The energy levels of the harmonic oscillator are

$\begin{align*}E_n = \left(n+ \frac{1}{2} \right) \hbar \omega\end{align*}$
where $n = 0,1,2,\ldots$ . Therefore, the ground-state energy is $E_0 = \boxed{\hbar \omega/2}$ . Therefore, answer (C) is correct.

Solution to 2008 Problem 11