The energy levels of the harmonic oscillator are
$\begin{align*}E_n = \left(n + \frac{1}{2} \right)\hbar \omega\end{align*}$
where $n = 0,1,2,3,\ldots$ . So, the difference in energy between adjacent levels is always $\hbar \omega$ . So I is correct.
The potential energy is $m \omega^2 x^2/2$ which is quadratic, not linear, in $x$ , so II is incorrect.

The ground-state corresponds to $n = 0$ , and it has energy $\hbar \omega/2$ . This is equal to the kinetic energy plus the potential energy. By the virial theorem, $\langle V\rangle = \langle T \rangle = \hbar \omega/4$ for this state, so the kinetic energy is not zero, and III is incorrect.

The wavefunction of the ground state of the harmonic oscillator is
$\begin{align*} %blah dog cat\psi(x) = a e^{-b x^2}\end{align*}$
where $a$ and $b$ are positive constants. So, there is a nonzero probability for the particle to be found at any $x \in \left(-\infty,\infty \right)$ . So, IV is correct.
So, answer (C) is correct.

Solution to 2008 Problem 65