Solution to 2008 Problem 65

 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.