Taking up the eigenvalue equation for $\hat{H}(\hat{a}\vert n \rangle)$
\begin{equation} \hat{H} \left(\hat{a} \vert n \rangle \right)= \left( E_n - \hbar\omega \right) \left( \hat{a} \vert n \rangle \right) \end{equation}
Applying the annihilation operator repeatedly the energy eigenvalue will decrease in integer multiples of $\hbar \omega$, the energy of harmonic oscillator must always be positive, therefore exists a lowest energy eigenvalue corresponding to eigenstate $\vert 0 \rangle$. The eigenvalue equation for the ground state is:
\begin{equation}\require{cancel} \hat{H}\vert 0 \rangle = \hbar\omega \left( \hat{a}^\dagger \hat{a} + \frac{1}{2} \right) \vert 0 \rangle = \cancelto{0}{ \hbar\omega \hat{n}\vert 0 \rangle} + \frac{1}{2}\hbar\omega \vert 0 \rangle = \frac{1}{2}\hbar\omega \vert 0 \rangle \end{equation}
So the lowest energy eigenvalue is $E_0 = \cfrac{\hbar\omega}{2}$. Since $E_{n+1}= E_n + \hbar\omega$, the energy eigenvalues are:
\begin{equation}E_n = \hbar\omega \left( n + \frac{1}{2} \right) \end{equation}
Acknowledgement
I would like to thank Dr. Carlos Herman Wiechers Medina for their support, this post is based on the notes from the Quantum Optics class.
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