Algebraic Solution of the Harmonic Oscillator in Quantum Mechanics (Operator Method)
Proposition
The ground state of the quantum harmonic oscillator and its energy are as follows.
$$ \psi_{0}(x) = \left(\textstyle \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{-\frac{m\omega}{2\hbar} x^{2}} $$
$$ E_{0} = \dfrac{1}{2}\hbar \omega $$
The excited states and their energies are as follows. For $n = 1, 2, \dots$,
$$ \psi_{n}(x) = \dfrac{1}{\sqrt{2^{n}n!}} \alpha^{-n/2} \left( \frac{\alpha}{\pi} \right)^{1/4} \left( \alpha x - \dfrac{\d }{\d x} \right)^{n} e^{-\frac{\alpha}{2} x^{2}} $$
$$ E_{n} = \left( n + \frac{1}{2} \right) \hbar\omega $$
Here, $\alpha = \frac{m \omega}{\hbar}$.
Derivation
For the quantum harmonic oscillator, the time-independent Schrödinger equation is as follows.
$$ H \psi = \left( -\frac{\hbar^{2}}{2m} \frac{ d^{2} }{ dx^{2} } + \dfrac{1}{2} m \omega^{2} x^{2} \right) \psi = E\psi $$
The ladder operators of the Hamiltonian $H$ are as follows.
$$ \begin{align*} a_{+} &= \dfrac{1}{\sqrt{2\hbar m \omega}}(- \i P + m\omega X) \\ a_{-} &= \dfrac{1}{\sqrt{2\hbar m \omega}}(+ \i P + m\omega X) \end{align*} $$
Let $\psi$ be an eigenfunction (wavefunction) of $H$, and let $E$ be the corresponding eigenvalue (energy). The role of the ladder operator $a_{-}$ is to act on an eigenfunction of $H$ to produce another eigenfunction whose eigenvalue is smaller by $\hbar\omega$. That is, it is an operator that satisfies the following.
$$ H (a_{-} \psi) = (E - \hbar\omega)(a_{-}\psi) $$
Ground State
By applying the lowering operator $a_{-}$ an arbitrary number of times, we can continue to obtain eigenfunctions with correspondingly lower energies.
$$ \begin{array}{c} H ((a_{-})^{2} \psi) = (E - 2\hbar\omega)((a_{-})^{2}\psi) \\ H ((a_{-})^{3} \psi) = (E - 3\hbar\omega)((a_{-})^{3}\psi) \\ \vdots \end{array} $$
Suppose that, while repeatedly applying the lowering operator, at some point the energy $E_{0}$ of the eigenfunction $\psi_{0}$ falls within the following range.
$$ 0 \lt E_{0} \lt \hbar\omega $$
Then applying $a_{-}$ once more to $\psi_{0}$ would make the energy negative, and (since the minimum value of the potential $V(x) = \frac{1}{2}m\omega^{2}x^{2}$ is $0$) in this case no physically possible solution exists. That is, it means the probability that the particle exists is $0$, so the wavefunction is $0$.
$$ a_{-}\psi_{0} = 0 $$
Since the momentum operator is $p = \frac{\hbar}{\i} \frac{\d }{\d x}$ (let us write it in lowercase for convenience), the above equation becomes the following differential equation.
$$ \dfrac{1}{\sqrt{2 \hbar m \omega}}\left( \hbar\dfrac{\d \psi_{0}}{\d x} + m \omega x \psi_{0} \right) = 0 \\[1em] \implies \dfrac{\d \psi_{0}}{\d x} = -\dfrac{m \omega x}{\hbar}\psi_{0} $$
Applying separation of variables and then integrating, we obtain $\psi_{0}$ as follows.
$$ \begin{align*} \implies && \dfrac{1}{\d \psi_{0}} \d \psi_{0} &= -\dfrac{m\omega}{\hbar} x \d x \\ \implies && \int \dfrac{1}{\d \psi_{0}} \d \psi_{0} &= -\dfrac{m\omega}{\hbar} \int x \d x \\ \implies && \ln \psi_{0} &= -\dfrac{m\omega}{2\hbar} x^{2} + \text{constant} \\ \implies && \psi_{0} &= e^{-\frac{m\omega}{2\hbar} x^{2} + \text{constant}} \\ \implies && \psi_{0} &= C_{0} e^{-\frac{m\omega}{2\hbar} x^{2}} \\ \end{align*} $$
Now let us determine the normalization constant $C_{0}$ exactly. $C_{0}$ is the constant that makes the square integral of the wavefunction equal to $1$.
$$ 1 = \int \psi_{0}^{\ast} \psi_{0} dx = |C_{0}|^{2} \int e^{-\frac{m\omega}{\hbar} x^{2}} dx $$
Here, the integral on the right-hand side is the Gaussian integral.
Gaussian integral:
$$ \int\limits_{-\infty}^{\infty} e^{-\alpha x^{2}} dx = \sqrt{\dfrac{\pi}{\alpha}} $$
Therefore, we obtain the following.
$$ 1 = |C_{0}|^{2} \sqrt{\dfrac{\pi \hbar}{m \omega}} \implies C_{0} = \left( \dfrac{m \omega}{\pi \hbar} \right)^{1/4} $$
Therefore, the wavefunction of the ground state with the lowest energy is as follows.
$$ \psi_{0}(x) = \left(\textstyle \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{-\frac{m\omega}{2\hbar} x^{2}} $$
Since the Hamiltonian of the harmonic oscillator is expressed in terms of the ladder operators as $H = \hbar \omega(a_{+}a_{-} + \frac{1}{2})$, the ground state energy is computed as follows.
$$ \begin{align*} \implies && H\psi_{0} &= E_{0}\psi_{0} \\ \implies && \hbar \omega \left( a_{+}a_{-} + \frac{1}{2} \right) \psi_{0} &= E_{0}\psi_{0} \\ \implies && \hbar \omega a_{+}a_{-}\psi_{0} + \frac{1}{2}\hbar \omega \psi_{0} &= E_{0}\psi_{0} \\ \implies && \frac{1}{2}\hbar \omega \psi_{0} &= E_{0}\psi_{0} \\ \implies && E_{0} &= \dfrac{1}{2}\hbar \omega \end{align*} $$
Excited States
Now let us find the wavefunction and energy of the excited states. First, since $a_{+}$ is the operator that increases the energy of a state by $\hbar\omega$, the energy for an arbitrary state $n$ is as follows.
$$ H(a_{+})^{n}\psi_{0} = \left( E_{0} + n\hbar\omega \right) (a_{+})^{n}\psi_{0} $$
$$ \implies E_{n} = \dfrac{1}{2}\hbar\omega + n\hbar\omega = \left( n + \frac{1}{2} \right) \hbar\omega $$
For the ladder operator $a_{+}$, $a_{+}\psi_{n}$ and the normalized $(n+1)$-th eigenfunction $\psi_{n+1}$ are proportional.
$$ a_{+} \psi_{n} = C_{+} \psi_{n+1} \tag{1} $$
Taking the square integral of the left-hand side gives the following. Since $\psi_{n}$ is a normalized wavefunction, using Dirac notation,
$$ \begin{align*} (a_{+} \ket{\psi_{n}})^{\ast}(a_{+} \ket{\psi_{n}}) &= \braket{\psi_{n} | a_{-}a_{+} | \psi_{n}} \\ &= \Braket{\psi_{n} | \frac{1}{\hbar \omega}H + \frac{1}{2} | \psi_{n}} \\ &= \frac{1}{\hbar \omega}\Braket{\psi_{n} | H | \psi_{n}} + \frac{1}{2} \braket{\psi_{n} | \psi_{n}} \\ &= \frac{1}{\hbar \omega}\Braket{\psi_{n} | E_{n} | \psi_{n}} + \frac{1}{2}\\ &= \left( n + \dfrac{1}{2} \right) + \frac{1}{2}\\ &= n + 1 \end{align*} $$
The second equality substitutes $a_{-}a_{+}=\frac{1}{\hbar \omega}H + \frac{1}{2}$. On the other hand, taking the square integral of the right-hand side of $(1)$,
$$ \begin{align*} (C_{+}\ket{\psi_{n+1}})^{\ast}(C_{+}\ket{\psi_{n+1}}) &= |C_{+}|^{2}\braket{\psi_{n+1} | \psi_{n+1}} \\ &= |C_{+}|^{2} \end{align*} $$
Therefore, we obtain the following.
$$ C_{+} = \sqrt{n+1} \implies \psi_{n+1} = \dfrac{1}{\sqrt{n+1}} a_{+}\psi_{n} $$
Finally, an arbitrary $n$-th state is expressed in terms of the ground state as follows.
$$ \begin{align*} \psi_{n} &= \dfrac{1}{\sqrt{n}} a_{+}\psi_{n-1} \\ &= \dfrac{1}{\sqrt{n}} a_{+}\dfrac{1}{\sqrt{n-1}} a_{+}\psi_{n-2} = \dfrac{1}{\sqrt{n(n-1)}} (a_{+})^{2}\psi_{n-2} \\ &= \dfrac{1}{\sqrt{n(n-1)}} (a_{+})^{2}\dfrac{1}{\sqrt{n-2}} a_{+}\psi_{n-3} = \dfrac{1}{\sqrt{n(n-1)(n-2)}} (a_{+})^{3}\psi_{n-3} \\ &= \vdots \\ &= \dfrac{1}{\sqrt{n!}} (a_{+})^{n} \psi_{0} \\ &= \dfrac{1}{\sqrt{n!}} \left( \dfrac{1}{\sqrt{2\hbar m \omega}}\left( m \omega x - \hbar \dfrac{\d }{\d x} \right) \right)^{n} \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{-\frac{m\omega}{2\hbar} x^{2}} \\ &= \dfrac{1}{\sqrt{2^{n}n!}} \left( \dfrac{1}{\sqrt{\hbar m \omega}} \right)^{n} \hbar^{n}\left( \dfrac{m \omega}{\hbar} x - \dfrac{\d }{\d x} \right)^{n} \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{-\frac{m\omega}{2\hbar} x^{2}} \\ &= \dfrac{1}{\sqrt{2^{n}n!}} \left( \dfrac{\sqrt{\hbar}}{\sqrt{ m \omega}} \right)^{n} \left( \dfrac{m \omega}{\hbar} x - \dfrac{\d }{\d x} \right)^{n} \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{-\frac{m\omega}{2\hbar} x^{2}} \end{align*} $$
Here, setting $\alpha = \frac{m \omega}{\hbar}$,
$$ \psi_{n}(x) = \dfrac{1}{\sqrt{2^{n}n!}} \alpha^{-n/2} \left( \frac{\alpha}{\pi} \right)^{1/4} \left( \alpha x - \dfrac{\d }{\d x} \right)^{n} e^{-\frac{\alpha}{2} x^{2}} $$
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