiltonian, there is a nice algebra trick called the ladder operators.
We define them as: $$ a = \sqrt{\frac{m\omega}{2\hbar}}(x + \frac{ip}{m\omega})$$
$$ a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}(x - \frac{ip}{m\omega})$$
and we call $a$ the annihilation operator, and $a^\dagger$ the creation operator. Also note that they are hermitian conjugate of one another. It is also useful to think about their commutators, and perhaps we can get insight to how their eigenvectors relate.
$$
\begin{gather}
[a, a^\dagger] = \frac{m\omega}{2\hbar}([x+\frac{ip}{m\omega}, x-\frac{ip}{m\omega}]) \\
= \frac{m\omega}{2\hbar}([x, x] - [x,\frac{ip}{m\omega}] + [\frac{ip}{m\omega}, x]-[\frac{ip}{m\omega}, \frac{ip}{m\omega}]) \\
= \frac{m\omega}{2\hbar}(0 - \frac{i}{m\omega}(i\hbar)+\frac{i}{m\omega}(-i\hbar)-0) \\
= \frac{m\omega}{2\hbar}(\frac{\hbar}{m\omega} + \frac{\hbar}{m\omega}) \\
= 1
\end{gather}
$$
Here I used the fact that $[x, p] = i\hbar$, an important commutator you should always keep in mind!
Now let’s further define a new operator
$$ N \equiv a^\dagger a$$
which we will call it the number operator.
Now we wish to rewrite the Hamiltonian in terms of these newly defined quantities. First, notice that $N=a^\dagger a$ is hermitian, i.e. $N = N^\dagger$.
Evaluating $a^\dagger a$, we obtain
$$a^\dagger a = \frac{m\omega}{2\hbar}(x^2 + \frac{p^2}{m^2\omega^2}) - 1/2$$
Reminding ourselves of the Hamiltonian of the system, we see that
$$a^\dagger a = \frac{H}{\hbar \omega} - 1/2$$
and
$$H = \hbar\omega(N + 1/2)$$
This means that if we can find the eigenvalues of $N$, we find the eigenvalues of $H$; they are just shifted by $(E(N) + 1/2)\hbar\omega$ ! You can justify this easily by arbitrarily defining the eigenvalues and eigenvectors of 2 operators related by linear operations. Explicitly:
$$\text{Let} \quad N\ket{n} = n \ket{n}$$
Then,
$$H\ket{n} = \hbar\omega(n+1/2)\ket{n}$$
What does $a, a^\dagger$ do? They actually can raise or lower our energy states!
In particular, $$Na^\dagger \ket{n} = (n+1) a^\dagger \ket{n}$$
$$Na\ket{n} = (n-1)a\ket{n}$$
To obtain the above relations, substitute the fact that $aa^\dagger - a^\dagger a=1$, shown previously by the commutator. Importantly, if we look at the second line above, this suggests that $a\ket{n}$ is an eigenstate of N with a lower eigenvalue by 1. And since we are denoting the eigenstates by a ket with their eigenvalues inside, then $a\ket{n}$ and $\ket{n-1}$ are related by some constant! i.e.
$$ a\ket{n} = c\ket{n-1}$$
$$ a^\dagger \ket{n} = \tilde{c}\ket{n+1}$$
By requiring that each state must be normalized, we can show:
$$ a \ket{n} = \sqrt{n}\ket{n - 1}$$
$$a^\dagger \ket{n} = \sqrt{n+1}\ket{n+1}$$
Now, suppose we start with some state $\ket{n}$,
and we, for some reason, continuously apply the annihilation operator $a$:
$$a\ket{n} = \sqrt{n}\ket{n-1}$$
$$a^2\ket{n} = \sqrt{n(n-1)}\ket{n-2}$$
$$a^3\ket{n} = \sqrt{n(n-1)(n-2)}\ket{n-3}$$
$$\vdots$$
Eventually, the sequence will reach some $\tilde{n}=0$, and it will terminate! We can justify this by the fact that
$$
\begin{gather}
n = \braket{n|N|n} \\
= \braket{n|a^\dagger a | n} \\
= \lVert a \ket{n} \rVert \\
\geq 0
\end{gather}
$$
With such requirement in mind, there must be a lowest state such that when applying the annihilation operator essentially does nothing on the state. i.e.
$$a\ket{0} = 0$$
This also means that $n$ must be an integer such that when repeatedly applying the annihilation operator yields 0 (otherwise, it bypasses 0 and goes to a negative number). This essentially completes our eigenvalue spectrum of $N$, which is the set of integers. Recall our relation between $N$ and $H$, we have
$$E_n = (n + 1/2)\hbar \omega$$
with the lowest energy being
$$E_0 = 1/2\hbar \omega$$
To obtain the eigenvector spectrum, we solve the differential equation in the position representation
$$\braket{x|a|0} = 0$$
by plugging in the definition of $a$. Sparing the mathematical trouble, we will arrive at a solution of the form
$$\psi_0 = (\frac{m\omega}{\pi\hbar})^{1/4}\exp{(\frac{-m\omega}{2\hbar}x^2)}$$
Theoretically, one could repeatedly apply the creation operator from here and obtain as many eigenstates as desired. Luckily, there is a recursion relation that is defined between the states, and they are the hermite polynomials. Rarely in an exam setting, however, you will be asked to toy with the explicit form of the solutions. Rather, it is more worth to play around with the states in the ket notation as eigenkets of the number operator $N$.
With that in mind, note that $$ x = \sqrt{\frac{\hbar}{2m\omega}} (a^\dagger + a)$$ $$ p = i\sqrt{\frac{\hbar m \omega}{2}} (a^\dagger - a)$$ We can then evaluate matrix elements of $\braket{x}$, $\braket{p}$, $\braket{x^2}$, $\braket{p^2}$ easily. From here, we can also arrive at the uncertainty principle in the basis of harmonic oscillator solutions, $\Delta x\Delta p = \hbar/2$. The matrix elements $\braket{x}$ and $\braket{x^2}$ are also related to the selection rules of infrared spectroscopy, as the harmonic oscillator potential near the origin resembles the potential energy of a linear molecule, and the molecular dipole moment can be approximated as the position operator.